In this blog post, RJ. Lipton mentions an example of common mathematical traps. In particular, that ``square root is not a function''. He shows the following trap:

Start with: $\frac{-1}{1}=\frac{1}{-1}$, then take the square root of both sides: $$ \frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}} $$ hence $$ \frac{i}{1} = \frac{1}{i} \\ i^2=1 \enspace , $$ which contradicts the definition that $i^2=-1$.

Question 1: I know that the square root is not a function because it is multi-valued, but I still can not wrap my head around this example. Where was the problem exactly? Was is that we

  • can not convert $\sqrt{1/-1}$ to $\sqrt{1}/\sqrt{-1}$?
  • both the RHS and LHS are unordered sets?
  • both?

Question 2: Also, does this problem only arise in equalities or in general algebraic manipulation? Because it would be a nightmare when manipulating an expression with fractional powers! Are there easy rules to determine what is safe to do with fractional powers? To see what I mean, there is another example of a similar trap:

One might easily think that $\sqrt[4]{16x^2y^7}$ is equivalent to $2x^{1/2}y^{7/4}$, which is not true for $x=-1$ and $y=1$.

  • $\begingroup$ Related: math.stackexchange.com/questions/445585/… . $\endgroup$ – Mohammad Alaggan Aug 7 '13 at 7:26
  • $\begingroup$ The condition that you need is that the base of all exponents are non-negative. $\endgroup$ – Baby Dragon Aug 7 '13 at 7:27
  • $\begingroup$ Where the question is "can not convert...", the answer is no . With a little complex analysis is reasonably simple to explain this by means of branches, multivalued function and stuff. The fact is that the rules of exponents can be applied only to non-negative reals...for the time being. $\endgroup$ – DonAntonio Aug 7 '13 at 7:31
  • $\begingroup$ Exact duplicate of math.stackexchange.com/q/185502/264 $\endgroup$ – Zev Chonoles Aug 7 '13 at 7:40
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    $\begingroup$ Here's another "proof": $-1 = \sqrt{-1}\cdot\sqrt{-1} = \sqrt{(-1)\cdot(-1)} = \sqrt{1} = 1$. And to debunk the misconception that this type of wrong reasoning is somehow special to square roots of negative quantities, here's one that "works" completely in the real numbers: $-1 = (-1)^1 = (-1)^{2\cdot\frac12} = ((-1)^2)^{\frac12} = \sqrt{(-1)^2} = \sqrt{1} = 1$ $\endgroup$ – celtschk Aug 7 '13 at 8:22

Without going into complex analysis, I think this is the simplest way I can explain this. Let $f(x) = \sqrt{x}$. Note that the (maximal) domain of $f$ is the set of all non-negative numbers. And how is this defined? $f(x) = \sqrt{x}$ is equal to a non-negative number $y$ such that $y^2=x$. In this sense, square root is a function! It is called the principal square root.

In contrast, the following correspondence is not a function: the relation $g$ takes in a value $x$ and returns a value $y$ such that $y^2=x$. For example, under $g$, 1 corresponds to two values, $1,-1$.

Now, the property of distributing a square root over a product is only proven (at least in precalculus) over the domain of the principal square root, that is, only for non-negative numbers. Given this, there is no basis for the step

$$\frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}.$$

As to why this property is not true, the best explanation for now is that because $-1$ is not in the domain of the principal square root. Hence, $\sqrt{-1}$ does not actually make sense, as far as our definition of square root is concerned. In complex analysis, more can be said. As a commenter mentioned, this has something to do with branches of logarithms.

For your second question, I think it is safe that you always keep in mind what the domain of the function is. If you will get a negative number inside an even root, then you can't distribute the even root over products or quotients.

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    $\begingroup$ And by the way, one rule you can keep in mind is that $\sqrt{x^2} = |x|$. $\endgroup$ – Alex Strife Aug 7 '13 at 7:51
  • $\begingroup$ So is it always safe to distribute the root over products and quotients if ["the number insider the root is positive" $\vee$ "the root is odd"]? Is that the rule? $\endgroup$ – Mohammad Alaggan Aug 8 '13 at 12:10
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    $\begingroup$ Yes, that would be the rule. :) $\endgroup$ – Alex Strife Aug 8 '13 at 14:59
  • $\begingroup$ But in this example, $e$, the base of the natural logarithm, is positive: $e^{z} \overset{!}{=} (e^{2\pi i})^{z/2\pi i} = 1^{z/2\pi i} = 1$ for all $z$. Is this an exception to the rule? $\endgroup$ – Mohammad Alaggan Aug 10 '13 at 13:55
  • $\begingroup$ @M.Alaggan: the thing you're doing there is not "distributing the root over a product or a quotient". Transforming $(ab)^c$ to $a^cb^c$ is distributing a root over a product (when $c$ is the reciprocal of an integer). That's not the same as transforming $a^{(bc)}$ to $(a^b)^c$, which is what you're doing here. $\endgroup$ – Steve Jessop Feb 6 '16 at 16:48

at real analysis

$$\sqrt{\frac{x}{y}}\ = \frac{\sqrt{x}}{\sqrt{y}}$$

only for $y,x>0$

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The problem lies in the conversion of: $$\sqrt{\frac{x}{y}}\ \to \ \frac{\sqrt{x}}{\sqrt{y}}$$

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  • $\begingroup$ Can you elaborate on why is this a problem? $\endgroup$ – Mohammad Alaggan Aug 7 '13 at 7:31

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