Question

For any real number $x$ prove that $x^2\ge 0$ and hence claim that $0<1$

Answer 1

Consider two cases:

$x\ge 0$ and $x<0$

Case$I$: $x^2=x.x\ge 0.0=0$

Case$II$: $x<0$ we have $-x\ge 0$

Hence $x^2=x.x=(-x).(-x)\ge 0.0=0$

Hence for any real number $x$ we have $x^2\ge 0$

Claim:First we will show that $0\le 1$

Note that $1=1.1\ge 0$

Now it suffices to show that $0\ne 1$

Suppose for a contradiction that $0=1$.

Choose any real number $x\ne 0$ then

$x=x.1$

$=x.0$ since we have assumed for a contradiction that $0=1$

$=0$

This contradicts the fact that we choose $x\ne 0$ and hence it must be the case that $0\ne 1$

Hence $0\le 1$ and $0\ne 1$

$\therefore 0<1$