I'm working my way through the videos on the Khan Academy, and have a hit a road block. I can't understand why the following is true: $$\frac{6}{\quad\frac{6\sqrt{85}}{85}\quad} = \sqrt{85}$$

  • $\begingroup$ I'm not even sure how to read this fraction. If you know some $\LaTeX$ please use this to enhance the equation; otherwise please at least use parenthesis to make the readability well-defined. $\endgroup$
    – Asaf Karagila
    Jun 26, 2011 at 9:26
  • $\begingroup$ At the moment what you've written is $6/(6\sqrt{85}/85) = 6\cdot 85/(6\sqrt{85}) = 85/\sqrt{85} = \sqrt{85}$, which is clearly not equal to 6. Can you copy down the fraction exactly? $\endgroup$ Jun 26, 2011 at 9:34
  • $\begingroup$ As it's currently written: $\dfrac{6}{6 \sqrt {85}} = \sqrt {85}$ is not true. But I am struggling to come up with what was really meant. $\endgroup$
    – davidlowryduda
    Jun 26, 2011 at 22:20

3 Answers 3


With respect to handling square roots in the denominator:

Say, for example, you have $\displaystyle \frac {a}{\sqrt{b}}$.

If you want a "square-root free" denominator, remember that you can multiply any fraction by the number $1$ (or its equivalent) without changing its value. In this case, we want to multiply the given fraction by $\displaystyle \frac{\sqrt{b}}{\sqrt{b}} = 1$. Can you see why we would want to choose this particular representation of 1?

Now, $$\frac{\sqrt{b}}{\sqrt{b}}\cdot \frac {a}{\sqrt{b}} = \frac{a\cdot \sqrt{b}}{(\sqrt{b})^2} = \frac{a\cdot \sqrt{b}}{b}$$

And if the greatest common divisor of a and b > 1, then we can reduce $\displaystyle \frac ab$ by dividing each of $a$ and $b$ by their greatest common divisor (i.e., reducing the fraction $\displaystyle \frac ab$.

Now, we take a look at a slightly more complicated situation, as in your problem, where the denominator is itself a fraction given by $\displaystyle \frac{6}{\frac{6\sqrt{85}}{85}}$

We can first multiply numerator and denominator by 85, then proceed as we did in the case above: $$\frac{6}{\frac{6\sqrt{85}}{85}}\cdot \frac{85}{85} = \frac{6\cdot 85}{6\cdot\sqrt{85}} = \frac {85}{\sqrt{85}} = \frac {85}{\sqrt{85}}\cdot \frac{\sqrt{85}}{\sqrt{85}} = \frac{85\cdot \sqrt{85}}{85} = \sqrt{85} $$

More simply, once we have reduced the expression to $\displaystyle \frac{85}{\sqrt{85}}$, we notice that $85 = (\sqrt{85})^2$, and so, this, together with the fact that we can then cancel a common factor of both the numerator and the denominator, we have:

$$\frac{85}{\sqrt{85}} = \frac{(\sqrt{85})^2}{\sqrt{85}} = \sqrt{85}$$

Note: When you have enough experience simplifying fractions and fractions with fractions with roots, etc., many of the steps which I've displayed explicitly above will be "second nature" for you, and you'll be able to manipulate such expressions with far less in the way of computations than I've shown above.


No one seems to have posted the really simple way to do this yet:

$$ \frac{85}{\sqrt{85}} = \frac{\sqrt{85}\sqrt{85}}{\sqrt{85}} $$

and then cancel the common factor.

  • $\begingroup$ No one? You're blind-sighted, my friend: $$\frac{85}{\sqrt{85}} = \frac{(\sqrt{85})^2}{\sqrt{85}} = \sqrt{85}$$ $\endgroup$
    – amWhy
    Jul 14, 2014 at 16:13

Are you sure you entered that right? It looks like you have $$(6) \over ({6 \sqrt{85}\over 85})$$ which would be equal to $$\sqrt{85}.$$

Note: this answer was written before the original question was edited. The original question was asking why (6)/(6√85/85) was equal to 6.

  • $\begingroup$ Yeh, sorry thats correct. I am not 'why' though. $\endgroup$
    – Ray
    Jun 26, 2011 at 9:56
  • $\begingroup$ Ohh. I think I understand now.I was getting confused with how to handle sqrt in the denominator $\endgroup$
    – Ray
    Jun 26, 2011 at 10:03

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