# Simple Contrapositive Proof

Printable View

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

I can prove this by substituting n for 2m and proving $n^2$ 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 $n \in \mathbb{Z}$ be an integer with $n \ge 0$. State and prove the contrapositive of the
following statement: If $n^2$ is an odd number, then n is odd.

I can prove this by substituting n for 2m and proving $n^2$ 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 $n^2$ is not odd (even) then $n$ 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 $n^2$ is not odd (even) then $n$ 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:

$n$ must be even for $n^2$ to be even,
let $n = 2m+1$
$(2m+1)^2 = 4m^2+4m+1$
$= 2Q + 1$ which is odd, therefore $n$ must be even for $n^2$ to be even and hence $n$ must be odd for $n^2$ 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 $n^2$ is not odd (even) then $n$ is not odd (even). Prove that (try contradiction).

I dont think that is right. Read this.

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

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