
# 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 $$b \notin A$$.

##### 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.