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?

Does this answer your question? What are some common pitfalls when squaring both sides of an equation? — , Jan 03 '22 at 09:09

score 7 · Answer 1 · answered Sep 09 '19 at 05:32

7

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

answered Sep 09 '19 at 05:32

José Carlos Santos

427,504

Yeah but isn’t it that we don’t really know if $a<b$ in the first place, cuz that’s what we’re trying to prove? So how would we know that $a^2<b^2$? Would you have to do like a proof by contradiction instead and assume that it’s greater than or equal to $sqrt26$, and then square it/show the negation leads to a contradiction? – Sep 09 '19 at 06:02
2

@gughy The implication is both ways here if the terms are non-negative. That means if you prove it for the squares it is true for the square roots - which is what you want. – Mark Bennet Sep 09 '19 at 06:08
3

@gughy No. You do it as follows:\begin{align}\sqrt2+\sqrt3<\sqrt{26}&\iff\left(\sqrt2+\sqrt3\right)^2<26\&\iff2+3+2\sqrt6<26\&\iff2\sqrt6<21.\end{align}Can you take it from here? – José Carlos Santos Sep 09 '19 at 06:20

score 6 · Answer 2 · answered Sep 09 '19 at 05:33

6

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

answered Sep 09 '19 at 05:33

DeepSea

77,651

5

Probably, but it doesn't address OP's immediate concern. :) – Allawonder Sep 09 '19 at 09:09

score 0 · Answer 3 · answered Sep 09 '19 at 05:37

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.

achille hui · Answer 4 · 2019-09-09T05:49:35.160

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.

Can you square both sides in a proof?

4 Answers4