Prove or disprove that if $\frac ab$ ($a$, $b\in\Bbb Z$) is in upper bound of $\{x\in\Bbb Q\mid x^2<2\}$, then $a^2>2b^2$. Do not introduce real numbers yet.
I suppose this is true, but it seems hard to prove it using only knowledge prior to the definition of rational number set. I thought of proving by contradiction, say $\frac ab$ is an upper bound (namely, for all $x\in\Bbb Q$, $x^2<2$, $x\le\frac ab$), and assume $a^2\le2b^2$. Since $a^2\ne2b^2$ by number theory, then $a^2\le2b^2-1$.
I don't know how to continue. Maybe we need to find $x\in\Bbb Q$, $x^2<2$ such that $x>\frac ab$. But how to do this?