# Write this surd in its simplest form.

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.

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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.
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.