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Apparently $\sqrt{ab}$ = $\sqrt{a}\sqrt{b}$ is only true if a and b are both positive or if a is negative and b is positive or if a is positive and b is negative. In other words, a and b can't both be negative.

Is it possible to algebraically prove this? Or is it just a result of the way the square root function is defined?

I know of 1 way to prove this radical property, but I'm still not sure why it won't work for negative numbers.

Let x = $\sqrt{ab}$.

Let y = $\sqrt{a}\sqrt{b}$

Square both sides for both equations.

$x^2 = (\sqrt{ab})^2 = ab$

$y^2 = (\sqrt{a}\sqrt{b})^2 = (\sqrt{a}\sqrt{b})(\sqrt{a}\sqrt{b}) = (\sqrt{a})^2(\sqrt{b})^2 = ab$

$\therefore x^2 = y^2$

$x^2-y^2=0$

$(x+y)(x-y)=0$

$\therefore x = y$ or $x = -y$

Or

$\therefore y = x$ or $y = -x$

A lot of people will go through this line of reasoning (shown below) in order to justify why a and b can't both be negative.

Considering that mathematicians define $i^2=-1$ or $i = \sqrt{-1}$

$1 = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = (i)(i) = i^2 = -1 $

But this is only a specific instance where this property fails us. This isn't a rigorous or at least satisfying proof of why $\sqrt{ab}$ = $\sqrt{a}\sqrt{b}$ can only be true if a and b are not both negative.

Note: I just started learning about complex and imaginary numbers and I am no means an expert in mathematical proofs, so if you do know the answer to this question please try (if possible) your best to answer the question without using too much complex or high-order math that I won't be able to understand.

Dom Turner
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    What is your definition of $\sqrt a$? For which $a$ does it apply? – lhf Jan 30 '20 at 13:14
  • See also https://math.stackexchange.com/a/49224/589 – lhf Jan 30 '20 at 13:15
  • The principal square root, where $\sqrt{a}$ (assuming a > 0) means 'the positive root of a'. Isn't that how the square root function is defined to be? – Dom Turner Jan 30 '20 at 13:18
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    Does this answer your question? Why $\sqrt{-1 \times -1} \neq \sqrt{-1}^2$? – Blueyedaisy Jan 30 '20 at 13:20
  • hold on, let me read the post. – Dom Turner Jan 30 '20 at 13:21
  • hmmmm..., I'm still confused to be honest after reading that post. – Dom Turner Jan 30 '20 at 13:26
  • @DomTurner What is the "positive square root of $a$" when $a$ is a negative number? Is the positive square root of $-1$ equal to $i$ or to $-i$? Because none of those numbers is positive, or negative, for that matter. – 5xum Jan 30 '20 at 13:30
  • You say that the principal square root means the positive root. This only makes sense within the real numbers. There is no standard or consistent definition of "positive" for complex numbers. You can set an arbitrary rule to pick out one square root as the principle but not in a way such that $\sqrt a \sqrt b = \sqrt{a b}$ is always true. – badjohn Jan 30 '20 at 13:33
  • I don't know, to be honest, I've always been taught that by convention $\sqrt{a}$ means the principle square root of a. Is this only true for the real number system? What happens if we use the definition i^2 = -1? Is the rule still true? This is what i'm confused about. – Dom Turner Jan 30 '20 at 13:36
  • When the real numbers are extended to the complex numbers we gain some things, most obviously square roots of negative numbers (and all of the new numbers). However, we also lose something which is that it is not possible to define positive and retain familiar behaviour such as positive x positive is positive. Later, you might learn of the quaternions which are a further extension, this time you lose commutativity of multiplication so not even $a \times b = b \times a$ is always true. – badjohn Jan 30 '20 at 13:58
  • That's not directly related to my question, but I read the post and no I'm still confused. – Dom Turner Jan 31 '20 at 06:58
  • To be clear, I'm not confused about why $\sqrt{-1 \times -1} \neq (\sqrt{-1})^2$, I'm confused about why $\sqrt{ab} = \sqrt{a}\sqrt{b}$ is not true when both b and a are negative in a general sense. – Dom Turner Jan 31 '20 at 07:21

1 Answers1

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This is perhaps rather a case of imprecise language than an issue with proof. The square root function is defined to have it's domain in positive real numbers (including zero). Whenever people say "square root of $x$", I must assume $x$ is a positive real number.

There is a completely different relation (and not a function) that I'll call $R(x)$, which relates any complex number with two other complex numbers. If $(x = r e^{\theta i})$ with $r$ a positive real number, then I have that $R(x) = \{\sqrt{r}e^{\theta/2 i}, \sqrt{r}e^{(\theta/2 +\pi)i}\}$.

Note that, in the particular case $\theta=0$, we have $R(x) = \{\sqrt{r}, -\sqrt{r}\}$.

That being said, if someone explicitly asks for the square root of $-1$, I'll forgive this person's abuse of language, and understand that he/she asked for $R(-1)$, which is $\{ i, -i \}$, and in some cases I might even consider that the person wanted to hear "$i$" only.

In your case, if both $a$ and $b$ are negative, at the left-hand side, I'll have a traditional square root written, whereas, on the right hand-side I should have a product of $R(a)$ and $R(b)$, which is undefined.

Mefitico
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  • Would it be incorrect to say $i = \sqrt{-1}$? I've heard some people say that it's wrong and $i^2 = -1$ is the correct definition of i while others claim that it is perfectly fine. – Dom Turner Jan 30 '20 at 13:30
  • Would it be incorrect to say $i=\sqrt{-1}$? Yes. Do I say it? Yes. I consider pedantic to say it is wrong unless I'm asked. But notice that $i^2=-1$ is correct, bona fide truth. The point is, while the square root is a function, and thus has an inverse, the relation $R$ isn't, so you should take care manipulating expressions with it. There is actually a definition of complex numbers based on matrices, and the $i$ symbol is just a simpler notation for it, so $i$ is not defined with $i^2=-1$. $i$ is defined by a 2x2 matrix. – Mefitico Jan 30 '20 at 13:35
  • Why do you consider it to be incorrect? Doesn't $\sqrt{a}$ denote 'the principle square' root by convention and therefore you are taking the positive version of the imaginary number -----> $+i = \sqrt{-1}$. If you have ${i^2=-1}$ then by definition of what the term, square root, means 'A "square root" of a real number a is a number y such that ${y^2=a}$' (could be both negative and positive). $\therefore i = \pm\sqrt{-1}$ – Dom Turner Jan 30 '20 at 22:48
  • Because saying "square root" instead of "principle square root" (as you said it now) is an abuse of language. And as I've said, I overlook language abuses in general, but if asked specifically, I answer that an abuse of language is "incorrect" (or at least "imprecise"). And no, the "square root" of a real positive number cannot be a negative number, because it is a function defined with domain and image in the positive reals. What you present is the kind of sloppy language that leads students (and professionals tbh) to make sloppy mistakes. – Mefitico Jan 31 '20 at 13:21
  • Note that $y^2=a$ is an algebraic equation of second degree, and thus it has two (not necessarily distinct) solutions. If you claim that the solutions to this equation are the result of an operation, then this operation has domain on complex numbers ($\mathcal{C}$) and domain in $\mathcal{C}^2$). And depending on the case, more care is needed to order the pair of solutions (otherwise, it wouldn't even be a function). – Mefitico Jan 31 '20 at 13:25