# Introduction

• Aug 21st 2012, 12:39 AM
Introduction
Hi,

I came to MHF through google, looking for a *direct* proof that for all integers x it holds that if $\displaystyle x^2$ is even, then $\displaystyle x$ is even. There's a post on this here: http://mathhelpforum.com/number-theo...-root-too.html, 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
• Aug 21st 2012, 12:57 AM
Prove It
Re: Introduction
The contrapositive proof is more direct than the direct proof in this case...
• Aug 21st 2012, 01:01 AM
Re: Introduction
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 $\displaystyle a$ s.t. $\displaystyle x=2a$ from the given that $\displaystyle x^2=2b$ for some integer $\displaystyle b$.
• Aug 21st 2012, 01:09 AM
Prove It
Re: Introduction
See what happens when x is odd and you square it.
See what happens when x is even and you square it.
• Aug 21st 2012, 03:23 AM
Re: Introduction
Well, of course if $\displaystyle x$ is odd, then $\displaystyle x^2$ is odd, and if $\displaystyle x$ is even, then $\displaystyle x^2$ 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 $\displaystyle x^2$ and end at $\displaystyle x$.

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

By the fundamental theorem of arithmetic, $\displaystyle x^2$ has a unique prime factorization, so $\displaystyle x^2=p_1^{c_1}\cdot p_2^{c_2}\cdot\ldots\cdot p_n^{c_n}$, for certain primes $\displaystyle p_i$, and positive whole numbers $\displaystyle c_i$. (Assume these primes are listed in ascending order.) Because $\displaystyle x^2$ is even, $\displaystyle p_1=2$ and because it is a square, we have that for all $\displaystyle i$, $\displaystyle 2\mid c_i$. (This shows why, if $\displaystyle x^2$ is even, $\displaystyle 4\mid x^2$.) So $\displaystyle x=2^{c_1/2}\cdot p_2^{c_2/2}\cdot\ldots\cdot p_n^{c_n/2}$, which means $\displaystyle x$ is even. QED.
• Aug 21st 2012, 03:52 AM
Prove It
Re: Introduction
Quote:

Originally Posted by tomasklos
Well, of course if $\displaystyle x$ is odd, then $\displaystyle x^2$ is odd, and if $\displaystyle x$ is even, then $\displaystyle x^2$ 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 $\displaystyle x^2$ and end at $\displaystyle x$.

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

By the fundamental theorem of arithmetic, $\displaystyle x^2$ has a unique prime factorization, so $\displaystyle x^2=p_1^{c_1}\cdot p_2^{c_2}\cdot\ldots\cdot p_n^{c_n}$, for certain primes $\displaystyle p_i$, and positive whole numbers $\displaystyle c_i$. (Assume these primes are listed in ascending order.) Because $\displaystyle x^2$ is even, $\displaystyle p_1=2$ and because it is a square, we have that for all $\displaystyle i$, $\displaystyle 2\mid c_i$. (This shows why, if $\displaystyle x^2$ is even, $\displaystyle 4\mid x^2$.) So $\displaystyle x=2^{c_1/2}\cdot p_2^{c_2/2}\cdot\ldots\cdot p_n^{c_n/2}$, which means $\displaystyle x$ 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.
• Aug 21st 2012, 04:04 AM
Re: Introduction
OK, thank you. What I meant was "direct" in the sense that I wanted to reason from $\displaystyle x^2$ being even to $\displaystyle x$ 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.
• Aug 21st 2012, 04:22 AM
Prove It
Re: Introduction
Quote:

Originally Posted by tomasklos
OK, thank you. What I meant was "direct" in the sense that I wanted to reason from $\displaystyle x^2$ being even to $\displaystyle x$ 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.
• Jan 30th 2013, 06:37 PM
EhtaChan
Re: Introduction
@thomasklos that's a hard 1. i dont even know the answer. help please!
• Jan 30th 2013, 11:52 PM
Re: Introduction
Quote:

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?
• Jan 31st 2013, 02:32 AM
pssingh1001
Re: Introduction
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...........
• Jan 31st 2013, 02:34 AM
pssingh1001
Re: Introduction
Can u suggest me how to solve pie chart questions very quickly??????????????

thanks
• Jan 31st 2013, 04:30 AM
emakarov
Re: Introduction
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.
• Feb 5th 2013, 02:56 AM
pssingh1001
Re: Introduction
Thanks for your suggestion ...........
I will remember in future