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Through the textbook, I've been taught the rule $\frac{\sqrt a}{\sqrt b} = \sqrt\frac{a}{b}$, however I realized that if all numbers are assumed to be real, and $a<0 ,b<0$, then the rule is not true as $\frac{\sqrt{-a}}{\sqrt{-b}} = \sqrt\frac{a}{b}$, whereas in $\frac{\sqrt{-a}}{\sqrt{-b}}$, the process makes it invalid as square rooting negative numbers is impossible. So does this mean that the surds rule (multiplication of surds included) is invalid for negative numbers? Or does it mean complex numbers have to be introduced in all cases?

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    $\begingroup$ If you look closely at the textbook source for this rule, it is likely that it specified $a$ and $b$ must be greater than or equal to $0$. An instructor might not always stress that restriction. And in many application scenarios, the values of $a$ and $b$ would naturally be positive, so the restriction wouldn't be important to stress. $\endgroup$ – alex.jordan Dec 31 '17 at 3:21
  • $\begingroup$ @alex.jordan: You are wrong about that. I have seen countless textbooks actually used in schools that don't care one bit about getting mathematics correct. $\endgroup$ – user21820 Dec 31 '17 at 8:08
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For all nonnegative real numbers $a$ and $b$ (with $b > 0$), we have $$ \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}. $$ After that, all bets are off. The basic problem is that the function $x \mapsto x^2$ is not injective (one-to-one) over the real numbers. Thus to define the principal square root, which is what the surd represents, we must restrict the domain of this function to the nonnegative real numbers. Thus if we are working with real variables, then, for example, $\sqrt{a}$ is nonsense for $a < 0$.

Okay, so real variables cause problems. What if we work with complex variables, instead? Unfortunately, even if we are careful, we generally cannot get the "surd rule" to work as you want. For example, if we could extend it, then we might have $$ 1 = \sqrt{1} = \sqrt{(-1)^2} = \sqrt{-1}^2 = i^2 = -1, $$ which is clear nonsense. The problem again comes down to how we define the square root function, and on what domain that definition holds. There are generally two possible choices, and these choices are more or less arbitrary.

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  • $\begingroup$ Principal square root aka positive square root. $\endgroup$ – user5402 Dec 31 '17 at 12:38
  • $\begingroup$ @inéquation Indeed. Thank you. $\endgroup$ – Xander Henderson Dec 31 '17 at 14:59
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The main problem is $\sqrt{\cdot}$ isn't defined for negative values as you've pointed out. While you can find numbers that square to negative numbers using the complex numbers, the problem is that there's no natural way to choose them. That is to say, $i$ and $-i$ are essentially indistinguishable (I mean mathematically. Our notation allows us to tell them apart, but we essentially just picked one of the two at random and called that one $i$ and the other one $-i$. It didn't matter which one we named which.) Thus while you can define a continuous function, $s: \Bbb{R}\to\Bbb{C}$, such that $s(x)^2 = x$ for all $x\in \Bbb{R}$, say by $$s(x) := \begin{cases}\sqrt{x} & x\ge 0 \\ i\sqrt{|x|} & x < 0, \end{cases}$$ if you check to see whether this satisfies $s(xy)=s(x)s(y)$, or $s(x/y)=s(x)/s(y)$, you'll see that it fails to satisfy either except in special cases.

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  • $\begingroup$ @infiniteaccelerator0643 I was in the middle of editing my answer to give the "real answer" given here. Since this has been posted, I've deleted my answer. My one qualm with jgon is that I would say that $\sqrt{x}$ is defined everywhere, it's just multivalued and not a function. We pretend it's a function by looking at it on positive values and declaring that we want to get a positive output. We can do this because the order on $\mathbb R$ lets us distinguish $-1$ and $1$. But as jgon says, the same can't be said of $-i$ and $i$. $\endgroup$ – Stella Biderman Dec 31 '17 at 3:03
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    $\begingroup$ @StellaBiderman If I were being more technical, I would just say the issue is that $z\mapsto z^2$ is a 2-sheeted ramified covering map, I would still say that $\sqrt{z}$ isn't defined. Though we can find local inverses of course. $\endgroup$ – jgon Dec 31 '17 at 3:06
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You should start with the definition of the square root and the set of numbers you are working in. In the reals $\sqrt x$ is only defined for $x \ge 0$ and the result is defined to be nonnegative. That makes it a well defined function and the rule you cite depends on that. If you are working in the reals and $a,b \gt 0$, $\frac {\sqrt {-a}}{\sqrt {-b}}$ is undefined. Whether you can make sense of it by going to the complex field is immaterial. There is a temptation to say $\sqrt {-a}=i\sqrt a$ but there is no way to distinguish that from $-i\sqrt a$. This can be useful for exploration of problems, because it does not always cause a problem. If you use it, you need to get rid of it in your final answer. If you are working in the complex numbers $\sqrt a$ has an infinite set of values until you choose a branch of the square root. The identity you quote does not hold any more because of the branch choice.

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Even if you define the square-root on all real numbers such that $\sqrt{-1} = i$, you will still not have any such property, otherwise:

(WRONG) $\dfrac{1}{i} = \dfrac{\sqrt{1}}{\sqrt{-1}} \overset{???}= \sqrt{\dfrac{1}{-1}} = \sqrt{-1} = i$ and hence $1 \overset{???}= i^2 = -1$.

Bottom-line: You should make sure your textbook is a mathematically proper one, and also try not to simply memorize mathematical 'facts'. Every theorem in mathematics has a proper proof, and the more rigorous mathematics you learn the more you will understand that there is no such thing as a random 'law' to be followed.

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