A clean proof of $x^2 \geq x$, for any integer $x$ I am trying to prove that $x^2 \geq x$ for any integer $x$.
Since we know that for any number $n$, $n^2 \geq 0$ we conclude that if $x \leq 0$ the proposition will hold.
Next we must prove that the proposition holds for $x > 0$.
First we note that $x^2=xx$, and apply this to the inequality in the proposition.
$xx \geq x$.
Divide each side by $x$ and arrive at the solution set $x \geq 1$. Since there is no integer $p$; $0 < p < 1$ the proposition must be true.
The problem I am having is the second part for $x > 0$. I feel like my argument is circular. If anyone has any advice I'd appreciate it!

Since $x \geq 1$, you have $x^{2} = x \cdot x \geq x \cdot 1 = x$.
