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I know there must be something unmathematical in the following but I don't know where it is:

\begin{align} \sqrt{-1} &= i \\ \\ \frac1{\sqrt{-1}} &= \frac1i \\ \\ \frac{\sqrt1}{\sqrt{-1}} &= \frac1i \\ \\ \sqrt{\frac1{-1}} &= \frac1i \\ \\ \sqrt{\frac{-1}1} &= \frac1i \\ \\ \sqrt{-1} &= \frac1i \\ \\ i &= \frac1i \\ \\ i^2 &= 1 \\ \\ -1 &= 1 \quad !!! \end{align}

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When dealing with square roots, it is important to remember that each number apart from 0 naturally has 2 different roots. This means that you have to be very careful when dealing with non-positive numbers – Casebash Jul 22 '10 at 4:22
@Casebash: I believe it's fairly standard to take sqrt(x) to mean the principal square root function of x--that is, if x is a nonnegative real number, the nonnegative square root. Defining which root is the principal root of nonreal complex numbers can be a little trickier (some texts use the one with argument in [0,π) while other texts and many calculators use the one with argument in (-π/2,π/2] ). – Isaac Jul 22 '10 at 4:48
$i^2=(-i)^2$ doesn't imply $i=-i$. Simple but complex :-). – copper.hat Jan 21 '14 at 7:49
How $\sqrt{-1}=i$ if and only if $i^2=-1$? Or how can we conclude that $\sqrt{-9}=\pm 3i$? – daulomb Jan 21 '14 at 16:34
$\sqrt{-1}\neq \frac{1}{i} = -i$ – Alvin Lepik Mar 30 at 6:54
up vote 87 down vote accepted

Between your third and fourth lines, you use $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$. This is only (guaranteed to be) true when $a\ge 0$ and $b>0$.

edit: As pointed out in the comments, what I meant was that the identity $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$ has domain $a\ge 0$ and $b>0$. Outside that domain, applying the identity is inappropriate, whether or not it "works."

In general (and this is the crux of most "fake" proofs involving square roots of negative numbers), $\sqrt{x}$ where $x$ is a negative real number ($x<0$) must first be rewritten as $i\sqrt{|x|}$ before any other algebraic manipulations can be applied (because the identities relating to manipulation of square roots [perhaps exponentiation with non-integer exponents in general] require nonnegative numbers).

This similar question, focused on $-1=i^2=(\sqrt{-1})^2=\sqrt{-1}\sqrt{-1}\overset{!}{=}\sqrt{-1\cdot-1}=\sqrt{1}=1$, is using the similar identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$, which has domain $a\ge 0$ and $b\ge 0$, so applying it when $a=b=-1$ is invalid.

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Well sqrt(-25) / sqrt(-1) = 5, so the explanation doesn't seem quite complete. – Wilhelm Jul 22 '10 at 4:21
Just because something is true for some cases, doesn't mean that it is true in other cases. The square root division law described by Isaac is the only defined way. sqrt(-25)/sqrt(-1) might = 5, but that's just a coincidence. There is only one rigorously defined way that keeps everything in order. – Justin L. Jul 22 '10 at 4:38
Thanks Justin and Himadri for clarifying my answer. Hopefully my edit will help as well. – Isaac Jul 22 '10 at 5:05
@workmad3: It's true that it works in both those cases, but it's misleading. It's comparable to saying that it's okay to cancel the 6s in 16/64 to get 1/4--it works, but for the wrong reasons. The identity sqrt(a) * sqrt(b) = sqrt(a * b) really does only work when a and b are nonnegative. – Isaac Jul 22 '10 at 13:33
@Isaac, I believe the key here originates from unit signs. Every number can be expressed in the form $N = e^{i \theta}*|N|$ where the $|N|$ gives the absolute value of the number and must strictly be greater than or equal to positive 0 while: $e^{i \theta}$ is the angle or orientation of the number in the complex number plane. The identity $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ holds true whenever a,b both share the same orientation ($e^{i\theta}$) (ie exist on the same line from the origin). – frogeyedpeas Jun 17 '14 at 23:14

Isaac's answer is correct, but it can be hard to see if you don't have a strong knowledge of your laws. These problems are generally easy to solve if you examine it line by line and simplify both sides.

$$\begin{align*} \sqrt{-1} &= \hat\imath & \mathrm{LHS}&=i, \mathrm{RHS}=i \\ 1/\sqrt{-1} &= 1/\hat\imath & \mathrm{LHS}&=1/i=-i, \mathrm{RHS}=-i \\ \sqrt{1}/\sqrt{-1} &= 1/\hat\imath & \mathrm{LHS}&=1/i=-i, \mathrm{RHS}=-i \\ \textstyle\sqrt{1/-1} &= 1/\hat\imath & \mathrm{LHS}&=\sqrt{-1}=i, \mathrm{RHS}=-i \end{align*}$$

We can then see that the error must be assuming $\textstyle\sqrt{1}/\sqrt{-1}=\sqrt{1/-1}$.

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Only problem I see with these, is that when one is shaky on the laws simplifying might be a problem. – Wilhelm Jul 23 '10 at 0:28
@Wilhelm: It works pretty well if you use the most certain laws first – Casebash Jul 23 '10 at 0:31

The simple reason this,

$\sqrt{\frac{-1}{1}} =\sqrt{\frac{1}{-1}}$

Is not valid is because of a branch cut that must be taken as a result of the square root function.

Using exponentials:

$$ \tag{1} e^{i\pi} = -1 $$ and thus, $$ \tag{2} e^{-i\pi} = \frac{1}{-1} = -1 $$

However, in the complex plane, these represent a semicircle rotation from $+1 \to -1$ in both anticlockwise and clockwise directions. The fact that both end up at $-1$ on the Real Axis simply conceals the path taken from $+1 \to -1$ in the complex plane.

Now if we think of both (1) and (2) they are two opposite semi-circles that make a circle in the complex plane of radius 1 around the origin: enter image description here

However, from complex analysis, we know that there must be a branch cut from $0 \to \infty$ somewhere on this circle. Thus, you are crossing a branch cut on the Riemann Surface by doing,

$$ \sqrt{\frac{-1}{1}} \to \sqrt{\frac{1}{-1}} $$

To correctly, dodge the Branch Cut issues, we can use the exponential form ensuring that we simplify the exponentials inside the square root before taking the square root, $$ \sqrt{\frac{e^{i\pi}}{1}\frac{e^{-i\pi}}{e^{-i\pi}}} = \sqrt{\frac{e^{0}}{e^{-i\pi}}} = \sqrt{\frac{e^{i\pi}}{1}} = i $$ or similarly, $$ \sqrt{\frac{e^{-i\pi}}{1}\frac{e^{+i\pi}}{e^{+i\pi}}} = \sqrt{\frac{e^{0}}{e^{+i\pi}}} = \sqrt{\frac{e^{-i\pi}}{1}} = -i $$ The fact we get two different answers from using (1) and then using (2) to invert the terms inside the square root demonstrate the chaos we give ourselves when crossing Branch Cuts. Morale of Story: Don't Cross Branch Cuts as "Here be multivalued-Dragons" see bottom for picture.

You can also be really sneaky and do two full rotations, which if you look at the picture below, will put you where you started on the Riemann sheet.

See this image for a pictorial representation of the manifold which you are attempting to abuse (This is 3D version of the above and the branch cut can be seen where the sheet intersects itself). This picture also shows why we can take the Branch Cut anywhere as long as it is within one rotation as we can spin the shape in the vertical axis and it will remain the same.

enter image description here

A demonstration of what happens to an unprepared Math Student, (Here called 'Bob') who unknowingly crossed a Branch Cut without being prepared. enter image description here (credit also to @marios-bounakis for some exciting discussions about Euler identities)

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+1 for "Multivalues in Fire Format" :-) – Bungo Mar 30 at 0:54

You use the rule $\sqrt{ab}=\sqrt a\sqrt b$, which indeed holds for $a,b\ge 0$: If $u,v$ are nonnegative real numbers with $u^2=a$ and $v^2=b$ then clearly $uv$ is a nonnegative ral number with $(uv)^2=ab$. However, without the assumption $a,b\ge0$, this rule does no longer hold in general.

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and what happens to it when a or b is negative? Is there another formula or is this a dead end? – Nick Jan 21 '14 at 7:36
@Nick Simply separate the roots in terms of positive ones and i. Also, it is not incorrect to say that $\sqrt{\frac{1,-1}}=1/i=-i$, as $(-i)^{2}=-1=\frac{1,-1}}=1$, it's just that square roots have two possible values always. – Anonymous Pi Apr 12 '15 at 19:05
@HagenvonEitzen, adding "only" to the condition makes it necessary however it is a sufficient condition. Do you see the Isaac's answer contains a statement "This is only(guarenteed to be true) when ..."? This sentence show that it is a sufficient condition and not necessary. In particular, addition of the term "only" made your answer incorrect. – Sufyan Naeem May 10 '15 at 11:35
@Hagen von Eitzen) I think your answer is wrong...Statement is TRUE when at least one of $a$ and $b$ is $\ge 0$. – S.Panja-1729 Oct 5 '15 at 19:00
As said in the comments, the phrasing makes this answer wrong. – YoTengoUnLCD Mar 30 at 3:30

There is always a danger with dealing with roots of any kind, that one might not be dealing with the same number all the time. This comes in part from both $1^2=1$ and $(-1)^2=1$.

Writing these into an equation as $\sqrt{1}=1$ and $\sqrt{1}=-1$, gives a result that $a=-a$. Such might be true in some mantissa-space (mantissa here is a multiplication form of modulus: ie just as $a + bn = a \pmod{n}$, so would $a * n^b \operatorname{man} n$).

By taking square roots, one is effectively dealing in a potential mantissa-space where $+x=-x$, and some subtly is needed to distingiush the two. This is one of the reasons that $\sqrt{x} \ge 0$ is taken as convention.

The actual mistake in the calculations, is that $\sqrt{x}$ is let to vary by sign.

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As others have mentioned, one approach is to be consistent about using principal roots, and make sure that the identities you're using are actually applicable for manipulating those principal roots. And your mistake there lies in equating the 3rd and 4th lines. But to me, that type of approach feels about as intuitive as memorizing a phone book.

Personally I find it much easier to think of square roots (or roots in general) as set-valued operators. Every number (other than zero) has two square roots, three cube roots, four cube roots, etc (which in general can be complex).

Define the $n^{th}$ root to be the set-valued mapping: $\sqrt[n]x \rightarrow \{ y : y^n = x \}$

Examples: \begin{gather} \sqrt{4} \rightarrow \{ +2 , -2 \} \\\\ \sqrt{1} \rightarrow \{ +1 , -1 \} \\\\ \sqrt{-1} \rightarrow \{ -i, +i \} \\\\ \sqrt[3]{1} \rightarrow \left\{ 1 , \tfrac{-1 + i \sqrt{3}}{2} , \tfrac{-1 - i\sqrt{3}}{2} \right\} \end{gather}

Then your question can be written as follows (treating multiplication and division as Kronecker products too see all the possible outcomes):

\begin{gather} \sqrt{-1} \rightarrow \{ i, -i \} \\\\ \frac{1}{\sqrt{-1}} \rightarrow \left\{ \frac{1}{i} , \frac{1}{-i} \right\} \\\\ \frac{1}{\sqrt{-1}} \rightarrow \left\{ \frac{1}{i} , \frac{1}{-i} \right\} \\\\ \frac{\sqrt{1}}{\sqrt{-1}} \rightarrow \frac{ \left\{ 1 , -1 \right\} }{ \left\{ i , -i \right\} } = \left\{ \frac{1}{i} , \frac{-1}{i} , \frac{1}{-i} , \frac{-1}{-i} \right\} = \left\{ \frac{1}{i} , -\frac{1}{i} \right\} = \left\{ -i , i \right\} \\\\ \sqrt{\frac{1}{-1}} = \sqrt{-1} \rightarrow \left\{ i , -i \right\} \\\\ \sqrt{-1} \rightarrow \left\{ i , -i \right\} \\\\ \left( \sqrt{-1} \right)^2 \rightarrow \left\{ i^2 , \left(-i\right)^2 \right\} = \left\{ -1 , -1 \right\} = \{ -1 \} \end{gather} so \begin{gather} \left( \sqrt{-1} \right)^2 = -1. \end{gather} Also note that \begin{gather} \left( \frac{\sqrt{1}}{\sqrt{-1}} \right)^2 \rightarrow \left\{ (-i)^2 , i^2 \right\} = \left\{ -1 , -1 \right\} = -1. \end{gather}

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I think that there have been a number of approaches used to find $\frac{1}{i}$. Here may be a few to consider:

The positive powers of $i$ run in a cycle $(i, -1, -i, 1)$. Projecting this back into zero and the negative exponents, $i^0 = 1$ and $i^{-1}$ should be $-i$.

If we "rationalize" $\frac{1}{i}$ by multiplying both sides by $i$, then $\frac{1}{i} * \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$.

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given $a,b \in \mathbb R $
Rule 1:
$\sqrt {a b} \iff \sqrt {a} \sqrt {b} $ is valid only if $a,b \geq0$
Similarly, $\sqrt { \frac{a}{b}} \iff \frac{\sqrt {a}}{ \sqrt {b}} $ is valid only if $a\geq 0,b>0$
Rule 2:
And,if $a>0$
Case 1: $\sqrt{a}$ will give only one positive value (more precise: you will get purely real complex number with positive real part) eg. $\sqrt{4}=2 $
Case 2: $\sqrt{-a} = i \sqrt{a}$ will give only one complex number with positive imaginary part. (more precise: you will get purely imaginary complex number with positive imaginary part) eg. $\sqrt{-4}= 2i $

So, you have done wrong in fourth line
Lets take a example, shows if you do not follows above rules you can create blunder. Take $a>0$
$\sqrt{-a} = i \sqrt{a} $
But if do not follow above rule you can do blunder as
$\sqrt{-a} = \sqrt{\frac{a}{-1}} $
$=>\sqrt{-a} = \frac{\sqrt{a}}{\sqrt{-1}} $
$=>\sqrt{-a} = -i \sqrt{a} $

All the above taking and discussion is done under the consideration that in general the notion of $\sqrt[n]{x}; x\in\mathbb C$ talking about only about principal root and principal root and it is unique.

But for all the roots of $\sqrt[n]{x}; x\in\mathbb C$ Some Mathematician generally uses $x^\frac{1}{n}; x\in\mathbb C$

For example:
$(4)^\frac{1}{2} = 2,-2$ { more precisely: $(4)^\frac{1}{2} \implies 2,-2$ but $2 \not\implies (4)^\frac{1}{2} $ or $(-2) \not\implies (4)^\frac{1}{2}$ }
Similarly, $(-1)^\frac{1}{2} = i,-i$ because $(-i)* (-i) = \sqrt{-1} = i *i$

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Since $e^{i\tau/2}=-1$ we have: $$\sqrt{e^{i\tau/2}}\sqrt{e^{i\tau/2}}=\sqrt{e^{i\tau/2+i\tau/2}}$$ $$e^{i\tau/2}=e^{i\tau^{1/2}}$$ which is obviously true. But what does it mean? The contradictory proofs all give $1$ for $\sqrt{1}$ but they could give $-1$, then there is no contradiction. We also know that $e^{i\tau}=1$, but calling the square root of that $1$ is just false. Why the difference? Well, if we use $e^{i\tau}$ we are explicitly stating that we are carrying out rotations so the square root must rotate it back. $1$ by itself in contrast may or may not be the result of a rotation so we have a choice for the square root.


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What does this add to this thread, other than shoehorning $\tau$ instead of $\pi$? – Asaf Karagila May 22 at 23:32
My argument is slightly different (but I think clear) and my source was that website. I always give credit where it is due. – selfawareuser May 22 at 23:34

protected by MJD May 2 '14 at 16:34

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