proof that if a square is even then the root is too.

**VonNemo19** · February 2nd 2011, 04:50 PM

Show that if $m^2=m\cdot{m}$ is even, then $m$ is even.

I think I've got it:

By definition, we can write $m^2=2n$ for some integer $n$ . This implies that $m\cdot{m}=2n$ . But, this means that $m\cdot{m}$ has a factor of 2. Therefore, $m$ has a factor of 2. So, by definition, $m$ is even.

Now, my gut instinct is that I butchered this. Can someone clean it up?

**dwsmith** · February 2nd 2011, 04:55 PM

Originally Posted by VonNemo19

Show that if $m^2=m\cdot{m}$ is even, then $m$ is even.

I think I've got it:

By definition, we can write $m^2=2n$ for some integer $n$ . This implies that $m\cdot{m}=2n$ . But, this means that $m\cdot{m}$ has a factor of 2. Therefore, $m$ has a factor of 2. So, by definition, $m$ is even.

Now, my gut instinct is that I butchered this. Can someone clean it up?

Since m = 2n, then m*m = 2n *2n = 4n^2 = 2(2n^2) which is of the form 2p

**TheEmptySet** · February 2nd 2011, 05:07 PM

Originally Posted by dwsmith

Since m = 2n, then m*m = 2n *2n = 4n^2 = 2(2n^2) which is of the form 2p

Isn't this circular reasoning. You are assuming m is even, that is what we want to prove. I think a cleaner way is to prove the logically equivalent statement the contra-positive.

Remember that if p then q is equivalent to not q implies not p. The reformulation is

If m is not even then $m^2$ is not even

Now just set $m=2k+1 \implies m^2=2(2k^2+2k)+1$ is odd. QED

**dwsmith** · February 2nd 2011, 05:10 PM

Originally Posted by TheEmptySet

Isn't this circular reasoning. You are assuming m is even, that is what we want to prove. I think a cleaner way is to prove the logically equivalent statement the contra-positive.

Remember that if p then q is equivalent to not q implies not p. The reformulation is

If m is not even then $m^2$ is not even

Now just set $m=2k+1 \implies m^2=2(2k^2+2k)+1$ is odd. QED

I thought m is even when I read it.

**Prove It** · February 2nd 2011, 06:06 PM

The Contrapositive is the quickest way, but if you wanted to prove it directly, you'd have to consider the cases " $\displaystyle m$ is even" and " $\displaystyle m$ is odd" and see what happens when you square each.

**LoblawsLawBlog** · February 2nd 2011, 08:25 PM

I think the original proof is fine, but could possibly use another sentence or two. You can always just say that if m is odd, then m and 2 are relatively prime. Since 2 divides the LHS but doesn't divide m, it must divide the other factor, which is m. Contradiction.

The contrapositive still seems fastest and most basic though.

**Drexel28** · February 2nd 2011, 08:52 PM

Originally Posted by VonNemo19

Show that if $m^2=m\cdot{m}$ is even, then $m$ is even.

I think I've got it:

By definition, we can write $m^2=2n$ for some integer $n$ . This implies that $m\cdot{m}=2n$ . But, this means that $m\cdot{m}$ has a factor of 2. Therefore, $m$ has a factor of 2. So, by definition, $m$ is even.

Now, my gut instinct is that I butchered this. Can someone clean it up?

Another way to do this is that if $2\nmid m$ then by the FTA $m=p_1^{\alpha_1}\cdots p_n^{\alpha_n}$ with $p_k\ne 2\;\; k\in[n]$ and thus $m^2=p_1^{2\alpha_1}\cdots p_n^{2\alpha_n}$ and since none of these primes are two we arrive at a contradiction.

**VonNemo19** · February 3rd 2011, 11:55 AM

Originally Posted by TheEmptySet

Isn't this circular reasoning. You are assuming m is even, that is what we want to prove.

This is what I was thinking, but as I am new to proof, I thought maybe I was just not seeing something.

Originally Posted by TheEmptySet

Remember that if p then q is equivalent to not q implies not p. The reformulation is

If m is not even then $m^2$ is not even

Now just set $m=2k+1 \implies m^2=2(2k^2+2k)+1$ is odd. QED

Thank you.

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