Some well-known sets of numbers

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 You should already know the following sets of numbers: natural numbers, integers, rational numbers, real numbers, and complex numbers. We review these sets of numbers and study some of their properties.



Natural numbers

The set of all natural numbers is usually denoted \(\mathbb{N}\). Of course, it is \[\mathbb{N} = \{1, 2, 3, 4, \cdots\}.\]

Some mathematicians prefer to consider 0 a natural number. In this blog, I will use \(\mathbb{N}_0\) to denote the set of natural numbers including 0: \[\mathbb{N}_0 = \{0, 1, 2, 3, \cdots\}.\]

Natural numbers are ordered in the usual sense:

\[1 < 2 < 3 < 4 < \cdots\]

The least element is 1. Let \(S\) be a non-empty subset of \(\mathbb{N}\). Now the following should be intuitively trivial: For a sufficiently large \(N\in \mathbb{N}\), we may select from \(1, 2,\cdots,N\) one that is least among these integers and also belongs to \(S\).

Let's make this an axiom:

Axiom (Principle of well-ordering in \(\mathbb{N}\))

Every non-empty subset \(S\) of \(\mathbb{N}\) has a least integer in \(S\).

Often, we have some statement (predicate), say \(P(n)\), that depends on a particular integer \(n\). Here are some examples.

Example. 

\[1 + 2 + \cdots + n = \frac{n(n+1)}{2} ~~ (n \in \mathbb{N}).\]

Here, the predicate \(P(n)\) is this equation itself. □

Example.

\[2n + 1 \leq 2^n~~(n\in\mathbb{N}).\] 

□

Note that these are just predicates that depend on \(n\in\mathbb{N}\), whether they are true or false. Nevertheless, we are often interested in whether a particular predicate is true for all \(n\in\mathbb{N}\), or for \(n\in\mathbb{N}\) greater than some specified value. This is done by using the Principle of Induction (i.e., mathematical induction), which is derived from the Principle of Well-ordering in \(\mathbb{N}\). This is actually a theorem, but, without proof, we list it here as an axiom for convenience.

Theorem Axiom (Principle of (Mathematical) Induction)

Let \(P(n)\) be a predicate depending on the integer \(n\). Suppose that

  1. \(P(1)\) is true, and
  2. if \(P(k)\) is true then \(P(k+1)\) is true.
Then, \(P(n)\) is true for all \(n\in\mathbb{N}\).

Example. Let us prove that 
\[1 + 2 + \cdots + n = \frac{n(n+1)}{2} ~~ (n \in \mathbb{N}).\tag{Eq:SumN}\]
by mathematical induction.
  1. For \(n = 1\), the left-hand side is 1, the right-hand side is \(1\times (1 + 1)/2 = 1\). Thus, (Eq:SumN) holds.
  2. Suppose that (Eq:SumN) holds for \(n = k\). Then, we have \[1 + 2 + \cdots + k = \frac{k(k+1)}{2}.\tag{Eq:IH}\] Now, \[\begin{eqnarray} 1 + 2 + \cdots + k + (k+1) &=& \frac{k(k+1)}{2} + (k+1) ~ (\text{by the induction hypothesis (Eq:IH)})\\ &=& (k+1)\left(\frac{k}{2} + 1\right) ~ (\text{factorization})\\ &=& \frac{(k+1)(k+2)}{2}.   \end{eqnarray}\] Therefore, (Eq:SumN) holds for \(n = k+1\). 
By the Principle of Induction, (Eq:SumN) holds for all \(n \in \mathbb{N}\). □

Integers

The set of all integers is usually denoted \(\mathbb{Z}\): \[\mathbb{Z} = \{\cdots, -2, -1, 0, 1, 2, \cdots \}.\]

By the way, the symbol \(\mathbb{Z}\) comes from the German word Zahl for number.

Rational numbers

The set of all rational numbers is usually denoted \(\mathbb{Q}\). It may be (informally) written as \[\mathbb{Q} = \{\frac{m}{n} \mid m, n \in \mathbb{Z}, n \neq 0\}.\] 

But what's the superset of \(\mathbb{Q}\) in this "definition"? What does the expression "\(\frac{m}{n}\)" mean to begin with? We don't dig into these details here, but just accept this set as given. 

By the way, the symbol \(\mathbb{Q}\) is after the word Quotient which is also used to mean rational numbers.

Real numbers

We know that \(\sqrt{2} = 1.141421356\cdots\) is not a rational number but an irrational number. \(\pi = 3.141592\cdots\) is also irrational. The set containing all rational and irrational numbers is the set of real numbers. In other words, a real number is either a rational or irrational number. We usually denote the set of real numbers by \(\mathbb{R}\). Clearly, \[\mathbb{Q} \subset \mathbb{R}.\] As you can see, our definition of real numbers is very informal. To begin with, irrational numbers are defined to be those numbers that are not rational. But what are those "numbers"? They can't be real numbers because we haven't defined real numbers when we are defining irrational numbers. The rigorous definition of real numbers is beyond the scope of this lecture. I might explain it in a future post. It's not that easy. 

Lemma

\(\sqrt{2}\) is irrational.

Proof. We prove this by contradiction. 

Suppose \(\sqrt{2}\) is rational, that is, \(\sqrt{2} = \frac{m}{n}\) for some \(m, n\in \mathbb{N}\), and \(m\) and \(n\) have no common divisor. Then, we have \(2 = m^2/n^2\), and hence \(m^2 = 2n^2\). Therefore, \(m^2\) is an even integer, which implies that \(m\) is also even (why?). So, let \(m = 2k\) for some \(k \in \mathbb{N}\). Then, we have \(4k^2 = 2n^2\), and hence \(2k^2 = n^2\). But this implies that \(n\) is also even so that \(m\) and \(n\) have the common divisor 2. This contradicts our assumption that \(m\) and \(n\) have no common divisor. ■

Anyway, we naively accept the set of real numbers as given. But we require the following as an axiom of real numbers:

Axiom (Archimedes' principle)

For an arbitrary positive real number \(a\) and an arbitrary real number \(b\), there exists a natural number \(n\) such that \(an > b\).

In other words, we assume that real numbers are those numbers that satisfy this axiom. Using this axiom, we can prove the following theorem.

The next theorem demonstrates that \(\mathbb{Q}\) and \(\mathbb{R}\) are dense: If we pick any two real numbers, we can always find some rational and irrational numbers in between.

Theorem

Let \(a\) and \(b\) be be real numbers such that \(a < b\). Then, there exist a rational number \(x\) such that \(a < x < b\) and an irrational number \(y\) such that \(a < y < b\).

Proof. By Archimedes' principle, we can choose a natural number \(N \in \mathbb{N}\) such that 

\[1/N < b - a. \tag{eq:N}\]

 Let \(z\in \mathbb{Z}\) be the least integer such that \(bN \leq z\). Accordingly, we have \(z - 1 < bN\). Together, we have

\[z-1 < bN \leq z,\]

or

\[\frac{z-1}{N} < b \leq \frac{z}{N}.\tag{eq:z}\]

Now,

\[\begin{eqnarray} a &=& b - (b - a)\\ &<& b - \frac{1}{N} ~~~ (\text{c.f. (eq:N)})\\ &\leq & \frac{z}{N} - \frac{1}{N} ~~ (\text{c.f., (eq:z)})\\ & = & \frac{z-1}{N}\\ & < & b. ~~ (\text{c.f., (eq:z)}) \end{eqnarray}\] In summary, we have

\[a < \frac{z-1}{N} < b,\]

and \[x = \frac{z-1}{N} \in \mathbb{Q}. \tag{eq:x}\]


Next, choose \(M \in \mathbb{N}\) sufficiently large so that

\[\sqrt{2} < M(b - x) \tag{eq:M}\]

where \(x\) is the rational number defined in (eq:x). Again, such a choice of \(M\) is possible due to Archimedes' principle. Let us define \(y\) by \[y = x + \frac{\sqrt{2}}{M}.\]

Then, \(y\) is irrational (otherwise, \((y - x)M = \sqrt{2}\) would be rational). By (eq:M), we have

\[x < y = x + \frac{\sqrt{2}}{M} < x + (b - x) = b.\]

Thus, 

\[a < y < b.\]

■

Complex numbers

\(i = \sqrt{-1}\) is called the imaginary unit. This is not a real number. For any real number, its square is always positive or zero. But, we have, by definition, \(i^2 = -1\). If we multiply the imaginary unit \(i\) by a real number \(b\), the result \(ib\) is called an imaginary number. If we add a real number \(a\) to an imaginary number \(ib\), the result \(a + ib\) is called a complex number. The set \(\mathbb{C}\) of complex numbers is (informally) defined as \[\mathbb{C} = \{a + ib | a, b\in \mathbb{R}\}.\] If we identify \(a + i0 \in \mathbb{C}\) with \(a \in \mathbb{R}\), that is, if we regard \(a + i0 = a\), we may assert that \[\mathbb{R} \subset \mathbb{C}\] holds. 

Of course, the above definition of complex numbers is very informal. What does \(a + ib\) mean, exactly? At this point, we don't have a precise definition of the multiplication between the imaginary unit and a real number; we don't have a precise definition between a real number and an imaginary number, either. What do they mean at all? In a later chapter, we will see how we can construct complex numbers from (pairs of) real numbers.



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