1.6: Applications of the Completeness Axiom

We prove here several fundamental properties of the real numbers that are direct consequences of the Completeness Axiom.

Theorem \(\PageIndex{1}\) - The Archimedean Property

The probabilities assigned to events by a distribution function on a sample space are given by.

Proof

Let us assume by contradiction that \(\mathbb{N}\) is bounded above. Since \(\mathbb{N}\) is nonempty,

\[\alpha=\sup \mathbb{N}\]

exists and is a real number. By Proposition 1.5.1 (with \(\varepsilon=1\)), there exists \(n \in \mathbb{N}\) such that

\[\alpha-1<n \leq \alpha.\]

But then \(n+1>\alpha\). This is a contradiction since \(n+1\) is a natural number. \(\square\)

The following theorem presents several immediate consequences.

Theorem \(\PageIndex{2}\)

The following hold:

1. For any \(x \in \mathbb{R}\), there exists \(n \in \mathbb{R}\) such that \(x<n\);
2. For any \(\varepsilon>0\), there exists \(n \in \mathbb{R}\) such that \(1 / n<\varepsilon\);
3. For any \(x>0\) and for any \(y \in \mathbb{R}\), there exists \(n \in \mathbb{N}\) such that \(y<nx)\);
4. For any \(x \in \mathbb{R}\), there exists \(m \in \mathbb{Z}\) such that \(m-1 \leq x<m\).

Proof

(a) Fix any \(x \in \mathbb{R}\). Since \(\mathbb{N}\) is not bounded above, \(x\) cannot be an upper bound of \(\mathbb{N}\). Thus there exists \(n \in \mathbb{N}\) such that \(x<n\).

(b) Fix any \(\varepsilon>0\). Then \(1 / \varepsilon\) is a real number. By (1), there exists \(n \in \mathbb{N}\) such that

\(1 / \varepsilon<n\)

This implies \(1 / n<\varepsilon\).

(c) We only need to apply (a) for the real number \(y / x\).

(d) First we consider the case where \(x>0\). Define the set

\(A=\{n \in \mathbb{N}: x<n\}\)

From part (a), \(A\) is nonempty. Since \(A\) is a subset of \(\mathbb{N}\), by the Well-Ordering Property of the natural numbers, \(A\) has a smallest element \(\ell\). In particular, \(x<\ell\) and \(\ell-1\) is not in \(A\). Since \(\ell \in \mathbb{N}\), either \(\ell-1 \in \mathbb{N}\) or \(\ell-1=0\). If \(\ell-1 \in \mathbb{N}\), since \(\ell-1 \notin A\) we get \(\ell-1 \leq x\). If \(\ell-1=0\), we have \(\ell-1=0<x\). Therfore, in both cases we have \(\ell-1 \leq x<\ell\) and the conclusion follows with \(m= \ell\).

In the case \(x \leq 0\), by part (1), there exists \(N \in \mathbb{N}\) such that

\(|x|<N\).

In this case, \(-N<x<N\), so \(x+N>0\). Then, by the result just obtained for positive numbers, there exists a natural number \(k\) such that \(k-1 \leq x+N<k\). This implies

\(k-N-1 \leq x<k-N\).

Setting \( m=k-N\), the conclusion follows. The proof is now complete. \(\square\)

Example \(\PageIndex{1}\)

Let \(A=\sup \left\{1-\frac{1}{n}: n \in \mathbb{N}\right\}\). We claim that \(\sup A=1\).

Solution

We use Proposition 1.5.1. Since \(1-1 / n<1\) for all \(n \in \mathbb{N}\), we obtain condition (1'). next, let \(\varepsilon>0\). From Theorem 1.6.2 (b) we can find \(n \in \mathbb{N}\) such that \(\frac{1}{n}<\varepsilon\). Then

\[1-\varepsilon<1-\frac{1}{n}.\]

This proves condition (2') with \(a=1-\frac{1}{n}\) and completes the proof.

Theorem \(\PageIndex{3}\) - The Density Property of \(\mathbb{Q}\)

If \(x\) and \(y\) are two real numbers such that \(x<y\), then there exists a rational number \(r\) such that

\[x<r<y.\]

Proof

We are going to prove that there exists an integer \(m\) and a positive integer \(n\) such that

\[x<m / n<y,\]

or, equivalently,

\[n x<m<n y=n x+n(y-x)\]

Since \(y-x>0\), by Theorem 1.6.2 (3), there exists \(n \in \mathbb{N}\) such that \(1<n(y-x)\). Then

\[n y=n x+n(y-x)>n x+1.\]

By Theorem 1.6.2 (4), one can choose \(m \in \mathbb{Z}\) such that

\[m-1 \leq n x<m.\]

Then \(n x<m \leq n x+1<n y\). Therefore,

\[x<m / n<y.\]

The proof is now complete. \(\square\)

We will prove in a later section (see Examples 3.4.2 and 4.3.1) that there exists a (unique) positive real number \(x\) such that \(x^{2}=2\). We denote that number by \(\sqrt{2}\). The following results shows, in particular, that \(\mathbb{R} \neq \mathbb{Q}\).

Proposition \(\PageIndex{4}\)

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

Proof

Suppose, by way of contradiction, that \(\sqrt{2} \in \mathbb{Q}\). Then there are integers \(r\) and \(s\) with \(s \neq 0\), such that

\[\sqrt{2}=\frac{r}{s}.\]

By canceling out the common factors of \(r\) and \(s\), we may assume that \(r\) and \(s\) have no common factors.

Now, by squaring both sides of the equation above, we get

\[2=\frac{r^{2}}{s^{2}},\]

and, hence,

\[2 s^{2}=r^{2}.\]

It follows that \(r^{2}\) is an even integer. Therefore, \(r\) is an even integer (see Exercise 1.4.1). We can then write \(r=2j\) for some integer \(j\). Hence \(r^{2}=4j^{2}\). Substituting in (1.3), we get \(s^{2}=2 j^{2}\). Therefore, \(s^{2}\) is even. We conclude as before that \(s\) is even. Thus, both \(r\) and \(s\) have a common factor, which is a contradiction. \(\square\)

The next theorem shows that irrational numbers are as ubiquitous as rational numbers.

Theorem \(\PageIndex{5}\)

Let \(x\) and \(y\) be two real numbers such that \(x<y\). Then there exists an irrational number \(t\) such that

\[x<t<y.\]

Proof

Since \(x<y\), one has

\[x-\sqrt{2}<y-\sqrt{2}\]

By Theorem 1.6.3, ther exists a rational number \(r\) such that

\[x-\sqrt{2}<r<y-\sqrt{2}\]

This implies

\[x<r+\sqrt{2}<y.\]

Since \(r\) is rational, the number \(t=r+\sqrt{2}\) is irrational (see Exercise 1.6.4) and \(x<t<y\). \(\square\)

Exercise \(\PageIndex{1}\)

For each sets below determine if it is bounded above, bounded below, or both. If it is bounded above (below) find the supremum (infimum). Justify all your conclusions.

1. \(\left\{\frac{3 n}{n+4}: n \in \mathbb{N}\right\}\)
2. \(\left\{(-1)^{n}+\frac{1}{n}: n \in \mathbb{N}\right\}\)
3. \(\left\{(-1)^{n}-\frac{(-1)^{n}}{n}: n \in \mathbb{N}\right\}\)

Answer

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Exercise \(\PageIndex{2}\)

Let \(r\) be a rational number such that \(0<r<1\). Prove that there is \(n \in \mathbb{N}\) such that

\[\frac{1}{n+1}<r \leq \frac{1}{n}.\]

Answer

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Exercise \(\PageIndex{3}\)

Let \(x \in \mathbb{R}\). Prove that for every \(n \in \mathbb{N}\), there is \(r \in \mathbb{Q}\) such that \(|x-r|<\frac{1}{n}\).

Answer

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Exercise \(\PageIndex{4}\)

Prove that if \(x\) is a rational number and \(y\) is an irrational number, then \(x+y\) is irrational. What can you say about \(xy\)?

Answer

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Exercise \(\PageIndex{5}\)

Prove that in between two real numbers \(a\) and \(b\) with \(a<b\), there are infinitely many rational numbers.

Answer

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Exercise \(\PageIndex{6}\)

Prove that in between two real numbers \(a\) and \(b\) with \(a<b\), there are infinitely many irrational numbers.

Answer

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