proof by contradiction

**hmmmm** · January 5th 2009, 01:42 AM

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

**craig** · January 5th 2009, 01:57 AM

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

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

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

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

Craig

ps does anyone know how to do the 'member of' sign in your maths formatting thing?

Edit: solved

**Last_Singularity** · January 5th 2009, 04:12 AM

Originally Posted by craig

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

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

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

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

Craig

ps does anyone know how to do the 'member of' sign in your maths formatting thing?

Edit: solved

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

**Chop Suey** · January 5th 2009, 04:35 AM

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 $n = 2m$ where $m \in \mathbb{Z}$ . But $n^2 = 4m^2 = 2(2m^2)$ , which is a contradiction.

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