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So basically, is there a set containing elements not in the complex numbers that squared is a negative real number?
$ \{ x \ \vert \ x^2 \in \mathbb{R} \land x^2<0 \} \stackrel{?}{=} \emptyset $

GNUSupporter 8964民主女神 地下教會
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  • Yes. The square of a real number is nonnegative. – saulspatz Nov 26 '20 at 13:09
  • excuse me when i read the bracket notation on the LHS, it's clear that it's complex, but from the context, it seems you're dealing with something real. in any case, this question has nothing to do with [tag:analysis], which is the logical foundations of calculus. – GNUSupporter 8964民主女神 地下教會 Nov 26 '20 at 13:12
  • However, $<$ is not usually defined on $\mathbb{C}$ and it cannot be defined to make $\mathbb{C}$ an ordered field. So, just using it pretty much says that $x^2$ is real. – badjohn Nov 26 '20 at 13:14
  • I realise that I'm being stupid. I guess that is what you get for asking a question on a site where everyone is smarter than you (not being sarcastic) ... –  Nov 26 '20 at 13:53

2 Answers2

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We can actually make a much stronger conclusion.

Let $z = re^{i\theta} \in \mathbb{C}$. We know $z^{2} = r^{2} e^{i 2 \theta} \in \mathbb{R}$ so $2\theta = k \pi$ so $\theta = k \pi/2$. We also know $z^2 < 0$ so $r > 0$ and $\theta = \pi/2$ or $\theta = 3\pi/2$. That is, we know $z$ is of the form $\pm b i$ where $b$ is a positive real number.

Matthew van Eerde
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For all $x \in \mathbb R$ we have that $x^2 \ge 0.$ Hence

$$\{ x \ \vert \ x^2 \in \mathbb{R} \land x^2<0 \} = \emptyset.$$

Fred
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