# 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 at 21:41
• I agree with @Fawkes4494d3; a purely imaginary complex number has no real part – J. W. Tanner Jul 24 at 21:41
• @Fawkes4494d3, oh, yes you're right. (>ლ), $\sqrt{-1}$ should delete it – UmbQbify Jul 24 at 21:43
• math educators community maybe of your (future). interests. – UmbQbify Jul 24 at 22:04
• I would simply avoid each potential misleading notation as $\mathbb C\setminus\mathbb R$. – Michael Hoppe Jul 25 at 15:20

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.
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 $$\Bbb{C}\setminus \Bbb{R}=\{a+bi:b \ne 0\:\text{and}\:a,b\in \Bbb{R}\}.$$
• 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 at 21:43
• 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 at 21:45