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Why is it in math, $\sqrt{ab}$=$\sqrt{a}$$\sqrt{b}$? I get why this is the case for any other power instead of $1/2$. For instance, if the power was for, then $(ab)^4$=$(a)^4$$(b)^4$ because on both sides, there will be 4 a's and 4 b's. But somehow, this proof doesn't seem as intuitive when used to prove that $\sqrt{ab}$=$\sqrt{a}$$\sqrt{b}$. So can someone please prove/explain to me why $\sqrt{ab}$=$\sqrt{a}$$\sqrt{b}$? Please don't do this too rigorously, just explain it at the level of a high school pre-calc student please.

Archer
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Ethan Chan
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3 Answers3

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Let $a,b\ge0$ (the version with complex numbers of this identity does not hold). By definition, you want to prove that $\alpha=\sqrt a\sqrt b$ is a non-negative real number such that $\alpha^2=ab$. Since $\sqrt a$ and $\sqrt b$ are non-negative real numbers, $\alpha$ is too. Finally, $\alpha^2=\left(\sqrt a\sqrt b\right)^2=\left(\sqrt a\right)^2\left(\sqrt b\right)^2=ab$.

  • But how does the fact that when they are squared, they yield the same results prove that they are the same? Because when -2 and 2 are squared, they are both equal to 2? But that doesn't mean they are the same. – Ethan Chan Jul 31 '18 at 08:09
  • @EthanChan That's why I kept writing "non-negative" all those times, and why the definition of square root - or $\sqrt[2n]{\bullet}$, for that matter - explicitly requires the root to be the non-negative solution to the equation $t^2=a$. Since the map $x\mapsto x^2$ (or $x\mapsto x^{2n}$ with $n\ge 1$) is strictly increasing on $[0,\infty)$, there is at most one such number in this interval. –  Jul 31 '18 at 08:13
  • Oh right, never mind. But then, how about for something like a cube root? – Ethan Chan Jul 31 '18 at 08:14
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    https://math.stackexchange.com/questions/1700731/prove-that-sqrtnx-cdot-sqrtny-sqrtnx-cdot-y?rq=1 –  Jul 31 '18 at 10:22
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The statement is only true if $a, b \geq 0$.

Let $a, b \geq 0$. Let $u = \sqrt{a}$; let $v = \sqrt{b}$. Since $\sqrt{x}$ is the principal (nonnegative) square root of $x$, $u, v \geq 0$, $u^2 = a$, $v^2 = b$. \begin{align*} \sqrt{ab} & = \sqrt{u^2v^2} && \text{substitution}\\ & = \sqrt{uuvv} && \text{by definition}\\ & = \sqrt{u(uv)v} && \text{associativity of multiplication}\\ & = \sqrt{u(vu)v} && \text{commutativity of multiplication}\\ & = \sqrt{(uv)(uv)} && \text{associativity of multiplication}\\ & = \sqrt{(uv)^2} && \text{by definition}\\ & = |uv| && \text{take principal square root}\\ & = uv && \text{$u,v \geq 0 \implies uv \geq 0 \implies |uv| = uv$}\\ & = \sqrt{a}\sqrt{b} && \text{substitution} \end{align*}

N. F. Taussig
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First of all: it is not always $\sqrt{ab}=\sqrt{a}\sqrt{b}$, there are some conditions you have to check first e.g. you want to have $a,b\geq 0$ or even more trivial $a,b\in\mathbb R$ as the root of complex numbers turns out to be...well, rather complex to handle.

Assuming that all of these conditions are met and you accept the fact that $(x\cdot y)^n=x^n\cdot y^n$ for $n\in\mathbb N$ maybe the following helps you:

$$\left(\sqrt[n]{x}\cdot\sqrt[n]{y}\right)^n = \sqrt[n]{x}\cdot\sqrt[n]{y} = x\cdot y=\left(\sqrt[n]{x\cdot y}\right)^n$$

Thus we have $$\left(\sqrt[n]{x}\cdot\sqrt[n]{y}\right)^n =\left(\sqrt[n]{x\cdot y}\right)^n$$ which yields $$\sqrt[n]{x}\cdot\sqrt[n]{y}=\sqrt[n]{x\cdot y}.$$

Hirshy
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