1. ## 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: 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

2. ## Re: Introduction

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

3. ## 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 $\displaystyle b$.

4. ## Re: Introduction

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

5. ## 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.

6. ## 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.
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.

7. ## 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.

8. ## 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.
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.

10. ## Re: Introduction

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

11. ## 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...........

12. ## Re: Introduction

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

thanks

13. ## 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.