My idea is that the two functions are not the same since for the first function, the domain of the function is only non negative reals for the numerator and positive reals for the denominator. However in the second function, the domain is any real provided $b$ isn't equal to $0$.

Am I correct in thinking so?

If they are indeed the same function, please do explain why their domains will be the same?

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    $\begingroup$ Letting $a=b=-1$ gives a problem with the first. Also: you are right to think the domains problematic. The first one takes a single number ${a\over b}>0$ and produces $\sqrt{a/b}$, so it has domain $\Bbb R_{\ge 0}$ and the second takes two numbers, $a,b\ge 0$ and produces the ratio of their square roots. Even if you make the $\sqrt{a/b}$ one to take two numbers, the domain is then $\{(a,b)\in\Bbb R^2 : ab>0\}\cup\{(0,b)\in \Bbb R^2 : b\ne 0\}$ $\endgroup$ – Adam Hughes Aug 28 '14 at 17:11
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    $\begingroup$ Yes, they are different, but where both are defined (i.e. on the intersection of domains) - the values will coincide. $\endgroup$ – gt6989b Aug 28 '14 at 17:13
  • $\begingroup$ @Yiorgos S. Smyrlis I see that you have added (roots) tag. The tag (roots) is for zeroes of functions, I think that (arithmetic) and (radicals) are better tags for questions about square roots. From roots tag-info: For questions about "square roots", "cube roots", and such, consider using the (arithmetic) tag. $\endgroup$ – Martin Sleziak Aug 29 '14 at 10:10
  • $\begingroup$ @MartinSleziak: You are right! $\endgroup$ – Yiorgos S. Smyrlis Aug 29 '14 at 13:43

Let us define the functions: $$ f(a,b)=\frac{\sqrt{a}}{\sqrt{b}}\quad\,\,\text{and}\,\,\quad g(a,b)=\sqrt{\frac{a}{b}}. $$ Then $f$ and $g$ AGREE on the intersection of their domains. However, they have different domains:

$$ \mathrm{Dom}(f)=\{(a,b): a\ge 0,\,\,b>0\}, $$ while $$ \mathrm{Dom}(g)=\{(a,b): a\ge 0,\,\,b>0\}\cup\{(a,b): a\le 0,\,\,b<0\}. $$

Strictly speaking, in order for two functions to be equal they need to have the same domain (and they same values for each element of their domain.) Hence, strictly speaking, these functions are not the equal.


Actually, you are right- as functions of two real variables, they have different domains. For example take $a=-1,b=-1$

  • $\begingroup$ Isn't $\dfrac{\sqrt{-1}}{\sqrt{-1}} = \dfrac{i}{i} = 1 = \sqrt{\dfrac{-1}{-1}} = \sqrt{1} = 1?$ $\endgroup$ – Warren Hill Aug 28 '14 at 17:43
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    $\begingroup$ I was only referring to the Real Number System, but if I extended this doubt to the set of imaginary numbers, would the functions then be same? $\endgroup$ – Niharika Aug 28 '14 at 17:48
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    $\begingroup$ @Niharika: You can run into trouble with imaginary numbers. For example, the principal square root of $i$ is $(1+i)/\sqrt{2}$, and the principal square root of $-i$ is $(1-i)/\sqrt{2}$. Then $\sqrt{-i}/\sqrt{i} = (1-i)/(1+i) = (1-i)^2/2 = -i$, but $\sqrt{-i/i} = \sqrt{-1} = i$. $\endgroup$ – Bungo Aug 28 '14 at 18:08
  • $\begingroup$ If we assume $a$ and $b$ are both real and allow the result to be imaginary then $\dfrac{\sqrt{1}}{\sqrt{-1}} = \dfrac{1}{i} = -i \ne \sqrt{\dfrac{1}{-1}} = i$ so it doesn't work. But if I require $a$ and $b$ to either both be positive or both be negative it looks like it works to me. I could be wrong however and I'm just asking for clarification. $\endgroup$ – Warren Hill Aug 28 '14 at 18:35
  • $\begingroup$ @WarrenHill: Both positive certainly will be fine- both negative would also be fine by your first comment. $\endgroup$ – voldemort Aug 28 '14 at 21:58

The question demonstrates the difference between expressions and functions. And the question of finding the maximal domain for a function determined of an expression depends on what one mean with the expressions in the formula. Either is $\sqrt x$ a function defined for $x\geq 0$ or it is an other function, defined for other domains. Anything goes but has to be declared.


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