How to prove $\frac{1}{\sqrt{a} + \sqrt{b}} = \sqrt{a} - \sqrt{b}$?

Question

My daughter is learning how to rationalise surds for her school exams. One example being worked through is the following:

$\frac{1}{\sqrt{6} + \sqrt{5}}$

In the tutorials she is following, the first step is rearranging to arrive at:

$\sqrt{6} - \sqrt{5}$

She doesn't know the reason for this first step - can hasn't been able to find anything to suggest why it works.

$\frac{1}{\sqrt{6} + \sqrt{5}} \to \sqrt{6} - \sqrt{5}$

Is this just something that just needs to be remembered? She doesn't like to do something without knowing why, hence posting this question.

Note that the equality would imply $1=(\sqrt a - \sqrt b)(\sqrt a +\sqrt b)=a-b$ but it is not always true that $a-b=1$. — lulu, May 14 '22 at 12:56
Even better, for $a=b$ we would obtain $\frac{1}{\sqrt{a} + \sqrt{b}} = 0$ for all $a$ and $b$. — Dietrich Burde, May 14 '22 at 12:59
@Joe Indeed, the usual homework is exactly what you said. Unfortunately the OP copied it wrongly. — Dietrich Burde, May 14 '22 at 13:09
@AnotherUser $\sqrt{4}-\sqrt{1}=1$, but this is neither equal to $\frac{1}{3}$ — Peter, May 14 '22 at 13:14

Joe · Accepted Answer · 2022-05-14T13:30:29.653

In general, $$ \frac{1}{\sqrt a+\sqrt b}=\frac{\sqrt a-\sqrt b}{(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)}=\frac{\sqrt{a}-\sqrt{b}}{(\sqrt a)^2-(\sqrt b)^2}=\frac{\sqrt a-\sqrt b}{a-b} \tag{*}\label{*} \, , $$ using the difference of two squares (since $(x+y)(x-y)=x^2-y^2$, we know that $(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=(\sqrt a)^2-(\sqrt{b})^2$). In the case $a=6,b=5$, we see that $$ \frac{1}{\sqrt{6}+\sqrt{5}}\stackrel{\eqref{*}}{=}\frac{\sqrt{6}-\sqrt{5}}{6-5}=\frac{\sqrt{6}-\sqrt{5}}{1}=\sqrt{6}-\sqrt{5} \, . $$ For a school student, it might be clearer if we derive this result directly: \begin{align} \frac{1}{\sqrt{6}+\sqrt{5}} &= \frac{\sqrt{6}-\sqrt{5}}{(\sqrt6+\sqrt5)(\sqrt6-\sqrt5)} \\[5pt] &= \frac{\sqrt6-\sqrt5}{(\sqrt6)^2-(\sqrt5)^2}\\[5pt] &= \frac{\sqrt6-\sqrt5}{6-5} \\[5pt] &= \frac{\sqrt6-\sqrt5}{1} \\[5pt] &= \sqrt6-\sqrt5 \, . \end{align} Finally, consider that the equality $$ \frac{1}{\sqrt6+\sqrt5}=\sqrt6-\sqrt5 $$ is equivalent to $$ 1=(\sqrt6+\sqrt5)(\sqrt6-\sqrt5) \, . $$ (In general, $\frac{a}{b}=c$ means $a=b \times c$.) This gives us a straightforward of checking that the result is indeed correct.

score 3 · Answer 2 · answered May 14 '22 at 13:23

Alright. I understand your question now, so please make sure to post 'full' questions like the one you posted later (otherwise it may get downvoted).

Now, here we are doing the following; We have the fraction; \begin{align*} \frac{1}{\sqrt{6}+\sqrt{5}} \end{align*} We can multiply both the numerator and the denominator by $\sqrt{6}-\sqrt{5}$ to get; \begin{align*} \frac{1}{\sqrt{6}+\sqrt{5}} &=\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\\ &=\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}\cdot \sqrt{6}-\sqrt{30}+\sqrt{30}-\sqrt{5}\cdot \sqrt{5}}\\ &=\frac{\sqrt{6}-\sqrt{5}}{6-5}\\ &=\frac{\sqrt{6}-\sqrt{5}}{1}\\ &=\sqrt{6}-\sqrt{5}. \end{align*}

So we are done. However, in general, we have; \begin{align*} \frac{1}{\sqrt{a}+\sqrt{b}} &=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\\ &=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}\cdot \sqrt{a}-\sqrt{ab}+\sqrt{ab}-\sqrt{b}\cdot \sqrt{b}}\\ &=\frac{\sqrt{a}-\sqrt{b}}{a-b}. \end{align*}

So in general it is; \begin{align*} \frac{1}{\sqrt{a}+\sqrt{b}} &=\frac{\sqrt{a}-\sqrt{b}}{a-b}. \end{align*}

How to prove $\frac{1}{\sqrt{a} + \sqrt{b}} = \sqrt{a} - \sqrt{b}$?

2 Answers2