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.
Best Answer
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 $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$.