For all positive real numbers $x$, if $x$ is irrational, at least one of the numbers $x+\sqrt{2} $ and $x^2 - 2$ is irrational. How to prove this theorem?
If I use contrapositive, the theorem becomes "If at least one of the numbers $x+\sqrt{2}$ and $x^2 - 2$ is rational, $x$ is rational". I have no idea how to prove the two numbers are rational.
If both $x+\sqrt{2}$ and $x^2-2$ are rational, then the quotient
$$\frac{x^2-2}{x+\sqrt{2}}=x-\sqrt{2}$$
is rational, and hence
$$x=\frac{1}{2}((x+\sqrt{2})+(x-\sqrt{2}))$$
is rational. Contradiction.
Well, $x+\sqrt{2}$ and $x-\sqrt{2}$ differ by $2\sqrt{2}$ so at most one of them can be rational.  If only one of them is rational, their product $x^2-2$ is irrational.  If $x^2-2$ is rational, both $x-\sqrt{2}$ and $x+\sqrt{2}$ must be irrational since they cannot both be rational.