How to simplify $\sqrt{3/2}$?

Question

My textbook says to do the following steps to simplify the $\sqrt{3/2}$:

$\sqrt{3/2}=\sqrt{((3/2)\times (2/2))} =\sqrt{6/4} = \sqrt{6}/\sqrt{4}=\sqrt{6}/2.$

However, if the fourth step is valid (1. please explain why and 2. I specifically mean where you "split" the square root into two parts), then why can't you simply do:

$\sqrt{3/2}=\sqrt{3}/\sqrt{2}$?

Why wouldn't that be valid?

My point is, are the two fractions equivalent?

Answer 1

It is true that $\sqrt{3/2} $ is equal to $\sqrt{3}/\sqrt{2}$.

However, the point of the transformations you give it to have a denominator without root.

More generally, for positive $a,b$ it is always true that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/ \sqrt{b}$.

To see that note $(\sqrt{a}\sqrt{b})^2 =(\sqrt{a})^2(\sqrt{b})^2 =ab$, thus $\sqrt{a}\sqrt{b}$ is the square-root of $ab$, that is it is $\sqrt{ab}$. And likewise for the second.

Answer 2

• $$\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{2^2}}=\frac{\sqrt{3}}{\sqrt{4}}=\sqrt{\frac{3}{4}}$$
• $$\sqrt{\frac{3}{2}}=\sqrt{\frac{2\cdot3}{2\cdot2}}=\sqrt{\frac{6}{4}}=\frac{\sqrt{6}}{\sqrt{4}}=\frac{\sqrt{6}}{\sqrt{2^2}}=\frac{\sqrt{6}}{2}$$

And if I read it right:

$$\color{red}{\frac{\sqrt{3}}{2}\ne\frac{\sqrt{6}}{2}}$$

Answer 3

It is true that $$\sqrt{\frac32} = \frac{\sqrt{3}}{\sqrt{2}}. $$

Generally, when one simplifies fractions involving square roots, you want to remove all square roots in the demoninator so

$$\sqrt{\frac32} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{3}\sqrt{2}}{2} $$

You also have that $$\frac{\sqrt{6}}{2} = \frac{\sqrt{3\cdot 2}}{2} = \frac{\sqrt{3}\sqrt{2}}{2}, $$ so you can write the same things in different ways: $$\sqrt{\frac32} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} = \frac{\sqrt{3}\sqrt{2}}{2}. $$ It just depends on what is considered "most simplified."