Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Express $\dfrac{1}{2+ \sqrt3}$ in its simplest form.

NB: The textbook has the answer as $2 - \sqrt3$ but I can't see how that was achieved.

I tried $\dfrac{1}{2} + \dfrac{1}{\sqrt3}$ and multiplying the top and bottom by $\sqrt3 $ to get $\dfrac{1}{2} + \dfrac{\sqrt3}{3}$ so far.

share|cite|improve this question
In general, $\dfrac{1}{a+b} \neq \dfrac{1}{a} + \dfrac{1}{b}.$ – Brad Jun 2 '14 at 14:10
For example, notice that $$\frac1{2+2}\ne \frac12+\frac12.$$ Similarly, it is also the case that $$\frac1{2+\sqrt3}\ne \frac12 + \frac1{\sqrt3}.$$ – MJD Jun 2 '14 at 14:11
How you split the fraction is a classic and very wrong move. – orion Jun 2 '14 at 14:11

Multiply by $\displaystyle 1 = \frac{2-\sqrt 3}{2-\sqrt 3}$. Use $(a-b)(a+b) = a^2-b^2$ on the denominator. It's called "rationalising the denominator" by multiplying the denominator by the "conjugate surd". You should look up the key phrases in quotes.

share|cite|improve this answer

Any time you are simplifying an expression like $$\frac{c}{a \pm \sqrt{b}},$$ multiply it with $$\frac{a\mp \sqrt{b}}{a\mp \sqrt{b}}$$ which gives you $$\frac{c(a\mp \sqrt{b})}{(a\pm \sqrt{b})(a\mp \sqrt{b})} = \frac{c(a\mp \sqrt{b})}{a^2 - b}$$ which has no more square roots in the denominator.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.