(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

if n is a natural number and n^{2}is odd, then n is odd

2. Relevant equations

odd numbers: 2k+1, where k is an integer

even numbers: 2K, where k is an integer

3. The attempt at a solution

ok so take the opposite to be true, or n^{2}is odd and n is even. Then we would have n^{2}= 2p+1 and n=2k, where p and k are both integers. then take n^{2}= (2k)^{2}= (2^{2})(k^{2})= 4k^{2}

Then set 2p+1= 4k^{2}→

4k^{2}-2p=1

2(2k^{2}-p)=1

(2k^{2}-p)=1/2

but since we know 2k^{2}is an integer because it is just the product of 3 integers, 2*k*k, and we also know p is an integer by definition, we also know that (2k^{2}-p) is an integer because it is the subtraction of two integers. This leaves us with a contradiction since we know (2k^{2}-p) can not equal 1/2, hence proof is complete.

**Physics Forums - The Fusion of Science and Community**

# If n is a natural number and n^2 is odd, then n is odd

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: If n is a natural number and n^2 is odd, then n is odd

Loading...

**Physics Forums - The Fusion of Science and Community**