Theorem 4.2

If $n$ is an odd integer, then $n^2$ is odd.
$\text{Note that the statement of the}$ theorem $\text{is of the form}:$

Where $P(n)$ is $\text{‘}n\text{ is odd}\text{’}$, $Q(n)$ is $\text{‘}{n}^2\text{ is odd}\text{’}$, and the domain being integers

Remark: For proving that something is true $\text{for all}$ elements of the domain, it is sufficient to prove that the statement is true for an $\text{arbitrary}$ element of the domain.
For this we pick a random variable, say, $n$, of the domain, and without making any additional assumptions on $n$, we show that $P(n)\to Q(n)$.
Then, since $n$ could have been any element, we will known that $P(n)\to Q(n)$ is true $\text{for all}$ elements of the domain.

Proof

Let $n$ be an arbitrary integer, and assume that $n$ is odd. We have to prove that $n^2$ is odd.
Since $n$ is odd, there exists an integer $k$ such that $n=2k+1$. Squaring both sides of the equation, we obtain

Hence, $n^2$ can be written as $n^2=2\ell +1$, for an integer $ell$, namely, for the integer $\ell =(2k^2+2k)$. Thus, by the definition of an odd integer, we can conclude that $n^2$ is an odd integer.
$\text{Note: we started with the assumption ( that }n\text{ is odd ), then made a number}$$\text{of logical inferences ( }\text{‘}\text{Since... we obtain...}\text{’}\text{, }\text{‘}\text{Hence...}\text{’}\text{‘}\text{Thus...}\text{’}\text{ ), until we}$
$\text{arrived at the conclusion}$