A direct proof is one of the most familiar forms of proof. It is used to prove statements of the form “if p then q” or “p implies q”. The method of the proof is to take an original statement p, which we assume to be true, and use it to show directly that another statement q is true.
Directly prove that if $n$ is an odd integer then $n^2$ is also an odd integer.
Let $p$ be the statement that $n$ is an odd integer and $q$ be a statement that $n^2$ is an odd integer.
Assume that $n$ is an odd integer, then by definition $n=2k+1$ for some integer $k$. We will now use this to show that $n^2$ is also an odd integer.
\begin{align} n^2 &= (2k+1)^2\\ &=(2k+1)(2k+1)\\ &=4k^2+2k+2k+1\\ &=4k^2+4k+1\\ &=2(2k^2+2k)+1 \end{align}
Hence, we have shown that $n^2$ has the form of an odd integer since $2k^2+2k$ is an integer. Therefore we have shown that $p\Rightarrow q$ and so we have completed the proof.
Directly prove that if $m$ and $n$ are odd integers then $mn$ is also an odd integer.
Assume than $m$ and $n$ are odd integers. Then by definition $m=2k+1$ and $n=2l+1$ for some integers $k$ and $l$.
Note: that different integers $k$ and $l$ are used in the definitions of $m$ and $n$.
Now we will use this to show that $mn$ is also an odd integer.
\begin{align} mn &= (2k+1)(2l+1)\\ &=4kl+2k+2l+1\\ &=2(2kl+k+l)+1 \end{align}
Hence we have shown that $mn$ has the form of an odd integer since $2kl +k+l$ is an integer.