# Quick Sum

Is this surd sum exactly 3?

## Problem

According to my calculator, I see that

$$

\frac{1}{\sqrt{1}+ \sqrt{2}}+ \frac{1}{\sqrt{2}+ \sqrt{3}} + \text{and so on up to}+\frac{1}{ \sqrt {15}+ \sqrt{16}} = 3.0000000000

$$

How could I prove that the answer is, indeed, exactly 3?

Did you know ... ?

Mathematicians often use calculating aids to help form an opinion concerning the likely numerical answer to a problem involving irrational numbers, but will always seek a full proof which does not rely on the calculator to complete the problem.

Mathematicians often use calculating aids to help form an opinion concerning the likely numerical answer to a problem involving irrational numbers, but will always seek a full proof which does not rely on the calculator to complete the problem.

## Getting Started

Try rationalising the denominators.

## Student Solutions

Let

$$X= \frac{1}{\sqrt{1}+ \sqrt{2}}+ \frac{1}{\sqrt{2}+ \sqrt{3}} + \text{and so on up to}+\frac{1}{ \sqrt {15}+ \sqrt{16}} $$

When surds appear in the denominator the first step is almost always to rationalise the denominator. We see that

$$

X= \frac{1}{\sqrt{1}+ \sqrt{2}}\left(\frac{\sqrt{1}- \sqrt{2}}{\sqrt{1}- \sqrt{2}}\right)+\frac{1}{\sqrt{2}+ \sqrt{3}}\left(\frac{\sqrt{2} -\sqrt{3}}{\sqrt{2}- \sqrt{3}}\right) + \dots+\frac{1}{ \sqrt {15}+ \sqrt{16}}\left( \frac{\sqrt {15}-\sqrt{16}}{ \sqrt {15}-\sqrt{16}}\right)$$

For each term the denominators multiply to $-1$. For example

$$

(\sqrt{2}+ \sqrt{3})(\sqrt{2}- \sqrt{3}) = (\sqrt{2})^2-(\sqrt{3})^2= 2-3= -1

$$

Thus we have

$$

X= -(\sqrt{1}- \sqrt{2})-(\sqrt{2}+ \sqrt{3}) -\dots-(\sqrt {15}-\sqrt{16})= -\sqrt{1}+\sqrt{16} = 3

$$