# Simple Contrapositive Proof

• Dec 19th 2009, 03:36 AM
Stylis10
Simple Contrapositive Proof
Let $\displaystyle n \in \mathbb{Z}$ be an integer with $\displaystyle n \ge 0$. State and prove the contrapositive of the
following statement: If $\displaystyle n^2$ is an odd number, then n is odd.

I can prove this by substituting n for 2m and proving $\displaystyle n^2$ is even. but am unsure if this is the final extent to what i need to proof, how far do i need to go to prove this, should i give more examples?
• Dec 19th 2009, 03:41 AM
Drexel28
Quote:

Originally Posted by Stylis10
Let $\displaystyle n \in \mathbb{Z}$ be an integer with $\displaystyle n \ge 0$. State and prove the contrapositive of the
following statement: If $\displaystyle n^2$ is an odd number, then n is odd.

I can prove this by substituting n for 2m and proving $\displaystyle n^2$ is even. but am unsure if this is the final extent to what i need to proof, how far do i need to go to prove this, should i give more examples?

The contrapositive is that if $\displaystyle n^2$ is not odd (even) then $\displaystyle n$ is not odd (even). Prove that (try contradiction).
• Dec 19th 2009, 03:46 AM
Stylis10
Quote:

Originally Posted by Drexel28
The contrapositive is that if $\displaystyle n^2$ is not odd (even) then $\displaystyle n$ is not odd (even). Prove that (try contradiction).

But then am i not just going back and provign the original statement and therfore not actually proving it through contrapositive?
• Dec 19th 2009, 03:48 AM
Drexel28
Quote:

Originally Posted by Stylis10
But then am i not just going back and provign the original statement and therfore not actually proving it through contrapositive?

No. It is quite different.
• Dec 19th 2009, 04:19 AM
Stylis10
ok, i think i understand, i have come up with this:

$\displaystyle n$ must be even for $\displaystyle n^2$ to be even,
let $\displaystyle n = 2m+1$
$\displaystyle (2m+1)^2 = 4m^2+4m+1$
$\displaystyle = 2Q + 1$ which is odd, therefore $\displaystyle n$ must be even for $\displaystyle n^2$ to be even and hence $\displaystyle n$ must be odd for $\displaystyle n^2$ to be odd.

or have i just done my own thing here and gone off on a tangent?
• Dec 19th 2009, 05:03 AM
Isomorphism
Quote:

Originally Posted by Drexel28
The contrapositive is that if $\displaystyle n^2$ is not odd (even) then $\displaystyle n$ is not odd (even). Prove that (try contradiction).

I dont think that is right. Read this.

The contrapositive of "If $\displaystyle n^2$ is an odd number, then $\displaystyle n$ is odd" is "If $\displaystyle n$ is even then $\displaystyle n^2$ is even".

So Stylis10, you were right in your first post when you explained the proof in the last line.