2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ You can't, unless you regard $2\sqrt2$ as simpler. $\endgroup$ – almagest Jun 22 '16 at 13:31
  • 1
    $\begingroup$ Assuming you mean $i\sqrt8$. Incidentally, $-i\sqrt8$ also works. The convention about $\sqrt a$ denoting the positive value does not really apply for $a<0$. $\endgroup$ – almagest Jun 22 '16 at 13:33
  • 2
    $\begingroup$ @BloodDrunk Depends on what you think is simpler... Is $i\cdot\sqrt{8}$ simpler than $i\cdot 2\cdot \sqrt{2}$? $\endgroup$ – 5xum Jun 22 '16 at 13:35
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jun 22 '16 at 13:35
  • $\begingroup$ 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. $\endgroup$ – BloodDrunk Jun 22 '16 at 13:37
1
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ The regular square root is always conventionally read as the positive root. It is only when you do algebra and take the square root that you consider negative values. $\endgroup$ – The Great Duck Jun 22 '16 at 13:55
  • 1
    $\begingroup$ @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". $\endgroup$ – Najib Idrissi Jun 22 '16 at 13:56
  • $\begingroup$ Yeah there is. Positive complex numbers are one's that lie within the first quadrant of the complex number line. Of course, i itself is neither positive nor negative, but I would imagine the convention to follow would be to presume positivity to mean the first quadrant if it is a possible solution. $\endgroup$ – The Great Duck Jun 22 '16 at 18:01
  • $\begingroup$ @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 $\mathbb{C}$). $\endgroup$ – Najib Idrissi Jun 22 '16 at 18:37
  • $\begingroup$ I shall do that. Edit: O.O $\endgroup$ – The Great Duck Jun 22 '16 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.