Introduction

**tomasklos** · August 21st 2012, 12:39 AM

Hi,

I came to MHF through google, looking for a *direct* proof that for all integers x it holds that if $x^2$ is even, then $x$ is even. There's a post on this here: proof that if a square is even then the root is too., but it doesn't give a direct proof, unfortunately. Anyway, MHF seemed like a nice and helpful site, so I registered, and here I am.

Tomas

**Prove It** · August 21st 2012, 12:57 AM

The contrapositive proof is more direct than the direct proof in this case...

**tomasklos** · August 21st 2012, 01:01 AM

Yes, it's quite neat, that's precisely the point: I'm looking at this as an example of a case where proving the contrapositive is a lot easier than giving a direct proof. Of course, I need the direct proof to justify this claim. (I'm talking about deriving that there exists an $a$ s.t. $x=2a$ from the given that $x^2=2b$ for some integer $b$ .

**Prove It** · August 21st 2012, 01:09 AM

See what happens when x is odd and you square it.
See what happens when x is even and you square it.

**tomasklos** · August 21st 2012, 03:23 AM

Well, of course if $x$ is odd, then $x^2$ is odd, and if $x$ is even, then $x^2$ is even. But without wanting to sound pedantic, I'd say that's a contrapositive proof in disguise. I'd like to start reasoning about $x^2$ and end at $x$ .

I just discussed it with a colleague, and we came up with this.

By the fundamental theorem of arithmetic, $x^2$ has a unique prime factorization, so $x^2=p_1^{c_1}\cdot p_2^{c_2}\cdot\ldots\cdot p_n^{c_n}$ , for certain primes $p_i$ , and positive whole numbers $c_i$ . (Assume these primes are listed in ascending order.) Because $x^2$ is even, $p_1=2$ and because it is a square, we have that for all $i$ , $2\mid c_i$ . (This shows why, if $x^2$ is even, $4\mid x^2$ .) So $x=2^{c_1/2}\cdot p_2^{c_2/2}\cdot\ldots\cdot p_n^{c_n/2}$ , which means $x$ is even. QED.

**Prove It** · August 21st 2012, 03:52 AM

Originally Posted by tomasklos

Well, of course if $x$ is odd, then $x^2$ is odd, and if $x$ is even, then $x^2$ is even. But without wanting to sound pedantic, I'd say that's a contrapositive proof in disguise. I'd like to start reasoning about $x^2$ and end at $x$ .

I just discussed it with a colleague, and we came up with this.

By the fundamental theorem of arithmetic, $x^2$ has a unique prime factorization, so $x^2=p_1^{c_1}\cdot p_2^{c_2}\cdot\ldots\cdot p_n^{c_n}$ , for certain primes $p_i$ , and positive whole numbers $c_i$ . (Assume these primes are listed in ascending order.) Because $x^2$ is even, $p_1=2$ and because it is a square, we have that for all $i$ , $2\mid c_i$ . (This shows why, if $x^2$ is even, $4\mid x^2$ .) So $x=2^{c_1/2}\cdot p_2^{c_2/2}\cdot\ldots\cdot p_n^{c_n/2}$ , which means $x$ is even. QED.

It's not a contrapositive proof. You are seeing what the possible ways are to get even-valued x^2. It's a proof by exhaustion, which is direct.

**tomasklos** · August 21st 2012, 04:04 AM

OK, thank you. What I meant was "direct" in the sense that I wanted to reason from $x^2$ being even to $x$ being even. What I meant by "contrapositive in disguise" is that just one of those cases (the one of x being odd) already provides the contrapositive proof.

I'm sorry, I'm new here; still learning the proper way of expressing myself on the forum. Thanks.

**Prove It** · August 21st 2012, 04:22 AM

Originally Posted by tomasklos

OK, thank you. What I meant was "direct" in the sense that I wanted to reason from $x^2$ being even to $x$ being even. What I meant by "contrapositive in disguise" is that just one of those cases (the one of x being odd) already provides the contrapositive proof.

I'm sorry, I'm new here; still learning the proper way of expressing myself on the forum. Thanks.

Logically speaking, a direct proof is an argument that can show that statement 1 implies statement 2. You can start wherever you like, as long as you get to an argument that shows this implication. Exhaustive proofs like what I've shown you are direct.

**EhtaChan** · January 30th 2013, 06:37 PM

@thomasklos that's a hard 1. i dont even know the answer. help please!

**tomasklos** · January 30th 2013, 11:52 PM

Originally Posted by EhtaChan

@thomasklos that's a hard 1. i dont even know the answer. help please!

Hi, there are several (ideas for) proofs on this page. Can you try to explain what you don't understand?

**pssingh1001** · January 31st 2013, 02:32 AM

Hi Friends!
I am the new member of the forum ...........
So first of all I would like say ".............HELLO........"
to all of u Dear..................

Thanks...........

**pssingh1001** · January 31st 2013, 02:34 AM

Can u suggest me how to solve pie chart questions very quickly??????????????

thanks

**emakarov** · January 31st 2013, 04:30 AM

Hi, pssingh1001,

Two things. First, start a new thread for a new issue (introduction, new question, etc.). Second, you are more likely to receive help if you post a concrete question. A forum is not a replacement for textbooks.

**pssingh1001** · February 5th 2013, 02:56 AM

Thanks for your suggestion ...........
I will remember in future

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