why is $\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$ correct?

Question

Im trying to understand how the equation below was generated. A step-by-step solution would be awesome. $$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$

Answer 1

$(a+b)(a-b) = a^2 - b^2.$

Put $\ a = \sqrt{n+1},\quad b = \sqrt{n}\ .$

$\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right) = \left(\sqrt{n+1}\right)^2 - \left(\sqrt{n}\right)^2 = n+1 - n =1.$

Therefore, starting with the RHS and rationalising the denominator,

\begin{align}\frac{1}{\sqrt{n+1}+\sqrt{n}} =\frac{1}{\sqrt{n+1}+\sqrt{n}} \times 1\\ \\ =\frac{1}{\sqrt{n+1}+\sqrt{n}} \times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}\\ \\ = \frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ \\ = \frac{\sqrt{n+1}-\sqrt{n}}{1}\\ \\ = \sqrt{n+1}-\sqrt{n}\ .\\ \end{align}

Answer 2

Recall that $$\frac{1}{x}$$ means something so that, when you multiply it by $x$, you get $1$. So check that $\sqrt{n+1}-\sqrt{n}$ is, indeed, something so that, when you multiply it by $\sqrt{n+1}+\sqrt{n}$, you get $1$.