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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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    You have exactly $50$ stars, $128 = 2^7$ upvotes, and $777$ reputation. You are a lucky human being...2018-05-03

10 Answers 10

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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, there might be those texts, but the standard is definitely$a$precise notion of quantification. This is especially important since the "domain" is not all that decides the truth value. It is a great difference between $\forall a\in A$ and $\exists a\in A$. It changes the whole meaning of the equation that follows the quantification (yes, follows! That's another pet peeve of mine: Writing quantors at the end of a formula, sometimes multiple quantors!)2016-01-08
66

The simple reason this,

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

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 \frac{\sqrt{1}}{\sqrt{-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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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 $\textstyle\sqrt{1}/\sqrt{-1}=\sqrt{1/-1}$.

18

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 real number with $(uv)^2=ab$. However, without the assumption $a,b\ge0$, this rule no longer holds in general.

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    I just stumbled over this from$a$duplicate and I find it confusingly phrased as it is **true** that when $u,v$ are (complex) numbers with $u^2 =a $ and $v^2 = b$ then $uv$ is a number with $(uv)^2 = ab$. Moreover taken strictly what you write under "If" is perfectly true without the assumption $a,b \ge 0$, though vacuously.2016-09-13
16

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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    A very nice step by step illustration.2018-02-08
14

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.

8

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 $\frac{1}{i} * \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$.

7

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 $(-i)* (-i) = \sqrt{-1} = i *i$

3

Some of the confusion here is because there are two types of roots: roots and principal roots. The principal roots always return only one value. However symbolically they are almost never different. You can very, very rarely encounter $_+\sqrt{\;}\;$for the principal root. So confusion is often created because people confuse these two types of roots, as they are represented by exactly the same symbols. $_+\sqrt1=1$ $\sqrt1=-1,+1$ $\sqrt{-1}=-i,+j$ If we are doing something like this $\sqrt{-1}=\frac{1}{i}$ it means we have to take the correct roots; otherwise, we can have even this: $\sqrt{1}=-1\;\ and \;\sqrt{1}=1$ Hence, $1=-1\,.\;$ That's not the way to go. We have multiple roots when dealing with complex numbers, even square roots from real numbers have two answers. To avoid this, the principal root is used but it doesn't differ symbolically from just square root. People extremely rarely write $_+\sqrt{\;}\;$. Therefore there's a lot of confusion. The rules of extracting roots from complex numbers don't strictly follow the rules used with principal roots, so you may easily arrive at a wrong answer. So when we see, for example, the formula: $\sqrt[n]{z}=\sqrt[n]{p}\, \bigg(\cos \frac{φ+2πk}{n}+i\sin\frac{φ+2πk}{n}\bigg)$ We have to understand that what is meant is this $\sqrt[n]{z}=\,_+\sqrt[n]{p}\, \bigg(\cos \frac{φ+2πk}{n}+i\sin{\frac{φ+2πk}{n}}\bigg)$ Another example. We can't reduce a multiple root $\;\sqrt[nk]{z^k}\;$ to $\;\sqrt[n]{z}\;$ because the first one has $nk$ different root values and the second--only $n\,$.

Formulas such as $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$ will work but not always. Generally, we can't use rules for principal roots for real numbers when we are dealing with complex numbers.

To recap:

First, $\sqrt{-1}=-i,+i\;\;$ ( NOT JUST $\;i\,$)

Second, when dealing with complex numbers, $\sqrt1=-1,+1\;\;$ ( NOT JUST $\;1\;$)

And third, $\sqrt{-1}\cdot\sqrt{-1}=1\;$ if we take $\;i,-i\;$ or $\;-i,i\;$ as roots, and $\sqrt{-1}\cdot\sqrt{-1}=-1\;$ if we take either $\;i,i$ or $\;-i,-i\;$ as roots. And even if we deal with real numbers and extract roots (not principal roots) we can still arrive at different values.

In very old books $\sqrt{-1}$ is sometimes used instead of $i$. It can only add to confusion.

Most crucial here: Roots must be distinguished from principal roots. When dealing with complex numbers we don't extract some principal root of them, but we have a set of different root values. So, almost every single line is wrong in the OP's 'proof'. Kevin Holt provided a very nice step by step illustration.