Is there another way of solving $\frac{\frac{\sqrt{a}}{\sqrt{2}}}{\frac{a}{2}}$. Without the method of multiplying by the inverse?

Question

So, with this: $$\frac{\frac{\sqrt{a}}{\sqrt{2}}}{\frac{a}{2}}$$

I could just do $${\frac{\sqrt{a}}{\sqrt{2}}}\times{\frac{2}{a}}$$

But I was wondering if there was a way of simplifying the $\sqrt{a}$ with $a$ or the $\sqrt{2}$ and $2$. I don't know, it seems there should exist a way of doing it. Another way of looking at a division of fractions, a more fundamental way, and using these similar expressions in the numerator and denominator to simplify it.

Answer 1

$$\frac{\frac{\sqrt a}{\sqrt 2}}{\frac{a}{2}}=\frac{\left(\frac{a}{2}\right)^{1/2}}{\frac{a}{2}}=\left(\frac{a}{2}\right)^{1/2-1}=\sqrt{\frac{2}{a}}$$

First step:

$$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}=\left(\frac{a}{b}\right)^{1/2}$$

Second step:

$$\frac{a^m}{a^n}=a^{m-n}$$

Third step:

$$\left(\frac{a}{2}\right)^{1/2-1}=\left(\frac{a}{2}\right)^{-1/2}=\left(\frac{2}{a}\right)^{1/2}=\sqrt{\frac{2}{a}}$$

Answer 2

Since $\dfrac{\sqrt a}{\sqrt2}\times\dfrac{\sqrt a}{\sqrt 2}=\dfrac{a}{2}$, we have $\dfrac{\left(\frac{a}{2}\right)}{\left(\frac{\sqrt{a}}{2}\right)}=\dfrac{\sqrt{a}}{\sqrt2}$. By taking reciprocals of both sides, we see that $$ \dfrac{\left(\frac{\sqrt{a}}{2}\right)}{\left(\frac{a}{2}\right)}=\dfrac{\sqrt2}{\sqrt a}=\sqrt{\dfrac{2}{a}}\, . $$