What type of numbers are roots of $x^{2} = -1$ themselves?

Question

Are the roots $i, -i$ themselves irrational numbers or complex numbers or left auxiliary so that undefined?

Answer 1

The usage varies. Quoting Ivan Niven (Irrational Numbers, ch. 7, sec. 1):

It is customary to separate the complex numbers into the types algebraic and transcendental, whereas it is the real numbers that are classified as rational and irrational. These usages are readily extended, of course. We would say that $a+bi$ is a complex rational if both $a$ and $b$ are rational. Likewise a real algebraic number is simply one that is simultaneously real and algebraic.

A couple of pages earlier Niven writes of complex integers (or Gaussian integers), which are complex numbers of the form $x=x_1 + x_2i$ where $x_1$ and $x_2$ are rational integers (and in particular, real).

So, to answer your question, the two numbers you mention could be called complex rationals or complex integers, but probably not just rationals or integers without the qualifier "complex". You can, of course, make your own definition and stick to it, as long as it is clear in the context what is meant.

Answer 2

Just $i$ and $-i$. The fundamental theorem of algebra tells us that an equation like $x^2 + 1 = 0$ has two roots. In this particular case, $i$ is the principal square root of $-1$, and $-i$ is the other root.

The idea that $i$ could be irrational seems ridiculous to me, but I have to admit that I have never stopped to think why I would consider one complex number rational and another one irrational. The thing about the purely imaginary numbers is that in some ways they are just like the purely real numbers, only the axis of the former is perpendicular to the axis of the latter.

Let me ask you this: do you consider the number $-21$ to be rational or irrational? What about $-\sqrt{7}$? It seems to me that if you consider $-21$ to be a rational number, you should also consider $21i$ and $-21i$ to be rational, and if you consider $-\sqrt{7}$ irrational, you should also consider $\sqrt{-7}$ and $-\sqrt{-7}$ irrational. (And I'm using the square root symbol to mean the "principal" square root, but contrarians will still find a reason to howl, and to push the limits of serial downvoting).

But I would still stop short of redefining the symbol $\mathbb{Q}$; I recommend you only use that to mean purely real rational numbers.

Answer 3

$i$ and $-i$ are purely imaginary numbers. In symbols, we can write $\Re(x) = 0$.

But the heart of your question is whether these numbers are rational or irrational. Most smart amateurs would say that of course these numbers are rational. Most professional mathematicians would hesitate, and a few would even declare that these numbers are neither rational nor irrational.

To understand that hesitation to extend the definition you need to remember that professional mathematicians don't like to extend definitions unless they can be sure that it's done in a rigorous way and that it does not "break" other definitions.

So how are rational and irrational numbers in $\mathbb{R}$ defined? I suppose first we need to define the integers of $\mathbb{Z}$. But since I don't want to take all day with this, I'm going to assume that you and I agree on the meaning of $1$ and the basic operations of arithmetic. We agree that $1 \in \mathbb{Z}$. Set $m = 1$ and $n = m + 1$, we agree that $n \in \mathbb{Z}$. Reset $m$ to $n$ and do $n = m + 1$ again, we agree that $n \in \mathbb{Z}$. Repeating this procedure forever, we obtain the positive integers. And by doing $n = m - 1$ similarly, we obtain $0$ and the negative integers.

Alright then, the rational numbers. If $x \in \mathbb{R}$ can be expressed as $x = \frac{a}{b}$ such that both $a$ and $b$ are integers from $\mathbb{Z}$, then $x$ is rational, but if not, then it is irrational. (By the way, $0$ is rational, just make sure you don't do $b = 0$).

Moving on to $x \in \mathbb{C}$: we define $i$ so that $i^2 = -1$. But can we express $i$ as $i = \frac{a}{b}$ such that both $a$ and $b$ are integers from $\mathbb{Z}$? We can't, and from that we can come to the "ridiculous" conclusion that $i$ is irrational.

This suggests that we need to extend the definition of rational and irrational to include purely imaginary numbers. Robert Soupe seems to suggest that we consider a number rational if it can be expressed as $\frac{a}{b}$ or $\frac{a}{b}i$. Then $i$ is rational, just as the amateur's intuition indicated.

Extending the definition this way does not seem to create problems, and by problems I mean that it does not suddenly turn some irrational real number into a rational real number.

But it does not address complex numbers with both real and imaginary parts, numbers like $\frac{1}{2} + 14i$. To address such numbers, we could say that a number is rational if it can be expressed as $\frac{a}{b}$ or $\frac{c}{d}i$ or $\frac{a}{b} + \frac{c}{d}i$, with $a$, $b$, $c$, $d$ all drawn from $\mathbb{Z}$. Obviously these are closed under addition, subtraction and multiplication. But are they closed under division? I believe that they are, but I don't want to hold off posting this answer much longer while I slowly verify that, I can come back and edit later.