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Mendicant_Bias
 one year ago
(Introductory Real Analysis) I'm trying to show by contraposition that if the square of a number has a certain number as a factor, its root will also have that certain number as a factor. The way I'm tempted to solve this (after taking the contrapositive of the above statement) is by using a not equal sign, but I'm not sure I understand the mechanics of it. Math below shortly.
Mendicant_Bias
 one year ago
(Introductory Real Analysis) I'm trying to show by contraposition that if the square of a number has a certain number as a factor, its root will also have that certain number as a factor. The way I'm tempted to solve this (after taking the contrapositive of the above statement) is by using a not equal sign, but I'm not sure I understand the mechanics of it. Math below shortly.

This Question is Closed

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Contrapositive of the above: \[\text{If n is not a factor of} \ a, \ \text{then n is also not a factor of} \ a^2.\] \[a \neq nk;\]\[a^2 \neq n^2k^2\]

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0I'd like to treat a not equals sign as a relationship that is maintained under normal algebraic operations, e.g. if you manipulate both sides in the same way, the relationship of "not holding"...will still hold. I don't know if I can do this, though.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Are we assuming \(a\in \mathbb{Z}\)?

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to do this specifically by contraposition, but we can do that, too, if we're doing what I think we're doing. And I don't think it should matter for this problem, I think a is an element of either Z or R. I don't totally understand the implications of what you wrote, I'm taking a minute to think about it.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Well, to be factorable...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the term, "factor", tells us that \(a\in \mathbb{Z}\)

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, this is actually just a lemma I'm having to prove in the middle of another problem to complete the whole problem; the original prompt doesn't specify the set the value is in, but the original problem itself is proving that sqrt(3) is irrational; I did it by contradiction, so within the context of a proof by contradiction, I should be assuming that sqrt(something) is rational, if that's sensible/correct?

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0And yeah, a in this context is supposed to be part of a ratio of two integers in lowest terms, so it definitely belongs to Z. Had to think that through for a minute, sorry, lol.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3There is an easier way.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(\mathbb{Z}\) is the set of integers \(\mathbb{Q}\) is the set of ratio numbers (rational)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the phrase "factor of a rational number" makes no sense

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2factor must refer to integer \(a\) must be an integer, then only we can talk about its factors/divisors

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Well, at least for proving the a^2 factor part (I brought it to my prof), I have to do that by contraposition, but I'd ideally like to learn any and all ways to solve this. In any case, I hate to be like, "bluh, this is semantics", but I think we're on the same page regarding what the rationals/integers are, I'm just being really imprecise with my language and throwing stuff together quickly to try and solve this.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Yes, and a is an integer. No disagreement there. But it's both an integer and rational, right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2all integers are rational whats exciting about that haha

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0And you have to assume both for the problemright? You have to assume that a and b are integers since it's a ratio of two integers that makes a rational number that you're assumign sqrt(something) is equal to for the sake of contradiction. Nothing, but you said that "factor of a rational number" doesn't make sense, but it does, doesn't it? Integers are rational numbers, and they have factors. Even in lowest terms. It's trivial, but it's not wrong, and that wasn't the point, in any case. People were asking me questions regarding what set the variable we're concerned with was in, and I was answering them. It's not really that important right this very moment, but I was just trying to answer questions I was asked so we could move on.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Suppse \(\sqrt{3}=\frac{a}{b}\) where \((a,b)=1\). Then \(3=\frac{a^2}{b^2}\implies b^2*3=a^2\) from here you may conclude that \(a^2,b^2\) are both odd. Suppose \(b^2\) was even, then so is \(3b^2\) and then so is \(a^2\) so that we must have \(a,b\) are both even contradiction \((a,b)=1\) So \(a^2,b^2\) are both odd, so that \(a,b\) are both odd. So \(a=2k+1\\ b=2k_0+1\) now plug this in, \(3b^2=a^2\implies 3(4k^2+4k+1)=4k_0^2+4k_0+1\\\implies 6k^2 + 6k + 1 = 2(k_0^2 + k_0).\) The left is odd, the right is even. Yikes!!!

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3\((a,b)=1\) just means that \(\frac{a}{b}\) CAN'T be reduced any further.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0^Part of what I'm not on board with with this/would need to be proved is this: http://i.imgur.com/8kyErQo.png I'm aware that the square of an odd number is odd, and the square of an even number is even, but I don't know how we can conclude that both a^2 and b^2 themselves are odd.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0What I'd conclude from 3b^2=a^2 is that 3 is a factor of a^2. I don't yet know, however, how I can demonstrate that both a^2 and b^2 are odd.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Ok well you also must conclude that \(a^2\) has the same parity as \(b^2\) right?

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Wait wait wait, sorry, lol. One minute. I scanned it the first time you posted it, my bad.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3forget what I just posted, I thought you had another problem.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3back to your issue. You agree that if \(b^2\) is even then so is \(a^2\) right?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3because \(a^2=3*\)even

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Ok so let us assume they are both even, then you also agree that if \(a^2\) is even, then so is \(a\)?

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0But we don't know that b^2 is even. (Nevermind, assumption) And yeah, agreed that if a^2 is even, so is a.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3They must either be both even or both odd. I will assume they are both even and show you that that cant be the case. there for they must both be odd. Make sense?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3OK so if \(a^2,b^2\) is even, then so is both \(a,b\) and then re can reduce our fraction \(\frac{a}{b}\) right?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3But we assumed at the start that our fraction was reduced so this is a contradiction.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3So \(a^2,b^2\) are both odd.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3p.s. this proof got a dude killed :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Maybe the smiley was too much

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I heard about that! What's his name, the dude who the Pythagoreans killed for disclosing it when they wanted to keep it a secret. And yeah, I get it now, at least I certainly get the proof that a and b must both be odd. Now I'm going to reread the above, one sec.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Well they did not want him to show the world that there were irrational numbers because they thought the world of integers and really did not like that idea. So death ... they also did not allow an open flame next to a mirror...goes to show what they know.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0I'm just figuring out the last bit of algebra, but it at the very least conceptually makes sense to me...thanks very much.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3the algebra will be easy. np.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I don't understand what the heck you did at the very end, it literally just looks wrong, so let me write this out step by step and confirm I'm not crazy:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0the starting question is not proved yet :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0since n can be any natural number, and not n=3

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3lol omg. actually what we proved has nothing to do with the starting question. The starting question is a result of going a different rout that is not needed.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0\[3(4k^2+4k+1)=4k_0^2+4k_0+1\](I'm on board with this) \[12k^2+12k+3=4k_0^2+4k_0+1\]\[12k^2+12k+2=4k_0^2+4k_0\]\[6k^2+6k+1=2k_0^2+2k_0\] Alright, so this is effectively the same resultsI just don't know...how you did your wizard algebra to end up with a different simplification.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3I did the exact same thing. I just skip steps...

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0You had a slightly different written end result, promise. You had plus one on the RHS, which is why I was really confused. Both sides had a plus one, guessing it was a typo. Either way, this makes sense now, heh.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Whoah, hey there, lol. I like hearing everybody here talk. ;_;

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3It was a personal jab at him, just playin.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0a good idea would be if you answer to the actual question :) @zzr0ck3r

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3read man, we covered all of this...

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0ooooooOooo shots fired

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Or tell us again how \(n\) might not be \(3\)....

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I tell you that chemistry is not physics, as you wrote yesterday, lol! :) @zzr0ck3r

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3This dude follows me to other questions and trys to but in and ALWAYS says the wrong thing. Then I show him why he was wrong, as I did here, he then moves on to something else. What you are referring to is someone asking a chem question in the math section and saying the reason was because people would not answer it is physics. I told the user that people were not answering because his question was chem, not physics. Just read @Michele_Laino , you might then actually know what you are talking about more.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0So at least for the original question, for n=3, we're good, right? We proved all the lemmas necessary in between about how both a and b must be odd, and provided a contradiction showing that sqrt(3) is irrational, but how could we show that if a^2 has a factor alpha, then a also has a factor alpha. The last thing I'm concerned with (and I'll open up a different question if people would like) is just figuring that out.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Ahh I did not read that post correcty. I did not know the number was a square... sorry. this is true then.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I like one line proofs in number theory below is a legitimate proof for showing\(\sqrt{3} \not\in \mathbb{Z}\) : suppose \(\sqrt{3}\) is rational, then for some \(a,b\in \mathbb{Z}\) we have \[\frac{a}{b} = \sqrt{~3~} \implies \left(\frac{a}{b}\right)^2 = 3 \implies 3b^2 = a^2 \] By euclid lemma, \((a,b)=1 \implies 3\mid 1\), contradiction \(\blacksquare\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I think we still need to show \((a^2, b^2)=1 \iff (a,b)=1 \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2** I like one line proofs in number theory below is a legitimate proof for showing\(\sqrt{3} \not\in \color{red}{\mathbb{Q}}\) : suppose \(\sqrt{3}\) is rational, then for some \(a,b\in \mathbb{Z}\) we have \[\frac{a}{b} = \sqrt{~3~} \implies \left(\frac{a}{b}\right)^2 = 3 \implies 3b^2 = a^2 \] By euclid lemma, \((a,b)=1 \implies 3\mid 1\), contradiction \(\blacksquare\)

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Wait a minute, I don't understand, what did we still need to show? Up to the point I thought I was done, what else did we need to prove?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we're done!im talking about my proof in the end... it has nothing to do with zzr his proof is complete

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3you are done, this is a different method. But you need euclids lemma for this version

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Number theorist like to make things easy.... pfft

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3jk :) this one is much "nicer"

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Alright. The last thing I'd like to double down on again is that I want to prove that a factor of a^2 is also a factor of a, and I'd like to do that doing contraposition, but I don't know how to go about doing that. Solving one problem in as many ways possible is right now pretty important to me so that I develop a sense of the different ways of solving problems, not just getting the solution.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3The best proofs are one liners. Here is my favorite proof that the identity element is unique in a group. \(e_1=e_1*e_2=e_2\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Can I see how you are using this in the proof that \(\sqrt{3}\) is irrational?

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0(Hell, that *was* my original question. I was focusing on that lemma which ended up never getting answered, but the larger question getting answered in a different way.) @zzr0ck3r , sure. One minute.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0How do you guys put (not LaTeX text) plaintext in line with math in openstudy comments, by the way?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3`\(\text{like this}\)`

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Actually I am using that in my proof above while making the jump : \((a^2,b^2)=1 \implies (a,b)=1\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3This is some number theory stuff, @ganeshie8 will love to do it :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Ah, I would go a different rout for that.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3if some number divides a and b then surely it divides a^2 and b^2...

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.01.) Assume sqrt(3) is rational (proof by contradiction.) \[\text{If} \ \sqrt{3} \ \text {is rational, then} \ \sqrt{3}=\frac{a}{b}, \ \ \ a,b,\in \mathbb{Z}\]\[3=\frac{a^2}{b^2}\]\[3b^2=a^2\] @zzr0ck3r Yeah, but I want that to be proven deductively, lol, not just going "lol it's obvious". If a^2=3b^2, 3 is a factor of a^2. If b is already assumed as an integer, its square will also be an integer (here's where this lemma is necessary).\[3b^2=a^2 \rightarrow 3b^2=(3 \gamma)^2=9 \gamma^2\]

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0\[3b^2=9 \gamma^2\]By the same logic, 3 is a factor of b^2. By the same lemma as before, if 3 is a factor of b^2, it must be a factor of b. Thus, b=3(delta), where delta is some integer. Putting these into the original expression a/b shows you that they were not, as assumed, in lowest terms. This is the contradiction that shows you that sqrt(3) is not rational.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0And Ganeshie, I actually have no idea what that notation means in your last post/how would that be read in words?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\((a, b)\) is read as \(\gcd\) of \(a\) and \(b\)

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0But yeah, that's why I'm trying to proveby contrapositionthat if a^2 has some root, a must also have it in common. Ah.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0first: I don't follow you, since I have no reason second: what I write is correct, furthermore, what I saw is that you are very adept at distorting the truth and in offending others, and less capable in doing mathematics, it is clear that your parents have not been able to teach good manners. @zzr0ck3r

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2i smell the mathematician vs physicist fight

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0@ganesh he started to offend me as you can see from the above replies

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0I promised to Mrs. Preetha that I would not fight with that person, but he continues to offend me

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0...I just want to get my original question answered. Could this please be done somewhere else. I still haven't figured out the question I first asked and was first and foremost was concerned with, everyone sort of ignored that, went, "No, I have a better method" and did whatever they felt like. Which is cool, but I still don't understand how to do this in this specific way that I'm concerned with.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0right! In fact I wrote that the starting question is not proved yet @Mendicant_Bias and @zzr0ck3r started to offend me

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2` if the square of a number has a certain number as a factor, its root will also have that certain number as a factor.` is equivalent to showing : if the \(\gcd(a^2,b^2)\) is \(d^2\), then the \(\gcd(a,b)\) is \(d\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2lets try proving the latter statement..

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0...What? Alright, hold off, I didn't try to offend you at all. I didn't say a word to you. I don't have an issue with you.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Don't bring me into this. Anyways, yeah, let's try proving that.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3@Michele_Laino again, simply read " but I'd ideally like to learn any and all ways to solve this." TheQuestionAsker just read man, read.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Alright, so I have no idea how to compare those two statements off the top of my head, hmm...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2do you have teamviewer/skype ?

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0And well wait a minute, I know it's in effect the same, but I don't need both a^2 and b^2, just doing the same thing for one number in the same set should apply to both right? So If\[\text{If} \ \gcd(a^2,b^2)=d^2, \\ \text{then}\gcd(a,b)=d\]Is effectively the same as\[\text{If} \ \gcd(a^2)=d^2, \ \text{then} \ \gcd(a)=d.\]

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0I do have skype where I am, but I wouldn't want to talk over the micit's pretty late where I am and my roommates are sleeping. In any case, I might call it a night unless I can figure this out soon, I'm getting pretty tired. Thank you guys for all the help in any case.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2gcd = greatest "common" dividsor gcd requires at least two operands

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0(Nevermind, googled it)

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I'm too sleepy to think well, I'm going to call it a night and come back to this in the morning. Goodnight and thanks.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2gnite! have good sleep :)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1I am sorry that I did not follow the whole thread in detail, but I wish I was there at the beginning. So please point it out if I am repeating some ideas presented before. As @michelelaino stated, the original statement is not proved yet. In fact, it is false, namely "that if the square of a number has a certain number as a factor, its root will also have that certain number as a factor." if ab^2k^2 then abk is false. (counter example: if "16144, then 1612 " ) So it's contrapositive is also false. As @ganeshie8 said, we should only be dealing with Z, for example: Perhaps the original statement meant to be. "that if the square of a number \(n\in Z\) has a certain \(prime~k \in Z\)as a factor, the (root of n) \(\in Z\) will also have k as a factor."

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0thanks! :) @mathmate
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