0

I have trouble to comprehend what my mistake is in the following calculation:

If we set $\sqrt{-1}$ to be the new number with the property that $(\sqrt{-1})^2 = -1$ then I can write $$\frac{1}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}.$$

But we also have (and I know this is the correct result) $$ \frac{1}{\sqrt{-1}}\cdot\frac{\sqrt{-1}}{\sqrt{-1}}=\frac{\sqrt{-1}}{-1}=-\sqrt{-1}$$

What am I missing? Thanks.

Max
  • 1,039
laguna
  • 169
  • 2
    Every nonzero complex number has two square roots. – Angina Seng Feb 13 '19 at 20:15
  • 3
    The identity $\sqrt{a} \sqrt{b} = \sqrt{ab}$ is not true in general for complex numbers. – Robert Israel Feb 13 '19 at 20:17
  • 1
    $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is a little white lie we tell our selves when we assume we are only dealing with real numbers. Well one of two. In truth we only know that $|\sqrt{|ab|}|=|\sqrt{|a|}||\sqrt{|b|}|$. this is a white lie for reals as only positive numbers have square roots and we only consider that positive square root. – fleablood Feb 13 '19 at 21:53
  • 1
    For real numbers only $x\ge 0$ have square roots. So if we see $\sqrt{ab}$ we assume $ab\ge 0$ which means either $a\ge 0; b\ge 0$ or $a<0;b<0$ but in either case $ab = (-a)(-b) = |a||b|$ so we assume $a\ge 0;b\ge 0$. And in that case $\sqrt{ab}=\sqrt{a}\sqrt{b}$ that assumes only real numbers $\sqrt{ab}=\sqrt{(-a)(-b)}\ne\sqrt{-a}\sqrt{-b}$ because negatives don't have square roots. In reality we have $\sqrt{ab}=\sqrt{|a|}\sqrt{|b|}$ but ONLY with the assumption $ab \ge 0$. Now if we assume that we can have square roots of negative numbers LOTS of things go out the window..... – fleablood Feb 13 '19 at 22:06
  • 1
    To begin with if $x = a^2$ then $x=(-a)^2$ so ever number has two square roots. One is $\sqrt{x}$, the other is $-\sqrt{x}$. That's true even if $x=a^2$ if $a$ isn't real or $x$ isn't positive. If $\sqrt{-1}=i$ exists then we can't say that $\sqrt{ab}=\sqrt{a}\sqrt{b};\sqrt{ab}=\sqrt{-a*-b}=\sqrt{-a}\sqrt{-b}=i\sqrt{a}i\sqrt{b}=i^2\sqrt{ab}=-\sqrt{ab}$. It isn't true that $\sqrt{ab}=\sqrt{a}\sqrt{b}$. Instead it's only $|\sqrt{ab}|=|\sqrt{a}||\sqrt{b}|$. – fleablood Feb 13 '19 at 22:15
  • So the second way works. But the first way doesn't. It's interesting but. $\frac 1{\sqrt{-1}} = -\sqrt{-1}$. – fleablood Feb 13 '19 at 22:16
  • And it's interesting that $i^1 = i; i^2 = -1; i^3 = -i = \frac 1i$ and $i^4 = (i^2)^2 = i^3*i =1$. – fleablood Feb 13 '19 at 22:21

2 Answers2

4

The incorrect step is $$\frac{1}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}$$ This kind of relations that you learned very soon in your life are so anchored in your thinking that you don't even think that they need to be re-checked whenever the settings are different. Complex numbers behave differently to positive real numbers in that respect.

Arnaud Mortier
  • 27,276
-1

The problem is due to the use of the same symbol $\sqrt{\cdot}$ with different meanings.

Let's separate the meanings by distinguishing:

  • $\sqrt[\text{real}]{\cdot}$ is an operator defined on non-negative real numbers, that associates to $x$ the unique non-negative real number $y$ such that $y \cdot y = x$;

  • $\sqrt[\text{new}]{-1}$ is a symbol used to indicate a new number, that satisfies the property $\sqrt[\text{new}]{-1}\cdot \sqrt[\text{new}]{-1} = -1$.

The first chain of equations you wrote translates to:

$$\displaystyle\frac{1}{\sqrt[\text{new}]{-1}}=\frac{\sqrt[\text{real}]{1}}{\sqrt[\text{new}]{-1}} \overset{!}{=} \frac{\sqrt[\text{real}]{1}}{\sqrt[\text{real}]{-1}} \overset{!}{=} \sqrt[\text{real}]{\frac{1}{-1}} \overset{!}{=} \sqrt[\text{real}]{-1},$$

but the steps marked with $\overset{!}{=}$ are not valid since $\sqrt[\text{real}]{-1}$ is not defined.

simonet
  • 467
  • This would be almost perfect if you added the word positive when necessary. Also, the equations at the end make no sense, because $\sqrt[\text{real}]{-1}$ doesn't exist. – Arnaud Mortier Feb 13 '19 at 21:30
  • 1
    I have to point out that this misses the point. $\sqrt[\text{real}]{x}$ is NOT unique. If $y*y=x$ then $(-y)(-y)=x$. Nor is $\sqrt[\text{new}]{-1}$. – fleablood Feb 13 '19 at 21:58
  • Thank you for suggestions. I tried to fix my answer. – simonet Feb 14 '19 at 08:35