3

My maths teacher claims that $(\sqrt{-4})^2$ is not defined if we consider only those numbers in the real number plane.

Even if we exclude imaginary numbers, IMO the statement can be written as

$$-4^{(2 \times \frac{1}{2})}$$

Then the powers obviously cancel and we are left with $-4$. Even using WolframAlpha results in $-4$.

According to Wikipedia, $(b^n)^m = b^{nm}$ for all $b \neq 0$. This is what makes me not believe my teacher's claim, even if we are not considering complex numbers.

My teacher's logic is that we need to first evaluate $\sqrt{-4}$ before we can proceed with squaring the result. He also claims that the $(b^n)^m = b^{nm}$ identity is true only for $b > 0$, contradicting what Wikipedia states.

So, my two questions, summarized:

  • Is $(\sqrt{-4})^2$ defined if we do not consider complex numbers?
  • Is $(b^n)^m = b^{nm}$ for only $b > 0$ or $\forall$ $b \neq 0$?
Box Box Box Box
  • 284
  • 4
    If you only consider real numbers, $\sqrt{-4}$ is not defined, neither is any function of that, so the square of $\sqrt{-4}$ cannot be defined either. The operations have to make sense in the order they are to be performed. – Paul May 01 '20 at 14:30
  • 1
    For the second question, cf. this – J. W. Tanner May 01 '20 at 14:37
  • @Paul ok, so, if I understand what you're saying, $\frac{\sqrt{-4}}{\sqrt{-4}}$ will be undefined as well, right? – Box Box Box Box May 02 '20 at 04:28
  • 1
    @J.W.Tanner Hmm, ok, I think that answers my whole question, because my complete reasoning behind $(\sqrt{-4})^2$ being defined is the exponents cancelling. If that cannot be done, then, well, I understand where I went wrong. Thanks. – Box Box Box Box May 02 '20 at 04:43
  • Yes, that expression will also be undefined. – Paul May 02 '20 at 10:46

1 Answers1

3

  • $\sqrt{-4}$ is not defined in real numbers, so neither is $\sqrt{-4}^2$

  • That Wikipedia page says $(b^n)^m=b^{nm}$ for integer exponents $m$ and $n$.

When $b<0$, it may not hold for fractional exponents such as $\frac12$ (as in your case).

J. W. Tanner
  • 60,406