How do you rationalize $1/(1+sqrt3-sqrt5)$ ?

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Answer 1 · 2015-05-24T17:26:13

The quick answer is: multiply both numerator (top) and denominator by:

$(1+sqrt(3)+sqrt(5))(1-sqrt(3)-sqrt(5))(1-sqrt(3)+sqrt(5))$

Quick to say, but a little slow to do...

Let's take it one step at a time...

$(1+sqrt(3)-sqrt(5))(1+sqrt(3)+sqrt(5))$

$=(1+sqrt(3))^2-sqrt(5)^2$

$=1+2sqrt(3)+3-5$

$=-1+2sqrt(3)$

For slightly complex reasons, we can reverse the sign on all occurences of $sqrt(3)$ to deduce:

$(1-sqrt(3)-sqrt(5))(1-sqrt(3)+sqrt(5))=-1-2sqrt(3)$

So

$(1+sqrt(3)-sqrt(5))(1+sqrt(3)+sqrt(5))(1-sqrt(3)-sqrt(5))(1-sqrt(3)+sqrt(5))$

$=(-1+2sqrt(3))(-1-2sqrt(3))$

$=(1-2sqrt(3))(1+2sqrt(3))$

$=1^2-(2sqrt(3))^2$

$=1-12=-11$

So

$1/(1+sqrt(3)-sqrt(5)) = -((1+sqrt(3)+sqrt(5))(1-sqrt(3)-sqrt(5)) (1-sqrt(3)+sqrt(5)))/11$

How do you rationalize 1/(1+sqrt3-sqrt5)1/(1+sqrt3-sqrt5)?