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I have a few problems proving stuff with integers, basically the most basic stuff is driving mad right now.

Basically, there's two exercices.

1) If 3x is even, then x is even

Which is true (I think, at least). The method of resolution that I have here assumes x is even, and then proves that 3x is also even. But that's completely the opposite of what I need. Using some methods that I saw in other questions, the only way I could think of is doing something like this:

"Assuming 3x is even, then 3x = 2a and x = $\frac{{2}}{3}a$, so x = $2(\frac{{1}}{3}a)$ and $(\frac{{1}}{3}a) = b$. Finally, x = 2b proves that x is even.

Naturally, both a and b are integers.

1) If 4x is even, then x is even

This is clearly false, and it's pretty easy to find a counterexample.

Except that if I use the method above, I can easily prove a FALSE statement.

Doing a quick copy paste:

"Assuming 4x is even, then 4x = 2a and x = $\frac{{1}}{2}a$, so x = $2(\frac{{1}}{4}a)$ and $(\frac{{1}}{4}a) = b$. Finally, x = 2b proves that x is even.

I am of course doing something very wrong somewhere, but I can't find a recipe for this.

Thank you very much!

Tricolor
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  • But I don't think your first argument works either. $x=\frac 23 a$ certainly does not imply that $x= 2(\frac 23 a)$. – lulu Aug 17 '16 at 00:24
  • I edited my first argument, that was what I meant. – Tricolor Aug 17 '16 at 00:26
  • You might want to read about The Fundamental Theorem of Arithmetic. As it appears in Euclid, it asserts that, if a prime divides the product of two integers then it divides one or the other of them. – lulu Aug 17 '16 at 00:27
  • Your argument (post edit) is incomplete. To complete it you would have to prove that $\frac 13 a$ was an integer, which is not obvious (though it is in fact true). – lulu Aug 17 '16 at 00:29
  • Try this: if $x$ is odd then we can write $x=2m+1$ for some integer $m$. In that case $3x=6m+3=6m+2+1=2(3m+1)+1$. Thus $3x$ is also odd. If, to the contrary, we are told that $3x$ is even then it must be the case that $x$ was even. – lulu Aug 17 '16 at 00:32
  • In your second "proof" you are assuming that $a/4$ is a half-integer, which it need not be.:) An analogous qualification (that $a$ is divisible by $3$) is also missing from your first proof.

    Instead, in your first proof, I would say: let $3x = 2a$. Then $a$ is divisible by $3$, so $a = 3n$ for some integer $n$. Consequently, $x = 2n$, even.

    I think you would find it useful to learn (if you haven't yet) the proof that $\sqrt{2}$ is irrational.

     – avs Aug 17 '16 at 00:34

1 Answers1

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Since the mistake in your reasoning has been pointed out in the comments, let me provide you with the correct argument.

1) Assume $3x$ is even. For contradiction assume that $x$ is odd. Then $x=2n+1$ for some $n \in \mathbb{Z}$. Then $$3x=3(2n+1)=6n+3=2(3n+1)+1$$ where $3n+1 \in \mathbb{Z}$. Thus, we expressed $3x$ as $2m+1$ for some $m \in \mathbb{Z}$ (namely, $m=3n+1$). Thus, $3x$ is odd, contrary to the assumption that $3x$ is even. Thus, by contradiction, $x$ has to be even, as desired.

Pawel
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  • Actually, I was just missing the "contradiction part". This makes things much easier and also makes very easy to prove the second one wrong. Thank you very much! – Tricolor Aug 17 '16 at 00:46