how would you prove by contradiction that if n is even then n^2 is even??

2. To prove by contradiction you assume the opposite of what is stated, ie that $\displaystyle n^2$ is odd.

As you are starting with an even number, then instead of $\displaystyle n$, lets use $\displaystyle 2n$.

$\displaystyle (2n)^2 = 4N$, therefore if $\displaystyle n \in Z$(intergers), $\displaystyle n^2$ is always even.

This is how I would do it anyway, I am sure there are different methods out there.

Craig

Edit: solved

3. Originally Posted by craig
To prove by contradiction you assume the opposite of what is stated, ie that $\displaystyle n^2$ is odd.

As you are starting with an even number, then instead of $\displaystyle n$, lets use $\displaystyle 2n$.

$\displaystyle (2n)^2 = 4N$, therefore if $\displaystyle n \in Z$(intergers), $\displaystyle n^2$ is always even.

This is how I would do it anyway, I am sure there are different methods out there.

Craig

Edit: solved
And of course, once you assume that $\displaystyle n^2$ is odd, the only way that is true is if $\displaystyle n$ itself is also odd. But that contradicts the original assumption that you made about it being even. So that completes the proof.

4. Originally Posted by hmmmm
how would you prove by contradiction that if n is even then n^2 is even??
This statement is equivalent to:
If n^2 is odd, then n is odd

By law of excluded middle: n is odd or n is even. Assume n is even, then $\displaystyle n = 2m$ where $\displaystyle m \in \mathbb{Z}$. But $\displaystyle n^2 = 4m^2 = 2(2m^2)$, which is a contradiction.