• Jan 5th 2009, 01:42 AM
hmmmm
how would you prove by contradiction that if n is even then n^2 is even??
• Jan 5th 2009, 01:57 AM
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 (Happy)

Edit: solved ;)
• Jan 5th 2009, 04:12 AM
Last_Singularity
Quote:

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 (Happy)

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.
• Jan 5th 2009, 04:35 AM
Chop Suey
Quote:

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.