Question

Prove that the square of real number is always non-negative.

Answer 1

Suppose $a$ is real number.

Then either $a>0$ or $a=0$ or $a<0$ and only one of these is true.

Suppose $a>0$.

Then $a\times a>0\times a$, since we are multiplying by a positive real number.

Hence $a^2>0$.

If $a=0$, then $a^2=0$.

Suppose $a<0$.

Then $a\times a>0\times a$, as the inequality changes when multiplied by a negative number.

Again we obtain $a^2>0$.

We can conclude that $a^2>0$ when $a$ is either positive or negative; and $a^2=0$ when $a=0$.

Thus $a^2\ge 0$ for any real number $a$.

So, the square of a real number is always non-negative.