i am having trouble proving the statement below using Proof by Contrapositive. I have negated the statements as required and then i prove that $n$ is odd if and only if $7n+4$ is odd. However, from there i am stuck and am not sure how to prove the 1st part that if $n$ is negative integer, then $n$ is odd. My work is show below.
Original Statement: if n is a positive integer then n is even if and only if is 7n+4 is even.
Contrapositive: If n is negative integer then n is odd if and only if 7n+4 is odd.
Therefore by definition of odd: n = 2k+1
Substitute n:
=7(2k+1)+4
=14k+7+4
=14k+11
=2(7k)+11
Therefore, n is odd and 7n+4 is odd.
Thats as far as i got and i dont even know if what i did above is even right though. Thanks.
Best Answer
There really is no need to prove by contrapositive (unless your teacher is requiring you to do that for some reason). The following direct proof works just fine.
Problem: Suppose $n\in\mathbb{Z}$ ($n$ being positive is really not necessary). Then $n$ is even if and only $7n+4$ is even.
Proof.
($\to$): Suppose $n$ is even. Then $n=2\ell$, where $\ell\in\mathbb{Z}$. Thus, we have that $$ 7n+4=7(2\ell)+4=14\ell+4=2(7\ell+2)=2m, $$ where $m=7\ell+2$ and $m\in\mathbb{Z}$. Thus, the forward direction is true.
($\leftarrow$): Suppose $7n+4$ is even. Consider what happens when $n$ is odd or even.
The above analysis shows that $n$ is even if and only if $7n+4$ is even, as desired.