I'm not Even sure I'm using the correct terminology here, but I'm helping out my high school daughter with her algebra and was presented with the following rule:

$\sqrt3/\sqrt2 = \sqrt{3/2}$

Accepting that this is true (which a calculator did demonstrate), I should be able to step through a proof using the process for simplifying radicals in the denominator. So, given:


I then multiply by the value of the denominator divided by itself:

$\sqrt3/\sqrt2 * \sqrt2/\sqrt2$

to get:

$(\sqrt3 * \sqrt2)/2$

which equals:


And that's where I get stuck. What can I do to get to:


Other than using a calculator of course. :)

  • 2
    $\begingroup$ Maybe it's easier in this case just to say that $\frac{\sqrt{3}}{\sqrt{2}}.\frac{\sqrt{3}}{\sqrt{2}} = \frac{3}{2}$. $\endgroup$ Jul 24, 2011 at 16:02

2 Answers 2


From my perspective (this is probably not the only way to arrive at the property you're asking about):

$$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$$ (with $a,b\ge 0$) is a special case of $$\left(\frac{a}{b}\right)^x=\frac{a^x}{b^x}$$ (or, equivalently, $(ab)^x=a^xb^x$), the Distributive Property of Exponentiation over Division (or Multiplication).

The Distributive Property of Exponentiation over Multiplication, for integer exponents, comes from associativity and commutativity of multiplication: $$(ab)^n=\underset{n\text{ factors of }ab}{\underbrace{(ab)\cdots(ab)}}=\left(\underset{n\text{ factors of }a}{\underbrace{a\cdots a}}\right)\left(\underset{n\text{ factors of }b}{\underbrace{b\cdots b}}\right)=a^nb^n$$

The property extends naturally to non-integer exponents.

original answer:


Is that what you were looking for?

  • $\begingroup$ Not quite. You used the same rule that I'm trying to prove. :) I need to know how to prove $\sqrt6/\sqrt4 = \sqrt{6/4}$ $\endgroup$ Feb 4, 2011 at 2:40
  • 1
    $\begingroup$ @Brian: Then I'm not clear on what you're trying to prove. If you're trying to justify that $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$, it's a special case of $(\frac{a}{b})^x=\frac{a^x}{b^x}$, which is the distributive property of exponents over division (or multiplication), and follows (for integer exponents) from commutativity and associativity of multiplication (and then gets extended to non-integer exponents). $\endgroup$
    – Isaac
    Feb 4, 2011 at 2:44
  • $\begingroup$ That's exactly what I needed. Thanks. $\endgroup$ Feb 4, 2011 at 2:46
  • $\begingroup$ @Brian: I've edited my answer to better fit what you were looking for, I think. $\endgroup$
    – Isaac
    Feb 4, 2011 at 2:55
  • $\begingroup$ Yep. You definitely got me on the right track. My solution simply replaced $\sqrt{x}$ with $x^{1/2}$ and the distributive property application did the trick. Thanks! $\endgroup$ Feb 4, 2011 at 3:03

While it's a simple consequence of the distributive law for exponents, one can view it from a more general perspective. Namely, both $\rm\ \sqrt{3/2}\ $ and $\rm\ \sqrt{3}/\sqrt{2}\ $ are roots of $\rm\ x^2 - 3/2 $. But this polynomial has a unique positive root since it is increasing from $\rm\: -3/2\ $ to $\: +\infty\ $ on $\ [0,\infty)\:.\ $ Therefore this uniqueness theorem implies that the two roots are equal. One can frequently apply analogous techniques to more much complicated expressions involving messy nested radicals - inferences far removed from the laws of exponents. Thus I point out once again what I have emphasized in many varied posts here: uniqueness theorems provide powerful tools for proving equalities.

  • $\begingroup$ How do you know that both forms are roots of $x^2-3/2$ without using properties of exponents? $\endgroup$
    – Isaac
    Feb 4, 2011 at 4:59
  • $\begingroup$ His method only requires using the rule for integer exponents which most people accept without difficulty as it follows easily from the associativity and commutativity of multiplication. To justify the rules for arbitrary real,as you surely know, one would first have to define such an expression and complete a few short proofs. Granted, if the exponent is rational defining what the symbol means isn't too difficult. $\endgroup$
    – Benji
    Feb 4, 2011 at 5:21

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