I am currently in high school and we are studying radicals. I had asked my math teacher why $\sqrt{ab}=\sqrt{a}\sqrt{b}$ (for all a,b>0) and he tries to prove it by arguing that $a^{1/2}*b^{1/2}=(ab)^{1/2}$ (an exponent law). However, I find this proof problematic since $x^{1/2}$ is simply defined as $\sqrt{x}$, so the reasoning is circular.

My view is that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ because once we square both sides we get $ab=ab$. Since we're obviously referring to the positive root, and the function $f(x) = \sqrt{x}$ is injective, it necessarily follows that the original expressions $\sqrt{ab}$ and $\sqrt{a}\sqrt{b}$ are equivalent because for injective functions it is not possible to map distinct elements in the domain to the same element in the range ($ab$). Hence they are equivalent expressions.

My question therefore is, is my proof valid and/or rigorous (I find it convincing but maybe it's wrong; I just want to be clear) and secondly was my teacher's proof correct?

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    $\begingroup$ Your proof is more conceptual, that is, better. The proof offered by your teacher is fine if the exponent law has been proved, and not merely asserted. $\endgroup$ – André Nicolas May 10 '15 at 1:06
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    $\begingroup$ Supposing $\sqrt x$ rigourously defined, your proof is fine. However it relies on the definition of $\sqrt x$. There are are various definitions. The handiest one is $$\sqrt x=\mathrm e^{\tfrac 12\ln x}.$$ $\endgroup$ – Bernard May 10 '15 at 1:08
  • $\begingroup$ Squaring both sides still requires law of exponents because you still have to use the law that permits us to do this : $(\sqrt{a} \sqrt{b})^2=(\sqrt{a})^2(\sqrt{b})^2 $ @AndréNicolas $\endgroup$ – randomgirl May 10 '15 at 1:20
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    $\begingroup$ @randomgirl: It is integer exponents, which is a little easier to establish. $\endgroup$ – André Nicolas May 10 '15 at 1:23
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    $\begingroup$ @randomgirl I believe that's just the associativity law for multiplying real numbers, not an exponent rule. $\endgroup$ – MathematicsStudent1122 May 10 '15 at 1:28

Could we proceed along the following lines, as long as we restrict ourselves to the positive reals? We observe $\sqrt{x}$ to be the unique positive real number $r$ such that $r \cdot r = x$.

Let $m = \sqrt{a}, n = \sqrt{b}$. Then $m \cdot m = a, n \cdot n = b$, and

$$ \begin{align} a \cdot b & = (m \cdot m) \cdot b \\ & = m \cdot (m \cdot b) \\ & = m \cdot (m \cdot (n \cdot n)) \\ & = m \cdot ((m \cdot n) \cdot n) \\ & = m \cdot ((n \cdot m) \cdot n) \\ & = m \cdot (n \cdot (m \cdot n)) \\ & = (m \cdot n) \cdot (m \cdot n) \end{align} $$

where we rely on the associativity and commutativity of multiplication. Hence $\sqrt{ab} = m \cdot n = \sqrt{a} \cdot \sqrt{b}$.

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  • $\begingroup$ Okay, I guess that works as well! Thanks :-) $\endgroup$ – MathematicsStudent1122 May 10 '15 at 2:34
  • $\begingroup$ That's basically your idea, put symbolically (and a couple of steps spelled out a bit more). $\endgroup$ – Brian Tung May 10 '15 at 4:53

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