# Radical question simplify?

How would you simplify the following radical.

$$\sqrt\frac1B$$

I am kind of confused do I multiply the numerator and denominator by B.

-

$\sqrt{\frac{1}{B}} = \sqrt{\frac{1}{B}\frac{B}{B}} = \sqrt{\frac{B}{B^2}} = \frac{\sqrt{B}}{\sqrt{B^2}} = \frac{\sqrt{B}}{B}$

of course assuming that $B \neq 0$.

Although I am not sure which is more simple or appealing.

-
You need $B>0$ because it's inside the root, merely $B\neq 0$ is not enough. – Asaf Karagila Jun 13 '12 at 20:43
@William: Agree with your comment on simplification. The "unsimplified" version can be approximated on the calculator by entering $B$, pressing reciprocal, pressing square root. The "simplified" version takes more keystrokes. A form that is simple for one purpose may not be so simple for another. – André Nicolas Jun 13 '12 at 20:51

$$\sqrt\frac1B=\frac1{\sqrt B}=\frac1{\sqrt B}\cdot\frac{\sqrt B}{\sqrt B}=\frac{\sqrt B}{(\sqrt B)^2}=\frac{\sqrt B}B$$

Of course that we have to have $B>0$ for this to be meaningful in the context of real numbers.

-
You mention both of what I would consider "simpler" forms: $\dfrac{1}{\sqrt{B}}$ and $\dfrac{\sqrt{B}}{B}$; however, I see nothing complicated about $\sqrt{\dfrac1B}$. (+1) – robjohn Jun 13 '12 at 21:03
@robjohn: I completely agree. However at a precalc level I could see why $\sqrt\frac1B$ may seem a bit complicated. – Asaf Karagila Jun 13 '12 at 21:52