In this post, we review different types of numbers classified based on how they are represented or their properties. We discuss natural, whole, integer, rational, real, and complex numbers.

#### Types of numbers: natural numbers, prime numbers, and whole numbers

The first type of numbers introduced and discussed since childhood is the class of counting numbers, i.e., one, two, three, and so on. We represent these numbers by the following numerals: $$1, 2, 3, \dots.$$ We usually collect the natural numbers in a set denoted by \(\mathbb{N}\). The “blackboard bold” letter \(\mathbb{N}\) stands for natural numbers: $$\mathbb{N} = \{1,2,3,\dots\}.$$

A natural number is prime if it has exactly two positive divisors, i.e. itself and 1. We denote the set of prime numbers by \(\mathbb{P}\). Examples of prime numbers include 2,3, 5, and many others, in particular, the following number: $$12345678910987654321.$$

In many references, they call the natural numbers together with the zero (denoted by 0) whole numbers. In some references, they denote the set of whole numbers by \(\mathbb{N}_0\): $$\mathbb{N}_0 = \{0,1,2,3,\dots \}.$$

We collect the first \(n\) natural numbers in the following set: $$\mathbb{N}_n = \{1,2,\dots,n\}.$$

#### Types of numbers: Integer numbers, even numbers, and odd numbers

The integer numbers are the whole numbers together with their additive inverses. The standard notation for the set of integer numbers is the blackboard bold letter \(\mathbb{Z}\). The set of integer numbers is $$\mathbb{Z} = \{ \dots,-2,-1,0,1,2, \dots \}.$$

The blackboard bold letter \(\mathbb{Z}\) stands for the German word “Zahl” literally meaning “number”.

An integer number is even if it is a multiple of 2. We denote the set of even numbers by \(\mathbb{E}\): $$\mathbb{E} = \{ \dots,-4,-2,0,2,4, \dots \}.$$

An integer number is odd if it is not even. We denote the set of even numbers by \(\mathbb{O}\): $$\mathbb{O} = \{ \dots,-3,-1,1,3, \dots \}.$$

It is clear that the union of the set of even numbers with the set of odd numbers is the set of integer numbers: $$\mathbb{Z} = \mathbb{E} \cup \mathbb{O}.$$

#### Types of numbers: rational numbers and irrational numbers

A number is a rational number if it is expressed by a ratio of an integer number to a natural number. Examples of rational numbers are $$1/2, -1/3, 2/9, -5/25.$$ The standard notation for the set of rational numbers is the blackboard bold letter \(\mathbb{Q}\). In the language of set theory, the definition of rational numbers is as follows: $$\mathbb{Q} = \{m/n: m \in \mathbb{Z}, n \in \mathbb{N} \}.$$

It is evident that each integer number is a rational number. In other words, the set of integer numbers is a subset of the set of rational numbers. However, there are many rational numbers that are not integer numbers. For example, \(1/3\) and \(-2/9\) are rational but not integer numbers.

Note that the blackboard bold letter \(\mathbb{Q}\) stands for quotient. Some references refer to rational numbers as quotient numbers.

Let \(x\) be a positive number such that its multiplication by itself is 2, i.e., \(x^2 = 2\). The interested reader can try to check the proof of this point that such a number, no matter what it is, is not a rational number. Many famous numbers such as \(e\) and \(\pi\) are not rational numbers. A number is irrational if it is not rational (quotient).

**Exercise**. Prove that \(\log^{3}_{2}\) is not rational.

**Solution**. Let the positive number \(\log^{3}_{2}\) be rational. So, there are natural numbers \(m\) and \(n\) such that $$\log^{3}_{2} = m / n.$$ This implies that \(3 = 2^{m/n}\) and this leads to the following contradiction: $$3^n = 2^n.$$ Why is this a contradiction? Because \(3^m\) is an odd number and \(2^n\) is an even number and a number cannot be an even and an odd number simultaneously.

#### Types of numbers: real numbers

There is no short algebraic definition of real numbers. However, one may geometrically define real numbers as numbers that correspond to points along a line.

It is obvious that rational numbers are real numbers and the rest are irrational numbers. However, the set theoretic gap between rational and real numbers is huge. We will not go through this discussion since it is off-topic, but the interested reader may refer to any standard textbook on set theory to learn more about the relevant topic known as “cardinal numbers”.

The standard notation for the set of real numbers is the blackboard bold letter \(\mathbb{R}\). It is obvious that \(\mathbb{Q} \subset \mathbb{R}\). Note that the set of irrational numbers is the same as the set-theoretic difference of the set of rational numbers from the set of real numbers, i.e. \(\mathbb{R} \setminus \mathbb{Q}\).

The real numbers are the foundation of calculus and other branches of mathematics. For more on real numbers see the post on the properties of real numbers.

According to the history of mathematics, Pythagorean theorem in geometry led to the discovery of irrational numbers and a blow to the philosophy of Pythagoreans who believed that the world is built up by numbers in the sense of natural numbers and their fractions (i.e., rational numbers).

Recall that Pythagorean theorem states that the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides. Therefore, the square root of 2 is equal to the length of the hypotenuse of an isosceles right triangle with legs of length 1. On the other hand, Pythagoreans proved that the square root of 2 cannot be a rational number. Therefore, they discovered an object not being built up by fractions of natural numbers!

#### Imaginary numbers

As we have already explained the equation \(x^2 = 2\), implicitly, led to the discovery of irrational numbers. On the other hand, it is natural to ask about a number that is a solution to the equation \(x^2 = -1\).

One of the most important properties of real numbers is that real numbers are totally ordered by the inequality relation. Therefore, if we take an arbitrary real number \(r\), then either \(r \geq 0\) or \(r \leq 0\). Evidently, in each case, we have \(r^2 \geq 0\). This means that if \(r\) is a real number, then its square \(r^2\) cannot be a negative number. Therefore, if \(x^2 = -1\), no matter what type of number it is, \(x\) cannot be real.

Now, if \(a\) is a positive real number, then the solutions of the equation \(x^2 = -a^2\) cannot be real numbers. This is perhaps why scientists called such numbers imaginary. An imaginary number is a real number multiplied by the imaginary unit \(i \). The imaginary unit \(i\) has this property that \(i^2 = -1\). Therefore, if \(a\) is an arbitrary real number, then the square of the imaginary number \(ai\) is computed as follows: $$(ai)^2 = a^2 i^2 = -a^2$$ which is a solution of the equation \(x^2 = -a^2\).

#### Types of numbers: complex numbers

A complex number is an addition of a real number and an imaginary one. In other words, a complex number is a two-dimensional real vector denoted by \(a+bi\). The addition of complex numbers is component-wise (similar to the addition of vectors): $$(a+bi) + (c+di) = $$ $$ (a+c) + (b+d)i.$$ For example, $$(2+3i) + (2-i) = $$ $$ 4+2i.$$ The subtraction of complex numbers is also component-wise. The multiplication of complex numbers is computed as follows: $$ (a+bi) (c+di) = $$ $$ac + adi + bci + bdi^2 = $$ $$(ac – bd) + (ad+bc)i.$$ In particular, $$(a+bi) (a-bi) = a^2 + b^2.$$

We use this to find a multiplicative inverse of a complex number. Let \(a+bi\) be a nonzero complex number. Observe that $$\frac{1}{a+bi} = $$ $$ \frac{1}{a+bi} \cdot \frac{a-bi}{a-bi} = $$ $$\frac{a-bi}{a^2+b^2} = $$ $$\frac{a}{a^2+b^2} + \frac{-b}{a^2+b^2} i.$$ This shows that each nonzero complex number is invertible.

**Exercise**. Compute the multiplication of the following complex numbers: $$ 2+3i \text{ and } 3-i.$$

**Exercise**. Find the multiplicative inverse of the complex number \(1 + i\sqrt{2}\).