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This page was last updated on Friday, August 02, 2002 07:42:46 PM. © Ken Monks
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FAQ - Modern Algebra I Dr. Monks This isn't really a FAQ, but rather is just an AQ... it is just a compilation of questions I have answered about modern algebra by email. I have not edited these email messages much and they are usually replies to questions asked by students, so keep that in mind when reading through them. They are in roughly chronological order. I hope they help. -------------------------------------------------------------------- Messages from the Fall 2001 course -------------------------------------------------------------------- ------------------------------------------------------------------------ Laaaaaadiess... and Geeeentleman... the first FAQ of 2001 is here! Let the fun begin!!! But first a word from your sponsor... be sure you read the previous year's FAQ's. I don't want to answer questions that are already answered in there. > -----Original Message----- > Sent: Sunday, September 09, 2001 5:39 PM > To: monks@UofS.edu > Subject: Math 448 Assignment #2, Problem #3 > > Dr. Monks, > For problem #3, (P<=>Q)<=>(Q<=>P), is it acceptable to use the > results for P<=>Q to prove Q<=>P? In other words, I would prove P<=>Q > by showing that Q=>P and P=>Q. Must I reprove those two things if I > want to prove Q<=>P or can I just use the previous results? > Thanks so much. You cannot prove P<=>Q because it is not a tautology (check it's truth table). You also can't prove P=>Q and Q=>P for the same reason. As I mentioned in class, only tautologies can be proven with our recipies. So this is not a good approach. -tautological tutor ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, September 12, 2001 11:03 PM > To: monks@UofS.edu > Subject: Contradictions > > Lets say you make a contradiction, then use that to show > something. If you need another contradiction later in the same proof, do > you have to make another contradiction, or can you use the previous one? While technically possible, it is extremely rare that you would be able to "reuse" a previous contradiction in any effective way in a proof. Technically, you can use a previous contradiction IF it is still available at the line you are referring to it from (i.e. if it is still an active line in your proof... i.e. if you haven't ended the assumption world it was based on with an end assumption (<-) yet). But in actual practice, whenever we get a -><- into a proof, it is always in an indented "Assume" world, and it is almost always because we want to do a proof by contradiction (~- or ~+) so that when we get a -><- in our proof, we almost always follow it with a <- and then with a conclusion using ~+ or ~-, so that it is no longer available in the proof at any line after that. So remember, once you end an Assume world with an <-, all the lines between the Assume and the <- are no longer "active" lines and cannot be referred to in later lines in your proof (because we have stopped "pretending" that the Assumption is true, and therefore anything we deduced from that assumption is no longer valid). So the technical answer is that you can in theory refer to a previous contradiction IF that contradiction is an active line, but in practice the answer is usually NO, because there is almost never a situation in a proof where you don't deactivate a contradiction immediately after it appears. --teaches contradictions ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, September 12, 2001 4:01 PM > To: monks@epix.net > Subject: assignment #2 problem #3 > > > Dr. Monks, > > just a quick question... > > i found this in the FAQ for '97: > Pf. > 1. Assume P=>Q > 2. Assume ~Q > 3. Assume P > 4. Q =>- ; 1,3 > > so i began my proof of number 3 as follows: > > Pf. > 1. Assume (P<=>Q) > 2. Assume P > 3. Q =>-;2,1 > 4. <- > 5. P=>Q =>+;2,3,4 > > it seems as though the =>- on line 3 is incorrect, but i think i followed > the pattern of the example from the FAQ '97. > > please let me know if i can do this. You can't. In the first example, line 1 has a => statement, so you can use =>-, but in the second example, the first line isn't a => statement so you can't use =>- on line 3 because there is no => to -. In addition your indentation is incorrect... line 3 and 4 in your second example should be lined up with line 2 and line 5 should be lined up with line 1. --gives indentation education ------------------------------------------------------------------------ > -----Original Message----- > Sent: Thursday, September 13, 2001 10:12 AM > To: monks@UofS.edu > Subject: Use of Previous Theorems > > Are these two examples correct uses of previous theorems? If not, how > should I implement the previous theorems into my proofs? > > Ex1: > 20. P & ~P from a previous line > 21. Q (P & ~P) => Q; 20 No, you can only use a previous theorem by putting it as a line in your proof, i.e. it can only go in the statement column, not in the reason column. The reason would then just be the name or reference to the theorem (like "Lemma A" or "Problem #2, Assignment #4" etc). > Ex2: > 2. ~(P or ~P) from a previous line > 3. ~P and ~(~P) ~(P or Q) <=> (~P and ~Q); 2 (using ~P as Q, of > course) > > Thanks! > > --user of previous theorems Same problem here. You can only insert the previous theorem directly as a line in your proof. --previous instructor ------------------------------------------------------------------------ > -----Original Message----- > Sent: Saturday, September 15, 2001 3:23 PM > To: monks@UofS.edu > Subject: contradictions > > Dr Monks, > > I was wondering if the contradiction for (~@x,P(x)) would be > ~(~@x,P(x)) or @x,P(x)and if the contradiction for (#x,~P(x)) would be > ~(#x,~P(x)) or (#x,P(x)). Thanks for your help. A contradiction is a statement of the form W and ~W where W is any statement. In our formal math language we have defined a special symbol for this, namely -><-, which is a symbol which can be derived from any such statement, or also (as seen in the recipes) anytime we have both a statement and its negation in a proof. Thus, you can't ask what the "contradiction for" a statement is, because a statement doesn't *have* a contradiction, it either *is* a contradiction or it isn't. That having been said, I think what you meant to ask is: "What is the negation of (~@x,P(x))? Since the negation of W is ~W, then if W is (~@x,P(x)) then clearly the negation is ~(~@x,P(x)). However, it is also true that the negation of @x,P(x) is ~@x,P(x) for the same reason. So for the purpose of *deducing* a contradiction from such statements, either pair would work. On the other hand, the negation of (#x,~P(x)) is clearly ~(#x,~P(x)), but is certainly NOT #x,P(x). You can't negate some sub-part of a statement and call that the negation of the whole statement. You simply put a ~ in front of the whole statement to form its negation. DeMorgan's Laws (one of which you are proving for homework) tell us how to "move a ~ past a quantifier", but you can't use them on this assignment, since that is what you are being asked to prove. -- ~~monks ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, September 17, 2001 11:21 AM > To: monks@scranton.edu; monks@epix.net > Subject: > > > Dr. Monks i got a question for you right quick. > > Does the negation of a quantifier (eg.(~@x,P(x)) affect the rules of > inference for that quantifier? For example: > 1-Assume (~@x,P(x)) > 2-Let t be arbitrary > 3-P(t) @-;1 (is this true despite the negation of @?) A statement of the form ~W is not the same as a statement of the form @x,W. The statement ~@x,P(x) is of the form ~W (where W is @x,P(x)). You can't find any W so that ~@x,P(x) is of the form @x,W because of the ~ in front. Since the @- rule requires as its first input a statement of the form @something and the statement you have on line one is of the form ~something, you can't use @- in line three. So yes, the negation of the quantifier completely prevents you from using the @- rule. Think of it this way... if there was NO ~ in the statement of line 1, then line three would be correct, so if it were also correct in your proof above it would mean that you would just be ignoring the ~ completely. [But there is a BIG difference between @x,P(x) and ~@x,P(x)!!! :)] -- ~ completely ignorant ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, September 17, 2001 4:10 PM > To: monks@UofS.edu > Subject: Assignment 3 Number 5 > > Dr Monks, > > In regards to #5 > Proof > 1. Assume x = y and y = z > 2. x = z Arithmetic; 1 > : > > Is that proper usage of the Arithmetic reason? > > Any words of wisdom would be appreciated. > > Thanks, No. The ONLY TIME YOU CAN USE "Arithmetic" as a reason is if the statement on that line refers to a fact about the natural, integer, rational, real, or complex numbers which the author assumes you know, i.e. that is not proven or a consequence of theorems proven in the course. Since this problem does not necessarily deal with those specific number systems, but rather is about "=" in its most general form, you cannot use arithmetic for ANY line in the proof of this theorem. In fact, you can't use "arithmetic" for any line in any proof in any problem that is in the first three assignments, because none of them deal with the number systems N,Z,Q,R, or C. For further examples, you cannot use "arithmetic" in problems like #15 or #21 in Appendix B because these have nothing to do with N,Z,Q,R, or C, but you could use it in problems like #25b and #25d because they clearly have to do with the real numbers. So it is ONLY problems that directly need to use facts about those SPECIFIC number systems, N,Z,Q,R,C that can use "arithmetic" as a reason. EVERY OTHER reason must be from a recipe (or from a recipe that is derived from a definition as discussed in the shortcuts). -still teaching arithmetic ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, September 19, 2001 10:47 AM > To: Monks@UofS.edu > Subject: Assignment #3 Appendix B > > Wise and Powerful Teacher, > > In the following: > > 1. @x in B, @y in C, f(x,y)=(y,x) Given > 2. z=(x,y) for some x in B and y in C > 3. f(z)=f(x,y) > > First, is line 2 okay or do we have to say "Define z=..."? No, line 2 declares x and y, but it does NOT declare z. You must declare z somehow. There are three ways to declare a free variable in your proof. 1. Let x be arbitrary (used for a @+) 2. blah-blah-blah for some x (only from #-) 3. Define x=blah (anytime you want, but never needed) The only other way a free variable can get in your proof is if it is in a Given line, i.e. if it is a free variable that is in the statement of the theorem. There is further discussion about this in the FAQ's! --Define monks=me ------------------------------------------------------------------------ "> -----Original Message----- > Sent: Wednesday, September 19, 2001 3:51 PM > To: monks@epix.net > Subject: Assignment 4 Problem 24 page 517 > > Dr Monks, > The question asks: "Determine whether the given operation on R is > commutitive or associative. > > If the operation is commutitive and associative do we need to prove both > parts as 1 proof or can we make it 2 seperate proofs, proving both parts > seperately? You can write it as one proof or two, your choice. Each part (a,b,c,d etc) will be graded as one problem. > If an operation is not commutitive and not associative > do we need to show anything in our proof or can we simply write "It > is not commutitive and not associative? Read the FAQs. > If a funtion only satisfies one condition (say associative) do we need to > prove that it doesn't satisfy the other condition (commutitive)? Yes. -grades each numbered (lettered) part of each question as one question ------------------------------------------------------------------------ So here is the example I promised to send you that we didn't get to in today's lecture. Ex: Show that the cube of any integer can be written in the form 9k, 9k+1 or 9k+8 for some integer k. OK, first lets see what this really says: @n in Z,#k in Z,#e in {0,1,8},n^3=9k+e
Ok, now on with the proof: Pf. 1. Let n in Z Given 2. n=9q+r and 0<=r<9 for some q,r in Z Div Alg Thm;1 3. n^3=(9q+r)^3 subst;2 4. =9^3*q^3+3*9^2*r+3*9*r^2+r^3 arith;3 5. =9(9^2*q^3+27*r+3*r^2)+r^3 arith;4 6. Define m=(9^2*q^3+27*r+3*r^2) - 7. m in Z arith;6,2 8. n^3=9m+r^3 subst;3,5,6 9. r in {0,1,2,3,4,5,6,7,8} arith;2
10. r=0 or r=1 or ... or r=8 Def in;3 11. Assume r=0 -
13. n^3=9m+0^3 subst;11,8
14. =9m+0 arith;13
15. #k in Z,#e in {0,1,8},n^3=9k+e #+;13,14
<-
16. Assume r=1 -
17. n^3=9m+1^3 subst;16,8
18. =9m+1 arith;17
19. #k in Z,#e in {0,1,8},n^3=9k+e #+;17,18
<-
20. Assume r=2 -
21. n^3=9m+2^3 subst;20,8
22. =9m+8 arith;21
23. #k in Z,#e in {0,1,8},n^3=9k+e #+;21,22
<-
24. Assume r=3 -
25. n^3=9m+3^3 subst;24,8
26. =9m+27 arith;25
27. =9(m+3)+0 arith;26
28. m+3 in Z arith;7
29. #k in Z,#e in {0,1,8},n^3=9k+e #+;25,27,28
<-
30. Assume r=4 -
31. n^3=9m+4^3 subst;30,8
32. =9m+64 arith;31
33. =9(m+7)+1 arith;32
34. m+7 in Z arith;7
35. #k in Z,#e in {0,1,8},n^3=9k+e #+;31,33,34
<-
36. Assume r=5 -
37. n^3=9m+5^3 subst;36,8
38. =9m+125 arith;37
39. =9(m+13)+8 arith;38
40. m+13 in Z arith;7
41. #k in Z,#e in {0,1,8},n^3=9k+e #+;37,39,40
<-
42. Assume r=6 -
43. n^3=9m+6^3 subst;42,8
44. =9m+216 arith;43
45. =9(m+24)+0 arith;44
46. m+24 in Z arith;7
47. #k in Z,#e in {0,1,8},n^3=9k+e #+;43,45,46
<-
48. Assume r=7 -
49. n^3=9m+7^3 subst;48,8
50. =9m+343 arith;49
51. =9(m+38)+1 arith;50
52. m+38 in Z arith;7
53. #k in Z,#e in {0,1,8},n^3=9k+e #+;49,51,52
<-
54. Assume r=8 -
55. n^3=9m+8^3 subst;54,8
56. =9m+512 arith;55
57. =9(m+56)+1 arith;57
58. m+56 in Z arith;7
59. #k in Z,#e in {0,1,8},n^3=9k+e #+;55,57,58
<-
60. #k in Z,#e in {0,1,8},n^3=9k+e or- (pf by cases);10,11,
15,16,19,20,23,24,29,30,
35,36,41,42,47,48,53,54,
59
61. @n in Z,#k in Z,#e in {0,1,8},n^3=9k+e @+;1,60
QED
(I didn't check the line numbering in the reasons carefully, so let me know if there are typos) Comments: First you might notice that I didn't number or refer to the lines that have the <-'s on them. That is because they are really just a kind of "punctuation" which show us where the assumption ends, but aren't saying anything about the topic of the proof itself. You can use this abbreviation too if you like. (Still show where your assumptions end, but you don't need to refer to <- lines in your reasons). Second, you might notice that there is a lot of "repetition" in checking the nine cases in lines 11-59. Since it is very tedious to check and write so many cases, there is a shortcut used in mathematics to condense all of this information... namely to use a table. So if we use a table in our proof above, it would look something like this: Pf (short version): 1. Let n in Z Given 2. n=9q+r and 0<=r<9 for some q,r in Z Div Alg Thm;1 3. n^3=(9q+r)^3 subst;2 4. =9^3*q^3+3*9^2*r+3*9*r^2+r^3 arith;3 5. =9(9^2*q^3+27*r+3*r^2)+r^3 arith;4 6. Define m=(9^2*q^3+27*r+3*r^2) - 7. m in Z arith;6,2 8. n^3=9m+r^3 subst;3,5,6 9. r in {0,1,2,3,4,5,6,7,8} arith;2
10. r=0 or r=1 or ... or r=8 Def in;3 11. r r^3 9m+r^3 arith
-- ---- ------
0 0 9m+0
1 1 9m+1
2 8 9m+8
3 27 9(m+3)+0
4 64 9(m+7)+1
5 125 9(m+13)+8
6 216 9(m+24)+0
7 343 9(m+38)+1
8 512 9(m+56)+8
12. m+3,m+7,m+13,m+24,m+56 in Z arith;7
13. #k in Z,#e in {0,1,8},n^3=9k+e or-;10,11,12
14. @n in Z,#k in Z,#e in {0,1,8},n^3=9k+e @+;1,13
QED
As you can see, this is much shorter. However, if you compare it to the proof above, you will see that it is exactly the same information, just formatted more compactly by using the table. So you can use table's like this in your proofs to make them shorter, when there are a lot of similar cases of information that needs to be checked. Note that both are still proofs, even thought the second one has a table in it. Its just a convenient shorthand for doing repetitive parts of our proofs. --cuttin proofs down to size ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Thursday, September 20, 2001 12:04 PM > To: monks@epix.net > Subject: assn 4 # 21 > > > Dr. Monks > > ... > n. (~x in A and x in A) or (~ x in A and x in B) (reason) > n+1. ~x in A and x in B (contradiction) > > is the line n+1 legal and if so where can i find the proper reason? > > thx Well, you can prove it by or-: :
n. (~x in A and x in A) or (~ x in A and x in B) (reason)
n+1. Assume ~x in A and x in A -
n+2. -><- -><-+;n+1
n+3. (~ x in A and x in B) -><-=>Q **
n+4. <- -
n+5. Assume ~x in A and x in B -
n+8. <- -
n+9. ~x in A and x in B or-;n,n+1,n+3,
n+4,n+5,n+5,n+8
:
**see the common tautologies list in the proof shortcuts handout However, now that we are starting to become *slightly* less formal, I would allow you to abbreviate this by saying: n. (~x in A and x in A) or (~ x in A and x in B) (reason) n+1. ~x in A and x in B logic;n The official meaning of the reason "logic" is this: If L(n) is the statement on line n, and L(n+1) is the statement on line n+1, then we can use the reason "logic" iff the statement "L(n)=>L(n+1)" is a tautology. So in this example, this would be a valid use of the reason "logic". -logical ------------------------------------------------------------------------ > -----Original Message----- > Sent: Thursday, September 20, 2001 10:38 AM > To: monks@UofS.edu > Subject: Assignment #4 Question #23 > > > Does (4)^(1/2) = +2 and -2? (I don't know how to write plus-minus 2 in > the e-mail) > > --trying to be positive and negative No, 4^(1/2)=2 by definition of rational exponents. x^(1/2) is defined to be the principal second root of x. -rational in principal ------------------------------------------------------------------------ The reason for line #10 in the proof I sent you in the last email message should be Def in;9 instead of Def in;3 --can't type at night -----------------------------------------
Ken Monks - Professor of Mathematics
University of Scranton
Scranton, PA 18510
email: mailto:monks@scranton.edu
web: http://www.scranton.edu/~monks
-----------------------------------------
------------------------------------------------------------------------
Cool... now we have a FAQ answering a question about a previous FAQ... :)
> -----Original Message----- > Sent: Saturday, September 22, 2001 6:44 PM > To: monks@UofS.edu > Subject: Appendix B, # 31,32,35, pg.517 > > > Dr. Monks, > > #31 > In the 1999 FAQ, you state that the Theorem we have to prove for > #31 should be stated: > > "Let f:B->C. Define Im(f)=f(B) and define g:B->Im(f) by g(x)=f(x) for > all x in B. Show g is surjective. > > How do you know to define g:B->Im(f) by g(x)=f(x) for all x in B? How do I know? (oh, no, not this again :)) That is what question #31 is asking. I am not "defining it". Question #31 is "defining it". I am just telling you (in the FAQ's) what the informal English sloppy wording of the textbook's question #31 actually means if you translate it into semi-formal mathematical language. > #32 > That leads to by next question. How do we define f:B->B in #32? You don't. f is an arbitrary function from B to B, i.e. f might be any function whatsoever. You can't pick one, you just have to treat it as an arbitrary function. > Thanks. > > --doesn't want to do more work than has to -answers FAQs with FAQs ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, September 23, 2001 10:28 AM > To: monks@epix.net > Subject: Assignment 6, #8, pg 7. > > Hello, > Is a proof needed for part A or can we work it out like we did the > examples in class? For part a you can just do the calculations. For part c you have to prove your conjecture from part b. Note that you can't get credit by making a ridiculously easy conjecture in part b like "Conj: n^2=n^2". :) -monks=monks ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, September 23, 2001 12:47 PM > To: monks@UofS.edu > Subject: Section 1.1, #5, pg. 6 > > > In the problems in Section 1.1, we have to show them both ways, right? > i.e. for number 5, we have to show that (a^2=3k or a^2=3k+1 for some > integer k) and that (3k=a^2 or 3k+1=a^2). No, you don't have to show equalities "both ways". We are assuming = is commutative and associative and transitive from now on since we proved those things in the previous assignment. If you read the Proof Shortcuts handout on the web site you would see that number in shortcut #IV is says that we will now identify x=y with y=x, i.e. we treat them as if they are the same statement from now on. READ THE HANDOUTS!! --the website-of-wisdom author ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, September 24, 2001 9:12 PM > To: monks@UofS.edu > Subject: Appendix B, #35, p.518 > > > Do we know what (gof)^-1(x) equals? Well, I know that we don't since > I am included in we. However, if we should know or if you know, I would > greatly appreciate you telling me. Thanks. > > - ~(likes inverses) Yes, *we* know what it equals... actually we know lots of things it equals. For example *we* know that (gof)^(-1)(x)=gof^(-1)(x). *We* also know that (gof)^(-1)(x) equals the element that (gof)^(-1) maps x to. *We* also know the definition of inverse function. *We* also know the definition of composition of functions. *We* also know that what it "equals" might not be as interesting as what those definitions are. --*we* ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, September 24, 2001 11:45 PM > To: monks@epix.net > Subject: 1.1, #4 Assignment 6 > > Dr. |Monks|, > > can i say, > > 450. c > 0 for a valid reason > 451. |c|>0 def of abs value, 450 > > if not, how do i get from 450 to 451, because 451 is obviously true > after 450. > > -- ~|sure| Yes, in fact you can use "arithmetic" on line 451 instead of "def of abs value" since that IS an example of a fact about the numbers in N,Z,Q,R, or C. The actual definition of absolute value for real numbers is: { x if x>0
|x|={
{ -x otherwise
But in this case I would accept either your reason or "arithmetic" for a reason for line 451 above. -- |sure| ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, September 26, 2001 10:42 PM > To: monks@epix.net > Subject: pg. 13, #10, sec. 1.1 > > > Dr. Monks, > > if my proof looks like: > > 60. n in Z for valid reason > 61. 1|n [reason] > > what should i use as a reason for line 61? arithmetic? the "1 divides > anything" rule? > thanks :O) Why not insert this line: 60 n in Z for a valid reason 60.5 1*n=n arith 61. 1|n def | ? --1*monks ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, September 26, 2001 11:16 PM > To: monks@UofS.edu > Subject: sec 1.2, #16 p 13 > > > For number 16, I have the whole proof, but now I am stuck on the last > part. In a FAQ, it said, after assuming c|(a/d) and c|(b/d), show that > c<=1... but in the recipe, it says show c|d (which is 1). So could I do > either? You can use either recipe... if you look at the recipe sheet you will see there are two recipes for showing something is a gcd. Either one is fine (because we proved in lecture that they are equivalent). --the recipe chef ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, September 26, 2001 11:25 PM > To: monks@epix.net > Subject: A new recipe for gcd? > > > While perusing the FAQ from '98, I came upon a message of interest. > (subject: more on Thm 1.3) Using the information in that message, would > it be possible to define a new recipe for finding the gcd as follows: > > To show d is the gcd(a,b) > 1. Show d = au + bv for some integers u, v > 2. Show @x, (x = au + bv) => x >= d > > I think that this recipe follows the second part of Thm 1.3 > faithfully. Of course, I wanted to confirm it with you prior to > staking my next homework grade on that fact. > > --trying out new recipes Aha! A fellow chef rises to the task of recipe creation!! Great!!! I like to see that kind of innovation. Now to the specific recipe at hand, it is unfortunately incorrect. Let's see why, then perhaps we can fix it. First, if you show the statement on step #1, then line 2 follows immediately from line 1 the way you have it written, because u,v are global variables in line 2, and hence since au+bv=d from line 1, we would have by substitution that line two is saying: @x, x=d => x>=d which is true in a silly way. Second, the theorem says that d is the smallest POSITIVE integer of the form an+bm, where n,m are integers. But its obviously not the SMALLEST integer of that form, just the smallest POSITIVE one. So if you want to make your recipe non-toxic, you should rewrite it this way: To show d is the gcd(a,b) 1. Show d = au + bv for some integers u, v 2. Show d>0. 3. Show @x in Z,@n in Z,@m in Z, (x = an + bm and x>0) => x >= d or if you prefer a more "expanded" version of this: To show d is the gcd(a,b) 1. Show d = au + bv for some integers u, v 2. Show d>0. 3. Let x,n,m in Z 4. Assume x = an + bm and x>0 5. Show x >= d 6. <- This skill at building recipes from Theorems is a VERY valuable one, and I am very glad to see you are trying to do it yourself. This is a basic proof-skill that we mathematicians need in order to turn our Theorems into useful recipes that we can use in our proofs. Note: If you do use one of the above (correct :)) recipes in your homework, the reason you should give when concluding that d=gcd(a,b) should be "Thm 1.3" or "u,v,gcd Thm" (which is the nickname I gave it in the lecture notes), but not "Def of gcd". This way we give credit where it is due (to the Theorem) instead of the definition of gcd (although it is a minor point to be sure). --detoxing poison recipes ------------------------------------------------------------------------ Based on the questions I am getting asked tonight about problem #16 on page 13, let me reword the question in a way that will avoid the evil fractions that are causing all the confusion. You may prove my reworded version below instead of the version in the book if you wish. #16 (revised) Let a,b,d,k,m in Z. If gcd(a,b)=d and a=dk and b=dm then gcd(k,m)=1. I HIGHLY recommend that you prove this alternate version. It is not mandatory (i.e. I will also accept proofs of the problem as worded in the book). But my experience at teaching modern algebra tells me that you will have a much better chance of getting the revised version above correct that the version in the book. -on a rescue mission ------------------------------------------------------------------------ > -----Original Message----- > Sent: Saturday, September 29, 2001 5:28 PM > To: monks@UofS.edu > Subject: appendix D prob. 2 > > > Dr Monks, > > I was wondering if this is the thm. for prob. 2: > > r~s <=> r - s in Z. NO! This is the definition of the relation ~ The theorem you are trying to prove is: Thm: Let ~ be the relation on Q defined by @r,s in Q, r~s <=>r-s in Z. Then ~ is an equivalence relation i.e. you are trying to prove that ~ is an equivalence relation. You can't prove that r~s <=> r-s in Z because that is the DEFINITION of ~. (Note: we usually use ~ to mean "not" in email, but for this message it refers to the relation defined in the problem under discussion. I was using R for the name of a generic relation in class whereas they are using ~ in Appendix D). > Also is the equivalence relation ~ defined as r - s. No. See above. > If these are true then do i show that r - s in Z is symmetric, reflexive and > transitive and then i would have to show that r - s is in fact in Z. No, you can't show r-s is symmetric or reflexive or transitive because r-s is an INTEGER... NOT a RELATION. You can only show ~ is reflexive, symmetric, transitive, etc. Perhaps reading Example 3* in Appendix D (and the whole Appendix) will help you with this problem. -knows when ~ is not ~ ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, September 30, 2001 1:21 PM > To: monks@UofS.edu > Subject: Math 448 - Appendix D #15 > > > Dr Monks: > In the above referenced problem, we are asked to prove that the > relation on Z defined by a~b iff (a^2) is equivalent to (b^2) (mod 6) is > an equivalence relation. My question to you is what is meant by mod 6? > > -Confused by the language For this problem you can use the following definition: Def: a^2 is equivalent to b^2 (mod 6) <=> 6|(a^2-b^2) ---mod -ern algebra prof ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, September 30, 2001 3:30 PM > To: monks@UofS.edu > Subject: Section 1.3, #11, p.18 > > I'm having a bit of a problem going between you and the book. I > think I get what The Fundamental Theorem of Algebra is saying, but I'm > having problem applying it. In problem #11, it states: > > a=p1^r1*p2^r2*p3^r3***pk^rk > > Could this be written as : > > a=2^r1*3^r2*5^r3***pk^rk ?? > > If it can't, how can I use the FToA in problems like 11 or 14? Well, its "technically" not the same (because the pi's and ri's are given in the problem and you are redefining them to be something else. But it doesn't change the content of the question at all if you assume p1=2, p2=3, p3=5, etc, which is what you are suggesting above, so I will allow it. You can assume that pi represents the ith prime in problem #11. > Also, when it states in Theorem 1.10 that every integer n except 0,1,-1 > is a product of primes, can n be written: > n=p1^r1*p2^r2*p3^r3***pk^rk? Yes, if all the pi's are prime. :) > Thanks. > > --apparently not in his prime --likes prime rib ------------------------------------------------------------------------ > -----Original Message-----
> Sent: Sunday, September 30, 2001 5:59 PM
> To: monks@epix.net
> Subject: def. of prime
>
>
> Dr. Monks,
>
> just a quick question about what i can conclude from:
>
> p is prime
>
> 1. p|b or p|c
>
> and
>
> 2. c e {+-1,+-p}
>
> can we conclude everything to the right of the last implication?
>
> thx
No. If you know p is prime then from the definition you can conclude: 1. ~ p in {0,1,-1}
and 2. @c in Z, c|p => c in {+-1,+-p}
and from the Alt Def of Prim Thm (see Lecture Notes) we can conclude 3. @b,c in Z, p|bc => p|b or p|c But you cannot conclude the things you want to prove without "breaking apart"
my statements #2,3. i.e. if you show c is an integer AND IF YOU SHOW that
c|p, THEN you can conclude that c in {+-1,+-p}. Similarly, if you show that
b,c are integers AND IF YOU SHOW that p|bc, THEN you can conclude that p|b or
p|c. You can't get the "stuff on the right of the implication" without
knowing the "stuff on the left". That is the =>- rule.
--teaches at the primary level ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message-----
> Sent: Sunday, September 30, 2001 8:43 PM
> To: monks@UofS.edu
> Subject: Section 1.3, #2, p.18
>
>
> Dr. Monks
>
> I am having a problem with #2 and I know why. I don't believe it.
> It states:
>
> @p in Z-{1,-1,0},(p is prime <=> (@a in Z,gcd(a,p)=1 or p|a))
>
> What I don't believe is this part:
>
> @p in Z-{1,-1,0},((@a in Z,gcd(a,p)=1 or p|a) => p is prime)
>
> Take a=16 and p=4. gcd(16,4)=4 and 4|16.
> Since 4 divides 16, 4 must be prime, but 4 is not prime. How am I
> screwing this up?
>
> --primetime
You are "screwing it up" because you aren't reading the parentheses correctly. Your "example" is interpreting the statement as saying: @p in Z-{1,-1,0}, @a in Z, ( gcd(a,p)=1 or p|a => p is prime )
which in English says, "if p isn't 0,1, or -1 then FOR ANY NUMBER a, IF either gcd(a,p)=1 or p|a, then p is prime". Clearly your example shows that this statement is false. But it isn't what problem #2 says or implies, or even what you have written above namely @p in Z-{1,-1,0}, ( @a in Z,gcd(a,p)=1 or p|a ) => p is prime
(notice the parentheses) This correct statement says in English: "if p isn't 0,1, or -1 and IF FOR EVERY NUMBER a, either gcd(a,p)=1 or p|a, then p is prime". So in order for you to "prove" p=4 is "prime" from this statement (assuming its a true statement, which it is but you have to prove that of course), you would have to show that for EVERY number a, either gcd(a,4)=1 or 4|a. But clearly if we take a=2, then ~(gcd(2,4)=1 or 4|2). So 4 isn't prime. So it boils down to reading the parentheses correctly. :) --( :) ) ------------------------------------------------------------------------ > How then would I show that pi^ri|p1^r1*p2^r2***pk^rk? I know that > pi^ri*(the rest of the primes)=p1^r1*p2^r2***pk^rk, but I don't know how > I would represent "the rest of the primes." Well, p1^r1*p2^r2*...p[i-1]^r[i-1]*p[i+1]^r[i+1]*...*pk^rk is a standard way to write "the rest of the primes". There is a more formal way to do it using the big Pi (product) symbol, but the above notation is fine. --master of notation ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, October 01, 2001 9:13 PM > To: Ken Monks > Subject: appendix D prob. 2 again > > > Dr. Monks, > your faq states that r~s <=> r - s in Z is the definition of ~. So is > our given that r~s and we must prove that r - s in Z is in fact an > equivalence relation? No, you are given the definition of ~ and you must prove ~ is an equivalence relation. Each mathematical symbol has a type... like creatures in a zoo. Some are amphibians, some are mammals, some are birds, etc. You can't ask "What color are a tiger's wings?" because tigers aren't the kind of animal that even have wings in the first place. Similarly the kind of mathematical creature that can be an equivalence relation is a "relation". You can't prove a number is an equivalence relation, and you can't prove a statement is an equivalence relation. You can only try to prove a relation is an equivalence relation. What is a relation? Well, as we said in lecture and as it says in the notes, a relation on a set X is a subset of XxX. What is XxX? It's the set of ordered pairs whose first and second entries (coordinates) are both elements of the set X. We also said that if R is a relation on a set X, and a,b in X, then we write "aRb" as another notation for the statement "(a,b) in R" (see the notes). So you can't prove that "r - s in Z" is an equivalence relation because "r - s in Z" is a statement, not a relation, and a statement can't be proven to be an equivalence relation anymore than you can prove a tiger's wings are green. In this problem you are given a relation, ~, which is defined as 1. ~ is the set of ordered pairs of rational numbers such that the difference between the coordinates is an integer or in other words: 2. for all rational numbers r,s, r~s if and only if r-s in Z or in even other words 3. ~={(r,s) : r,s in Q and r-s in Z }
So as you can see clearly by the third equivalent definition, ~ is indeed a relation on QxQ, so you can try to show ~ is an equivalence relation, because its the correct "species". -mathematical zoologist ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 03, 2001 2:39 PM > To: monks@epix.net > Subject: assn 10, #5 > > > Dr. Monks, > > if we know that something divides something; > > namely that nk | a-b for some integer k > > can we conlude that (nk)^2 | (a-b)^2? > > thx No. You would have to prove it. -math skeptic ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 03, 2001 6:00 PM > To: monks@epix.net > Subject: assignment 10, prob. 6, pg 29 > > > Dr. Monks, > The problem states: > if a, b are integers such that a=b(mod p) for every positive prime p, > prove that a = b. If i let > a=15 > b=12 > then 15=12(mod 3), where 3 is a positive prime. Yet 15 doesn't equal > 12. Am i reading the question wrong? Yes. The question says that if a=b(mod p) FOR EVERY positive prime p, then a=b. For your example, a=15 and b=12 THERE EXISTS a positive prime p, namely 3, such that a=b(mod p). But THERE EXISTS and FOR EVERY are two completely different things. It ISN'T true that 15=12(mod p) FOR EVERY positive prime p. For example, 15~=12(mod 7). So the question asks you to show that if they are congruent mod p FOR EVERY p (not FOR SOME p) then they are equal. > Don't they HAVE to equal each > other in order to prove a=b? > thanks :) Yes. If they are not equal to each other you can't prove they are equal to each other. :) -- #monks ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 03, 2001 8:01 PM > To: monks@epix.net > Subject: assignment #10, question #22 > > > Dr. Monks, > In question 22, if I have n*k = something, for some K in the integers > and 2k is an element of the integers...can I substute 2k for k? Is > this legel: n. 2k=a^2 - 1 > n+1. 2k in Z > n+2. 2(2k)= a^2 - 1 No. You can't substitute 2k for k unless k=2k, but this only happens if k=0. So in general you can't do it. - ~=2monks ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 03, 2001 8:21 PM > To: monks@epix.net > Subject: General Question > > > In our proofs can we assign a variable to be a certain value, for > example, can we: > > Let k = 1 > > or > > Define k=1 > > Thanks You can Define k=1 anytime you like, as long as k has not already been declared in the proof (i.e. as long as k is a new free variable in the proof, not used on a previous line. You usually can Let k=1 only if it is a "Given", i.e. the first line and given in the problem. Technically you could use this as an abbreviation for: Let k be arbitrary Assume k=1 as long as k is not already declared in the proof, but I doubt this is what you have in mind. -mindreader ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 03, 2001 11:03 PM > To: monks@epix.net > Subject: define k = 2m > > > Dr. Monks, > If we've worked out a proof and we get to something like, > 34. 2k=a^2-1 for some k in Z for some valid reason > 35. 2k in Z def in No, this line is by "arith" from line 34. > 36. define 2f=k - Nope nope nope. You can't do that. You can only DEFINE a variable as a shortcut name for a messy expression. You NEVER EVER need to use a Define in a proof, and a lot of people seem to be looking to it for some kind of mathematical help. But Define is of NO HELP WHATSOEVER. All it does is make you proof easier to read and write. So if you have a big expression like: sqrt(ln(17+sin(45)/3)) and you don't want to keep typing this 50 times in your proof you can give it a name by Define: Define w=sqrt(ln(17+sin(45)/3)) so then later in the proof you can just refer to w instead of that big messy expression. But it DOES NOTHING MATHEMATICALLY. Repeat after me "DEFINE IS MATHEMATICALLY USELESS". You NEVER EVER really need to define anything and it DOESN'T HELP mathematically with the proof. Its just formatting, like changing the font. If you change the font on an essay you typed or change it to boldface or italics it doesn't improve the writing or detract from the content of what you have written, it just makes it look different. That is ALL that DEFINE does. It CAN'T EVER HELP YOU TO GET A PROOF. --thinking about BANNING DEFINE ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 03, 2001 9:26 PM > To: monks@UofS.edu > Subject: Section 2.1, #7, p.29 > > > > Do we know that a^m is odd if a is odd and even if a is even? I > know it, but do we know it for the sake of our proofs? No. > --hopes we know odds who multiply only create other odds We don't. -doesn't know much ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 03, 2001 9:39 PM > To: monks@UofS.edu > Subject: Section 2.1, #6, p.29 > > > If we know that x|a for some reason, we know xk=a. But, k can not > ONLY be equal to zero, right? Thanks. Wrong. 3|0 and 3*k=0 is only true for k=0. > --preparing more questions for the answerer -answering many questions in a row ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 03, 2001 10:13 PM > To: monks@UofS.edu > Subject: Section 2.1, #7 and other stuff, p.29 > > > I'm not sure if my interpretation of the notation is correct or if I > am messing up. > > Theorem 1 from 1.3 states: > Let RCXxX (R subset of X cross X) be an equivalence relation and a,b in > X. Then > [a]=[b] <=> aRb > > For problem #7, given a,m,n in Z, m>0 and n>0, we have to prove that > [a^m]=[a^n] in Z(2). I KNOW we can't just show that = is an equivalence > relation. What I don't completely understand is why. > > Also, it hints that we use Theorem 2.3: > a congruent to c(modn)<=> [a]=[c] > > Could you explain what exactly [a]=[c] means again? It is with respect > to a certain number system I am guessing due to the "modn", but in the > right hand side, how do we know what mod it should be? > > --[confused] In problem #7, the "mod n" is "mod 2" because we are working in Z2. [a]=[c] means the sets [a] and [c] are equal, i.e. that they contain the same elements. Thm 2.3 that you state above is the special case of Theorem 1 from 1.3 that you state above if you take R to be congruence mod n. --teaches a modern (equivalence) class ------------------------------------------------------------------------ > If we know that if n is odd then n^2 is on, could you give me a clue on > how to expand this to n^m is odd for all m in Z+? Thanks. How about the binomial theorem from discrete? Induction would work too, but we didn't learn that yet. :( -running out of signatures tonight ------------------------------------------------------------------------ > -----Original Message----- > Sent: Tuesday, October 09, 2001 11:15 PM > To: monks@UofS.edu > Subject: Math 448: page 39 questions > > > Dr. Monks: > Hello! Hope you had a nice break. I was working on the assignment > for page 39 over break and I ran into some problems...so I have some > questions. First, for #4, is it sufficient to prove that a=b or is > there something else we must also show? I don't understand your question. You are not asked to show a=b, you are asked to show: @n in N, n is composite => #a,b in Zn, a~=0 and b~=0 and ab=0 In other words if you assume n is an arbitrary composite integer, then you must show that there exists two numbers a,b in Zn, neither of which is zero, but whose product is zero. Note that an example doesn't suffice here, in other words you can't prove it by giving some particular values of n, a, and b as an example. You have to prove it for an arbitrary composite natural number. > My second question comes in at > #11. The questions asks that we use exercise 9 to slove the equation. > My trouble is that I do not understand exactly what exercise 9 is > saying. Do you think you could help translate? Well, #9 is pretty clearly explained already in the text... I think you need to read it more carefully and slowly... break it down into tiny bits and digest each one. Given a,b,n they define d to be the gcd(a,n) and they assume d|b. Under that assumption they define the values of u, v, a1, b1, and n1 in part 9(a), and then they use those values to explicitly list all the solutions to [a]x=[b] in 9(b). So #9 is giving you the solution set explicitly. This brings up an interesting point, that I would like to philosophize about briefly. I have noticed that many students judge the difficulty of a problem by its "visual complexity" rather than its "mathematical complexity". If I look at which homework problems most students attempt vs the ones most students avoid, I am surprised to see that they often tackle the hardest problems because the wording of the questions is very simple, yet avoid a very easy to prove question simply because there are lots of symbols in the question. If a question has variables with subscripts, for example, the likelihood that students will not attempt that question seems to increase dramatically! If there are LOTS of symbols and variables or if the question is very long to read, students will often avoid that question even if the question is quite easy. Thus I would encourage you to do the following when facing your homework problems (1) Do not judge the question by how long it is or how many variables are in it or how many subscripts are in it (2) For lengthy questions, don't get intimidated by their appearance... perhaps they are just a long winded way to say something trivial (3) Learn to read mathematics SLOWLY... break it down into tiny parts... read one expression at a time and digest it... formulate the meaning of the question in your mind... be patient... math can't be rushed (4) after you understand the question then decide if it is mathematically difficult or easy. Problem #11 is just such a problem. THE ANSWER TO #11 IS GIVEN IN #9! So mathematically this is a really easy question. #9 explicitly is giving you the solutions to [a]x=[b] in Zn, so its just a matter of reading it carefully and using it to answer #11. > Thank you very much for all your help. Hey, my pleasure! > Sincerely, > Confused again :( --not afraid of subscripts ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 10, 2001 11:40 AM > To: monks@UofS.edu > Subject: P 36 equivalence classes vs. integers > > > Dr. Monks, > I am confused about when things are integers and when they are > meant to be equivalence classes. For example in number 8, we have > x^2 + x = 0. > Are these equivalence classes? How do you know? Yes, they are equivalence classes. The way we know is because of the phrase "in Z5". If we say "3" we mean the integer 3, but if we say "3 in Z5" we mean the equivalence class of 3 in Z5. Similarly, if we say "3=7" that is a false statement about integers, but if we say "3=7 in Z4" that is a true statement about equivalence classes. So the following statements all say the same thing: 3=7 in Z4 3=7(mod 4) [3]=[7] (where the equiv class is wrt congruence mod 4) So keep an eye out for the phrase "in Zn" and you should be ok. > #10 ax = 1 > I'm guessing that a and 1 are equivalence classes but x is > supposed to be an integer? Is this correct? No, they are all equivalence classes. You can't multiply an equivalence class by an integer, so that would tip you off that if a and 1 are equivalence classes, so is x. --classifies math animals ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, October 14, 2001 4:19 PM > To: monks@epix.net > Subject: small ring question > > > Dr. Monks, > Hi. > Living by the recipes, i would of course let a, b, c in R as one of my > first steps in proving something is a ring. You are Wise, young mathematician! > So it's true then to say
>
> 1. let R = {ab=1|a,b in Z} -
Uh, I doubt this is what you want... you don't want R to be a set of
statements, right? The expression, "ab=1", is a statement... its either true
or false. So the way you have this set R defined it consists of all
statements of the form "ab=1" where a,b in Z. i.e. you are defining R to be
a set which contains {1*2=1,3*5=1,1*1=1,-12*135=1,... etc }. I highly doubt
there are any problems in the book that ask about such a set.
> 2. let a, b, c, in R - > (showing closure of addition) > 3. a+b in R property of Z Not with your set R as defined above, because you can't add two equations without defining the addition for two elements of the set first. > (showing closure of multiplication) > 4. ab=1, 1 in R def R But 1 is not in your set R, so you really need to redefine the set R. > But should i be saying "in R" for line 3 and 4? or "in Z" by property > of Z? I think it's "in R" but i just wanted to make sure. Neither. You should define R correctly and then ask the question again. :) > thanks :) You're welcome! -can't give correct reasons for incorrect statements ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, October 14, 2001 8:10 PM > To: monks@UofS.edu > Subject: Math 448: page 52, #18 > > > Dr Monks: > For number 18 on page 52, may we use the fact that Z is a ring, or > must we prove it? Ahh, you have hit an important point right on the head!!! Its not as clearly stated in the book, but if you look in the lecture notes you will see that a ring is a TRIPLE, (R,+,*), not a SET, R. However, as it says in the lecture notes (and as is done in the book), we often abbreviate the ring (R,+,*) by just referring to it as simply "R". Usually this abbreviation is not a problem, because USUALLY once you name the set that you are looking at it is clear what the addition and multiplication operators are on that set. So when we talk about the "ring of integers" or "the ring Z", we mean the SET of integers Z, with the ORDINARY addition and multiplication, i.e. we mean the ring (Z,+,*) where + and * are ordinary addition and multiplication of integers. However this abbreviation of using the name of the set to refer to the whole ring gets us into trouble if the addition and multiplication on the set are something OTHER than what we expect it to be. That is the case in this problem. In other words, if I write (+) for the circled plus then define in the problem, and (*) for the circled dot they define in the problem, then this question is asking you to show that (Z,(+),(*)) is an integral domain (and hence a ring). So when you say "may we use the fact that Z is a ring", the fact that you are referring to is the fact that (Z,+,*) is a ring. We do know that (Z,+,*) is a ring... but that fact doesn't help us here because we are not asked to show (Z,+,*) is a ring... we are asked to show (Z,(+),(*)) is a ring... a fact which certainly we don't know without proving it. So the point is that they are defining a completely different addition and multiplication on the same SET, and you need to follow the recipe to see if indeed that is a ring (and an integral domain). > Using the recipies you gave us, in order to prove > that Z is an integral domain, we must show that it is a commutative > ring, which involves showing that Z is a ring. You are correct! > So can we use the fact that Z is a ring? No.. because of what I explained above. > Thanks :) > -Trying to decipher the recipies You are deciphering the recipes correctly... its the problem you needed to decipher differently. (Uh, oh, he's going to philosophize now...) This makes me realize one thing that is definitely true about trying to teach abstract mathematics, and proofs in particular to new mathematicians... In teaching this for several years now and seeing what repeatedly causes problems for students I am convinced of one thing... One of the main sources of confusion for students is using ambiguous notation or abbreviations. For example, using Z as a shorthand for (Z,+,*) in the above situation is often confusing to students. The Shortcuts and notational abuse that mathematicians and textbook authors regularly use are some of the most common sources for confusion. However, there is only so much we can do about it. As a teacher I try to invent new notation for REALLY bad situations where the notation always causes horrendous problems for students (which is why my notation and my proof format is so different from the text book in many cases). But there is only so much you can change, because, on the other hand, the students need to be familiar wit h standard notation used in mathematics books and articles so they can read them after they leave the course. So until they designate me as "Math Notation Czar" so I can eliminate all the confusing notation and abbreviations in mathematics, here is what I recommend... Read, memorize, and study carefully, all the shortcuts and abbreviations I explain in the Proof Shortcuts handout and also in the Lecture notes (which I usually label either as "Notation" or as "Remark" in the notes). This will help you to understand what is being asked in the problems in the book, since the book uses exceedingly informal wordy English and *lots* of notational shortcuts (and this is a very good book!). It isn't the fault of the book... for an experienced mathematician it is very easy to read our book... but for beginners you need to have your "hands held" a little more until you memorize an learn all the "secret handshakes". That is why I try to expose them in the lecture notes and handouts. There is a lot of material, but if you put that Frito chip on your head for a long enough time, it will eventually soak in. :) -math philosopher ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, October 15, 2001 11:43 AM > To: monks@epix.net > Subject: arithmetic question > > You said that we would have to use the algebra theorems from now on > instead of arithmetic for the reasons in our proofs. Does this apply > to section 3.1, since the algebra theorem is introduced in section 3.2. > > -I sure hope not According to the "Rule of Allowability" link on the course home page, you cannot use any theorems from a section of the book which comes AFTER the section containing the homework problem you are working on. So in particular you cannot refer to ANY of the algebra theorems from section 3.2 in your proofs for problems in section 3.1. The only exception to this rule is that I will allow you to use the Subring Theorem, which is in section 3.2 in the book but which is in section 3.1 of the Lecture Notes (so it replaces Theorem 3.2, which we don't need because the subring thm is stronger). Remember that you still can use the reason "arithmetic" for properties of the specific number systems Z,Q,R (the real numbers, not an arbitrary ring R!), and C if you need to. So for example, in section 3.1 in problems #1,4,5,6,9,10,12, etc you COULD use some statements with the reason "arith" if you needed them because those questions do involve Z,Q,Reals, or C. But in problems like #2,3,7b,8b,c,11,24,25 etc y ou can't use ANY (useful) statements whose reason is "arith" because those questions don't deal with the number systems Z,Q,Reals,C at all. "Arith" refers to arithmetic facts that you learned in elementary school ... certainly nothing about abstract rings! :) -has an abstract ringing in his ear ------------------------------------------------------------------------ > -----Original Message-----
> Sent: Monday, October 15, 2001 9:40 PM
> To: monks@epix.net
> Subject: two more questions
>
>
> Dr. Monks,
> Hi. I have two questions. If i'm trying to define a new
> multiplication, is
>
> 1. let a, b, c in Z -
> 2. Define R={ a(*)b | a(*)b=0 for all a, b in R+} -
> (where R+ denotes positive reals)
> one way to define it?
No. You are confusing several different notations. I think what you meant is: 1. *:ZxZ->Z and @a,b in Z, a(*)b=0 Given > Secondly, would i have to prove a rule of arithmetic that i learned in > calc, such as: > > log b log a > a^ = b^ > where a, b are positive reals? You can use any common properties of logarithms you like with the reason "arith" or "calc" since that is not a topic we learned in this course and it is a property of the real numbers. However, I don't think the above logarithm property is a really common one, but if you say you learned it in calc then that's ok with me. I'm more interested in what you learned in modern. :) - ~your calc teacher ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Tuesday, October 16, 2001 12:10 AM > To: monks@epix.net > Subject: help with 0 sub R > > > dr. monks, > > im having a little trouble with proving that there exists an additive > identity on a set and that it is an element of the set when the > operations > are something other than the normal mult. and addition. could you do an > example without giving away the H.W.? > > thx OK, lets try this: Let R be the set of all subsets of {a,b,c}, i.e. R={{},{},{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}.
Define (+):RxR->R by
@x,y in R, x(+)y = x intersect y (It doesn't matter what we define (*) as since you are asking about 0R). Thm: (R,(+)) has a zero element. Pf:
1. Let R={S:S subset {a,b,c}},(+):RxR->R,
and @x,y in R, x(+)y = x intersect y Given
2. Let x in R
3. Define t={a,b,c}
4. x subset t def R;subst
5. x intersect t=x an easy Lemma (omitted)
6. t intersect x=x an easy Lemma (omitted)
7. x(+)t= x intersect t @-;1 (or def (+);1)
8. = x subst;5
9. = t intersect x subst;6
10. = t(+)x @-;1 (or def (+);1)
11. @x in R, x(+)t=x=t(+)x @+;2,7,8,10
12. #t in R,@x in R, x(+)t=x=t(+)x #+;11
13. 0R=t def 0R;11
QED
I omitted the set theory lemma that we need here so it wouldn't clutter the proof, namely that is A is a subset of B then A intersect B equals A, but that's easy to prove. - hopes you get a 1R, not a 0R ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, October 15, 2001 11:16 PM > To: monks@UofS.edu > Subject: Subrings > > > If we are told that S is a subring of R, do we know that all of the > following are true: > S is a subset of R > 0R is in S > a+b in S > ab in S > -a in S > S is a nonempty set > a-b in S > > --search for subring properties If S is a subring of R then its a subset, and its also a ring in its own right. So all of your properties above hold as long as a,b are elements of S. If they are elements of R then they might not be in S, so then many of your properties above need not hold. So once again, you can't talk about math symbols without saying what they represent. We must classify all expressions in our mathematical zoo. -zookeeper ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, October 15, 2001 7:29 PM > To: monks@UofS.edu > Subject: Section 3.1 ;#22 > > > > Dr. Monks, > A question right quick: > for > > a(+)b=ab a(*)b=a^log b > > are we to treat the operations on the right hand side of the equal sign as > ordinary multiplication and what not as we did in problem 18 of the same > section. > > a fellow hassletoid Yes, the operations on the right hand side of the equals signs are ordinary multiplication, ordinary ^, and ordinary log that you learned in elementary school. -learned about log cabins in elementary school ------------------------------------------------------------------------ Mathletes, There is a MAJOR logic error that many of you made on the homework you handed in yesterday that I want to tell you about so you don't continue to make the same error on the homework that is due tomorrow night. That is that both Theorem 3.2, Theorem 3.6, and the Subring Theorem from the lecture notes ALL are IF-THEN statements, NOT IF-AND-ONLY-IF! Each of these theorems says: "IF BLAH THEN S is a subring". But its logically wrong to try to use these to show that S is NOT a subring. But what many of you tried to say was that if ~BLAH then S is NOT a subring! But that is not logically correct... you can't conclude ~Q from (P=>Q and ~P)! Now it turns out that all three of the above theorems can be reproven so that they ARE indeed <=> statements, in which case you could then use ~BLAH to conclude S is not a subring. But you would have to prove that first since it the theorems in the book and in the lecture notes are only IF-THEN. Moral: You can't show a subset of a ring is NOT a subring by using Thm 3.2, 3.6, or the Subring thm. Its an incorrect logical argument (even though the conclusion will be true). The sad part of this is that 3/4 of the class lost a point on this problem, even though it explicitly tells you that this is mistake and how to avoid it in the 1999 FAQ's (Subject: Problem #6 in Section 3.1 (Page 51)). If you read the FAQ's you can avoid losing all these points. If you don't read them they can't help you. -- ~happy and (+1's)=>happy -----------------------------------------
Ken Monks - Professor of Mathematics
University of Scranton
Scranton, PA 18510
email: mailto:monks@scranton.edu
web: http://www.scranton.edu/~monks
-----------------------------------------
------------------------------------------------------------------------
Note: This is a VERY IMPORTANT EMAIL MESSAGE. If you read it carefully and
learn what it says, you will be blessed with a shower of +1's on future
homework assignments.
> -----Original Message----- > Sent: Wednesday, October 17, 2001 9:35 PM > To: monks@epix.net > Subject: Section 3.3, #5, page 76 > > Dr. Monks, > I'm good with the entire proof, until i have to prove it's surjective. > I looked at the recipe, and i know it would go something like > 21. let y in R(bar) - > 22. f ( ? ) = y ? > Here's my question... > whatever i put into f, i'm going to get an ordered pair out because f > is defined by f:R->R(bar) given by f(a) = (a,0s). So how could i find > something for line 22 that will give me y, and not (y,0s)? > If you define y = (y,0s), then wouldn't you get ((y,0s),0s), which > still isn't y? how can you get y? i'm stuck > thanks :) These are the SAME TWO QUESTIONS that are at the root of at least half of all the mistakes everyone is making, so I will take this opportunity to try to clear them up. Here are the two biggest misunderstandings that most of you are having: THE TWO BIGGEST SOURCES OF CONFUSION CURRENTLY IN OUR COURSE ------------------------------------------------------------ by Ken Monks - Fearless Leader I. Not knowing what KIND OF OBJECT the expressions and symbols you use represent. This is the ZOO CLASSIFICATION I keep talking about, but nobody seems to be understanding. You MUST KNOW WHAT KIND OF OBJECT EVERY SINGLE SYMBOL AND EXPRESSION REPRESENTS! Fact 1: Every mathematical expression and variable has a TYPE which is a set or class of objects that it can represent. It is imperative that you know the TYPE of every object you discuss! Examples of common TYPES: set, function, ordered pair, real number, integer, matrix, polynomial, sequence, ring, binary operation, relation, equivalence relation, statement, n-tuple, complex number, equivalence class, etc Notice that just as an element can be a member of more than one set, an expression can have more than one TYPE. For example, every ordered pair is also a n-tuple, every function is also a relation, every binary operation is also a function, every equivalence class is also a set, every ring is also an ordered triple, every integer is also a real number. So in your example above, since Rbar is a set of ordered pairs (that is ITS TYPE), then if we let y in Rbar, then we know that the y must be a variable WHICH REPRESENTS AN ORDERED PAIR! The TYPE of the variable y is ORDERED PAIR. Just because it doesn't "look" like an ordered pair doesn't mean it isn't one. If I say "Let x be an ordered pair of real numbers" then x is an ordered pair of real numbers, no matter what it looks like. Similarly, if I say, "Let z be a complex number" then z is a complex number, even if it doesn't look like one (i.e. its not a+bi). So a *variable* can be ANY TYPE. Get it? ANY TYPE. Let me repeat that one more time. FACT 2: A VARIABLE CAN REPRESENT ANY TYPE OF MATHEMATICAL OBJECT!!! So your variable y above IS AN ORDERED PAIR! Yes, it is! Really! I'm not kidding! Just like n can be an integer even though it doesn't LOOK like one (e.g. its not 3 or 17 etc). Which brings us to the second most common error that all of you are making in the course: II. Not understanding Shortcut XI on the Proof Shortcuts handout. This shortcut is used A ZILLION TIMES in the homework, so if you don't figure it out soon you ARE DOOMED. Now, everyone go to the Proof Shortcut page right now and read carefully Shortcut XI and in particular be sure to read the example using complex numbers in the same section. Go ahead, I'll wait... tick.. tick.. tick.. OK, did you read that? Hey, who are you kidding? You didn't! Now go and read it for real this time! tick.. tick.. tick.. OK, I see that some of you did read it that time. Now here is what the gist of it is. If you have a set like: { E(a) : P(a) }
where E(x) is an expression containing x that isn't a variable, and P(x) is a statement about x, then that is an abbreviation for: { x : #a, x=E(a) and P(a)}
So in your above example, you have: Rbar={(a,0S) : a in R }
Here E(a) is (a,0S) and P(a) is "a in R". So this REALLY MEANS: Rbar = { x : #a, x=(a,0S) and a in R }
Thus if we have: 21. Let y in Rbar We know that "#a, y=(a,0S) and a in R" (because y is an element of { x : P(x)
} iff P(y)). So we can just skip this step and go directly to the #-:
22. y=(w,0R) for some w in R So as you can see, y is indeed an ordered pair, just as we claimed, and I think you can easily complete the proof of surjectivity using line 22 (assuming w is a new variable in your proof). But the lesson learned here is a VERY important one, because it is the source of MOST of the problems people have when trying to do their proofs. Fix this and earn oodles of +1's! --debunks false proof-ettes ------------------------------------------------------------------------ One easy error to avoid when doing your proofs involving the polynomial ring R[x] or F[x]... DON'T use the variable x in your proof! x is already used as the indeterminate of the polynomials, so use other letters for your variables. In other words don't do something like this: Let x,y in R[x] That is wrong, because x is already declared in the proof as the indeterminate in R[x], i.e. x is the polynomial "x", so you can't declare it as an arbitrary element of R[x]... it is a SPECIFIC element of R[x]. Some of the recipes want you to "Let x in something", so be sure to pick some other variable besides x. -trying to grade your assignments before you do them -----------------------------------------
Ken Monks - Professor of Mathematics
University of Scranton
Scranton, PA 18510
email: mailto:monks@scranton.edu
web: http://www.scranton.edu/~monks
-----------------------------------------
------------------------------------------------------------------------
> -----Original Message----- > Sent: Saturday, October 20, 2001 1:05 PM > To: monks@UofS.edu > Subject: Appendix C pg. 525 prob. 6 > > > Dr Monks, > > Problem 6 says to prove 3 is a factor or (4^n)-1. I was > wondering if that is the same saying 3|(4^n)-1 b/c for 3 to be a factor > it would have to divide it evenly. thanks for your help. > > trying to factor it out Yes, you are correct! -can see the Frito grease glistening on your head ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, October 21, 2001 7:56 PM > To: monks@UofS.edu > Subject: Sec. 4.1, #7,11,16, p.89 > > > Dr. Monks! > > #7 > If we have, c, d in R[x] should we say c= a0 + a1x + a2x^2+...+akx^k > and d = b0 + b1x +...bmx^m for some m,k in N and for some a0,a1,...,ak > in R and for some b0,b1,...,bm in R? Yes. But you can shorten the last bit a little by saying "for some a0,...,ak,b0,...bm in R" > And must we say that m has to be greater than or equal to k (if so, > why)? No. You mustn't. However... You do know this: m>=k or k<=m arith Case 1: Assume m>=k (insert some proof here) <- Case 2: Assume k>=m (insert the same proof here, except that the roles of c and d are interchanged) <- Now, since there is really no mathematical difference between c and d, since they were both selected to be arbitrary elements of R[x], we do know that any proof we do in Case 1, can also be done in Case 2, with the roles of c,d interchanged. So if there is such a statement, whose proof depends only on which is bigger, m or k, we can often get around this duplication (writing the same proof twice, but a common proof trick which I haven not taught you yet... so here is a good place to teach it to you (and those of you who actually read these email messages). The proof trick is called: WLOG This stands for "Without Loss Of Generality What this means is that, since we know that either m is bigger than or equal to k or k is bigger than or equal to m, and we know that we can prove the same thing in either case, except with the symbols reversed, then it doesn't harm anything to assume that, say, m>=k. So an abbreviation for the above proof snippet would be: Let c, d in R[x] c= a0 + a1x + a2x^2+...+akx^k and d = b0 + b1x +...bmx^m for some m,k in N and for some a0,a1,...,ak,b0,b1,...,bm in R WLOG assume m>=k (now do the proof for case1 here) The nice thing about the WLOG assume line as opposed to the Assume line in a proof, is that we don't have to indent the WLOG, because we are saying that even if we didn't assume that m>=k, we would then know that k>=m and we are claiming that the same proof would work in that case, but we are just not writing it. In this senses it is like our abbreviation "Let z in Z" which has a hidden "assume" but doesn't get indented. Moral: While you mustn't, in some cases you could. > #11 > We know that x is not a unit (assuming we are referring to the > indeterminate x in a polynomial). What I am not sure of is how we know > this. Could you help to enlighten me? Yes, I could. But I won't. When you say "We know P but I am not sure of how we know this." what you really mean is "I don't know P.". That is whole point of a proof... to explain why you know something. In mathematics, if a statement isn't proven, i.e. if we don't know WHY its true, then it isn't considered to be known. Consider the Twin Prime conjecture... it says that "there are infinitly many primes whose difference is 2". Do we know this? It seems true enough... here are some examples: [3, 5] [5, 7] [11, 13] [17, 19] [29, 31] [41, 43] [59, 61] [71, 73] [101, 103] [107, 109] [137, 139] [149, 151] [179, 181] [191, 193] [197, 199] [227, 229] [239, 241] [269, 271] [281, 283] [311, 313] [347, 349] [419, 421] [431, 433] [461, 463] [521, 523] [569, 571] [599, 601] [617, 619] [641, 643] [659, 661] [809, 811] [821, 823] [827, 829] [857, 859] [881, 883] [1019, 1021] [1031, 1033] [1049, 1051] [1061, 1063] [1091, 1093] [1151, 1153] [1229, 1231] [1277, 1279] [1289, 1291] [1301, 1303] [1319, 1321] [1427, 1429] [1451, 1453] [1481, 1483] [1487, 1489] [1607, 1609] [1619, 1621] [1667, 1669] [1697, 1699] [1721, 1723] [1787, 1789] [1871, 1873] [1877, 1879] [1931, 1933] [1949, 1951] [1997, 1999] [2027, 2029] [2081, 2083] [2087, 2089] [2111, 2113] [2129, 2131] [2141, 2143] [2237, 2239] [2267, 2269] [2309, 2311] [2339, 2341] [2381, 2383] [2549, 2551] [2591, 2593] [2657, 2659] [2687, 2689] [2711, 2713] [2729, 2731] [2789, 2791] [2801, 2803] [2969, 2971] [2999, 3001] [3119, 3121] [3167, 3169] [3251, 3253] [3257, 3259] [3299, 3301] [3329, 3331] [3359, 3361] [3371, 3373] [3389, 3391] [3461, 3463] [3467, 3469] [3527, 3529] [3539, 3541] [3557, 3559] [3581, 3583] [3671, 3673] [3767, 3769] [3821, 3823] [3851, 3853] [3917, 3919] [3929, 3931] [4001, 4003] [4019, 4021] [4049, 4051] [4091, 4093] [4127, 4129] [4157, 4159] [4217, 4219] [4229, 4231] [4241, 4243] [4259, 4261] [4271, 4273] [4337, 4339] [4421, 4423] [4481, 4483] [4517, 4519] [4547, 4549] [4637, 4639] [4649, 4651] [4721, 4723] [4787, 4789] [4799, 4801] [4931, 4933] [4967, 4969] [5009, 5011] [5021, 5023] [5099, 5101] [5231, 5233] [5279, 5281] [5417, 5419] [5441, 5443] [5477, 5479] [5501, 5503] [5519, 5521] [5639, 5641] [5651, 5653] [5657, 5659] [5741, 5743] [5849, 5851] [5867, 5869] [5879, 5881] [6089, 6091] [6131, 6133] [6197, 6199] [6269, 6271] [6299, 6301] [6359, 6361] [6449, 6451] [6551, 6553] [6569, 6571] [6659, 6661] [6689, 6691] [6701, 6703] [6761, 6763] [6779, 6781] [6791, 6793] [6827, 6829] [6869, 6871] [6947, 6949] [6959, 6961] [7127, 7129] [7211, 7213] [7307, 7309] [7331, 7333] [7349, 7351] [7457, 7459] [7487, 7489] [7547, 7549] [7559, 7561] [7589, 7591] [7757, 7759] [7877, 7879] Yet nobody has proven that there are infinitely many such pairs. So do we "know" this "fact"? No. To know something in mathematics MEANS you can prove it (or at least someone you trust can :)). >- waiting for the Frito grease -hasn't eaten a Frito since 1979 ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, October 22, 2001 9:23 PM > To: monks@UofS.edu > Subject: Sec. 4.1, #7,16,18, p.89 > > > Dr. Monks, > > In your definition of polynomials you use sigma notation. In #7, it > appears that we have to turn (I really don't know how to use sigma > notation through e-mail, but I'll give it a shot) (Sigma a[j]b[i-j] from > j=0 to i) to (Sigma b[j]a[i-j] from j=0 to i). I'm not quite sure how > to do this, so would it be okay if I just used the polynomial notation > in the book. Also, when using Sigma notation, how do we alter it. What > I mean is what reasons would we use for placing a constant inside of the > sigma? (i.e. 1*Sigma (something) = Sigma (1*something)) We didn't define or discuss the use of Sigma notation formally, i.e. I never defined it and we have no recipes for it. You don't have to use it at all and can just use the a0+a1*x+...+an*x^n notation if you prefer. I only mentioned it because you already learned Sigma notation in other courses like Calculus, so if you want to use it in your proofs, that is ok with me, but use it at your own risk. If you want to change the indexing variable in a Sigma expression, you can just go ahead and do it, because the variable is a dummy variable anyway (just say "change of dummy variable" as your reason). But if you want to say something like: 1*Sigma (something) = Sigma (1*something) then something like this is not a property of Sigma at all, but rather the distributive law property of a ring. Remember Sigma is just a fancy way to write +, so if you want to know why 1*Sigma(a_i,i=0..n)=Sigma(1*a_i,i=0..n) then just think of it as: 1*(a0+a1+a2+...+an)=1*a0+1*a1+1*a2+...+1*an and ask yourself why this is true (ans: distributive law). So I would say if you are not comfortable with knowing which properties of Sigma you can use and which you can't, just don't use it at all, since it shouldn't be necessary. But if you want to use it, think of it as a messy + and you should be able to get the correct reasons. And if you want to change the dummy variable go right ahead. --your dummy instructor ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 24, 2001 2:35 PM > To: monks@epix.net > Subject: Section 4.2, #5, p. 94 > > > Dr. Monks, > > Problem #5 asks to use the Euclidean Algorithm to find the gcds. It > doesn't say show or prove, so we find the gcds without using the format > of a proof, correct? :) > > --promises answers will be neat and clear Correct, with one caveat: you have to "show your work", like the examples I did in class. You can't just give me the answer, or only use Maple to printout the answers (note you can use Maple to check your answers if you want to though, however I am not responsible for any mistakes you make if you use Maple incorrectly). -knows calculations are just proofs w/o reasons anyway ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 24, 2001 6:40 PM > To: Dr. Monks > Subject: Sec. 4.2, #3, p.94 > > > Dr. Monks, > > For #3, would this be valid: > > a-b in F Closure of Subtr. in F (assuming a and b are in F) > (a-b)^-1 in F Def. Mult. Inv. only if a-b ~= 0F... > Also, can we use 0F and 0F[x] interchangeably, along with 1F and > 1F[x], yet, or do we still have to go through all of the motions of > wrinting that if something equals one then it equals the other? Yes, you can use them interchangeably, although I was a little anal about it in the lecture notes proofs for a couple of proofs so I emphasized there was a distinction. > Also, if we know that f*k~=0 and f and k are in F[x], do we know > that f~=0F and k~=0F by the Sign Theorem? I'm not sure if I'm > interpreting it correctly. Thanks. No, you know it by the new Secret Corollary to the Sign Thm that I snuck into the Lecture notes recently in the Sign Thm section when I realized that you need to use that fact about ten thousand times in proofs. > --likes teachers who like students -likes students who like Fritos ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 24, 2001 1:30 PM > To: monks@UofS.edu > Subject: Sec. 4.2, #9, p.94 > > > Dr. Monks, > > I was going to ask you this yesterday, but figured it was in the FAQs > and I did not have time to look in them. What exactly is #9 looking > for? > > --frightened by simple questions Its asking you to make the strongest possible mathematical statement you can about f, and then prove it is a theorem (ok, I admit, I added the last bit... :) ). So state your best conclusion as a Theorem and prove it. Weeny Thms won't be accepted even if true. --wants proofs whenever possible ------------------------------------------------------------------------ > -----Original Message----- > Sent: Thursday, October 25, 2001 10:54 AM > To: monks@epix.net > Subject: Polynomials of degree 1 > > > Is this valid in a proof? > > 1. Let a in F[x] Given > . > . > . > n. deg(a) = 0 -Some reason- > n+1. a in F ??? > > If this is legitimate, what reason should I put? Def of F? Def of > deg? Something else? > > --mathematician in the 0th degree That's fine. You can say "def of degree" because if a=b0+b1*x+...+bn*x^n and
deg(a)=0 then the leading coefficient is the coefficient of x^0, i.e.
LC(a)=b0, so that all the other coefficients must be 0F, i.e. a=b0 and b0 in
F by definition of F[x] so a in F by substitution (but we can skip all those
trivial steps and just say its by definition of degree. Since degree isn't
defined for the zero polynomial you can even say something more... that a is
in F-{0F}.
-wonders if degree is in Centigrade or Fahrenheit ------------------------------------------------------------------------ On this problem you need to assume that r,s are distinct roots, or else the problem is false as stated. So consider it given that r is not equal to s. -sending email from class -----------------------------------------
Ken Monks - Professor of Mathematics
University of Scranton
Scranton, PA 18510
email: mailto:monks@scranton.edu
web: http://www.scranton.edu/~monks
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> -----Original Message-----
> Sent: Sunday, October 28, 2001 1:23 PM
> To: monks@epix.net
> Subject: Assignment #19 Section 4.3 #6
>
>
> Hey Dr. Monks I was just wondering if I could use the quadratic equation
> in number 6 to prove that the two roots of x^2 + 1 are imaginary making
> the equation irreducible. Thanks
That is a very good question. The simple answer is NO you can't use the quadratic equation. You should prove it the way they suggest in the problem. There are several reasons for this. First, we can only use arithmetic facts about N,Z,Q,R, and C that we didn't prove or define or study in the course... for example, in chapter 1 we couldn't use the fact that an odd number is of the form 2k+1 for some k in Z without proving it from the division algorithm, even though everyone knows this as a fact of arithmetic. That's because we did study divisibility in the integers, so we need to prove any facts that involve divisibility. Similarly, we did study divisibility and factorization of polynomials (and roots), so we need to prove any facts about factoring polynomials (or roots) that we use. Second, there are some fine points that we would need to investigate to make this argument work. Namely we need to know things like Q[x] is a subset of C[x] and that if a polynomial facts in C[x] as a prod uct of two polynomials in C[x]-Q[x] then there are no associates of the two polynomials which are are in Q[x] so that the factorization can be carried out in Q[x] anyway (for example, the polynomial ix+i is in C[x] and not in Q[x], but is an associate in C[x] of x+1 which is in Q[x]). Plus we didn't even discuss square roots yet in the course, which is an algebraic topic in its own right. So I think it would be easier to prove it the way the book suggests, and for the first reason above you can't use the quadratic formula anyway. Its not a hard proof the way the book suggests. --works on Sundays ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, October 28, 2001 10:21 PM > To: monks@epix.net > Subject: Section 4.4, #8, p. 105 > > > Dr. Monks, > Do we really have to PROVE 8? It just says determine, and i can > determine things without actually proving them... > > -likes an extra hour to do modern You must prove everything. I can simply ask Maple if they are irreducible. But I can only ask a mathematician WHY they are irreducible or not. Remember, a "calculation" is nothing more than a proof that you didn't give any reasons for. Also I will say in general... no matter WHAT a question asks you to do in modern algebra, you have to prove your answer. Every if the question is a Yes or No answer or a True/False question you MUST prove your answer. Sometimes your proof will consist of giving a counterexample, sometimes it will involved doing a calculation with reasons for each step, sometimes it will involve a semi-formal line by line proof like we usually do. However, in every case you should justify your answer. A proof is just a mathematical explanation if what you are doing, and this course is about explaining everything. Maple can tell me if a polynomial is irreducible, but it can't explain WHY. That is your job. -likes to prove with determination ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, October 31, 2001 6:45 PM > To: monks@epix.net > Subject: General Allowability > > > Dr. Monks, > Is it acceptable to refer to an example in the text as a reason which > states completely verbatim what you want to say? > > -dressing up as a polynomial for Halloween If the example PROVES what you are trying to say, then there is no need to recopy the whole thing from the book to the homework, as long as the example is not located in a future section of the book (i.e. beyond the one you are doing the assignment for) and as long as the proof of the statement given in the example does not depend on a the homework problem you are currently working on either directly or indirectly (e.g. by depending on a thm which in turn depends on the problem you are working on), i.e. no circular arguments. In that case an example in the text can be treated like a proven theorem in the text in the Rule of Allowability. -going to test a Hazleton superstitious Halloween legend at midnight tonight ------------------------------------------------------------------------ Just to further clarify what I said below... if the example PROVES the statement you want to use, then you can use it, but if it only makes a statement without proving it, and then the author asks you to prove that statement in the homework problems, then OF COURSE you can't refer to the example... you have to prove it. In that situation it would be similar to the author stating a theorem without proof and then asking you to prove it as a homework problem.. you couldn't refer to that theorem in your proof in that case. --needs the proof to be SOMEWHERE > > -----Original Message----- > > Sent: Wednesday, October 31, 2001 6:45 PM > > To: monks@epix.net > > Subject: General Allowability > > > > > > Dr. Monks, > > Is it acceptable to refer to an example in the text as a reason which > > states completely verbatim what you want to say? > > > > -dressing up as a polynomial for Halloween > > If the example PROVES what you are trying to say, then there is > no need to recopy the whole thing from the book to the homework, > as long as the example is not located in a future section of the > book (i.e. beyond the one you are doing the assignment for) and > as long as the proof of the statement given in the example does > not depend on a the homework problem you are currently working on > either directly or indirectly (e.g. by depending on a thm which > in turn depends on the problem you are working on), i.e. no > circular arguments. In that case an example in the text can be > treated like a proven theorem in the text in the Rule of Allowability. > > -going to test a Hazleton superstitious Halloween legend at > midnight tonight ------------------------------------------------------------------------ > -----Original Message----- > Sent: Thursday, November 01, 2001 12:54 AM > To: monks@UofS.edu > Subject: 5.1 #2,7 > > > First the easy question, which I'm afraid to ask. Do we need to > prove how many distinct classes there are and/or prove the steps > needed to get to that result? Since the question involves counting something (combinatorics) and we don't have recipes for combinatorics in the course, you should just state the number of classes AND explain why that is the correct number, i.e. a very informal proof. You don't need to do a line-numbered proof, but simply giving the answer is not acceptable either. So give an answer and also the explanation of WHY it is the answer. -explains explanations ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, November 04, 2001 2:46 PM > To: monks@UofS.edu > Subject: Math 448 - page 128 #2 > > > Dr Monks: > I have a quick question regarding the calculations in #2. I am > working with the multiplication tables for Z3/(p(x)). I understand how > to figure out the equivalence classes for this, but there is one thing I > am not sure of. When multiplying, if I end up with an (x^2) term, does > this translate to 0 in Z3/(p(x))? Thanks a lot :) > -Having fun with math :) No, since p-0 is divisible by p, p is congruent to zero in Zn/(p), so if, for example, we had p=x^2+2x+1 and n=3, then x^2+2x+2 is equal to zero in Z3[x]/(x^2+2x+2), so x^2=-2x-2 which is equal to x+1 in Z3[x]. So if I wanted to multiply [2x+1][x+2] in Z3[x]/(x^2+2x+2) we would get [2x+1][x+2]=[2x^2+5x+2]
=[2x^2+2x+2]
=[2(x+1)+2x+2]
=[2x+2+2x+2]
=[4x+4]
=[x+1]
So x^2 isn't replaced by zero... its replaced by what its equal to which in my example is x+1... in your problem you can figure out the appropriate thing to substitute. On a related matter everyone should notice that in problem number 4 the multiplication table is... shall we say... a tad bit large. If you want to use Maple to do problem #4, that is ok with me. There is an example of the appropriate Maple code on the course home page which you can modify to do problem number 4. You can check you answer for problem number 2 using Mapel as well if you like, but I want you to do problem number 2 by hand... that means explain how you did the computations in addition to just giving me the table. Number 4 you can do entirely on Maple. --makes some recipes with Maple syrup ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, November 04, 2001 3:27 PM > To: Ken Monks > Subject: Re: [FAQ] RE: Math 448 - page 128 #2 > > > What did you mean in this email when you said that we should explain our > computations for #2? What should we write in addition to the two tables? I mean that you should, in addition to giving the two tables, explain briefly how you computed the entries in the table... give some example hand calculations showing where the entries came from (i.e. how you did them) and explain briefly why your calculation gives the correct value (without doing a semiformal proof, though that is fine too of course :) ). In other words, you can't just say "I did it this way because that's the way you did your calculation in class." because that isn't a math explanation. You should explain what theorems you are using that tell you that the calculation you are doing is correct. I want the explanation because I don't want people to use Maple to solve #2. -doesn't believe his own calculations > Ken Monks wrote: > > > > -----Original Message----- > > > Sent: Sunday, November 04, 2001 2:46 PM > > > To: monks@UofS.edu > > > Subject: Math 448 - page 128 #2 > > > > > > > > > Dr Monks: > > > I have a quick question regarding the calculations in #2. I am > > > working with the multiplication tables for Z3/(p(x)). I > understand how > > > to figure out the equivalence classes for this, but there is > one thing I > > > am not sure of. When multiplying, if I end up with an (x^2) > term, does > > > this translate to 0 in Z3/(p(x))? Thanks a lot :) > > > -Having fun with math :) > > > > No, since p-0 is divisible by p, p is congruent to zero in > Zn/(p), so if, > > for example, we had p=x^2+2x+1 and n=3, then x^2+2x+2 is equal > to zero in > > Z3[x]/(x^2+2x+2), so x^2=-2x-2 which is equal to x+1 in Z3[x]. So if I > > wanted to multiply > > > > [2x+1][x+2] > > > > in Z3[x]/(x^2+2x+2) we would get > > > > [2x+1][x+2]=[2x^2+5x+2] > > =[2x^2+2x+2] > > =[2(x+1)+2x+2] > > =[2x+2+2x+2] > > =[4x+4] > > =[x+1] > > > > So x^2 isn't replaced by zero... its replaced by what its equal > to which in > > my example is x+1... in your problem you can figure out the appropriate > > thing to substitute. > > > > On a related matter everyone should notice that in problem number 4 the > > multiplication table is... shall we say... a tad bit large. If > you want to > > use Maple to do problem #4, that is ok with me. There is an > example of the > > appropriate Maple code on the course home page which you can > modify to do > > problem number 4. You can check you answer for problem number > 2 using Mapel > > as well if you like, but I want you to do problem number 2 by > hand... that > > means explain how you did the computations in addition to just > giving me the > > table. Number 4 you can do entirely on Maple. > > > > --makes some recipes with Maple syrup > ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, November 04, 2001 3:46 PM > To: monks@UofS.edu > Subject: sec 5.2 #6 > > > What form do we need to answer #6 in? If it is a proof, what is the thm > we are tryin to prove (ie- could you translate the question so i can > figure out the first line of the proof!) Sure... proof format leaves no room for error, and this one is easy enough, so here's how I would like you to state it. First, do some calculations to figure out what the formula is for the product [ax+b][cx+d] in the given ring... let's make up an answer for the sake of this discussion (notice THE FOLLOWING FORMULA IS THE WRONG FORMULA, I'M MAKING IT UP FOR THE SAKE OF THE DISCUSSION!). Suppose you calculate that: [ax+b][cx+d]=[(abc+d)x+(2cad+5d^2+a)] (That's not the answer! I made it up!) Then the theorem you want to prove is: Thm: Let s,t in Q[x]/(x^2-2),a,b,c,d in Q, s=[ax+b], and t=[cx+d]. Then [ax+b][cx+d]=[(abc+d)x+(2cad+5d^2+a)]. [NOTE THE ABOVE THM IS WRONG!! ITS JUST AN EXAMPLE! DID I MENTION ITS FALSE?!] Of course to make the theorem correct you have to work out the correct formula for the product. Then you can start your proof the way we always do namely: Pf: 1. Let s,t in Q[x]/(x^2-2),a,b,c,d in Q, s=[ax+b], and t=[cx+d] Given 2. and let the fun begin!!! Yeeeeehaaaaah! But the proof is just if not easier than the calculation because all you have to do is prove that the formula you got is correct. -loves proofs ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, November 04, 2001 11:39 PM > To: Ken Monks > Subject: Re: [FAQ] RE: sec 5.2 #6 > > > Dr. Monks, > > What?!!?? Why wouldn't the thm be: > let R, S in Q[x]/(x^2-2), A, B, C, D in Q, [Ax+B][Cx+D] > then R = AD+BC and S=2AC+BD? > I thought we were supposed to be defining the rules for addition and > multiplication for these congruence classes... aren't the rules that R > = AD+BC and S=2AC+BD? > :) Math humor... that's a sure sign you have been corrupted completely now. You are a true math geek. :) -math geek -------------------------------------------------------------- > First, a quick quest about #6. It says in the directions for this one > "in other words, [rx+s]=[ax+b][cx+d]" , but when i write that line in my > proof, should i write it as a let, or is there a thm that proves it (if > so, could you please tell me which one bc this is the only line im > having trouble with!) You don't need that line at all, in fact you don't have to use r,s in your proof at all... you can just prove it the way I stated it in the previous email message where you give the formula directly. The reason they state the question with the r,s thing is only to explain what form they want the product formula to be in. So you don't need to actually do it in you prof. However, if you wanted to use it in your proof you could do it something like this example (note this isn't the same question): Let V,W in Q[x]/(x^2+x+2)
Q[x]/(x^2+x+2)={[ux+v] : u,v in Q}
V=[ax+b] and W=[cx+d] for some a,b,c,d in Q
VW in Q[x]/(x^2+x+2)
VW=[rx+s] for some r,s in Q
:
then go ahead and derive your formula for the product as usual. Once you get it you can say that [rx+s]=[your formula] and thus rx+s=your formula (why?!). I'll let you figure out the reasons for all the lines above. -unreasonable -------------------------------------------------------------------- > can you show us an ex like #1, where the ring is a > field? ... So now, could you? Lets say Z3/x^3+x^2+3x+1 > - apparantly likes to bombard you with emails First, in your example: Z3/(x^3+x^2+3x+1) its sort of silly to have the 3x term because 3 is zero in Z3, so its the same polynomial as x^3+x^2+1. Second we can't quotient Z3 by a polynomial, we quotient Z3[x] by a polynomial. Z3 only contains 0,1,2, not general polynomials. Third, your polynomial is not irreducible in Z3 because 1 is a root: 1^3+1^2+1=3=0 mod 3. So that means the quotient in your example is not a field. If you want to see an example that IS a field let's do something like this: THM: Z3[x]/(x^3+x^2+2) is a field. Pf:
Define p=x^3+x^2+2 in Z3[x]
Z3={0,1,2}
Let r in Z3
r=0 or r=1 or r=2
(case 1:)
Assume r=0
_ _
p(r)=p(0)
=0^3+0^2+2
=2
~=0
<-
(case 2:)
Assume r=1
_ _
p(r)=p(1)
=1^3+1^2+2
=1
~=0
<-
(case 3:)
Assume r=2
_ _
p(r)=p(2)
=2^3+2^2+2
=2
~=0
<-
@r in Z3, r is not a root of p
deg(p)=3
p is irreducible
Z3[x]/(x^3+x^2+2) is a field.
QED
I'll let you supply the reasons. -still un-reason-able ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, November 11, 2001 2:35 PM > To: Ken Monks > Subject: Sec 7.1 Prob 16 pg. 172 > > > Dr Monks, > > For prob 16 do I have to show the composition for the negatives > of the functions or do I just show the compositions of the functions > given. Thanks. > > Praying for double jeopardy points You can't "show the composition"... you can only "show" a statement and "the composition" is not true or false, therefore it is not a statement. However, if I understand what you are trying to ask, the answer is that the set G in the problem contains ONLY those six functions listed and NO other functions. -dis-functional ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, November 11, 2001 4:57 PM > To: Ken Monks > Subject: Sec 7.1 Prob 16 and 19 pg. 172 > > > Could you give me an example of how you prove something > is closed under some given operation. Thanks > > missing football for modern Sure. Let's prove the operation * in problem #4b is closed. 1. Let G={2^x : x in Q} and @a,b in G, a*b=ab Given
2. Let a,b in G
3. a=2^x and b=2^y for some x,y in Q Def G
4. a*b=ab Def *;2 (or @-;2)
5. =(2^x)(2^y) Substitution
6. =2^(x+y) arith (yes!)
7. x+y in Q arith (Q is closed under +)
8. a*b in G Def G;1,4,6,7
9. G is closed under * Def closed;2,8
There you go... --closed for now ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, November 11, 2001 9:56 PM > To: Ken Monks > Subject: Sec 6.2 prob 6 > > > Dr Monks, > > For prob 6b can you show me an example of how to write out > the addition and multiplication tables. I found the ideals but don't > know how to write up the tables. There is an example of writing out the tables on the bottom of page 146 in the textbook. If you would rather use the [a] notation instead of the a+I notation the book uses, that's fine with me. > Another question I have is will you be able to give me the > reasons for lines 5 and 8 for thereom 1 in the lecture notes for 7.1. Line 5: Def of 0R (or def of ring);1 Line 8: Def of -a;1,3 You can see Section 3.1 in the Fun Facts Lecture notes for the definition of 0R and -a in a ring to confirm these reasons to yourself. > Also can you give me the reasons for thereom 4 in sec 7.1. Thanks for > all your help. You should be able to figure out the ones I omitted yourself, they are easy. I already filled in all the reasons for the hard ones. Is there a particular line or lines you are confused about? > Give me one good reason --def of monks ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Sunday, November 11, 2001 11:47 PM > To: monks@UofS.edu > Subject: Thm 4 - 7.1 > > > Dr. Monks, I am trying to figure out the reasons for Thm 4 in the > lecture notes. I understand the notation but I am not sure about > some of the reasons. Could you tell me if these are correct? > > line 2 - def permutation Yes. > line 3 - def bijection, comp. of functions? Def of composition of functions > lines 5,10,15 - def Sn Yes, because we showed they are bijections from In to In on lines 3,4,8,9,13,14. > line 6 - closure of mult. with composition? No, its "Def of function". We are saying that o is a function from SnxSn to Sn because we showed that for any a,b in Sn, a o b in Sn (on lines 1,5). We had to check this in case, e.g. the compositions of two bijections wasn't defined or wasn't itself a bijection, etc in which case we wouldn't have closure. So o is a function from Sn to Sn, which is just another way to say we have closure. > line 8 - def identity? def of the identity map... see the set theory definitions sheet. > lines 12,17 - #+ Yep. > lines 13 AND 14 - thm B.1 No. Line 13 is by Thm B.1 which says that a function has an inverse function iff it is a bijection. Line 14 says that the inverse of a bijection is a bijection, which is what it says on the bottom of pg 514 (last paragraph). They use Thm B.1 to prove it, so you might say its a corollary to Thm B.1, even though it isn't labled as a corollary in the book. > line 16 - def inverse? Def of inverse function. > I think I understand what these statements are saying but I am not sure that I am using the right reasons. You were mostly correct. --mostly reasonable ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, November 12, 2001 9:42 PM > To: monks@UofS.edu > Subject: p 173 #24 > > > Dr. Monks, > > After we prove that GxH is a group in thm. 7.4, do we have to also prove > if G and H are abelian, then so is GxH and > if G and H are finite, then so is GxH and #(GxH) = #(G)#(H)? > > or are these automatic results (by def. abelian and finite)once > we know GxH is a group? They are NOT automatic results!! They must be proven. The question asks you to prove Thm 7.4, and these statements are part of the theorem, so they must be proven. NOTE: They (the statements about abelian and finite) are not proven in the lecture notes! NOTE #2: You can refer to the fact from discrete math about what the order of a Cartesian product of two sets is in order to prove the latter statement. NOTE #3: The former statement is a piece of cake if you follow the recipe for showing a group is Abelian. --wonders if the GNP is Abelian ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Monday, November 12, 2001 10:15 PM > To: monks@UofS.edu > Subject: page 172 #16 > > > Dr Monks: > For #16 on page 172, it ask whether G is agroup under the operation > of function composition. I know that in order for something to be a > group there must be closure under the function. For this problem, must > we show that f o f, f o g, f o h, etc. are all closed under the function > composition? If this is not a group, would it suffice to say that one > of the compostions is not within the set of the six functions listed? > Thanks for the help! > -Confused Yes. -man of few words ------------------------------------------------------------------------ ------------------------------------------------------------------------ > Do we actually have to write out all of the compositions or can > we just say generally that it would work for all of them? Ha, ha, very funny. You are kidding with me, right? :) How could you possibly know that it will work for all of them if you don't check them all? What if it doesn't work and the pair you don't check is the counterexample? > Would it suffice to say that the composite of two bijective > functions is bijective by thm. B1 in appendix B, conclude that > there is closure under the operation of composition, and show a > few examples? No, no, no! Just because the composition of two bijections is a bijection, doesn't mean that the composition of any two of the six functions listed is another one of the six! You are not proving that the operation of composition is closed on the set of all bijections, you are trying to determine if it is closed on the set containing the six functions listed (and who said they are bijections anyway? Did you prove that all six of them are?). So the whole bijection thing is irrelevant. There is no easy way around it... there are 36 pairs of functions to check and one way or another you have to check them all if you want to prove closure. -can compose music too ------------------------------------------------------------------------ ------------------------------------------------------------------------ > >> -----Original Message----- > >> Sent: Monday, November 12, 2001 9:42 PM > I still don't understand how to prove: > > if G and H are finite, then so is GxH and #(GxH) = #(G)#(H)? > > Dr. Monks wrote: > You can refer to the fact from discrete math about what the order > of a Cartesian product of two sets is. > > What does this mean? Can you refresh my memory. I don't have a > discrete structures book anymore. You are having a really bad day... this is not something you need to have a Discrete Structures textbook for... how many elements there are in GxH if G has n elements and H has m elements? You are not thinking. This is trivially easy. But we don't have any recipes for counting stuff in this course, so for your reason you can just say "combinatorics" or "fact from discrete" for anything that involves counting. BTW, I hope you didn't sell your Discrete Structures textbook. You should never sell a math textbook. You will regret it later on when you want to look something up because the books you learned out of are the ones you are most familiar with. --has all his old textbooks ------------------------------------------------------------------------ > -----Original Message----- > Sent: Tuesday, November 13, 2001 1:01 PM > To: monks@epix.net > Subject: section 7.1, #19 pg. 172 > > > Dr. Monks, > I want to use "def #' as a reason a lot in my proof, but i'm having > trouble defining it. Would i say > 2. # : GxG -> G Given > > or would > 2. let # be a new operation on G such that a#b=b*a Given > be okay? You should say: 1. Let (G,*) be a group and #:GxG->G by @a,b in G, a#b=b*a Given Then you are all set to use either "Def #" or "@-;1" as you reasons when you need to use it. -nks#Mo ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Tuesday, November 13, 2001 9:46 AM > To: "Ken Monks" > Subject: RE: RE: [FAQ] RE: page 172 #16 > > > Dr. Monks, > > Can we use the fact that the composition of functions is > associative by example on pg 509-510 to show that the set Q is > associative? Otherwise I'd have to show all the possibilities > like closure and that would take forever! I'm not even sure how > many there are supposed to be. fogoh, fogoi, fogoj, fogok, fohog, > foiog, fojog, fokog... it seems endless. Maybe I don't understand > how this works. Can you give me a clue? Yes, you can refer to the example on pg 509-510 because it is proven in that example. I used that in line #17 of Thm (6) in Section 7.1 of the lecture notes. -associates with functions ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, November 14, 2001 6:40 PM > To: monks@epix.net > Subject: Order of 1 > > > Dr. Monks, > > If G is a group, a in G and |a|=1, is a=eG? > > thanks :) Of course. |a|=1 => a^1=eG by def of order and a^1=a by def of exponents so by substitution |a|=1 => a=eG. -gives orders ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, November 14, 2001 9:32 PM > To: monks@epix.net > Subject: RE: 7.2 #14b Page 179 > > > Dr. Monks, > > For #14b, can we just make the conjecture based on the orders we got for > 14a as it says in the problem or do we have to prove the conjecture for > all elements and all orders? > > Thanks. Obviously you have to make a non-trivial conjecture AND also prove it! -doesn't give 143 points for conjectures ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Tuesday, November 20, 2001 7:32 AM > To: monks@UofS.edu > Subject: p 190 #46 > > > I am a little confused about the proof for this problem... I showed it > is a subgroup using the same method as the 3rd ex in this section. now > i am trying to prove it is cyclic, but i am confused about the recipe. > am i trying to find any k and a in Z such that b = ak? No, you are trying to find an element a in the subgroup such that for any b in the subgroup there exists a k in Z such that b=a^k. In other words you are trying to find an element, a, such that every element in the subgroup is a power of that element a. So a more detailed recipe for showing a group G is cyclic would be this: To show G is cyclic 1. Find some special element a in G (some elements will work and some will not) 2. Let b in G 3. Show b=a^k for some k in Z The reason I don't have it written this way in the recipe sheet is because we have no recipe command for "Find some special element". So the only way to say it in the recipe sheet is the way I did, but it doesn't break down the steps as much. -has a cyclic sense of humor ------------------------------------------------------------------------ ------------------------------------------------------------------------ > -----Original Message----- > Sent: Wednesday, November 28, 2001 5:36 PM > To: monks@UofS.edu > Subject: Sec. 7.7, #16 > > > Dr. Monks, > > For #16, do we just have to show that Q/Z is the set of finite > elements of R/Z or do we also have to show that Q/Z is a subgroup of > R/Z. The question seems to imply that we know Q/Z is a subgroup of R/Z, > but I do not plan to lose 2 points for a faulty assumption. Thank you! > > --doesn't like to lose 2 points You must show that the set Q/Z is equal to the set of elements of finite order in R/Z AND you must show that Q/Z is a subgroup of R/Z. This latter part is very important in this question because it is not even clear a-priori that Q/Z is even a SUBSET of R/Z, let alone a subgroup. So you must prove all of this. --thinks your point loss estimate is off by a factor of ten ------------------------------------------------------------------------
------------------------------------------------------------------------
> -----Original Message-----
> Sent: Wednesday, November 28, 2001 6:24 PM
> To: Ken Monks
> Subject: Section 7.7 #2
>
>
> Dr Monks,
>
> I was working on prob 2 and I was wondering if I was able to get
> to this point:
>
> Z6/N = {[0],[1],[2]}
> = Z3
>
> Is there any thereom that would tell me that since Z6/N = Z3 that Z6/N
> would be isomorphic to Z3 because I think I remember you saying that two
> things that are isomorphic are in actuallity the same things just
> written differently. If there is no thereom could you tell me if I am on
> the right track and give me a hint on where to go from here.
>
> --clueless in 7.7
OUCH!!! Here is one case where using a consistent notation for equivalence classes is causing confusion. The equivalence classes in Z6/N are with respect to congruence mod N, but the equivalence classes in Z3 are with respect to congruence mod 3... two entirely different relations! So when we abbreviate the equivalence classes in Z3 as just 0,1,2 instead of [0],[1],[2] and when we also abbreviate [x] R by writing simply [x] (see the Notation before Thm 1 in Appendix D of the Lecture Notes), what happens is you end up writing classes in Z6/N in the notation in the form [0], [1], etc which "looks" just like the notation for classes in Z3. But a class in Z6/N is really a class with respect to congruence mod N, and the element inside the brackets is really an element in Z3, which is a class wrt congruence mod 3, so if we didn't use the abbreviations, and assuming your set above is the correct set (I'm not saying if it is or if it isn't) then we would have to write your set this way: Z6/N = {[[0] ] ,[[1] ] ,[[2] ] }
R S R S R S
where R is the congruence mod 3 symbol and S is the congruence mod N symbol (I can't type them in email). On the other hand if we wrote Z3 without using abbreviations, we would have: Z3={[0] ,[1] ,[2] }
R R R
So as you can see, these two sets are HARDLY equal. So it would be wrong to claim that they are. --discusses abbreviations at length ------------------------------------------------------------------------ |
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