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

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

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

I think I've got it:

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

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

2. Originally Posted by VonNemo19
Show that if $\displaystyle m^2=m\cdot{m}$ is even, then $\displaystyle m$ is even.

I think I've got it:

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

3. 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 $\displaystyle m^2$ is not even

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

4. 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 $\displaystyle m^2$ is not even

Now just set $\displaystyle m=2k+1 \implies m^2=2(2k^2+2k)+1$ is odd. QED
I thought m is even when I read it.

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

6. 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.

7. Originally Posted by VonNemo19
Show that if $\displaystyle m^2=m\cdot{m}$ is even, then $\displaystyle m$ is even.

I think I've got it:

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

Now, my gut instinct is that I butchered this. Can someone clean it up?
Another way to do this is that if $\displaystyle 2\nmid m$ then by the FTA $\displaystyle m=p_1^{\alpha_1}\cdots p_n^{\alpha_n}$ with $\displaystyle p_k\ne 2\;\; k\in[n]$ and thus $\displaystyle m^2=p_1^{2\alpha_1}\cdots p_n^{2\alpha_n}$ and since none of these primes are two we arrive at a contradiction.

8. 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 $\displaystyle m^2$ is not even

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