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We know $$i^2=-1 $$then why does this happen? $$ i^2 = \sqrt{-1}\times\sqrt{-1} $$ $$ =\sqrt{-1\times-1} $$ $$ =\sqrt{1} $$ $$ = 1 $$

EDIT: I see this has been dealt with before but at least with this answer I'm not making the fundamental mistake of assuming an incorrect definition of $i^2$.

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Why is $\sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1}$? We know that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ but this is only true for $a,b \geq 0$... –  algebra_fan Jul 3 '11 at 9:17
Thank you algebra_fan. I've never seen restrictions placed on log/exponent laws before. Where/when do those get properly defined? Most math textbooks? –  Greg Jul 3 '11 at 9:24
Duplicate of math.stackexchange.com/q/438/7850 –  The Chaz 2.0 Jul 3 '11 at 11:49
In every math textbook! If there is a math textbook that says $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$ without providing restrictions on $a,b$, then you should write to the author immediately! –  GEdgar Jul 3 '11 at 12:43
$i^2=-1$ is the definition of $i$. –  Tpofofn Feb 8 '13 at 18:00
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10 Answers

up vote 27 down vote accepted

From $i^2=-1$ you cannot conclude that $i=\sqrt{-1}$, just like from $(-2)^2 = 4$ you cannot conclude that $-2=\sqrt 4$. The symbol $\sqrt a$ is by definition the positive square root of $a$ and is only defined for $a\ge0$.

It is said that even Euler got confused with $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$. Or did he? See Euler's "mistake''? The radical product rule in historical perspective (Amer. Math. Monthly 114 (2007), no. 4, 273–285).

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If I cannot conclude that $i=\sqrt{-1}$ then what good is it? Wouldn't that mean I couldn't use $i$ for what it's intended? Namely $\sqrt{-5} = \sqrt{-1\times 5} = i\sqrt{5}$ ? –  Greg Jul 4 '11 at 13:04
@Greg, then only property of $i$ that you need is that $i^2=-1$. The symbol $\sqrt{-1}$ is abuse of notation. As for $\sqrt{-5}$, you get $(i\sqrt{5})^2= (-1)\cdot 5 = -5 =$, as you expect. –  lhf Jul 4 '11 at 15:46
There's nothing wrong with defining(!) $\sqrt{-1}=i$, just as there's nothing wrong with defining $\sqrt{4}=2$ despite the fact that from $x^2=4$ you cannot conclude $x=2$. The difference to real numbers is that you cannot define the square root function so that it is continuous on all of $\mathbb C$. And the calculation in the question is just incorrectly applying power laws; you don't even need complex numbers to do something similar: $-1 = (-1)^1 = (-1)^{2\cdot1/2} = ((-1)^2)^{1/2} = 1^{1/2} = 1$ –  celtschk Jul 10 '13 at 12:40
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Any non zero number has two distinct square roots. There's an algebraic statement which is always true : "a square root of $a$ times a square root of of $b$ equals a square root of $ab$", but this does not tell you which square root of $ab$ you get.

Now if $a$ and $b$ are positive, then the positive square root of $a$ (denoted $\sqrt{a}$) times the positive square root of $b$ (denoted $\sqrt{b}$) is a positive number. Thus, it's the positive square root of $ab$ (denoted $\sqrt{ab}$). Which yields

$$\forall a,b \ge 0, \ \sqrt{a} \sqrt{b} = \sqrt{ab}$$

In your calculation, because $i$ is a square root of $-1$, then $i^2$ is indeed a square root of $1$, but not the positive one.

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Following Willie Wong's suggestion, I will put the comments I made before into an answer. The problem with this argument is that the formula $\sqrt{ab} = \sqrt{a} \sqrt{b}$ is valid for $a,b \geq 0$. In particular, it is not valid in the case when $a=b=-1$ so that the step $\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)}$ is incorrect.

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$\hspace{0.2in}$ The problem is this: the (polar) representation of a complex number depends on your choice of branch. Once you choose a branch of the square root, you cannot simultaneously represent $1$, and $-1$, because they are $\pi$ apart, so that $-1$ and $1$ will necessarily be in different branches of $z^{1/2}$. The choice of square root you make (if you want it to be well-defined) depends on the choice of branch you are working with. You are trying to combine numbers that live in different branches; it is as if you add $1+1$ and get $1-i\pi$, since 1 can also be represented as $i\pi$ ($~1$ can actually be represented as $ik\pi$, but when you operate, you are supposed to stay within a branch, and, in your case, you are not$~$).

$\hspace{0.2in}$ In complex numbers, from the perspective of polar representations, when you multiply $z_1 \dot z_2$, you multiply the respective lengths, and add the respective angles (but you have to make up for the fact that the sum of the angles may be larger than $2\pi$ (or whatever argument-system you are working with. So in this sense, when you multiply $i$ by itself, you multiply the length of $i$ by itself, and add the argument to itself; i.e., you double the argument. In complex variables, you have many possible polar representations for a given number; specifically, given $z=re^{it}$, then $z=re^{i(t+2\pi)}$ is also a valid representation.

$\hspace{15.0in}$ $\hspace{0.2in}$ You must then choose a specific representation for your $z$, specifically; you will need to specify the range of the argument you will be working with. So, say you work with the "standard" range of $[0,2\pi)$. Then the expression for $i$ (me) is as $i=1e^{i\pi/2}$, so that $i^2$= $(1)(1).e^{i(\pi/2+\pi/2)}=e^{i\pi}=-1$. But the way backwards from multiplying to taking roots is more complicated if your base is non-negative, and/or your exponent has non-zero imaginary part. When this last is the case, you define:

$z^{{1}/{2}} \:=e^{{\Large{log(z)}}},~$ where we define:

$\log(z):=\ln |z|+i arg(z)$

(This choice of definition has to see in part with wanting to have the complex log agree with the standard real $\log$--though this agreement is possible only for one choice of "branch", as we will see.) But because of the infinitely-many possible choicesfor the argument of a number, the $\log(z)$ itself--defined locally as the inverse of $e^z$ is somewhat-ambiguously-defined, since $e^z$ does not have a global inverse (since it is not $1-1$, for one thing, but $e^z$ does have local inverses, e.g., by using the inverse function theorem). So when we mention $\log$, we are referring just to one of (infinitely-) many possible local inverses of $e^z$ .Each possible local inverse to $e^z$ is called a "branch" of the $\log$. So once we choose a branch for the $\log$, which is a choice of an open set (technically, it is half-open) of width $2\pi$ from which we will choose the argument we will use. So, say we choose the standard branch $(0,2\pi)$, which we call Log(z). We then define :


But , in this branch , $(-1)^{1/2}$ is not even defined, because the argument for $(-1)$ is $0$ , which is outside of the allowable values $(0,2\pi)$. So, in this sense, the expression $(-1)^{1/2}$ is not well-defined, i.e, does not really make sense. ($~$Note that this particular branch of $\log$ reduces to the standard one when you select an argument of $t=0~$).

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Maybe I overdid it, trying to understand the log better. Could someone help with the spacing? I am saturated on formatting, sorry. –  gary Jul 3 '11 at 10:58
Help exactly with what spacing in your answer? –  night owl Jul 4 '11 at 2:34
Well, I try to space out the paragraphs, but they do not come out with the spacing of the original; specifically, in spreading the text out some more, and in separating paragraphs. –  gary Jul 5 '11 at 17:26
@ gary: Alright got it. –  night owl Jul 7 '11 at 13:48
Another downvote without a constructive comment. Is it personal? If there is something specific you do not like, it would be nice to hear; I may be able to learn from it. Still, it is up to you, whomever downvoted me without an explanation. –  gary Jul 12 '11 at 1:12
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What you can do is $a^x a^y = a^{x+y}$, so $(-1)^{1/2}(-1)^{1/2}=(-1)^{1/2+1/2}=(-1)^1=-1$.

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I don't follow. What is $(-1)^{1/2}$? Is it $i$ or $-i$? –  Hendrik Vogt Jul 3 '11 at 11:29
@Hendrik Vogt I think $i=(-1)^{1/2}$ –  Fan Zhang Jul 3 '11 at 11:47
And why not $(-1)^{1/2} = -i$? The problem is that you first have to specify which branch of the square root you want; cf. gary's answer. –  Hendrik Vogt Jul 3 '11 at 11:50
This answer can at best muddy waters. @upvoter Why the upvote? –  Did Mar 7 at 16:45
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There are many ways to show that your second equality is incorrect. Just for the sake of argument and a refreshing change of pace, suppose your point is true. That is, suppose $i^{2} = 1$. Then \begin{align} (x + i)(x-i) = x^2 - i^{2} = x^{2} - 1. \end{align} Use the Descartes Rule of signs to derive a contradiction. Hint: $i$ is not real. Can you finish the line of reasoning and derive an absurdity?

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Isn't a quicker absurdity $-1 = i^2 = 1$? –  user92843 Jul 3 '11 at 9:33
Hmm. if $1+i=0$ and $1-i=0$ then you can set them equal to each other. So $1+i=1-i$ which means $i=-i$. But I'm not sure why you conclude $(1+i)(1-i)=0$. –  Greg Jul 3 '11 at 9:33
@user02138: The OP is assuming $i$ is the square root of $-1$ in the derivation. If you're assuming nothing about $i$ then there's no contradiction, since for all you know $i$ could equal 1. –  user92843 Jul 3 '11 at 9:46
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You may only apply sqrt(a) sqrt(b) = sqrt (a*b) when a,b >=0

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Don't take me the wrong way, all of you are right from a certain point of view.

Let me show you the following. Let $a$ and $b$ be complex numbers. If $\sqrt a$ is a square root of $a$ and $\sqrt b$ is a square root of $b$, then $(\sqrt a\sqrt b)^2 = (\sqrt a)^2 (\sqrt b)^2 = ab$ by assumption, so indeed $\sqrt a\sqrt b$ is a square root of $ab$. Notice that it is some square root; there is no canonical choice over $\mathbb{C}$!

From this point of view, your arguments work until $i^2 = \sqrt1$. All you've shown is that $-1 = i^2$ is a square root of 1 and no one will disagree with that fact.

So what I'm trying to say is, if you want to write something like $\sqrt a\sqrt b = \sqrt{ab}$, which isn't the worst idea, you can't write something like $\sqrt 1 = 1$.

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In general,

$\forall a,b \in \mathbb{R} , ab\geq 0, (\sqrt{ab} =\sqrt{ |{a}| }\sqrt{|b|}) $

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@Greg: $i^2=-1$ is a matter of definition. It was introduced to handle a negative sign under the radical when solving polynomial equations. The best example is $x^2 +1 =0$, which yields $x = \pm\sqrt{-1}$. Using your manipulation above would yield $x^2 = 1$ which clearly does not solve the equation.

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