Simplifying fractions with radicals: $1/(2\sqrt{2})$ vs $\sqrt{2}/4$

Question

I just spent half an hour trying to figure out why I'm mathing wrong because a solution that I found was in the form $\frac{1}{2\sqrt{2}}$, but I was coming up with $\frac{\sqrt{2}}{4}$. This started off in the form $\frac{\sqrt{2}/2}{2}$.

Despite the fact that all three forms are equivalent, I can't think of any way to get to the all-in-the-denominator expression from... anywhere.

So, specifically, I was going $\frac{\sqrt{2}/2}{2} = \frac{\sqrt{2}}{4}$, but the solution was expressed as $\frac{1}{2\sqrt{2}}$. Is this the more obvious formulation of this fraction?

Answer 1

$$\frac{\sqrt{2}}{4} = \frac{\sqrt{2}}{2\cdot 2} = \frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{1}{2}\cdot \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{1}{2\sqrt{2}} $$

We rewrite the denominator three times to see that the square root of $2$ in the numerator cancels, leaving a factor of $2\sqrt{2}$ in the denominator and a $1$ up top.

Answer 2

Hint $\ $ Rationalize the $\, \color{#c00}{numerator}\ \ \dfrac{\color{#c00}{\sqrt 2} }4 = \dfrac{\sqrt{2}\cdot \sqrt 2 } {\ \ 4\cdot \sqrt 2} = \dfrac{2}{4\sqrt{2}} =\dfrac{1}{2\sqrt 2 }$

Remark $\ $ Though not used as frequently as rationalizing the denominator, this does prove useful in various contexts, e.g. specializing the quadratic formula when the leading coeff $\,a\to 0.$

$$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\ =\ \dfrac{2c}{-b \pm \sqrt{b^2-4\:a\:c}}$$

As $\,a\to 0,\,$ the latter gives the root $\,x = -b/c\,$ of $\,bx+c\,\ (= ax^2\!+bx+c\,$ when $\,a=0).$