I have a fraction -
$-\frac{1}{3}$
Which could either mean the value of fraction is $\frac{-1}{3}$ or $\frac{1}{-3}$ Note the minus sign
Now, what is the sqaure root of the fraction? I tried and I got this -
$\sqrt{-\frac{1}{3}}$
$=\frac{i}{\sqrt3}$ or $\frac{1}{\sqrt3i}$
Now what is the actual square root? Or is it both? Am I going wrong somewhere? Which one should I use in my calculations?
$\endgroup$ 34 Answers
$\begingroup$Good question! You have discovered that it's not possible to define a square root function in the complex numbers that obeys the rule $\sqrt{ab}=\sqrt{a}\,\sqrt{b}$ (or the equivalent $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$, with $b\ne0$).
You get the same dilemma, in an easier way, by considering $$ i=\sqrt{-1}=\sqrt{\frac{1}{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{i}=-i $$ Note that this is clearly wrong, which doesn't tell us that mathematics is contradictory, but that we have used an unproved (and unprovable) property, namely that we can define a square root function satisfying the rule above.
Note that the false argument produces both complex numbers whose square is $-1$, the same happens in your argument.
A suggestion: never use the symbol $\sqrt{-1}$, because it suggests the possibility to apply the wrong property. Neither use $\sqrt{z}$, for the same reason, unless $z$ is a real number with $z\ge0$.
$\endgroup$ 5 $\begingroup$In a sense it's both, because both $(\frac{i}{\sqrt{3}})^2$ and $(\frac{1}{\sqrt3 i})^2$ are equal to $-\frac{1}{3}$. So both could fairly be said to be square roots of $-\frac{1}{3}$.
However, we like $\sqrt x$ to be a function, so it can only give one value. This means we have to choose which of the two values above to call the square root (or principal square root) of $-\frac{1}{3}$.
The same problem arises with positive numbers: both $2$ and $-2$ square to $4$. With positive numbers, we're used make sure to always choose the positive root to call the square root, ie $\sqrt4 = 2$ and not $-2$.
Over the complex numbers, we can come up with a variety rules to choose the principal square root; one of them is "take the root that's is in the upper-half plane (with imaginary part positive)". Under this rule, which is the usual rule mathematicians use, $\frac{i}{\sqrt3}$ is the principal root. But under a different rule it might not be.
$\endgroup$ $\begingroup$As you could see taking imaginary unit on the numerator the two roots are symmetric: $r_1=\frac i{\sqrt3}$ and $r_2=-\frac i{\sqrt3}$. That's because every number has 2 square roots, 3 cubic roots and so on (at least in complex number) and they form a n-poligon in the complex plane (where n is the index of the root). So if you don't have other condition on your equation/exercise/formula you have to consider them all or, in the most case you can consider the one in the first quarter, where real and imaginary part are both positive.
$\endgroup$ $\begingroup$You cannot use the rule $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ when $a,b$ are not both positive.
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