How would I prove $x^2 \ge x $ for integral $x$?

Question

As the title says, how would I simply prove $x^2 \ge x$ for integral $x$?

Should it not already be assumed that a number multiplied by itself is simply larger or equal to the existing number it was multiplied by based upon integer values only.

I only get as far as knowing $x^2 \ge 0$ but I am unsure where to proceed from this point.

Answer 1

If $x \le 0 \implies x^2 \ge 0 \ge x \implies x^2 \ge x$, if $x > 0 \implies x \ge 1$ since its an integer. So $x^2 -x = x(x-1) \ge 0$ since $x \ge 0, x - 1\ge 0$. So $x^2 \ge x$.

Answer 2

It is sufficient to show for natural $x$. But $x \ge 1$ so $x^2 \ge 1\cdot x = x$.(or am I using something bad?)