How to simplify $\sqrt{-8}$

How would I go about simplifying square root of $-8$?

I know I can rewrite that as $\sqrt{(-1)(8)}$, and then I would get $i\sqrt{8}$, but how do I simplify that $8$ further?

Thanks for your help.

• You can't, unless you regard $2\sqrt2$ as simpler. – almagest Jun 22 '16 at 13:31
• Assuming you mean $i\sqrt8$. Incidentally, $-i\sqrt8$ also works. The convention about $\sqrt a$ denoting the positive value does not really apply for $a<0$. – almagest Jun 22 '16 at 13:33
• @BloodDrunk Depends on what you think is simpler... Is $i\cdot\sqrt{8}$ simpler than $i\cdot 2\cdot \sqrt{2}$? – 5xum Jun 22 '16 at 13:35
• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Jun 22 '16 at 13:35
• Thanks for your help, I was just curious about trying to simplify the root as much as possible. That expression will be good. :) @N. F. Taussig Will do. – BloodDrunk Jun 22 '16 at 13:37

The formula "$\sqrt{-8}$", although "grammatically correct", does not define a number. There are two complex numbers, $i \sqrt{8} = 2i \sqrt{2}$ and $-2 i \sqrt{2}$, that square to $-8$. Since there is no reasonable way to say that one is better than the other, "$\sqrt{-8}$" is not a well-defined object, pretty much on the same level as $0/0$ or $\infty-\infty$. Each of the number $2i\sqrt{2}$ and $-2i\sqrt{2}$ are both called "a square root" of $-8$, but there is no such thing as "the square root". The best you can get is probably $\pm i \sqrt{8}$ or $\pm 2i\sqrt{2}$, though you have to be careful with this notation and understand that it's not a number.
(When one uses real numbers, one has the convention that, for $x \ge 0$, $\sqrt{x}$ is the unique nonnegative number such that $\sqrt{x}^2 = x$, but of course there is always the other solution $-\sqrt{x}$. With real numbers we can decide to always choose the nonnegative solution, with complex numbers it's not so easy.)
• @TheGreatDuck I'm not sure what this comment is supposed to mean. If you're working in $\mathbb{R}$, then I think I already explained that in the parenthetical comment. If you're working in $\mathbb{C}$, there is no such thing as "positive root". – Najib Idrissi Jun 22 '16 at 13:56
• @TheGreatDuck That's an... odd definition. I can't say I've ever seen anyone use it. Anyway, look up "complex logarithm branches"; you can define $x^{1/2} = \exp(\ln(x)/2)$ but realize that this hinges on choosing a branch of the logarithm (something which is impossible to do on all of $\mathbb{C}$). – Najib Idrissi Jun 22 '16 at 18:37