proof by contradiction

[rsk098]

For all integers n, if n^2 is odd then n is odd.
What statement am I trying to prove false here?
Am I trying to prove that there exists at least one integer such if n^2 is even than n is odd,
or there exists at least one integer such if n^2 is odd than n is even? please help I'm really confused right now!

 

[Dr.Peterson]

There are several ways you could proceed, but a basic proof by contradiction would start with supposing that the condition is true (n^2 is odd), but then assuming that the conclusion is false: n is not odd (that is, n is even). If n is even, what does that tell you about n^2? You are expecting this to lead to a contradiction.

 

[JeffM]

Yes, that will do nicely.

[MATH]\exists \ k \in \mathbb Z \text { such that } k \text { and } k^2 \text { are both odd.}[/MATH]
[MATH]\therefore \exists \ p \in \mathbb Z \text { such that } 2p = k.[/MATH]
Finish it up.