# Can you find the square root of -1/2 ?

Oct 10, 2017

You can't technically, but it's $\frac{1}{2} i$

#### Explanation:

You can't square root a negative. Because we come up with a problem here it is. Remeber a when you square root a number you get a number when multiplied by itself gives us the number that we just square rooted

E. g. $\sqrt{16} = 4 \cdot 4 \mathmr{and} {4}^{2}$

Quick Example

When we try to square root a negative we realise this isn't possible. Because when you square root a number the root has to be the same number. So to complete the problem below we'll have to split the $- 16$ into $- 1$ and $16$ ($- 1 \cdot 16 = - 16$).

E. g. $\sqrt{- 16} = \sqrt{- 1 \cdot 16}$

Now we know that 16 is a square number, so when its square root is... 4 of course! However, we cannot square the negative one so we rewrite it as $i$ which is called an imaginary number as there is no square root of any negative.

Now onto the question
$\sqrt{- \frac{1}{2}}$

Let's split it into $- 1 \mathmr{and} \frac{1}{2}$

$\sqrt{- 1 \cdot \frac{1}{2}}$

This gives us $\sqrt{\frac{1}{2} i}$

Oct 13, 2017

$\sqrt{- \frac{1}{2}} = \frac{\sqrt{2}}{2} i$

#### Explanation:

Assuming you are asking about the square root of $\left(- \frac{1}{2}\right)$ ...

In common with all non-zero numbers, $- \frac{1}{2}$ has two square roots, but since $- \frac{1}{2}$ is negative, those square roots are non-real Complex numbers.

The imaginary unit $i$ satisfies ${i}^{2} = - 1$

Hence we find:

${\left(\frac{\sqrt{2}}{2} i\right)}^{2} = {\left(\frac{\sqrt{2}}{2}\right)}^{2} \cdot {i}^{2} = \frac{1}{2} \cdot - 1 = - \frac{1}{2}$

So one square root of $- \frac{1}{2}$ is $\frac{\sqrt{2}}{2} i$

We also have:

${\left(- \frac{\sqrt{2}}{2} i\right)}^{2} = {\left(- \frac{\sqrt{2}}{2}\right)}^{2} \cdot {i}^{2} = \frac{1}{2} \cdot - 1 = - \frac{1}{2}$

So the other square root is $- \frac{\sqrt{2}}{2} i$

By convention, when we write $\sqrt{- \frac{1}{2}}$, we mean $\frac{\sqrt{2}}{2} i$, which is known as the principal square root of $- \frac{1}{2}$.

Footnote

Why not simply write:

$\sqrt{- \frac{1}{2}} = \sqrt{- 1 \cdot \frac{1}{2}} = \sqrt{- 1} \cdot \sqrt{\frac{1}{2}} = i \sqrt{\frac{1}{2}}$

I tend to avoid being too hasty with the "rule" $\sqrt{a b} = \sqrt{a} \sqrt{b}$, because it does not always work, once you are starting to deal with square roots of negative and/or complex numbers.

For example:

$1 = \sqrt{1} = \sqrt{- 1 \cdot - 1} \ne \sqrt{- 1} \cdot \sqrt{- 1} = - 1$

On the other hand, the convention that if $x < 0$ then:

$\sqrt{x} = i \sqrt{- x}$

is safe.