1.3: Rational Numbers

Let $P=\{(p, q): p, q \in \mathbb{Z}, q \neq 0\} .$ We define an equivalence relation on $P$ by saying $(p, q) \sim(s, t)$ if $p t=q s$.

Exercise $\PageIndex{1}$

Show that the relation as just defined is indeed an equivalence relation.

We will denote the equivalence class of $(p, q) \in P$ by $p / q,$ or $\frac{p}{q} .$ We call the set of all equivalence classes of $P$ the rational numbers, which we denote by $\mathbb{Q}$. If $p \in \mathbb{Z},$ we will denote the equivalence class of $(p, 1)$ by $p ;$ that is, we let

$$\frac{p}{1}=p.$$

In this way, we may think of $\mathbb{Z}$ as a subset of $\mathbb{Q}$.

1.3.1 Field Properties

We wish to define operations of addition and multiplication on elements of $\mathbb{Q}$. We begin by defining operations on the elements of $P .$ Namely, given $(p, q) \in P$ and $(s, t) \in P,$ define

$$(p, q) \oplus(s, t)=(p t+s q, q t)$$

and

$$(p, q) \otimes(s, t)=(p s, q t).$$

Now suppose $(p, q) \sim(a, b)$ and $(s, t) \sim(c, d) .$ It follows that $(p, q) \oplus(s, t) \sim$ $(a, b) \oplus(c, d),$ that is, $(p t+s q, q t) \sim(a d+c b, b d),$ since

$$(p t+s q) b d=p b t d+s d q b=q a t d+t c q b=(a d+c b) q t.$$

Moreover, $(p, q) \otimes(s, t) \sim(a, b) \otimes(c, d),$ that is, $(p s, q t) \sim(a c, b d),$ since

$$p s b d=p b s d=q a t c=q t a c.$$

This shows that the equivalence class of a sum or product depends only on the equivalence classes of the elements being added or multiplied. Thus we may define addition and multiplication on $\mathbb{Q}$ by

$$\frac{p}{q}+\frac{s}{t}=\frac{p t+s q}{q t}$$

and

$$\frac{p}{q} \times \frac{s}{t}=\frac{p s}{q t},$$

and the results will not depend on which representatives we choose for each equivalence class. Of course, multiplication is often denoted using juxtaposition, that is,

$$\frac{p}{q} \times \frac{s}{t}=\frac{p}{q} \frac{s}{t},$$

and repeated multiplication may be denoted by exponentiation, that is, $a^{n},$ $a \in \mathbb{Q}$ and $n \in \mathbb{Z}^{+},$ represents the product of $a$ with itself $n$ times.

Note that if $(p, q) \in P,$ then $(-p, q) \sim(p,-q) .$ Hence, if $a=\frac{p}{q} \in \mathbb{Q},$ then we let

$$-a=\frac{-p}{q}=\frac{p}{-q}.$$

For any $a, b \in \mathbb{Q},$ we will write $a-b$ to denote $a+(-b)$.

$\quad$ If $a=\frac{p}{q} \in \mathbb{Q}$ with $p \neq 0,$ then we let

$$a^{-1}=\frac{q}{p}.$$

Moreover, we will write

$$\frac{1}{a}=a^{-1},$$

$$\frac{1}{a^{n}}=a^{-n}$$

for any $n \in \mathbb{Z}^{+},$ and, for any $b \in \mathbb{Q}$,

$$\frac{b}{a}=b a^{-1}.$$

It is now easy to show that

1. $a+b=b+a$ for all $a, b \in \mathbb{Q}$;

2. $(a+b)+c=a+(b+c)$ for all $a, b, c \in \mathbb{Q}$;

3. $a b=b a$ for all $a, b \in \mathbb{Q}$;

4. $(a b) c=a(b c)$ for all $a, b, c \in \mathbb{Q}$;

5. $a(b+c)=a b+a c$ for all $a, b, c \in \mathbb{Q}$;

6. $a+0=a$ for all $a \in \mathbb{Q}$;

7. $a+(-a)=0$ for all $a \in \mathbb{Q}$;

8. $1 a=a$ for all $a \in \mathbb{Q}$;

9. if $a \in \mathbb{Q}, a \neq 0,$ then $a a^{-1}=1$.

Taken together, these statements imply that $\mathbb{Q}$ is a field.

1.3.2 Order and Metric Properties

We say a rational number $a$ is positive if there exist $p, q \in \mathbb{Z}^{+}$ such that $a=\frac{p}{q}$. We denote the set of all positive elements of $\mathbb{Q}$ by $\mathbb{Q}^{+} .$

Given $a, b \in \mathbb{Q},$ we say $a$ is less than $b,$ or, equivalently, $b$ is greater than $a,$ denoted either by $a<b$ or $b>a$, if $b-a$ is positive. In particular, $a>0$ if and only if $a$ is positive. If $a<0,$ we say $a$ is negative. We write $a \leq b,$ or, equivalently, $b \geq a,$ if either $a<b$ or $a=b$.

Exercise $\PageIndex{2}$

Show that for any $a \in \mathbb{Q},$ one and only one of the following must hold: (a) $a<0,(b) a=0,(c) a>0$.

Exercise $\PageIndex{3}$

Show that if $a, b \in \mathbb{Q}^{+},$ then $a+b \in \mathbb{Q}^{+}$.

Exercise $\PageIndex{4}$

Suppose $a, b, c \in \mathbb{Q} .$ Show each of the following:

a. One, and only one, of the following must hold:

(i) $a<b$,

(ii) $a=b$,

(iii) $a>b$.

b. If $a<b$ and $b<c,$ then $a<c$.

c. If $a<b,$ then $a+c<b+c$.

d. If $a>0$ and $b>0,$ then $a b>0$.

Exercise $\PageIndex{5}$

Show that if $a, b \in \mathbb{Q}$ with $a>0$ and $b<0,$ then $a b<0$.

Exercise $\PageIndex{6}$

Show that if $a, b, c \in \mathbb{Q}$ with $a<b,$ then $a c<b c$ if $c>0$ and $a c>b c$ if $c<0$.

Exercise $\PageIndex{7}$

Show that if $a, b \in \mathbb{Q}$ with $a<b,$ then

$$a<\frac{a+b}{2}<b.$$

As a consequence of Exercise 1.3 .4 we say $\mathbb{Q}$ is an ordered field. For any $a \in \mathbb{Q},$ we call

$$|a|=\left\{\begin{aligned} a, & \text { if } a \geq 0, \\-a, & \text { if } a<0,\end{aligned}\right.$$

the absolute value of $a$.

Exercise $\PageIndex{8}$

Show that for any $a \in \mathbb{Q},-|a| \leq a \leq|a|$.

Proposition $\PageIndex{1}$

For any $a, b \in \mathbb{Q},|a+b| \leq|a|+|b|$.

Proof

If $a+b \geq 0,$ then

$$|a|+|b|-|a+b|=|a|+|b|-a-b=(|a|-a)+(|b|-b).$$

Both of the terms on the right are nonnegative by Exercise $1.3 .8 .$ Hence the sum is nonnegative and the proposition follows. If $a+b<0,$ then

$$|a|+|b|-|a+b|=|a|+|b|+a+b=(|a|+a)+(|b|+b).$$

Again, both of the terms on the right are nonnegative by Exercise $1.3 .8 .$ Hence the sum is nonnegative and the theorem follows. $\quad$ Q.E.D.

It is now easy to show that the absolute value satisfies

1. $|a-b| \geq 0$ for all $a, b \in \mathbb{Q},$ with $|a-b|=0$ if and only if $a=b$,

2. $|a-b|=|b-a|$ for all $a, b \in \mathbb{Q}$,

3. $|a-b| \leq|a-c|+|c-b|$ for all $a, b, c \in \mathbb{Q}$.

Note that the last statement, known as the triangle inequality, follows from writing

$$a-b=(a-c)+(c-b)$$

and applying the previous proposition. These properties show that the function

$$d(a, b)=|a-b|$$

is a metric, and we will call $|a-b|$ the distance from $a$ to $b .$

Suppose $a, b \in \mathbb{Q}^{+}$ with $a<b$ and let $p, q, r, s \in \mathbb{Z}^{+}$ such that $a=\frac{p}{q}$ and $b=\frac{r}{s} .$ For any $n \in \mathbb{Z}^{+},$ we have

$$n a-b=n \frac{p}{q}-\frac{r}{s}=\frac{n p s-r q}{q s}.$$

If we choose $n$ large enough so that $n p s-r q>0,$ it follows that $n a-b>0,$ that is, $n a>b .$ We say that the ordered field $\mathbb{Q}$ is archimedean. Note that it also follows that we may choose $n$ large enough to ensure that $\frac{b}{n}<a$.

1.3.3 Upper and Lower Bounds

Definition

Let $A \subset \mathbb{Q} .$ If $s \in \mathbb{Q}$ is such that $s \geq a$ for every $a \in A,$ then we call $s$ an upper bound for $A$. If $s$ is an upper bound for $A$ with the property that $s \leq t$ whenever $t$ is an upper bound for $A,$ then we call $s$ the supremum, or least upper bound, of $A,$ denoted $s=\sup A .$ Similarly, if $r \in \mathbb{Q}$ is such that $r \leq a$ for every $a \in A,$ then we call $r$ a lower bound for $A .$ If $r$ is a lower bound for $A$ with the property that $r \geq t$ whenever $t$ is a lower bound for $A,$ then we call $r$ the infimum, or greatest lower bound, of $A,$ denoted $r=\inf A .$

Exercise $\PageIndex{9}$

Show that the supremum of a set $A \subset \mathbb{Q},$ if it exists, is unique, and thus justify the use of the definite article in the previous definition.

A set which does not have an upper bound will not, $a$ fortiori, have a supremum. Moreover, even sets which have upper bounds need not have a supremum.

Example $\PageIndex{1}$

$\mathbb{Q}$ does not have an upper bound.

Example $\PageIndex{2}$

Consider the set

$$A=\left\{a: a \in \mathbb{Q}^{+}, a^{2}<2\right\}.$$

Note that if $a, b \in \mathbb{Q}^{+}$ with $a<b,$ then

$$b^{2}-a^{2}=(b-a)(b+a)>0,$$

from which it follows that $a^{2}<b^{2} .$ Hence if $a \in \mathbb{Q}^{+}$ with $a^{2}>2,$ then $a$ is an upper bound for $A .$ For example, 4 is an upper bound for $A .$

Now suppose $s \in \mathbb{Q}^{+}$ is the supremum of $A .$ We must have either $s^{2}<2,$ $s^{2}>2,$ or $s^{2}=2$.

Suppose $s^{2}<2$ and let $\epsilon=2-s^{2} .$ By the archimedean property of $\mathbb{Q},$ we may choose $n \in \mathbb{Z}^{+}$ such that

$$\frac{2 s+1}{n}<\epsilon,$$

from which it follows that

$$\frac{2 s}{n}+\frac{1}{n^{2}}=\frac{2 s+\frac{1}{n}}{n} \leq \frac{2 s+1}{n}<\epsilon.$$

Hence

$$\left(s+\frac{1}{n}\right)^{2}=s^{2}+\frac{2 s}{n}+\frac{1}{n^{2}}<s^{2}+\epsilon=2,$$

which implies that $s+\frac{1}{n} \in A .$ since $s<s+\frac{1}{n},$ this contradicts the assumption that $s$ is an upper bound for $A .$

So now suppose $s^{2}>2 .$ Again let $n \in \mathbb{Z}^{+}$ and note that

$$\left(s-\frac{1}{n}\right)^{2}=s^{2}-\frac{2 s}{n}+\frac{1}{n^{2}}.$$

If we let $\epsilon=s^{2}-2,$ then we may choose $n \in \mathbb{Z}^{+}$ so that

$$\frac{2 s}{n}<\epsilon.$$

It follows that

$$\left(s-\frac{1}{n}\right)^{2}>s^{2}-\epsilon+\frac{1}{n^{2}}=2+\frac{1}{n^{2}}>2.$$

Thus $s-\frac{1}{n}$ is an upper bound for $A$ and $s-\frac{1}{n}<s,$ contradicting the assumption that $s=\sup A$.

Thus we must have $s^{2}=2 .$ However, this is impossible in light of the following proposition. Hence we must conclude that $A$ does not have a supremum.

Proposition $\PageIndex{2}$

There does not exist a rational number $s$ with the property that $s^{2}=2$.

Proof

Suppose there exists $s \in \mathbb{Q}$ such that $s^{2}=2 .$ Choose $a, b \in \mathbb{Z}^{+}$ so that $a$ and $b$ are relatively prime (that is, they have no factor other than 1 in common) and $s=\frac{a}{b} .$ Then

$$\frac{a^{2}}{b^{2}}=2,$$

so $a^{2}=2 b^{2} .$ Thus $a^{2},$ and hence $a,$ is an even integer. So there exists $c \in \mathbb{Z}^{+}$ such that $a=2 c .$ Hence

$$2 b^{2}=a^{2}=4 c^{2},$$

from which it follows that $b^{2}=2 c,$ and so $b$ is also an even integer. But this contradicts the assumption that $a$ and $b$ are relatively prime. $\quad$ Q.E.D.

Exercise $\PageIndex{10}$

Show that there does not exist a rational number $s$ with the property that $s^{2}=3$.

Exercise $\PageIndex{11}$

Show that there does not exist a rational number $s$ with the property that $s^{2}=6$.

Exercise $\PageIndex{12}$

Let $A=\left\{a: a \in \mathbb{Q}, a^{3}<2\right\}$.

1. Show that if $a \in A$ and $b<a,$ then $b \in A$.

2. Show that if $a \notin A,$ and $b>a,$ then $b \notin A$.

1.3.4 Sequences

Definition

Suppose $n \in \mathbb{Z}, I=\{n, n+1, n+2, \ldots\},$ and $A$ is a set. We call a function $\varphi: I \rightarrow A$ a sequence with values in $A$.

Frequently, we will define a sequence $\varphi$ by specifying its values with notation such as, for example, $\{\varphi(i)\}_{i \in I},$ or $\{\varphi(i)\}_{i=n}^{\infty} .$ Thus, for example, $\left\{i^{2}\right\}_{i=1}^{\infty}$ denotes the sequence $\varphi: \mathbb{Z}^{+} \rightarrow \mathbb{Z}$ defined by $\varphi(i)=i^{2} .$ Moreover, it is customary to denote the values of a sequence using subscript notation. Thus if $a_{i}=\varphi(i),$ $i \in I,$ then $\left\{a_{i}\right\}_{i \in I}$ denotes the sequence $\varphi .$ For example, we may define the sequence of the previous example by writing $a_{i}=i^{2}, i=1,2,3, \ldots$.

Definition

Suppose $\left\{a_{i}\right\}_{i \in I}$ is a sequence with values in $\mathbb{Q} .$ We say that $\left\{a_{i}\right\}_{i \in I}$ converges, and has limit $L, L \in \mathbb{Q},$ if for every $\epsilon \in \mathbb{Q}^{+},$ there exists $N \in \mathbb{Z}$ such that

$$\left|a_{i}-L\right|<\epsilon \text { whenever } i>N.$$

If the sequence $\left\{a_{i}\right\}_{i \in I}$ converges to $L,$ we write

$$\lim _{i \rightarrow \infty} a_{i}=L.$$

Example $\PageIndex{3}$

We have

$$\lim _{i \rightarrow \infty} \frac{1}{i}=0,$$

since, for any rational number $\epsilon>0$,

$$\left|\frac{1}{i}-0\right|=\frac{1}{i}<\epsilon$$

for any $i>N,$ where $N$ is any integer larger than $\frac{1}{\epsilon}$.

Definition

Suppose $\left\{a_{i}\right\}_{i \in I}$ is a sequence with values in $\mathbb{Q} .$ We call $\left\{a_{i}\right\}_{i \in I}$ a Cauchy sequence if for every $\epsilon \in \mathbb{Q}^{+},$ there exists $N \in \mathbb{Z}$ such that

$$\left|a_{i}-a_{k}\right|<\epsilon \text { whenever both } i>N \text { and } k>N.$$

Proposition $\PageIndex{3}$

If $\left\{a_{i}\right\}_{i \in I}$ converges, then $\left\{a_{i}\right\}_{i \in I}$ is a Cauchy sequence.

Proof

Suppose $\lim _{i \rightarrow \infty} a_{i}=L .$ Given $\epsilon \in \mathbb{Q}^{+},$ choose an integer $N$ such that

$$\left|a_{i}-L\right|<\frac{\epsilon}{2}$$

for all $i>N .$ Then for any $i, k>N,$ we have

$$\left|a_{i}-a_{k}\right|=\left|\left(a_{i}-L\right)+\left(L-a_{k}\right)\right| \leq\left|a_{i}-L\right|+\left|a_{k}-L\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$

Hence $\left\{a_{i}\right\}_{i \in I}$ is a Cauchy sequence. $\quad$ Q.E.D.

The proposition shows that every convergent sequence in $\mathbb{Q}$ is a Cauchy sequence, but, as the next example shows, the converse does not hold.

Example $\PageIndex{1}$

Let

$$f(x)=x^{2}-2$$

and consider the sequence constructed as follows: Begin by setting $a_{1}=1$, $b_{1}=2,$ and $x_{1}=\frac{3}{2} .$ If $f\left(a_{1}\right) f\left(x_{1}\right)<0,$ set

$$x_{2}=\frac{a_{1}+x_{1}}{2},$$

$a_{2}=a_{1},$ and $b_{2}=x_{1} ;$ otherwise, set

$$x_{2}=\frac{x_{1}+b_{1}}{2},$$

$a_{2}=x_{1},$ and $b_{2}=b_{1} .$ In general, given $a_{n}, x_{n},$ and $b_{n},$ if $f\left(a_{n}\right) f\left(x_{n}\right)<0,$ set

$$x_{n+1}=\frac{a_{n}+x_{n}}{2},$$

$a_{n+1}=a_{n},$ and $b_{n+1}=x_{n} ;$ otherwise, set

$$x_{n+1}=\frac{x_{n}+b_{n}}{2},$$

$a_{n+1}=x_{n},$ and $b_{n+1}=b_{n} .$ Note that for any positive integer $N, f\left(a_{N}\right)<0,$ $f\left(b_{N}\right)>0,$ and

$$a_{N}<x_{i}<b_{N}$$

for all $i>N .$ Moreover,

$$\left|b_{N}-a_{N}\right|=\frac{1}{2^{N-1}},$$

so

$$\left|x_{i}-x_{k}\right|<\frac{1}{2^{N-1}}$$

for all $i, k>N .$ Hence given any $\epsilon \in \mathbb{Q}^{+},$ if we choose an integer $N$ such that $2^{N-1}>\frac{1}{\epsilon},$ then

$$\left|x_{i}-x_{k}\right|<\frac{1}{2^{N-1}}<\epsilon$$

for all $i, k>N,$ showing that $\left\{x_{i}\right\}_{i=1}^{\infty}$ is a Cauchy sequence. Now suppose $\left\{x_{i}\right\}_{i=1}^{\infty}$ converges to $s \in \mathbb{Q}$. Note hat we must have

$$a_{i} \leq s \leq b_{i}$$

for all $i \in \mathbb{Z}^{+} .$ If $f(s)<0,$ then, since the set $\left\{a: a \in \mathbb{Q}^{+}, a^{2}<2\right\}$ does not have a supremum, there exists $t \in \mathbb{Q}^{+}$ such that $s<t$ and $f(t)<0 .$ If we choose $N$ so that

$$\frac{1}{2^{N-1}}<t-s,$$

then

$$\left|s-b_{N}\right| \leq\left|a_{N}-b_{N}\right|=\frac{1}{2^{N-1}}<t-s.$$

Hence $b_{N}<t,$ which implies that $f\left(b_{N}\right)<0,$ contradicting the construction of $\left\{b_{i}\right\}_{i=1}^{\infty} .$ Hence we must have $f(s)>0 .$ But if $f(s)>0,$ then there exists $t \in \mathbb{Q}^{+}$ such that $t<s$ and $f(t)>0 .$ We can then choose $N$ so that $t<a_{N}$, implying that $f\left(a_{N}\right)>0,$ contradicting the construction of $\left\{a_{i}\right\}_{i=1}^{\infty} .$ Hence we must have $f(s)=0,$ which is not possible since $s \in \mathbb{Q}$. Thus we must conclude that $\left\{x_{i}\right\}_{i=1}^{\infty}$ does not converge.