Are $\mathbb{C}-\mathbb{R}$ imaginary numbers?

Question

Background

I am teaching senior high school students about the structure of numbers. Start from defining $\mathbb{Q}$ and $\mathbb{R}$ as the rational and real numbers respectively, we can define $\mathbb{R}-\mathbb{Q}$ as the irrational numbers.

I am trying to use the same logic to define imaginary numbers by making use of the relationship between $\mathbb{R}$ and $\mathbb{C}$. Another definition for imaginary numbers is

numbers that become negative under squaring operation.

Let $\mathbb{C}$ and $\mathbb{R}$ be the complex and real number sets respectively. Are $\mathbb{C}-\mathbb{R}$ imaginary numbers?

@UmbQbify-Key20- only the purely real numbers have been cast out, right, $\mathbb{C}\backslash \mathbb{R}$ contains all complex numbers except purely real numbers. For e.g $1+2\iota$ is still there. — Fawkes4494d3, Jul 24 '20 at 21:41
I agree with @Fawkes4494d3; a purely imaginary complex number has no real part — J. W. Tanner, Jul 24 '20 at 21:41
@Fawkes4494d3, oh, yes you're right. (>ლ), $\sqrt{-1}$ should delete it — UmbQbify, Jul 24 '20 at 21:43
I would simply avoid each potential misleading notation as $\mathbb C\setminus\mathbb R$. — Michael Hoppe, Jul 25 '20 at 15:20
@MichaelHoppe: Then, what notation should we use to represent imaginary numbers? — Display Name, Jul 25 '20 at 15:25
Just first define them ... You're "trying to use the same logic to define imaginary numbers", but in fact: one can't. — Michael Hoppe, Jul 25 '20 at 15:29

score 7 · Accepted Answer · answered Jul 24 '20 at 21:41

7

Imaginary numbers are real multiples of $\mathrm{i}$. So the complex number $1+\mathrm{i} \in \Bbb{C} \smallsetminus \Bbb{R}$ is neither real nor imaginary.

answered Jul 24 '20 at 21:41

Eric Towers

67,037

@Artificial Stupidity, R lies inside C, ofc. A complex number is made of real and imaginary part.. so I don't quite get you. – UmbQbify Jul 24 '20 at 22:02
@UmbQbify-Key20- : A real number has zero imaginary part and an imaginary number has zero real part. Any complex number with neither part zero is neither real nor imaginary. – Eric Towers Jul 27 '20 at 13:04

Alekos Robotis · Answer 2 · 2020-07-24T21:46:22.020

1

Depends what you mean by "imaginary." Perhaps you mean an element of $\Bbb{C}$ of the form $ai$ for $a\in \Bbb{R}$ in which case this is false. Indeed, in the complex plane you have removed only the "$x$-axis" so that $$\Bbb{C}\setminus \Bbb{R}=\{a+bi:b \ne 0\:\text{and}\:a,b\in \Bbb{R}\}.$$

edited Jul 24 '20 at 21:46

answered Jul 24 '20 at 21:41

Alekos Robotis

26,184

Shouldn’t this be the set of all $bi$’s ? Where $a = 0$? Since you’re taking out the “real part” of the complex number - giving a simply pure imaginary number. – Taylor Rendon Jul 24 '20 at 21:43
1

It's possible I'm misunderstanding, but for instance like in Eric Towers' answer, $1+i\in \Bbb{C}\setminus \Bbb{R}$, which is not of the form $bi$. (Edit: there was a mistake, but the mistake was that it should be $b\ne 0$ rather than $a\ne 0$). – Alekos Robotis Jul 24 '20 at 21:45
Gotcha! Your correction makes sense now. Thanks for your time. – Taylor Rendon Jul 24 '20 at 21:48

Are $\mathbb{C}-\mathbb{R}$ imaginary numbers?

Background

2 Answers2