Integers and Rationals

The natural numbers can be added and multiplied, but not always subtracted: $3 - 5$ has no answer in $\mathbb{N}$. Adding a missing answer is a pattern that runs right through mathematics, and this chapter does it twice. First we add the answers to subtractions, which gives the integers $\mathbb{Z}$. Then we add the answers to divisions, which gives the rationals $\mathbb{Q}$. Both times the tool is the one from 2A.2 Sets, Functions and Equivalence: take pairs of numbers you already have, declare when two pairs should count as the same, and pass to the quotient.

The rationals look complete: between any two of them there is another, and they can be added, subtracted, multiplied and divided freely. Yet they have holes. The chapter ends by finding one, at $\sqrt 2$. Filling those holes is the job of the next chapter, and the reason analysis exists.

By the end of this chapter you will be able to:

• construct $\mathbb{Z}$ and $\mathbb{Q}$ as quotients, and prove that their operations are well defined;
• explain why every familiar law of arithmetic for $\mathbb{Z}$ and $\mathbb{Q}$ now follows from the laws for $\mathbb{N}$;
• work with order, absolute value and distance on $\mathbb{Q}$, including the triangle inequality;
• prove that $\sqrt 2$ is irrational, and explain what that says about $\mathbb{Q}$.

3.1Why build the integers at all

Negative numbers are older than any axiom. The Indian mathematician Brahmagupta stated rules for calculating with them in 628 CE, thinking of positive numbers as fortunes and negative numbers as debts. European mathematics resisted them for many centuries more.11 Even in the 17th century, some European mathematicians called negative solutions of equations "false" or "absurd". Today every bank statement, thermometer and altimeter uses them without comment. The question here is not whether negative numbers make sense. It is how to define them so that their laws (why is $(-1)(-1) = 1$?) become theorems, proved from what we already have, rather than rules to be memorised.

The obvious approach is to take $\mathbb{N}$ and add a new symbol $-n$ for each positive $n$. That can be made to work, but every definition and proof then splits into cases: positive plus positive, positive plus negative with the positive one larger, and so on. The construction below avoids all the cases by a single idea.

3.2The integers as formal differences

An integer is going to be "what you get by subtracting one natural number from another". Since we can't subtract yet, we record the subtraction as a pair, and write it $a \mathbin{—} b$ (read "$a$ minus $b$", but at this stage just a pair of naturals). Different pairs can describe the same intended number: $3 \mathbin{—} 5$ and $10 \mathbin{—} 12$ should both be "$-2$". When should two pairs be the same? We'd like $a - b = c - d$, which, moving terms across, means $a + d = c + b$, and that involves only addition of naturals.

Before this definition can be trusted, we must check that "$=$" really is an equivalence relation, because equality has to be reflexive, symmetric and transitive (2A.2 Sets, Functions and Equivalence).

The proof of transitivity used cancellation, which in turn rested on Axiom 1.4 of 2A.1 The Natural Numbers. The axioms keep doing work long after they've been stated.

3.2.1Arithmetic, and checking it is well defined

The definitions of the operations come from what we want: $(a - b) + (c - d) = (a + c) - (b + d)$ and $(a - b)(c - d) = (ac + bd) - (ad + bc)$.

These are formulas in terms of representatives, so each one must be checked to be well defined: the answer must not depend on which pair we chose for each integer (2A.2 Sets, Functions and Equivalence). Here is the check for addition. The others are similar (Exercise 3.18).

The naturals sit inside the integers. The natural number $n$ is identified with the integer $n \mathbin{—} 0$. This is consistent with the operations: $(n \mathbin{—} 0) + (m \mathbin{—} 0) = (n + m) \mathbin{—} 0$ and $(n \mathbin{—} 0) \times (m \mathbin{—} 0) = nm \mathbin{—} 0$. We may therefore write $n$ for $n \mathbin{—} 0$, $-n$ for $0 \mathbin{—} n$, and $x - y$ for $x + (-y)$. Once this is done, the formal symbol $\mathbin{—}$ has served its purpose: $a \mathbin{—} b$ really is $a - b$.

So the quotient construction gives back exactly the picture everyone has of the integers, without our ever having to define them by cases.

Each law is proved by writing $x$, $y$, $z$ as formal differences and using the corresponding laws in $\mathbb{N}$. For example, $(a \mathbin{—} b) + (b \mathbin{—} a) = (a + b) \mathbin{—} (b + a) = 0$, because $(a + b) + 0 = 0 + (b + a)$. In particular, $(-1)(-1) = (0 \mathbin{—} 1)(0 \mathbin{—} 1) = (0 + 1) \mathbin{—} (0 + 0) = 1$. The sign rule that schoolchildren memorise is a two-line computation.

The first statement follows from Lemma 3.5 and the corresponding fact for $\mathbb{N}$ (2A.1 The Natural Numbers), by checking signs. The second follows from the first applied to $(x - y)z = 0$.

Order. For integers, $n \geq m$ means $n - m$ is a natural number, and $n > m$ means $n - m$ is a positive natural number. The order laws of 2A.1 The Natural Numbers carry over, with one important change: multiplying by a negative number reverses an inequality. If $a > b$ and $c < 0$ then $ac < bc$ (Exercise 3.19).

3.3The rationals as formal quotients

Now do the same for division. A rational number should be "$a$ divided by $b$", with $b \neq 0$, recorded as a formal quotient $a /\!\!/ b$. When should $a /\!\!/ b$ and $c /\!\!/ d$ be equal? We'd like $a/b = c/d$, which means $ad = cb$, an equation in the integers alone.

Showing this is an equivalence relation needs one more ingredient than for the integers. Transitivity requires cancelling a nonzero factor: from $ad = cb$ and $cf = ed$ we get $adf = cbf = ebd$, so $(af - eb)d = 0$, and since $d \neq 0$, Proposition 3.7 gives $af = eb$.

Each of these is well defined. For multiplication: if $a/b = a'/b'$, that is, $ab' = a'b$, then $(ac)(b'd) = (a'c)(bd)$ (multiply both sides by $cd$ and rearrange), which says $\tfrac{ac}{bd} = \tfrac{a'c}{b'd}$. The integer $n$ is identified with $n/1$, and the operations agree with those of $\mathbb{Z}$.

A set with an addition and multiplication satisfying these laws is called a field. The integers are not a field, because $2$ has no integer reciprocal; the rationals are. We can now define division, $x / y := x \times y^{-1}$ for $y \neq 0$, and powers with integer exponents, $x^{-n} := (x^n)^{-1}$, with the usual laws (Exercise 3.21).

3.3.1Order and absolute value

Every rational is exactly one of positive, zero or negative, by the integer version of trichotomy. With this order $\mathbb{Q}$ is an ordered field: sums and products of positive rationals are positive, and order is compatible with addition and with multiplication by positive numbers.

The triangle inequality is the most-used inequality in analysis. Every argument of the form "these two things are close because each is close to a third" is an application of it, and in 2B.1 Metric Spaces it becomes the defining property of a distance.

This is a deliberately modest notion: it is about a single tolerance $\varepsilon$, not about limits. But the next chapter builds the real numbers entirely out of statements of the form "for every $\varepsilon > 0$, eventually, things are $\varepsilon$-close". Two facts are worth having in hand: if $x$ is $\varepsilon$-close to $y$ and $y$ is $\delta$-close to $z$, then $x$ is $(\varepsilon + \delta)$-close to $z$ (triangle inequality); and if $x, y$ are $\varepsilon$-close then $xz, yz$ are $\varepsilon|z|$-close.

3.4Between any two rationals, another

The integers are spread out, one unit apart. The rationals are not: they crowd together everywhere. The first step is to show that the integers at least reach everywhere.

Applying Proposition 3.16 again and again, there are infinitely many rationals between any two. A picture of the rationals between $0$ and $1$ suggests a continuum with no room left over.

3.5A hole in the rationals

The square on a unit length has a diagonal of length $d$ with $d^2 = 1^2 + 1^2 = 2$, by Pythagoras. Is that length a rational number?

The proof uses exactly the descent form of well-ordering from 2A.1 The Natural Numbers. A shorter version takes $q$ to be the smallest possible denominator at the start, and reaches a contradiction in one step.

So the rationals, though dense, are full of holes. There are rationals whose squares are as close to $2$ as you please, from both sides, but none whose square is $2$. Calculus needs every such gap filled. The intermediate value theorem (2A.9 Continuous Functions) would be false over $\mathbb{Q}$: the function $x^2 - 2$ is negative at $1$ and positive at $2$, yet has no rational zero. Filling the gaps is the construction of the real numbers.

3.6Exercises

Hint

You are given $a + b' = a' + b$. Multiply by $c$ and by $d$, and add the results in the right combination.

Solution

We need $(ac + bd) + (a'd + b'c) = (a'c + b'd) + (ad + bc)$. Multiplying $a + b' = a' + b$ by $c$ gives $ac + b'c = a'c + bc$, and multiplying by $d$ (with the sides swapped) gives $a'd + bd = ad + b'd$. Adding these two equations gives exactly the required identity.

Solution

For $a \geq 0$ this is 2A.1 The Natural Numbers. For $a < 0$, divide $-a$: $-a = mb + s$ with $0 \leq s < b$. If $s = 0$ take $n = -m$, $r = 0$. Otherwise $a = -mb - s = (-m - 1)b + (b - s)$, with $0 < b - s < b$. Uniqueness is as in 2A.1 The Natural Numbers. For $-7/3$: $-7 = (-3)\cdot 3 + 2$, so the integer part is $-3$, not $-2$.

Solution

By the triangle inequality, $|x| = |(x - y) + y| \leq |x - y| + |y|$, so $|x| - |y| \leq |x - y|$. Swapping $x$ and $y$, $|y| - |x| \leq |x - y|$. Together these say $\big||x| - |y|\big| \leq |x - y|$. In words: two points that are close have nearly equal distances from $0$.

Hint

For general $n$, use unique factorisation: in $p^2 = nq^2$, each prime appears to an even power on the left. Or use the well-ordering principle directly: take the smallest $q$ with $q\sqrt n$ an integer $p$, and consider $q' = p - \lfloor\sqrt n\rfloor q$.

Solution

Any rational strictly between them has denominator at least $13$. With denominator $13$, the candidates near $0.38$–$0.4$ are $\tfrac{5}{13} \approx 0.3846$, which works, since $0.375 < 0.3846 < 0.4$. Checking denominators $1$ to $12$ shows none works: for example $\tfrac{3}{8} = 0.375$ and $\tfrac{2}{5} = 0.4$ are adjacent among fractions with denominators up to $8$, and the next fraction to appear between them is the mediant $\tfrac{3 + 2}{8 + 5} = \tfrac{5}{13}$.