Can you square both sides in a proof?

Question

Like say we had to prove that $\sqrt{2} + \sqrt{3} < \sqrt{26}$. Could you square both sides to prove it?

Answer 1

Yes, you can, since both numbers are non-negative. That is, if $a,b\in[0,\infty)$, then $a<b\iff a^2<b^2$.

Answer 2

We have: $\sqrt{2} + \sqrt{3} < 2\sqrt{3} = \sqrt{12} < \sqrt{26}$. That's probably faster.

Answer 3

Expanding on José Carlos Santos's answer: yes, in this case, but not generally. He shows the "yes, in this case" part. Generally, however, signs matter. From the true statement $-1 < 1$, squaring both sides seems to give $1 \overset{?}{<} 1$, which is not true.

Answer 4

In general, if $f(x)$ is any strictly increasing function over any interval $I = (a,b)$ (or $[a,b), (a,\infty), (-\infty,b),...$), then for any $u,v \in I$, $u < v \iff f(u) < f(v)$.

Taking square $x \mapsto x^2$ and square roots $x \mapsto \sqrt{x}$ are strictly increasing functions over $[0,\infty)$. So for any non-negative $u, v$, you have $$u < v \iff u^2 < v^2 \iff \sqrt{u} < \sqrt{v}$$

For your case, you know both sides are non-negative. This means $$\begin{align} & \sqrt{2} + \sqrt{3} \stackrel{?}{<} \sqrt{26}\\ \iff & 5 + 2\sqrt{6} = (\sqrt{2}+\sqrt{3})^2 \stackrel{?}{<} 26\\ \iff & 2\sqrt{6} \stackrel{?}{<} 21\\ \iff & 24 = (2\sqrt{6})^2 < 21^2 = 441 \end{align} $$ Since you know the last inequality is true, the inequality you want to show $\sqrt{2} + \sqrt{3} < \sqrt{26}$ is also true.