# 1.2: Rational and Irrational Numbers

- Page ID
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We start with a few results we need in the remainder of this subsection.

Theorem 1.7: Well-ordering Principle

Any non-empty set \(S\) in \(\mathbb{N} \cup \{0\}\) has a smallest element.

**Proof**-
Suppose this is false. Pick \(s_{1} \in S\). Then there is another natural number \(s_{2}\) in \(S\) such that \(s_{2} \le s_{1}-1\). After a finite number of steps, we pass zero, implying that \(S\) has elements less than 0 in it. This is a contradiction.

Note that any non-empty set \(S\) of integers with a lower bound can be transformed by addition of a integer \(b \in N_{0}\) into a non-empty \(S+b\) in \(N_{0}\). Then \(S + b\) has a lower bound, and therefore so does \(S\). Furthermore, a non-empty set \(S\) of integers with a upper bound can also be transformed into a non-empty \(-S+b\) in N0. Here, \(-S\) stands for the collection of elements of S multiplied by \(-1\). Thus we have the following corollary of the well-ordering principle.

Corollary 1.8

Let be a non-empty set \(S\) in \(\mathbb{Z}\) with a lower (upper) bound. Then \(S\) has a smallest (largest) element.

Definition 1.9

An element \(x \in \mathbb{R}\) is called rational if it satisfies \(qx-p = 0\) where \(p\) and \(q \ne 0\) are integers. Otherwise it is called an irrational number. The set of rational numbers is denoted by \(\mathbb{Q}\).

The usual way of expressing this, is that a rational number can be written as \(\frac{p}{q}\). The advantage of expressing a rational number as the solution of a degree 1 polynomial, however, is that it naturally leads to Definition 1.12.

Theorem 1.10

Any interval in \(\mathbb{R}\) contains an element of \(\mathbb{Q}\). We say that \(\mathbb{Q}\) is dense in \(\mathbb{R}\).

**Proof**-
Let \(I = (a, b)\) with \(b > a\) any interval in \(\mathbb{R}\). From Corollary 1.8 we see that there is an n such that \(n > \frac{1}{b-a}\). Indeed, if that weren’t the case, then \(\mathbb{N}\) would be bounded from above, and thus it would have a largest element \(n_{0}\). But if \(n_{0} \in \mathbb{N}\), then so is \(n_{0}+1\). This gives a contradiction and so the above inequality must hold.

It follows that \(nb-na > 1\). Thus the interval \((na, nb)\) contains an integer, say, \(p\). So we have that \(na < p < nb\). The theorem follows upon dividing by n.

The crux of the following proof is that we take an interval and scale it up until we know there is an integer in it, and then scale it back down.

Theorem 1.11

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

**Proof**-
Suppose \(\sqrt{2}\) can be expressed as the quotient of integers \(\frac{r}{s}\). We may assume that \(\gcd (r, s) = 1\) (otherwise just divide out the common factor). After squaring, we get

\[2s^2 = r^2 \nonumber\]

The right hand side is even, therefore the left hand side is even. But the square of an odd number is odd, so \(r\) is even. But then \(r^2\) is a multiple of 4. Thus \(s\) must be even. This contradicts the assumption that \(\gcd (r, s) = 1\).

It is pretty clear who the rational numbers are. But who or where are the others? We just saw that \(\sqrt{2}\) is irrational. It is not hard to see that the sum of any rational number plus \(\sqrt{2}\) is also irrational. Or that any rational non-zero multiple of \(\sqrt{2}\) is irrational. The same holds for \(\sqrt{2}, \sqrt{3}, \sqrt{5}\), etcetera. We look at this in exercise 1.7. From there, is it not hard to see that the irrational numbers are also dense (exercise 1.8). In exercise 1.15, we prove that the number \(e\) is irrational. The proof that \(\pi\) is irrational is a little harder and can be found in [**1**][section 11.17]. In Chapter 2, we will use the fundamental theorem of arithmetic, Theorem 2.14, to construct other irrational numbers. In conclusion, whereas rationality is seen at face value, irrationality of a number may take some effort to prove, even though they are much more numerous as we will see in Section 1.4.