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This page was last updated on Monday, August 21, 2000 10:40:19 AM. © 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. Each message is separated from the next by it's subject line. They are in roughly chronological order. I hope they help. -------------------------------------------------------------------- Messages from the Fall 1999 course -------------------------------------------------------------------- ------------------------------------------------------------------- OK, math-ites, a question from the Sunday in-bin... > -----Original Message----- [Note from the Coach: I am not saying whether the student's proof above is correct or not, so if you copy it, you do so at your own risk! :) ] OK, before I read and answer your question,
I need to mention one thing.... you need to put parentheses somewhere in Line
11. You either need to write P=>(~~P=>P) or (P=>~~P)=>P, because the
=> operator is not associative. > In line 11: P=>~~P=>P reflects
the =>- rule of inference which So this would say where the parentheses should go, right? > In line 18: I would like to make
(P=>~~P) my new A and changing my B from Aha! So it turns out that my point about parentheses is exactly the cause of your question. In general you want to know if you can write a statement like: X=>Y=>Z in a proof and then treat this statement as both X=>(Y=>Z) (*) and (X=>Y)=>Z (**) The answer is NO!!! Because => is not associative. To see that the two statements (*) and (**) are not equivalent, notice that if X is false, Y is true, and Z is false, then (*) is true but (**) is false. So they are not equivalent statements and you must use the parentheses here where they are needed. So the answer to your question is NO, you can't do that. > In other words, I AM sure. You can't do it. :) > Thanks for your time Dr. Monks and I
will be awaiting a No, it did help clarify your question and
make it easier for me to follow. Thanks. --usually dials non-associative phone operators ------------------------------------------------------------------- > -----Original Message----- Sounds like a question you would ask your dentist. Two comments: a. You left out the "end assume" which is fine by me, but I want to point it out in case the other students are wondering where it is. The end assumes are not part of the proof, but are just there to help the student know when to stop assuming and unindent. I just added them this year (next year I am adding a soda fountain and snack bar). If you put it in, your proof would look like: > 1. Assume ~(P or Q) b. Your proof is not correct as written. The conclusion in line 5 using the ~- rule would have to use lines 2 and 4, not lines 1 and 4 in order to follow the recipe correctly (because the -><- has to be shown in the SAME fantasy world as the Assume, not in a sub-fantasy world). Thus your conclusion on line 5 would be ~(P or Q), which of course does you no good whatsoever because you already have that on line 1. --inDentist ------------------------------------------------------------------- > -----Original Message----- Yes. If you put it in the non-imaginary
world then it must be true (so you must give a reason why it is, like if it is a
previous theorem or something). In the indented fantasy worlds the statements
you have do not have to be true in the real world, in fact they can be false
(and often are, as in proof by contradiction). > Question 2: I would accept either one. They are both
fine. For our purposes, -><- is just a convenient abbreviation for (Q and
~Q). This is not the or- rule. Check the recipe sheet. Or- is: ; or- ; or- This is the Or- rule I usually use. It is often called "proof by cases" in textbooks. It says that if we know "P or Q" and Case 1: If P is true then R and Case 2: If Q is true then R, then we can conclude that in every case R is true, so we can conclude R. But there is NOT an or- rule that says: To show P This is wrong! "P or Q" does not
imply that P is true. It also does not imply that Q is true. a. Let's use our email symbols consistently so everyone is reading the same thing. Look at the recipe sheet for a list of email symbols. In particular, => is implies and -><- is contradiction. b. You can't use =>+ on line three because you didn't show ~P in the same world as the Assume P in line 1. So you haven't shown ~P, you have just pretended it as far as Line 1's world is concerned. So this is not right. --fellow math-educator ------------------------------------------------------------------- Reminder: Please identify your question on the subject line. For example, you could say Subject: Clarifying or+ rule for this message instead of just
"Modern Algebra". This will make it easier for future students to find
answers in the FAQ list in the future. There is nothing to clarify. It is simply true, not too good to be true. If you know P then you know "P or Q" where Q can be anything whatsoever, including ~P if you like. To see why that is consider a truth table for or: "True or True" is True Thus in "P or Q" if P is true then so is the whole statement "P or Q", regardless of whether or not Q is true. --Monks or ~Monks
> -----Original Message----- Yes. > If so, can we insert it under any Yes. > Also, if the answer is yes, can we do that with proofs we already proved? I think you mean "with theorems we
already proved". In that case the answer is: yes. I hope so too! I am always open to suggestions for improving the course and the use of technology in the course as well, so feel free to pass along great ideas that I can then take credit for. :) --techno-prof ------------------------------------------------------------------- Thm: (@x,x is bloopy)=>(#y,y is bloopy) Pf: Note: Line 2 is not really used in the proof, but it is a good practice to declare all the variables we use in a proof so we know what each variable represents, so you should do this as well. --m@nks ------------------------------------------------------------------- You should think of a statement like @x,@y,P(x,y) as being of the form: @x,(@y,P(x,y)) We don't write the extra parentheses for an abbreviation. Thus this statement is of the form: @x, Q where Q is the statement: @y,P(x,y) So you handle the quantifiers one at a time. > Also, can you let an order pair be
arbitrary? For example use P(m,n) No, only one variable at a time in this particular situation. --arbitrator or arbitrariness ------------------------------------------------------------------- Please don't identify your modern questions in the subject line by "Assignment n, #x" because it is hard for the FAQ reader to know which Assignment number a particular problem belongs to, since the assignment numbers are not always the same from year to year, and even if they are the user has to first look up the assignment to see what page the problem is on. Instead use "Section n, #x, page pp" to identify the section and page in the book where the problem is found. Thanks. Now, on with the show... > -----Original Message----- No, f(x,y) is an abbreviation for f((x,y)). They drop the extra parentheses as an abbreviation because it looks kooky with the double parentheses. So the correct notation in the book should be: f((x,y))=(y,x). --((monks)) ------------------------------------------------------------------- > -----Original Message----- No. You are confusing the x and y in the definition of injective with the x and y in the definition of the function f in this problem. What you are given is: f:BxC->CxB and @x in B,@y in C,f(x,y)=(y,x) or equivalently f:BxC->CxB and @z in BxC,z=(x,y) => f(x,y)=(y,x) To show f is injective you want to follow the recipe, i.e. for any z,w in BxC, if f(z)=f(w) then z=w. But z and w are ordered pairs. So to show they are equal you must show their corresponding components are equal. Read the FAQ's. Look under the message entitled "Top Ten Bad Things Commonly Found in Monday's Proofs". There are lots of tips and hints there about this homework set. --FAQ finder ------------------------------------------------------------------- Hey, nice subject line!! Future modern students will thank you! > -----Original Message----- The definitions are explained on page 507, and are also in the set theory handout on the web page and I went over them in class. So I won't restate the definitions. Let's do an example instead. I will use U for the union symbol and I for the intersection symbol here Ex: Simplify U (n^2,infinity) Well, since Z = {...,-3,-2,-1,0,1,2,3,...} we can think of this as U (n^2,inf) =
...U((-2)^2,inf)U((-1)^2,inf)U(0^2,inf)U(1^2,inf)U(2^2,inf)... So the "intuition" is that you substitute every element n of Z into (n^2,inf) one at a time and then take the union of all those sets. On the other hand: I (n^2,inf) would equal the empty set, because for every real number x, there exists an integer N such that N^2>x (for example, take n=ceiling(x)) so that x would not be in (N^2,inf) and hence it is not in the intersection of all those sets. Thus, indexed unions and intersections are similar to sigma notation for series... if I write: SIGMA (1/2)^n then you know from calculus that means = 1/2 + (1/2)^2 + (1/2)^3 + (1/2)^4 + ... The idea is the same, except you are taking unions and intersections instead of sums. --sum teacher ------------------------------------------------------------------- You must prove Corollary 1.2. If it says q and r are unique, then you have to show that. If it says other things you must prove them too. Whatever it says, that is what you have to prove. If it doesn't say something, then you don't have to prove that thing. If it tells you to assume that c<>0 then you can assume that. If it doesn't tell you to assume that a>0 then you can't assume that. So first read the Corollary, then prove exactly what it says. --states the obvious ------------------------------------------------------------------- Thm: Let i:Z->Z and @x in Z, i(x)=x. Then i is surjective. Pf. Here is another one. Thm: Let f:RxR->R, and @x in R,@y in R, f(x,y)=x. Then f is surjective. Pf: -onto something ------------------------------------------------------------------- > how do you write g o f? This doesn't look formal (C-->D(B-->C)). To say: "g o f maps B to D" you simply write: g o f : B -> D I think that is the answer to what you are asking. > or would it be Thm. B-->D is injective ==> B-->C is injective ? No because B->D is not a defined expression. We defined: "f:B->C" to mean "f is a function from set B to set D" so f:B->C is a statement, but B->C doesn't mean anything at all by itself. It is like typing a square root symbol with nothing inside it... it doesn't mean anything because it is only a part of an expression. Notice that we also defined: f to mean the same thing as f:B->C. This notation is useful in some situations that will arise later on. Sometimes in mathematics as a shorthand we treat f:B->C as a noun as well as a statement, for example if I say: "f:R->R is differentiable" then that is a shorthand for "f:R->R and f is
differentiable". > Second Question: Number 31, page 517 No, f:B-->C is not always the same function as f:B-->Im(f) (and it is poor notation that they use the letter f for both functions) because they can have different codomains. Two functions are equal (i.e. the same) if and only if they have the same domain, the same codomain, and they map every element in the domain to the same element in the codomain. So they shouldn't use the same letter f for both functions above. A better statement of this question might be this: "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." By using different letters for the functions f and g it should be clear that g is always surjective while f may or may not be. This is another example of a mathematical
shorthand... we sometimes use the same letter for two different functions which
have different codomains as long as the two functions agree on all values of
their common domain. See the recipe sheet for a note to that effect. Well, this is certainly a true statement, but it is certainly not what they are asking you to prove... they are asking you to show that f:B->Im(f) (i.e. g) is surjective regardless of whether or not f:B->C is. --math translator ------------------------------------------------------------------- No. Because if this were legal you could then say this on line 4... 4. @x,y in B, x=y @+;2,3 which says that the set B has only one element. --math lawyer ------------------------------------------------------------------- Sure. Example: Let R be the set of real numbers and f:R->R by f(x)=x^2. Then Im(f)=[0..infinity). Pf: Note that I don't like the notation, Im(f), If f:A->B then I prefer to call it f(A) than Im(f). Also keep in mind that Im(f) is just another name for the range of the function f. --electronic Im(Monks)
------------------------------------------------------------------- > -----Original Message----- Right. --m0nks ------------------------------------------------------------------- > -----Original Message----- This is a true statement, but isn't what you mean to say. > 5. 1|n
;Def of Divides;2,4 See above. > 9. 1|(n+1)
;Def of Divides;2,8 You can write it... you just won't get any credit for it. > But isn't that the point of the whole proof? Yes. > What am I missing? The end of the proof. Well, when you are stuck, just try what comes naturally... Look at line 11. You have that as an assumption. What CAN you conclude from that? Just see what you CAN conclude and see where it leads you. If you are lucky it will lead you to what you are after. "You can't get always get what you
want, --Mick Monks ------------------------------------------------------------------- Don't use / for "divides". Use |
for "divides". / means "divided by" which is totally
different. For example 6|2 is false, whereas 6/2 is 3. Yes. > b. Is line 4 acceptable or does it
have to be done in a more piecewise The detail level is acceptable, but the
conclusion on line 4 is not correct. 3. Let s,t in Z be constants such that since one of the steps you are skipping is a #-. In your proof s,t are arbitrary and so you don't know that as=b for since all three variables in that equation are arbitrary integers in the proof. An alternative way to say it would be: 3. #s,t in Z, as=b and ct=d or 3. as=b and ct=d for some integers s,t instead of lines 3 and 4 in your pre-proof, and then skip the #- in the manner I described in lecture. I prefer this last one personally. --heartbreaker ------------------------------------------------------------------- > -----Original Message----- The rules for what you can use in your proof and what you can't are somewhat complicated, but mostly intuitive. I think there is a FAQ from one of the previous years in which I spell it all out in detail, so I won't try to type it all again. In this particular case, since the proof of the fact you want to use is (a) in the book and (b) occurs BEFORE problem #10 in Section 1.2, you may use it in your proof if you wish. --The Referee ------------------------------------------------------------------- > -----Original Message----- I would put a couple more steps like this: 1 Let a,b,d in Z Then continue from there. This way you have proven that a/d and b/d are integers, which is a MAJOR concern. So you should prove this as I have shown. Anytime in the proof that you say "we already know" or "because we know that" it is usually a good sign that you probably don't, and the BS meter should go off! :) --helping you get your BS degree ------------------------------------------------------------------- No. How can you justify this last step? The definition of divides is that c|1 iff ck=1 for some integer k. Where have you shown this? In two lines above you have shown that cj(d/a)=1 but you didn't show that j(d/a) is an integer. So no, this isn't legal. --math lawyer ------------------------------------------------------------------- No. You can't just "let rd/b be in Z". You might as well replace this line with: . I really wish that rd/b was in Z since that is what you really mean. --wishful thinker ------------------------------------------------------------------- > -----Original Message----- The deal is that in the recipe you must let c be an ARBITRARY integer, not one particular integer. So in your line 4, you are not following the recipe, you are just saying that 3 is an integer which is not the same as saying: Let c be arbitrary. or its abbreviation: Let c be an integer. So when we say "Let c be an integer" in a proof we are saying effectively "Let c be an ARBITRARY, UNSPECIFIED integer, so that whatever I prove about c will be true about ANY integer." But by saying that 3 is an integer and then proving 3|6 you are only showing that 3 is an integer which divides 12 and 24 and 3|6, not that ANY integer which divides 12 and 24 also divides 6. So you would have to show that ANY integer c which divides 12 and 24 also divides 6 in order to conclude that 6 is the gcd. But 12 divides both 12 and 24, but 12 does not divide 6, so 6 is NOT the gcd. --divisor advisor ------------------------------------------------------------------- > -----Original Message----- I wouldn't. --hates fractions ------------------------------------------------------------------- > -----Original Message----- This isn't really by "arith", its a Lemma that follows from "arith": Lemma: @n in Z, 1|n So there you go, now everyone can just say 1 divides whatever integer you like and the reason is the above Lemma. > 6) Let c be an element of Z No. In order to use the corollary to conclude that c|1 on line 8, you would have to already know that 1=gcd(a,p). But you don't know that. So you can't apply the Corollary to conclude c|1. In addition, I suspect that what you are trying to do here is prove that gcd(a,p)=1 using only the hypotheses on lines 1 and 2. But if that is your intention then your attempt is doomed to fail, because you can't possibly conclude that gcd(a,p)=1 from the mere fact that p is prime and a is an integer as can be seen by the example p=2 and a=4. So there is a helpful tip. --big tipper ------------------------------------------------------------------- > -----Original Message----- You can certainly SHOW that 1|p and 1|a, but you don't need the definition of gcd to do it. I just proved that 1 divides any integer in a Lemma in the previous email message! So if you want to say that 1|p and 1|a from now on, be my guest. You can also say 1|3 and 1|q+w (if q,w in Z) if you are in the mood. :) > Is OK to let 1 be your d (acc to
recipe) and show it like above intead of bzzt... bfft... I .. cannot ... understand ... this .. question ... > Can you let 1 be your integer d (acc to the recipe) and let 1=1? You don't have to "let 1=1", you KNOW 1=1 by the reflexive axiom for =. Let me see if I can guess what you are trying to ask here, but correct me if I am wrong. You are roughly asking if you can use the "recipe" that shows d=gcd(a,b) "backwards". The answer is yes in the following sense. The definition of gcd(a,b) (for a,b not both zero) is: gcd(a,b)=d <=> d|a and d|b and @c in Z, c|a and c|b => c<=d So this gives us two facts: I. gcd(a,b)=d => d|a and d|b and @c in Z, c|a and c|b => c<=d and II. d|a and d|b and (@c in Z, c|a and c|b
=> c<=d) => gcd(a,b)=d Remember that we can make LOTS of recipes from such statements. One such recipe uses Fact II to show gcd(a,b)=d To show gcd(a,b)=d (try to give reasons for each of the steps in the recipe to convince yourself why it is justified by FACT II.) BUT we can also make a recipe using FACT I. To show d|a and d|b and (@c in Z, c|a and
c|b => c<=d) and if we combine this with and- we could make several recipes out of this: To show d|a To show d|b To show @c in Z, c|a and c|b => c<=d But we can "break down" this last recipe even further if we wish: To show c<=d So as you can see, there are MANY recipes that can be obtained from a single definition. Thus the important thing is to understand the definitions, and then you can make your own proof recipes on the fly. But as you can see from the last few recipes above, yes, you can conclude lots of stuff by knowning that the gcd(a,b)=d. > Also, would you mind doing an example
of prime because there are only Thm: p is prime => -p is prime. Pf. --still in his prime ------------------------------------------------------------------- > -----Original Message----- Oh, I see. You want to know if d in the recipe for showing p is prime can be chosen to be some specific number, like 1? The answer is NO. The definition of prime is: p in Z-{0,1,-1} is prime <=> @d in Z, d|p => d in {1,-1,p,-p} So this gives us the recipe above (if you assume that p is not 0 or 1 or -1 in the recipe). Thus the d you Let on line 1 in the recipe is an ARBITRARY integer d. So you aren't following the recipe if you replace d by 1, because you can't "Let 1 be an arbitrary integer". This you should read line 1 in the recipe as saying "Let d be an arbitrary integer." > For the second step, where you are to
assume d|p, can you show with reason, No, because if you let d be an arbitrary integer, and then you SHOWED that d|p, then you could conclude that EVERY integer divides p by @+ rule. But 0 does not divide any integer, so this can't be. Thus you cannot show d|p for an arbitrary integer d. > And, given the gcd info for the third
step of the recipe, can you say 1=1 No because you can't use the recipe if you substitute 1 for d. You must leave d as a variable and let it be an arbitrary one. If the d in the recipe is causing you to confuse it with the variable d in the recipe for gcd, then you should change all the d's to c's or some other variable. The d in the recipe for showing p is prime has nothing at all to do with the d in the recipe for gcd, they are just variable names. --d prof ------------------------------------------------------------------- > -----Original Message----- Every example IS a proof if you explain it. For example if I want to show that x^2=3 mod 6 has a solution then if I say: ---------------------------------- 3^2=9 So x^2=3 has a solution, namely 3. Then that is just an informal way of doing the (informal) proof: ------------------------------------------------- Notice that the "Example" and the "Pf" both say exactly the same thing, it is only a matter of formatting. The proof is a little bit better explained, while the example is shorter and less formal. For problem #12 on page 29, I will accept either format. Comment: you can prove that a statement has a solution by giving a single example of such a solution. But if you are going to claim that a statement has NO solution then you have to prove it. But proving that an equation has no solution in Zn is easy because THERE ARE ONLY n NUMBERS in Zn! Thus to show a statement has no solution in Zn all you have to do is SUBSTITUTE EVERY NUMBER and show that none of them make the statement true. Gotta love finite number systems! --likes Z2 best ------------------------------------------------------------------- > -----Original Message----- It does, but only if you skip quite a few steps (too many for this point in the course). I assume your goal is to prove that "a is even" if you know "2|(a-2)". But that is easy: n. 2|(a-2) Right? --getting even ------------------------------------------------------------------- > -----Original Message----- No, if I understand what you are asking. There are two problems in what you have above. 1. If you know that n^2|blah^2 then you know that n^2k=blah^2 for some k in Z by def of divides, but you DON'T know that n^2k^2=blah^2 for some k in Z by definition of divides. 2. If you know blah^2=k^2 n^2 then you
can't conclude that blah=kn because blah might also be -km ------------------------------------------------------------------- > -----Original Message----- On the right. -right on ------------------------------------------------------------------- No. --knows 4|6^2 but ~(4|6) ------------------------------------------------------------------- > -----Original Message----- Right. This is the abbreviation I mentioned in lecture. It is also mentioned on the top of page 35. - [monks] ------------------------------------------------------------------- > -----Original Message----- Yes, that is the whole idea behind the #+ rule. If you show it is true for something, then there exists something for which it is true. Sounds almost humorous when we say it that way. :) > For example in Z4 2*2=0, therefore there exists a and b not equal to 0, but a*b=0. This is fine as far as proving the theorem when n=4. However, the question is asking you to prove it for an arbitrary composite integer n, not specifically n=4. --pondering the meaning of existence ------------------------------------------------------------------- It doesn't help at all. A student sent me an email asking if it was positive, and I replied in the FAQ that, yes, it is positive. I was just answering his question, not giving a hint. > Can we solve #11 without using #9 since we don't have to do #9 anyway? No, the whole point of the problem is to teach you the facts that are being explained in #9. While you can certainly solve these two equations just by trying all 18 numbers in Z18 (for part a) and all 65 numbers in Z65 (for part b) that is sort of inefficient, because #9 tells you exactly what the solutions are for ANY n. What if there was a part (c) to the problem that asks you to solve 23x=54 in Z1000000? Would you want to "try" every one of the one million numbers one by one? Or would you rather just use the answer that is give in problem #9. I didn't ask you to prove problem number 9 but it important to understand what it says. That is what problem 11 is getting you to do. So you must use the method of number 9 to solve number 11. Of course, you CAN prove number 9 too, if you feel guilty about using a fact that you didn't prove to solve a problem. :) --answers what he is asked ------------------------------------------------------------------- > -----Original Message----- Yes. So is 1. --likes short answers ------------------------------------------------------------------- > -----Original Message----- This question can be interpreted and answered on many different levels. Let's see: 1. What does DISPROVE mean? Well, "disprove P" means "prove ~P". So when they ask "Prove or disprove P" you can interpret that as "Prove P or Prove ~P, whichever you prefer.". Thus this is just one more way to ask you to do a proof in disguise. 2. There is a special case of this which comes up quite often, namely when you try to disprove a FORALL statement. To disprove @x,P(x), is by our convention in #1 above, to prove ~@x,P(x). But by DeMorgan's law ~@x,P(x) is logically equivalent to #x,~P(x). We know by the #+ rule that to prove #x,~P(x) it suffices to find one value c for which ~P(c) is true. This value c is called a COUNTEREXAMPLE to the statement @x,P(x). Thus to disprove @x,P(x), it suffices to find a single counterexample. 3. Of course, if the statement @x,P(x) is true, then it will not be possible to find a counterexample in which case a proof of @x,P(x) is what is being asked for. 4. The wording of your question implies that you think the statement is false. Perhaps others in your class believe it is true. If you think it is false then you should try to disprove it as described above (i.e. give a counterexample). If some other student thinks it is true then they should try to prove it. So when you say "we already know is false" in your question above you really mean "I am under the impression it is false". If you are under that impression, then you should try to find a counterexample. 5. One could ask a similar question for ANY homework question in the book. If you read the question and immediately know if it is true or false then "why should it ask you to prove something you already know?". Well, because it is one thing to know it and another thing to prove that you know it by proving it. --monks or ~monks ------------------------------------------------------------------- > -----Original Message----- So far so good. > 2. Let (Z, +, *) be a ring and S is a subset of Z Not so good. First, you can't "let (Z,+,*) be a ring" because (Z,+,*) IS a ring by the first example on page 43. Second, you don't want S to be any old subset of Z, you want it to be a specific subset, namely the set of all multiples of k. > Is there another example of proving a subring that you can give us? Thanks! You can't "prove a subring"... you mean "...proving that a subset is a subring". Sure. Thm: Z is a subring of Q. [Recall that we are now abbreviating (R,+,*) as just R and that for Z, Q, Reals, and C, the + and * are assumed to be the usual ones unless otherwise specified.] Pf: Or how about: Thm: {0,3} is a subring of Z6 Pf: and *|0 3 in Z6 by def of + and * in Z6 2. {0,3} is closed under + and *
def of
closed;1 Note that this is the same example as the one given on page 50, but I have explained it a little more carefully in the format we have been using. It would be beneficial for you to compare the "English" proof in the book with the slightly more formal proof above. --trying to set a good example ------------------------------------------------------------------- > -----Original Message----- Yes. 0+x=x for any x in the ring by
definition of the 0 element, so in particular if x is 0 then 0+0=0. Right, IF the answer above were no, then you would not have closure for addition and so you couldn't prove 8b and 8c. But the answer is yes, and indeed you can prove 8b and 8c. -making a closed call ------------------------------------------------------------------- > -----Original Message----- If you want to show that a subset S of a ring R is a subring, you can use Thm 3.2. But you cannot use Thm 3.2 to show that a subset S of a ring R is NOT a subring. To show a subset is not a subring, it suffices to show that it does not satisfy at least one of the 8 properties that define a ring, and closure of multiplication (which you mentioned in your example above) is one of those 8 properties (i.e. multiplication must be abinary operator on S). So it a subset S of a ring R is not closed under multiplication (again using your example above, not referring to problem #6) then S is not a subring by definition of subring, not by Thm 3.2. --running rings around you ------------------------------------------------------------------- > -----Original Message----- Well the set S is an infinite set, but the set you have listed here has only n elements and is therefore finite. I would have said: S={x : x=mk for some m in Z } > 1. Let k be a fixed integer in Z This is true, but not relevant. In fact a stronger statement is true, namely 2. @m in Z, mk is a multiple of k definition of multiple What you want to say is: 1. Let k in Z --km ------------------------------------------------------------------- Q: When can you use the reason "arithmetic" for a statement in your proof? A: ONLY if the statement is an elementary property of the integers Z (and sometimes Q, R, and C) that we did not prove in the course. For example, facts such as the fact that Z is a ring, or that 3+4=7 and so on. We allow this as a reason because we did not define the rings Z,Q,R, and C in this course from axioms... our book assumes we know what they are and are familiar with their properties from previous courses. So we give the reason "arithmetic" to describe facts about the integers, rational numbers, reals, and complex numbers that we know from previous courses (not counting new facts about them that we prove in this course). However, now that we have moved beyond these familiar rings of arithmetic, we have defined every other ring we use from the ground up, Zn, Mn(F), and now generic rings that we usually call R. These are defined in our course and therefore "arithmetic" is NEVER the reason for a statement involving elements of these rings. So if you are talking about anything other than normal everyday numbers you can't refer to "arithmetic" as a reason for a line. You need to invoke some property of the ring in question to do what you want. --arithmetic teacher ------------------------------------------------------------------- > -----Original Message----- If you mean by this question that you have 1. Let ... Then the reason for line n+2 is just substitution (and reflexive of =) using lines n,n+1 to substitute into the statement n+3/2. f(a)+f(b)=f(a)+f(b) which is true by reflexive of =. But you don't have to write the extra line n+3/2 at this point in the course, nor do you have to do the #- rule on lines n and n+1 (which is technically needed to remove the "for some c in T" part before using the equations to do the substitution). So if the above three lines were in a proof you only have to say "substitution;n,n+1" as the reason for line n+2. If that is not what you are asking in your question, then I can't figure out what you are asking from the wording of the question and you will have to be more specific. --only partially telepathic ------------------------------------------------------------------- > -----Original Message----- See Page 72, Thm 3.12, part (1). --maps some homework problems to +0 ------------------------------------------------------------------- > -----Original Message----- You do indeed understand correctly. > What happens when after doing the
Euclidean Algorithm you Since we only defined divisibility and gcd for polynomials in F[x] where F is a field, every nonzero element of F is a unit. Suppose you get, say, ax^2+bx+c from the Euclidean algorithm and want to make it monic, then since a is nonzero, a^(-1) exits in the ring F (NOTE: a^(-1).. not 1/a !!). Since multiplying a polynomial by a unit does not change its degree or change which polynomials it divides, you can multiply your polynomial by a^(-1) which will give you the monic polynomial: x^2+a^(-1)bx+a^(-1)c. For example, if we want the monic associate of 5x+4 in Z_11[x] we just need to find the element of Z_11 that is 5^(-1). Well, 9*5=1 in Z_11, so that means the inverse of 5 in Z_11 is 9. So we multiply our polynomial by 9 to get: 9(5x+4)=45x+36=x+3 in Z_11[x] So the unique monic associate of 5x+4 in Z_11[x] is x+3. Cool, huh?! :) --unique monkic associate professor ------------------------------------------------------------------- > -----Original Message----- For the ones that are subrings, yes, you should prove they are a subring using our "previous" and indeed our "current" definition of subring, unless you have some other definition of subring that I am not aware of. :) For the ones that are not subrings, you need to give a counterexample (which is a method of proving as I discussed previously) that shows they are not subrings. One comment regarding this: You can't prove something is not a subring by Theorem 3.2, because Thm 3.2 is not stated as <=> it is and =>. Thus Thm 3.2 does help you AT ALL when trying to prove something is not a subring of a ring. Instead you must show it does not satisfy some necessary conditions in the definition ("previous" and "current" :)) of a subring. --has no expiration dates on his defs ------------------------------------------------------------------- > -----Original Message----- The def is that g(x) is an associate of f(x) in R[x] iff g(x)=cf(x) for some unit c in R. So you can think of an associate of a polynomial f(x) in R[x] as being what you get if you multiply f(x) by a "constant", c, which is a unit in R (the ring of coefficients). Now the simplest case is when R=F is a field. In a field F, ALL of the nonzero elements, c in R are units (by definition of a field). Thus in that case you can multiply a given polynomial f(x) (in F[x]) by ANY nonzero constant in the field and you will get an associate. For example, if the field is Q (the rationals) and f(x)= x^2+2x+3 then the following are all associates of f(x) in Q[x]: 3x^2+6x+9
(this is 3*f(x)) but: sqrt(2)x^2+2sqrt(2)x+3sqrt(2) is not an associate of f(x) in Q[x] even though it is sqrt(2)*f(x) because sqrt(2) is not an element of Q. Slightly more complicated is the situation where R is not a field. In that case you can only multiply f(x) by units in R to get associates. For example, the only associates of f(x)= x^2+2x+3 in Z[x] are x^2+2x+3 and -x^2-2x-3 because the only units in Z are 1 and -1. Hope this helps! --conductor of the train of thought ------------------------------------------------------------------- > -----Original Message----- They work for me. --your mathematical associate ------------------------------------------------------------------- > -----Original Message----- There are three recipies on the Recipe Sheet for doing this: To show p(x) is irreducible in F[x] To show p(x) is irreducible in F[x] To show a f(x) is irreducible in F[x] where
deg(f(x))=2 or deg(f(x))=3 The first recipe is from the definition of
irreducible. Moral: Check the recipe sheet and book first when working on a homework problem. There's treasure there! > That is what we need to do for this problem, isn't it? Read Theorem 5.10 and then try to solve this mystery. --Mister Mystery
------------------------------------------------------------------- > -----Original Message----- By Corollary 4.5, the elements of Q[x]/(x^2-2) are exactly the congruence classes of the polynomials having degree less than the degree of x^2-2 (and [0]). The degree of x^2-2 is 2. So Q[x]/(x^2-2) is the set of equivalence classes of the polynomials with degree less than 2, i.e. polynomials of the form ax+b where a,b in Q. To contrast this with some other situation, every element of Q[x]/(x^3+2x-1) are can be written in the form [ax^2+bx+c] for some a,b,c in Q (i.e. they are the classes of polynomials of degree less than deg(x^3+2x-1)=3 and [0]). --Math Guidance Counselor ------------------------------------------------------------------- > -----Original Message----- No, your set: I={a^n| n>0, n in Z+} is the set of all positive powers of a fixed element a in R, i.e. what your set describes is: {a,a^2,a^3,a^4,a^5, ... } not the set of all nilpotent elements of R. > i have proven that I is a subring and
am stuck on the next step of the Well, you couldn't have shown it if you used the above definition of the set I. For example if R=Z and a=1 then your set I is just {1} which is not a subring of Z. > I let r in R and s in I. i have to now
show that rs in I and sr I think your first task is to find the
right way to describe the set I in mathematical notation, then you will have a
much easier time proving what you need. Certainly. > or is this saying that I would contain
positive elements, but not Aye! Correct again! --an aye for an I ------------------------------------------------------------------- > -----Original Message----- Yes, once again, it means isomorphic. It's defined on the top of page 69. --isomonkic ------------------------------------------------------------------- This question is not easy to understand. You are supposed to make the addition and multiplication tables for Z12/I. So the entries in the tables are equivalence classes which in turn are subsets of Z12. The product and sum of the equivalence classes is defined in terms of the product and sum in Z12 in accordance with the definition given in lecture (and in the book on page 146). Does this answer what you are asking? --Monks/I ------------------------------------------------------------------- > -----Original Message----- The symmetry groups consist of all symmetry operations on the given figure. A symmetry operation is one that maps the figure to itself without changing distances, i.e. without bending, stretching or tearing it. So if you want, you can photocopy the given figure, then cut it out, then see how many different ways you can put it back into the hole you cut out of the page. Not all four sided figures have D4 for a
symmetry group. For example, a trapezoid like: Has only one symmetry operation... the identity map e. There is no way to cut out that shape from a piece of paper and put it back into the hole you cut it out of other than to put it back in the same way it was in there originally. (Can you see why? i.e. can you prove it?) Thus the symmetry group for this figure is {e}, i.e. the trivial group. So not all four sided figures have D4 for a symmetry group. --group leader ------------------------------------------------------------------- > -----Original Message----- This is called permutation notation. (Though usually you write it without the commas, e.g. (123) instead of (1,2,3)). Permutation notation is a shorthand for the array notation used in the book in this section. The advantage to permutation notation is it is more compact than the array notation. The disadvantage is that it is much more complicated to understand than the array notation (hence easier to make mistakes). We DO learn permutation notation in section 7.9 in the book (take a look ahead and you will see what I mean). But the author feels that you should begin with the slightly messier but easier to understand notation in the beginning sections of Chapter 7 and then switch to the shorter but more complicated permutation notation later on. This doesn't answer your question though. Here is the answer: I don't recommend that you use permutation notation yet, because it is just one more thing you can screw up when doing the problems. However, if you are so in love with permutation notation that you can't possibly conceive of life without it, then feel free to use it on your homework. Don't use the commas though unless you are working in Sn for n>10. I can read (and grade) either notation so it doesn't matter to me which notation you use. -multilingual ------------------------------------------------------------------- > -----Original Message----- You have to make the multiplication table (use 1,i,j,k in the table, not the matrices!). But then you also have to indicate why it is a group. It is easy to see from a multiplication table if there is an identity element e (so you should point out what that is) and it is easy to see if every element has an inverse by looking at the table (why?) but to show that the product is associative is NOT EASY to see from a table. So you need to verify that in some other way. Of course once you have done these three things, then you have shown its a group. --math groupie ------------------------------------------------------------------- > -----Original Message----- No, you need to show all 4 aspects of a group for the set containing the 6 elements given. > If you do then you would need 30
calculations Two things about this: 1. If you intend to use brute force and check every single case by hand (as you seem to be suggesting) then for associativity alone you have to check a(bc)=(ab)c for every possible a,b,c. Since there are 6 choices for a, 6 for b, and 6 for c, there are 6^3=216 calculations needed for associativity, not 30. 2. You shouldn't do it by brute force. There are much more clever ways to prove it is associative without going through 216 calculations. So I would advise you not to do it by brute computation. --brutally honest ------------------------------------------------------------------- > -----Original Message----- Yes. --yes man ------------------------------------------------------------------- I assume you are asking about problems #22,23,35. In each of these problems you only have to show that G is Abelian. You are given that G is a group by the Note at the beginning of the Exercises section on page 178. So you don't have to prove that G is a group for these problems, you are given that. --thanks-given ------------------------------------------------------------------- > -----Original Message----- You can get the order based on the information given, though I won't say if it's infinite or finite. But I will say that you can figure it out using the other parts of Theorem 7.8. --doesn't like giving orders ------------------------------------------------------------------- > -----Original Message----- I assume you want blah and bleep to be elements of a group. In that case: 1. blah=bleep
Given Right? If you don't like formal proofs, then here is plain everyday intuition: If blah=bleep then blah and bleep are the SAME element of the group. Since every element has a unique inverse (by Theorem 7.5(3)), blah and bleep must have the same inverse because they are the same element! :) -formally intuitive ------------------------------------------------------------------- > -----Original Message----- Every ring is a group with addition. Every ideal is also a subgroup because it is a subring. However, in general, a subgroup of the additive group of a ring might not be an ideal because it might not satisfy the other condition of being an ideal (ra in I and ar in I for all r in R and all a in I). For example, S={(n,n) : n in Z} is a subgroup of (ZxZ,+) but it is not an ideal of (ZxZ,+,*) because (1,2) in ZxZ, (1,1) in S and (1,2)*(1,1)=(1,2) is not in S. So every ideal is a subgroup, but not every subgroup of the additive group of a ring is an ideal. Now in the special case of Zn, things are a bit better. The group Zn is a cyclic group, e.g. Zn = <1>. By Theorem 7.16, every subgroup of Zn is also cyclic. So if S is a subgroup of Zn, then S=<a> for some a in Zn. Thus S=<a> so that S is the principal ideal generated by a in the ring Zn. Since every ideal is also a subgroup we see that a S is a subgroup of Zn iff S is a principal ideal of Zn. --idealistic and principaled ------------------------------------------------------------------- > -----Original Message----- Every subgroup contains e, so e is in H intersect K, so the order of H intersect K is greater than or equal to 1. -- e-z ------------------------------------------------------------------- > -----Original Message----- Yes. -beside himself ------------------------------------------------------------------- No. You must show that EVERY group if order 33 has such an element. --nay sayer
------------------------------------------------------------------- > -----Original Message----- It's the image of N. --image of monks ------------------------------------------------------------------- > -----Original Message----- Well the case n=k is trivial, so whether
you can do it all in one shot or in two cases its no big deal either way. Multiplicative. --solving your identity crisis ------------------------------------------------------------------- > -----Original Message----- Sure. The Kernal is the set of all things that map to e. So all you have to do is figure out what e is in the codomain, and then find all the things in the domain that phi maps to e. --Kernal Klink ------------------------------------------------------------------- > -----Original Message----- Like it says on page 216, "G/N denotes the set of all right
cosets of N in G" and the group operation is defined to be So in your problem we define G/K to be the set of all right cosets of K in G and the group operation is defined to be (Ka)(Kb)=Kab, which we obtain by substituting K for N in the definition of quotient group. -substitute teacher ------------------------------------------------------------------- > -----Original Message----- No. -Dr. Nyet ------------------------------------------------------------------- > -----Original Message----- The order of a permutation is the same as the order of any element in any group... it is the smallest positive power of the element which gives you the identity (if such a power exists). For example, if you want to find the order of a=(13)(245) in S6 then you compute: a=(13)(245) so the order of (13)(245) is six. You might want to read and prove problem #13, since that is very relevant to finding the orders of elements (you don't have to prove #13 to do #14, but it is a nice fact to know anyway). --maintaining order |
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