The Problem with Rationals

As pleasant as the rational numbers are, we very quickly run into problems.

Theorem 1. Let $p$ be any prime number. There is no rational number $r$ such that $r^2 = p$ .

Proof. Suppose for a contradiction such an $r$ exists. Write $r = m/n$ , where $m,n$ are integers with $n > 0$ and $m,n$ having no common factors (in number-theoretic language, $\gcd(m, n) = 1$ ). By algebra,

$m^2 = (rn)^2 = r^2 n^2 = pn^2.$

This tells us $p \mid m^2$ . Using Euclid’s lemma, we have $p \mid m$ . Thus, find an integer $k$ such that $m =kp$ . This implies

$k^2 p^2 = m^2 = pn^2 \quad \Rightarrow \quad n^2 = p k^2.$

This yields $p \mid n$ , so that $p \mid \gcd(m, n) = 1 \Rightarrow p = 1$ , a contradiction.

What we have shown is that if we denote any positive $r$ that satisfies $r^2 = p$ by $r = \sqrt p$ , if $r$ exists, whatever it is, it cannot be a rational number. In particular, $\sqrt 2$ , if it exists, is not rational. This does not imply that $\sqrt p$ exists! However, how can we define $\sqrt p$ ? Intuitively, we want to “plug” the holes in the rational numbers (i.e. completing the rational numbers).

Let’s use $\sqrt 2$ as an example. Really, $r =\sqrt 2$ is meant to capture the idea that $r^2 = 2$ . Define the set $K \subseteq \mathbb Q$ by

$K := \{x \in \mathbb Q_{\geq 0} : x^2 < 2\}.$

We make several observations about this set which will motivate our defining property of a real number. We will need to use $\mathbb Q$ to create the set $\mathbb R$ of real numbers, so this property will guide our efforts in doing so.

Definition 1. Let $L \subseteq \mathbb Q$ .

We call $M \in \mathbb Q$ an upper bound of $L$ if for any $x \in L$ , $x \leq M$ . In this case, we say that $L$ is bounded above.
We call the upper bound $M$ a least upper bound of $L$ if for any upper bound $M'$ of $L$ , $M \leq M'$ .

Theorem 2. The set $K$ satisfies the following properties.

$K \neq \emptyset$ .
$K$ is bounded above.
Any upper bound for $K$ does not belong to $K$ .
$K$ has no least upper bound.

Proof. For the first claim, observe that $0^2 = 0 < 2$ , thus $0 \in K$ . For the second claim,

$x \in K \quad \Rightarrow \quad x^2 < 2 \leq 4 = 2^2 \quad \Rightarrow \quad x \leq 2.$

For the third claim, what we need to show is that for any $x \in K$ , there exists $y \in K$ such that $x < y$ . The idea is to define $y = x + 1/n$ for sufficiently large $n$ (so that the numbers differ but not too much) and show that $y^2 < 2$ so that $y \in K$ . To that end, let’s reverse-engineer our choice of $n$ :

$\displaystyle \left(x + \frac 1n \right)^2 = x^2 + \frac{2x}{n} + \frac 1{n^2} < 2.$

At this point in time, we need to be exceedingly careful, because we are touching on the main complication of real analysis. If $a \leq b$ and $b \leq c$ , then we can conclude that $a \leq c$ . Sadly, however, the converse does many not always hold. So we need to think very logically.

Now, if $n > 1$ , then it’s not hard to see that

$\displaystyle \frac 1{n^2} = \frac 1n \cdot \frac 1n < 1 \cdot \frac 1n = \frac 1n.$

Thus, whatever our choice of $n$ is, if we stipulate $n > 1$ , then

$\displaystyle x^2 + \frac{2x}{n} + \frac 1{n^2} < x^2 + \frac{2x}{n} + \frac 1{n} = x^2 + \frac{2x+1}{n}.$

What do we still know? Since $x \in K$ , $x^2 < 2$ . This means that $(2-x^2)/(2x+1) > 0$ . In particular, we would like to stipulate that we can “squeeze” a $1/n$ between them:

$\displaystyle 0 < \frac{1}{n} < \frac{2 - x^2}{2x+1} \quad \Rightarrow \quad x^2 + \frac{2x+1}{n} < 2.$

If we are allowed this squeezing property, then we are good to go in the following sense: choose $n_0$ such that $0 < 1/n_0 < (2-x^2)/(2x+1)$ and $n = 2 + n_0$ so that $n \geq 2 > 1$ implies

$\displaystyle \frac 1{n^2} < \frac 1n$

and $n \geq n_0$ implies

$\displaystyle 0 < \frac{1}{n} \leq \frac{1}{n_0} < \frac{ 2-x^2 }{ 2x+1 } \quad \Rightarrow \quad x^2 + \frac{2x+1}{n} < 2.$

Combining our results,

$\displaystyle y^2 = \left(x + \frac 1n\right)^2 = x^2 + \frac{2x}{n} + \frac 1{n^2} < x^2 + \frac{2x+1}{n} < 2.$

For the last property, we will employ a similar strategy. What we need to show is that for any $x \in \mathbb Q$ with $x^2 \geq 2$ , there exists $y \in \mathbb Q$ such that $y < x$ and $y^2 \geq 2$ . We will define $y := x-1/n$ —very similar to our previous proof. We’ll leave the following calculation as an exercise in choosing the correct $n$ :

$\begin{aligned} y^2 = \left(x - \frac 1n\right)^2 &= x^2 - \frac{2x}{n} + \frac{1}{n^2} \\ &> x^2 - \frac{2x}{n} - \frac{1}{n} \\ &> x^2 - \frac{2x+1}{n} \\ &> 2 + \frac{2x+1}{n} - \frac{2x+1}{n} = 2.\end{aligned}$

Now, in this proof, we used this “squeezing” property. We will now verify that this squeezing move is legitimate.

Theorem 3 (Archimedean Property). Let $x \in \mathbb Q$ with $x > 0$ . Then there exists $n \in \mathbb N$ such that

$\displaystyle 0 < \frac 1n < x.$

Proof. Write $x = M/N$ for positive integers $M, N$ . Then

$\displaystyle 0 < \frac 1n < \frac MN \quad \iff \quad N < nM.$

If $M \geq N$ then choosing $n = 2$ suffices. Suppose, therefore that $M < N$ . Thus, we need to show that there exists $n \in \mathbb N$ such that

$N < nM \quad \iff \quad N-nM < 0.$

Consider the set

$\{N - kM : k \in \mathbb N\} \cap \mathbb N,$

which is nonempty since $N - 1 \cdot M = N- M > 0$ . By the well-ordering principle, the set contains a minimal element $N - k_0 M$ . Suppose, for a contradiction, that for any $k \in \mathbb N$ , $kM \leq N$ . In particular, $(k_0 + 1)M \leq N$ . Since $N - k_0 M$ is the minimal element,

$N - k_0 M \leq N - (k_0 + 1) M = N - k_0 M - M < N - k_0 M,$

a contradiction. Therefore, there must exist $n \in \mathbb N$ such that $nM > N$ , as required.

Motivated by Theorem 2, we want to define the real numbers $\mathbb R$ that does satisfy some properties that $\mathbb Q$ does not: For any nonempty $K \subseteq \mathbb R$ that is bounded above, $K$ has a least upper bound. The definitions of bounds in the context of $\mathbb R$ remain the same, except replacing $\mathbb Q$ with $\mathbb R$ .

We will eventually construct this set laboriously using the method of Dedekind cuts, and this we will elaborate more in future posts.

For now, it should not be too hard to show that if $K \subseteq \mathbb R$ has a least upper bound, it must be unique. We call this the supremum of $K$ , denoted $\sup K$ .

Theorem 4. Let $L,M$ be least upper bounds for $K$ . Then $L = M$ .

Proof. Since $L$ be a least upper bound, then $M$ being an upper bound must satisfy $L \leq M$ . Symmetrically, since $M$ is a least upper bound, then $L$ being an upper bound must satisfy $M \leq L$ . Combining the inequalities yields $L = M$ .

In supremum notation, we will define $\mathbb R$ as the ordered field that satisfies the following property: For any nonempty $K \subseteq \mathbb R$ that is bounded above, $\sup K$ exists.

There are many more useful properties of the supremum which deserve our attention on its own post (or its exercise). Finally, once we construct the real numbers, we will be able to use the supremum freely in future discussions.

—Joel Kindiak, 17 Dec 24, 1730H

KindiakMath

The Problem with Rationals

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The Problem with Rationals

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