Prove the product of an even integer and an odd integer is even Is this proof correct?
Theorem:  The product of an even integer and an odd integer is even.
Proof:  Let $a$ and $b$ be integers.  Assume $a$ is even and $b$ is odd, so there exists an integer $p$ so that $a=2p$ and there exists an integer $q$ so that $b=2q+1$.  If $a \cdot b$ is even then by definition of even there exists an integer $r$ such that $a \cdot b = 2r$. So we have $a \cdot b = (2p) (2q+1) = 2r$, where $r$ is an integer.
Therefore, $a \cdot b$ is even.
 A: No.  You are assuming what you are trying to prove.
Consider this "proof" (the same as yours but I'm replacing the gray text with red text:
Theorem: The product of an even integer and an odd integer is $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$.
Proof: Let $a$
and $b$ be integers. Assume $a$ is even and $b$ is odd, so there exists an integer $k$ so that $a=2k$ and there exists an integer $q$ so that $b=2q+1$. If $a⋅b$ is $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$ then by definition of$\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$ there exists an integer $r$ such that $\color{gray}{a*b=2r}$ $\color{red}{a*b=2r+1}$. So we have $a⋅b=(2p)(2q+1)=\color{gray}{2r}$ $\color{red}{2r+1}$, where $r$
is an integer.
Therefore,$ a⋅b$
is $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$.
A: This proof is not clearly written, it could be interpreted as you proving your statement by assuming it first to be true (however not in the inductive sense. This could be a clearer way:
Let $a=2n$ (ie $a$ is even) and $b=2m+1$ (ie $b$ is odd). Then the product $ab$ is:
$$\begin{array}{rcl}
ab & = & (2n)(2m+1) \\
& = & 2(n(2m+1)) 
\end{array} $$
Letting $k=n(2m+1)$, this gives $ab=2k$, which implies $ab$ is even (since $k$ is an integer). 
Edit: having seen the updated question, your working is all correct, however it could be made more explicit what your $r$ is. 
A: The basic idea is correct, but it is not presented precisely. 

If ab is even then by definition of even there exists an integer r such that ab = 2r. So we have ab=(2p)(2q+1)=2r,where r is an integer. Therefore,ab is even.

Why "if"? What you need is rather the other direction, so: If $ab=2r$ for an integer $r$, then $ab$ is even. We have $ab = 2p(2q+1)= 2(p(2q+1))$. Since $p(2q+1)$ is an integer, the claim follows, setting $r = p(2q+1)$. 
As mentioned in a comment, it is not really needed to say $b=2q+1$. It is simpler to work with $b$ directly in this case. 
If $ab=2r$ for an integer $r$, then $ab$ is even. We have $ab = 2pb= 2(pb)$. Since $pb$ is an integers, the claim follows, chosing $r = pb$. 
An additional point, you use once $p$ and once $k$ for the same thing.
A: $a$ is even $\implies a=2n:n\in\mathbb{Z}$
$b$ is odd $\implies b=2m+1:m\in\mathbb{Z}$
$ab=2nm+2n=2n(m+1)$
$2n$ factors $2n(m+1)\implies ab$ is even.
A: I think the proof clarifies if you employ the Distributive Property fully:
ab = (2n)(2m + 1)
   = (2n2m) + 2n
   = 2( 2mn + n ) only ONE 2 pulls out of the first product
Since m and n are integers, ab will always have a factor of 2 among its Prime factors.
