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I have seen many people say that a and b can't be positive for example in this false proof :

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

Trust me, I understand that $1\neq -1$ and also by seeing this, I believe and accept that a and b should be positive or greater than 0 (for $\sqrt{ab}=\sqrt{a}\sqrt{b}$)

But I'm interested to know why is that, which is something I don't know? What is the proof ?

Thanks a lot.

M.S.E
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  • In general, $(ab)^x=a^xb^x$ for any $a\ge 0$ and $b,x\in\mathbb C$. So no, you don't need both to be positive. – user2345215 Jan 08 '15 at 20:27
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    If you work in $\mathbb R$ then the square root of a negative number is not defined. If you work in $\mathbb C$ then the square root is a "multi-valued" function, so $\sqrt{-1}$ could be $i$ or $-i$. – Martin R Jan 08 '15 at 20:33
  • The real underlying reason is that "the square root" has to be carefully defined when you try to generalize it. Notice that there are two square roots of any number. Because of that, we have to be careful in our definition for the square root to be well-defined. – Max Jan 08 '15 at 20:34
  • @MartinR really? I was taught that $\sqrt{-1}$ is $i$ – M.S.E Jan 08 '15 at 20:35
  • @MartinR That's your view. $\sqrt{-1}=i$ taking the principal value on $(-\pi,\pi]$. This way we can define $a^b\in\mathbb C$ for all $a,b\in\mathbb C$. – user2345215 Jan 08 '15 at 20:36
  • @user2345215: The reality is that $\sqrt a $ is defined only for $a\ge 0$, and you do not clarify this situation by introducing the more general notions of $a^x$ and $b^x$ for complex exponents. – hardmath Jan 08 '15 at 20:36
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    @user2345215: $a^b$ is so far from being well-defined that it usually has infinitely many possible values. The best one can do is to define it to be $e^{b \log(a)}$, but the complex valued logarithm function has infinitely many branches. – Lee Mosher Jan 08 '15 at 20:37
  • @LeeMosher I should have mentined it in the comment, I'm taking the $(-\pi,\pi]$ branch of the complex logarithm. Too late to edit now, why does this always creates so much confusion? – user2345215 Jan 08 '15 at 20:39
  • The notation $\sqrt{z}$ doesn't make sense for complex numbers, because it is multi-valued. That's why we have to write $i$ or $-i$ instead of $\sqrt{-1}$ – Dylan Jan 08 '15 at 21:05
  • To more carefully state a point user2345215 could have made, it is NOT TRUE that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ only when $a$ and $b$ are positive. For example, this identity holds when $a=b=0.$ – Dave L. Renfro Jan 08 '15 at 21:22
  • @user2345215: Choosing a specific domain, as you wish to do, does not help with questions such as the one asked here. For example, the formula $a^b \cdot a^c = a^{b+c}$ is false with your choice, or indeed any choice, of the branch. – Lee Mosher Jan 08 '15 at 22:57
  • @LeeMosher How so? $a^b\cdot a^c=e^{b\ln a}e^{c\ln a}=e^{(b+c)\ln a}=a^{b+c}$, the branch doesn't really play any role here, as long as stick to one. – user2345215 Jan 08 '15 at 23:26
  • @user2345215: Sorry, I wrote the wrong formula. I meant to write that the formula $a^c b^c = (ab)^c$ (which is the formula in the original question) is false with your choice of $(-\pi,\pi]$ branch of the complex logarithm. In fact it is false when substituting the values $a=b=-1$, $c=\frac{1}{2}$ (which are the values of $a,b,c$ in the original question): $a^c b^c = -1$ whereas $(ab)^c=1$. – Lee Mosher Jan 09 '15 at 02:40
  • @LeeMosher OMG. See the first comment in this question, there's a reasonable generalization (I should have added that $a,b\ne 0$). – user2345215 Jan 09 '15 at 07:05

2 Answers2

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If $a$ is a complex number and you write $\sqrt{a}$ to denote a specific number, you have introduced a problem, as there are two complex numbeers whose square is $a$.

The problem shows up when you need to choose one of the two alternatives for all complex numbers at the same time, and this is unavoidable in some situations. For example, notice that for a statement such as

for all complex numbers $a$ and $b$ we have $\sqrt a\sqrt b=\sqrt{ab}$

to mean anything, we need to give sense to $\sqrt{a}$ for all $a$s.

A good choice for $\sqrt{-1}$ is certainly $i$. A good choice for $\sqrt{1}$ is $1$, of course. Now $\sqrt{-1}\cdot\sqrt{-1}$ is, according to these choices, equal to $i\cdot i$ which is $-1$. On the other hand $\sqrt{(-1)(-1)}$ is $\sqrt{1}$ and we chose that to be $1$; we have a problem.

Ok. Maybe we should have chosen $\sqrt{1}$ to be $-1$? Let's see what happens. Now $\sqrt{1}\cdot\sqrt{1}$ is $(-1)\cdot(-1)$, which is $1$, and $\sqrt{1\cdot1}=\sqrt{1}=-1$: oh no!

And you can go on like this...

Indeed, theere are four choices in all for what $\sqrt{1}$ and $\sqrt{-1}$ can mean:

| sqrt{1} sqrt{-1}
|   1       i      
|   1      -i     
|  -1       i     
|  -1      -i

In each of these four options you can reach a problem.

MJD
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Mariano Suárez-Álvarez
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  • I don't think this is fully satisfactory. In the positive reals we also have the problem of $2$ roots, yet people define $\sqrt x$ without problems. In $\mathbb C$, we can still define $\sqrt x$ (or any power) using the principal branch of the complex logarithm. While the formula $\sqrt{a}\sqrt{b}=\sqrt{ab}$ doesn't hold for arbitrary $a,b\in\mathbb C$, it holds if we require just one of them to be nonnegative real. I think this is still useful. – user2345215 Jan 08 '15 at 20:48
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    When we only care about defininf square roots of positive real numbers, this does not arise: we can choose $\sqrt{a}$ for $a\geq0$ to denote the nonnegative real square root, and then we can actually prove our choices work out. But choosing square roots for a larger set than that is more demanding, because more equalities have to hold. – Mariano Suárez-Álvarez Jan 08 '15 at 20:49
  • @user2345215, yes , the formula has validity in several special situations. That is irrelevant, really. – Mariano Suárez-Álvarez Jan 08 '15 at 20:49
  • See also wikipedia https://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities – Lehs Jan 08 '15 at 22:38
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$\sqrt{x}$ is defined in $\mathbb{R}$ as long as $x \geq 0$. Thus in the first place for $\sqrt{ab}$ to be defined and make sense in $\mathbb{R}$ we would need $ab \geq 0.$ When $ab=0,$ the case is trivial. So let's consider the case when $ab>0$ which yeilds either $a>0,b>0$ or $a<0,b<0$. In the first case, both $\sqrt{a}$ and $\sqrt{b}$ are well defined and hence $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is true. However, in the second case, as both $a$ and $b$ are negative, neither of $\sqrt{a}$ and $\sqrt{b}$ are defined. Thus we cannot write $\sqrt{ab}=\sqrt{a}\sqrt{b}$ when $a$ and $b$ are negative.

As far as complex numbers are concerned, your question does not apply since complex numbers are not ordered in the usual sense. I mean, you cannot talk of a complex number being positive or negative. For example, if $i$ were positive, then $i^2 \geq 0$ which is not true as $-1 \ngeq 0$ and also if $i$ were negative, then again we get $i^2 \geq 0$ which is not true as $-1 \ngeq 0$.

Ralph Lazarus
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