Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$?

Question

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}

Answer 1

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.

Answer 2

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.

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

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

Answer 3

$1, -1$ and $\sqrt 1$ are all real numbers.

$\sqrt{-1}$ is not a real number.

There is no reason to assume that the rules for using the $\sqrt{\phantom x}$ symbol apply to non real numbers. In fact, you have found out that they don't.

If by $\sqrt{-1 \times -1}$ you mean $\sqrt{(-1 \times -1)}$, then $\sqrt{-1 \times -1} = \sqrt{(-1 \times -1)} = \sqrt 1 = 1$.

Complex number are based on the creation of a new non real number, $i$. The "definition" of $i$ is $i^2 = i \cdot i = -1$.

Side comment: This is often expressed as $i$ is the square root of $-1$. In mathematics the word "the" usually implies that there is exactly one such thing. However, since it turns out that $(-i)(-i)= -1$, then $-1$ seems to have more than one square root. Hence is it incorrect to refer to $i$ as **the ** square root of $-1$. Good luck trying to correct that particular abuse of the language.

If we want to use the notation $i = \sqrt{-1}$ to indicate that $i$ is the square root of $-1$, then it follows, by definition , that

$$\sqrt{-1} \cdot \sqrt{-1} = i^2 = -1$$

and we already know that

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

We can prove that there is no real number whose square is $-1$. We had to invent a not-real number, $i$, to make one. It shouldn't come as too big of a surprise that $i$ does not behave entirely like a real number.

Answer 4

Guaranteeing that all $a$ and $b$ is larger than $0$, $\sqrt{a}\sqrt{b} = \sqrt{ab}$. This also can be used when only one is larger than zero. However, if those are all negetive, it comes out to be $-\sqrt{ab}$. This can be proven by figuring out the square roots of each two negative numbers. For instance, think about the numbers $-2$ and $-3$. You'll see that $\sqrt{-2}\sqrt{-3} = i\sqrt 2 i \sqrt 2 = i^2\sqrt 6 = -\sqrt 6.$ The fact that $\frac{\sqrt a}{\sqrt b}=-\sqrt{\frac{a}{b}}$ when a is positive and b is negative can be proven in the same way.

Answer 5

$\def\cmlt{\operatorname{cmlt}}$ $\def\csqr{\operatorname{csqr}}$

In this case $-1$ is seen as a complex number, and the square root function over complex numbers is multi-valued, and mostly $\sqrt{e^{2\pi i}}=-1$, whereas $\sqrt{e^{0}}=1$.

Define multiplication over $\mathbb{C}$ as:

$$\cmlt:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$$ $$\cmlt(r_1e^{i \theta_1},r_2e^{i \theta_2})=r_1r_2e^{\theta_1+\theta_2}$$

We need to also specify the lower of $\theta_1$ and $\theta_2$ as the 'base' level for the multi-valued function, so that the multiplication is oriented in a positive fashion.

Define square root over $\mathbb{C}$ as:

$$\csqr:\mathbb{C}\to\mathbb{C}$$ $$\csqr(re^{i \theta})=\sqrt{r}e^{\frac{i\theta}{2}}$$

It is easy to see that, if $a,b$ are two complex numbers, then:

$$\csqr(\cmlt(a,b))=\cmlt(\csqr(a),\csqr(b))$$

Now, $-1\equiv e^\frac{i\pi}{2}$ (via base levels defined above), and so both definitions return $-1$, this being $\sqrt{e^{2\pi i}}$.

Answer 6

If $z$ and $w$ are two complex numbers, then it is not true in general $$ \sqrt{z}\sqrt{w}=\sqrt{zw} \tag{*}\label{*} \, . $$ The "rule" $\sqrt{\dfrac{z}{w}}=\dfrac{\sqrt{z}}{\sqrt{w}}$ also does not hold in general, as saying that $\sqrt{\dfrac{z}{w}}=\dfrac{\sqrt{z}}{\sqrt{w}}$ is equivalent to saying that $\sqrt{w}\sqrt{\dfrac{z}{w}}=\sqrt{z}$, which is a special case of $\eqref{*}$.

So why do these rules fail in the complex world, and what do we mean by the square root of a complex number in the first place? These are good questions, and the answers are more ... complex ... than one might expect.

When working in the real numbers, if $x$ is positive then the symbol "$\sqrt{x}$" refers to the positive number $a$ such that $a^2=x$. Although $2$ and $-2$ are both square roots of $4$, we designate $2$ as the "principal" root, and write $\sqrt{4}=2$. We adopt this notational convention so that $\sqrt{\cdot}$ may be regarded as a well-defined function from $[0,\infty)$ to $[0,\infty)$.

Things become trickier in the complex world, as there is no such thing as "positive" and "negative" complex numbers. (To be precise, there is no way of defining a total order $\leq$ on $\mathbb C$ so that it becomes an ordered field, and so any attempt to define the terms "positive" and "negative" will have limited success.) On what grounds can we say that $i$ is the principal square root of $-1$, then? Both $i$ and $-i$ are square roots of $-1$, but neither of them are positive. It is for this reason that some mathematicians would argue that the notation $i=\sqrt{-1}$ should be avoided entirely (see Ihf's answer to this question).

If we do insist upon defining the function $\sqrt{\cdot}$ in the complex world, then here is one possible approach. Every non-zero complex number $z$ can be uniquely written in the form $r\exp(i\theta)$, where $\theta \in (-\pi,\pi]$ (we say that $\theta$ is the principal argument of $z$) and $r>0$. If $z=r\exp(i\theta)$, then we could define $$ \sqrt{z} = \sqrt{r}\exp\left(\frac{i\theta}{2}\right) \, , $$ where $\sqrt{r}$ denotes the positive real root of $r$. This means that if $z=r_1\exp(i\theta_1)$ and $w=r_2\exp(i\theta_2)$ (where $\theta_1,\theta_2\in(-\pi,\pi]$ and $r_1,r_2>0$), then \begin{align} \sqrt{z}\sqrt{w} &= \sqrt{r_1}\exp\left(\frac{i\theta_1}{2}\right)\sqrt{r_2}\exp\left(\frac{i\theta_2}{2}\right) \\ &= \sqrt{r_1}\sqrt{r_2}\exp\left(\frac{i\theta_1}{2}\right)\exp\left(\frac{i\theta_2}{2}\right) \\ &= \sqrt{r_1r_2}\exp\left(\frac{i(\theta_1+\theta_2)}{2}\right) \, . \end{align} It might appear at first glance that the final line is the very definition of $\sqrt{zw}$, but this is not so. Notice that we defined $\sqrt{z}$ as $\sqrt{r}\exp\left(\frac{i\theta}{2}\right)$, we were careful to specify that $\theta$ is the principal argument of $z$. If $\theta_1+\theta_2\not\in(-\pi,\pi]$, then $\theta_1+\theta_2$ is not the principal argument of $zw$, and so we are not justified in writing $$ \sqrt{zw} = \sqrt{r_1r_2}\exp\left(\frac{i(\theta_1+\theta_2)}{2}\right) \, . $$ As it turns out, unless $\theta_1 + \theta_2$ is the principal argument of $zw$, the above equality does not hold, and so the "rule" $\sqrt{z}\sqrt{w}=\sqrt{zw}$ fails. In other words, for two non-zero complex numbers $z$ and $w$, the equality $\sqrt{z}\sqrt{w}=\sqrt{zw}$ holds if and only if the principal argument of $zw$ is the sum of the principal argument of $z$ and the principal argument of $w$. We can finally answer your question in detail. You wrote $$ \sqrt{\frac{1}{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} \, , $$ which is equivalent to saying that $$ \sqrt{-1}\sqrt{\frac{1}{-1}} = \sqrt{1} \, . $$ or $$ \sqrt{-1}\sqrt{-1} = \sqrt{1} \, . $$ We can easily verify that this is incorrect because the $\text{LHS}=i^2=-1$, while the $\text{RHS}=1$. The deeper reason for why equality does not hold is because the principal argument of $-1$ is $\pi$, but the principal argument of $1$ is not $\pi+\pi=2\pi$.

What do we learn from this fake proof? Well, the $\sqrt{\cdot}$ function is much less well-behaved in the complex numbers compared to the real numbers, to the point where some mathematicians abstain from defining it entirely. More generally, we are reminded that blindly applying "rules" without checking whether we have satisfied the conditions to apply them inevitably leads to trouble.