Prove every odd number is of the form 2k+5. Is every even number of the form 2k+6? I know even numbers are of the form $n=2k, k \in \mathbb Z$, and that odd numbers are of the form $n=2k+1, k \in \mathbb Z$. 

I'm being asked to prove that all odd numbers are of the form $n=2k+5, k \in \mathbb Z$ and asked if all even numbers are of the form $n=2k+6, k \in \mathbb Z$.  


I was wondering if my proofs are right, and if not, if you could please guide me in the right direction.
Proof 1. If $n$ is an odd integer then $n=2k+1, k \in \mathbb Z$. Let $n=2l+5, l \in  \mathbb Z$. Then, $n=2l+4+1 =2(l+2)+1$ Since $l+2$ is an integer, $n$ is odd.
Proof 2. If $n$ is an even integer, then $n=2k, k \in \mathbb Z$. Let $n=2l+6, k \in \mathbb Z$. Then, $n=2l+6=2(l+3)$. Since $(l+3)$ is an integer, then $n=2l+6$ is even.
Thank you all in advance.
Alex

You are trying to show that if $n$ is odd, then $\exists l \in \mathbb{Z}, n = 2l+5$ rather than the opposite.
If $n$ is an odd integer, then $ \exists k \in \mathbb{Z}, n = 2k+1,$.
$n = 2k+1 = 2(k-2)+4+1= 2(k-2) + 5$. 
Since $k-2 \in \mathbb{Z}$, $n$ can be written in the form of $2l+5$ by letting $l=k-2$.
Also note that an even number can be written as an odd number $+1$.
