# How does $\sqrt{n+1}-\sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1}+\sqrt{n}}$?

How does $\sqrt{n+1}-\sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1}+\sqrt{n}}$ ?

What are the exact steps to get to the right side from the left?

• You can't because they are not equal. Check your signs. It should be $+$ somewhere, perhaps in the denominator. – user147263 Feb 17 '15 at 5:01
• There should be a $+$ in the denominator – AvZ Feb 17 '15 at 5:03

$$\bigg( \sqrt{n+1}-\sqrt{n} \bigg) \cdot 1 = \bigg( \sqrt{n+1}-\sqrt{n} \bigg) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}$$
$$\bigg( \sqrt{a} - \sqrt{b} \bigg) \cdot \bigg( \sqrt{a} + \sqrt{b} \bigg) = a - b$$
hint: $1 = (n+1) - n = \left(\sqrt{n+1}\right)^2 - \left(\sqrt{n}\right)^2 = \left(\sqrt{n+1} + \sqrt{n}\right)(?-?)$
Hint: Evaluate $$\sqrt{n+1}-\sqrt{n} = (\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$$