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This is indeed an identity, but I cannot seem to understand how you would go about proving these two equations are equal algebraically? Do you see a method to prove it? $$\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}.$$

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The method in Brian's answer is called rationalizing the denominator. It is widely used when dealing with algebraic numbers. – Math Gems Apr 4 '13 at 14:08
up vote 7 down vote accepted

Just multiply the denominator by the righthand side and check that you get the numerator:

$$(\sqrt3-1)(\sqrt3+2)=3+2\sqrt3-\sqrt3-2=\sqrt3+1$$

To get the result without knowing it ahead of time, multiply the fraction by a very carefully chosen form of $1$:

$$\frac{\sqrt3+1}{\sqrt3-1}=\frac{\sqrt3+1}{\sqrt3-1}\cdot\frac{\sqrt3+1}{\sqrt3+1}=\frac{4+2\sqrt3}{3-1}=2+\sqrt3\;.$$

The key point is that $(a+\sqrt b)(a-\sqrt b)=a^2-b$, with no square root left.

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Multiply both sides by $\sqrt{3}-1$ and expand.

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Is this considered a proof? $$ \frac{\sqrt 3 + 1}{\sqrt 3 - 1} = \frac{(\sqrt 3 + 1)^2}{(\sqrt 3 - 1)(\sqrt 3 + 1)} = \frac{4 + 2\sqrt 3}{2} = 2 + \sqrt 3 $$

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