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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?

Arulx Z
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4 Answers4

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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 $z\ge0$.

egreg
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  • There is no need to explain that rule. The OP has simply asked that whether they are square roots of $\frac{-1}{3}$ or not? And, the answer is YES! – Sufyan Naeem May 09 '15 at 13:26
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    @SufyanNaeem The OP's problem is falling into a contradiction! It can't be $i/\sqrt{3}=1/(i\sqrt{3}$, can it? – egreg May 09 '15 at 13:35
  • I appreciate that you caught the miconception from the post and tried to explain it to OP but this is still not an answer. Better as a comment or to share the link where @Isaac has already given great answer over this matter. I repeat that OP asked that whether they are square roots of $\frac{-1}{3}$ or not??? – Sufyan Naeem May 09 '15 at 13:43
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    @SufyanNaeem The question reads “what is the actual square root?”, notice the definite article. The answer is that a square root function cannot be defined so that it allows that kind of computations. An abbreviated form of my answer could be “there is nothing that can be called the square root” and I gave an explanation why. – egreg May 09 '15 at 14:02
  • the question reads "What is the actual square root?" plus "or is it both?". The exact answer is "YES" which follows the definition of $n^{th}$ root. And of course, $$\frac{1}{\sqrt{3}i} \neq \frac{i}{3}$$ but, still they both are square roots of $-\frac{1}{3}$. – Sufyan Naeem May 09 '15 at 17:07
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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.

Zach Effman
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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.

AlienRem
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You cannot use the rule $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ when $a,b$ are not both positive.

wythagoras
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