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1.3: Rational Numbers

( \newcommand{\kernel}{\mathrm{null}\,}\)

Let P={(p,q):p,qZ,q0}. We define an equivalence relation on P by saying (p,q)(s,t) if pt=qs.

Exercise 1.3.1

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

We will denote the equivalence class of (p,q)P by p/q, or pq. We call the set of all equivalence classes of P the rational numbers, which we denote by Q. If pZ, we will denote the equivalence class of (p,1) by p; that is, we let

p1=p.

In this way, we may think of Z as a subset of Q.

1.3.1 Field Properties

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

(p,q)(s,t)=(pt+sq,qt)

and

(p,q)(s,t)=(ps,qt).

Now suppose (p,q)(a,b) and (s,t)(c,d). It follows that (p,q)(s,t) (a,b)(c,d), that is, (pt+sq,qt)(ad+cb,bd), since

(pt+sq)bd=pbtd+sdqb=qatd+tcqb=(ad+cb)qt.

Moreover, (p,q)(s,t)(a,b)(c,d), that is, (ps,qt)(ac,bd), since

psbd=pbsd=qatc=qtac.

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 Q by

pq+st=pt+sqqt

and

pq×st=psqt,

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,

pq×st=pqst,

and repeated multiplication may be denoted by exponentiation, that is, an, aQ and nZ+, represents the product of a with itself n times.

Note that if (p,q)P, then (p,q)(p,q). Hence, if a=pqQ, then we let

a=pq=pq.

For any a,bQ, we will write ab to denote a+(b).

If a=pqQ with p0, then we let

a1=qp.

Moreover, we will write

1a=a1,

1an=an

for any nZ+, and, for any bQ,

ba=ba1.

It is now easy to show that

1. a+b=b+a for all a,bQ;

2. (a+b)+c=a+(b+c) for all a,b,cQ;

3. ab=ba for all a,bQ;

4. (ab)c=a(bc) for all a,b,cQ;

5. a(b+c)=ab+ac for all a,b,cQ;

6. a+0=a for all aQ;

7. a+(a)=0 for all aQ;

8. 1a=a for all aQ;

9. if aQ,a0, then aa1=1.

Taken together, these statements imply that Q is a field.

1.3.2 Order and Metric Properties

We say a rational number a is positive if there exist p,qZ+ such that a=pq. We denote the set of all positive elements of Q by Q+.

Given a,bQ, we say a is less than b, or, equivalently, b is greater than a, denoted either by a<b or b>a, if ba is positive. In particular, a>0 if and only if a is positive. If a<0, we say a is negative. We write ab, or, equivalently, ba, if either a<b or a=b.

Exercise 1.3.2

Show that for any aQ, one and only one of the following must hold: (a) a<0,(b)a=0,(c)a>0.

Exercise 1.3.3

Show that if a,bQ+, then a+bQ+.

Exercise 1.3.4

Suppose a,b,cQ. 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 ab>0.

Exercise 1.3.5

Show that if a,bQ with a>0 and b<0, then ab<0.

Exercise 1.3.6

Show that if a,b,cQ with a<b, then ac<bc if c>0 and ac>bc if c<0.

Exercise 1.3.7

Show that if a,bQ with a<b, then

a<a+b2<b.

As a consequence of Exercise 1.3 .4 we say Q is an ordered field. For any aQ, we call

|a|={a, if a0,a, if a<0,

the absolute value of a.

Exercise 1.3.8

Show that for any aQ,|a|a|a|.

Proposition 1.3.1

For any a,bQ,|a+b||a|+|b|.

Proof

If a+b0, then

|a|+|b||a+b|=|a|+|b|ab=(|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. Q.E.D.

It is now easy to show that the absolute value satisfies

1. |ab|0 for all a,bQ, with |ab|=0 if and only if a=b,

2. |ab|=|ba| for all a,bQ,

3. |ab||ac|+|cb| for all a,b,cQ.

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

ab=(ac)+(cb)

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

d(a,b)=|ab|

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

Suppose a,bQ+ with a<b and let p,q,r,sZ+ such that a=pq and b=rs. For any nZ+, we have

nab=npqrs=npsrqqs.

If we choose n large enough so that npsrq>0, it follows that nab>0, that is, na>b. We say that the ordered field Q is archimedean. Note that it also follows that we may choose n large enough to ensure that bn<a.

1.3.3 Upper and Lower Bounds

Definition

Let AQ. If sQ is such that sa for every aA, then we call s an upper bound for A. If s is an upper bound for A with the property that st whenever t is an upper bound for A, then we call s the supremum, or least upper bound, of A, denoted s=supA. Similarly, if rQ is such that ra for every aA, then we call r a lower bound for A. If r is a lower bound for A with the property that rt whenever t is a lower bound for A, then we call r the infimum, or greatest lower bound, of A, denoted r=infA.

Exercise 1.3.9

Show that the supremum of a set AQ, 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 1.3.1

Q does not have an upper bound.

Example 1.3.2

Consider the set

A={a:aQ+,a2<2}.

Note that if a,bQ+ with a<b, then

b2a2=(ba)(b+a)>0,

from which it follows that a2<b2. Hence if aQ+ with a2>2, then a is an upper bound for A. For example, 4 is an upper bound for A.

Now suppose sQ+ is the supremum of A. We must have either s2<2, s2>2, or s2=2.

Suppose s2<2 and let ϵ=2s2. By the archimedean property of Q, we may choose nZ+ such that

2s+1n<ϵ,

from which it follows that

2sn+1n2=2s+1nn2s+1n<ϵ.

Hence

(s+1n)2=s2+2sn+1n2<s2+ϵ=2,

which implies that s+1nA. since s<s+1n, this contradicts the assumption that s is an upper bound for A.

So now suppose s2>2. Again let nZ+ and note that

(s1n)2=s22sn+1n2.

If we let ϵ=s22, then we may choose nZ+ so that

2sn<ϵ.

It follows that

(s1n)2>s2ϵ+1n2=2+1n2>2.

Thus s1n is an upper bound for A and s1n<s, contradicting the assumption that s=supA.

Thus we must have s2=2. However, this is impossible in light of the following proposition. Hence we must conclude that A does not have a supremum.

Proposition 1.3.2

There does not exist a rational number s with the property that s2=2.

Proof

Suppose there exists sQ such that s2=2. Choose a,bZ+ so that a and b are relatively prime (that is, they have no factor other than 1 in common) and s=ab. Then

a2b2=2,

so a2=2b2. Thus a2, and hence a, is an even integer. So there exists cZ+ such that a=2c. Hence

2b2=a2=4c2,

from which it follows that b2=2c, and so b is also an even integer. But this contradicts the assumption that a and b are relatively prime. Q.E.D.

Exercise 1.3.10

Show that there does not exist a rational number s with the property that s2=3.

Exercise 1.3.11

Show that there does not exist a rational number s with the property that s2=6.

Exercise 1.3.12

Let A={a:aQ,a3<2}.

1. Show that if aA and b<a, then bA.

2. Show that if aA, and b>a, then bA.

1.3.4 Sequences
Definition

Suppose nZ,I={n,n+1,n+2,}, and A is a set. We call a function φ:IA a sequence with values in A.

Frequently, we will define a sequence φ by specifying its values with notation such as, for example, {φ(i)}iI, or {φ(i)}i=n. Thus, for example, {i2}i=1 denotes the sequence φ:Z+Z defined by φ(i)=i2. Moreover, it is customary to denote the values of a sequence using subscript notation. Thus if ai=φ(i), iI, then {ai}iI denotes the sequence φ. For example, we may define the sequence of the previous example by writing ai=i2,i=1,2,3,.

Definition

Suppose {ai}iI is a sequence with values in Q. We say that {ai}iI converges, and has limit L,LQ, if for every ϵQ+, there exists NZ such that

|aiL|<ϵ whenever i>N.

If the sequence {ai}iI converges to L, we write

limiai=L.

Example 1.3.3

We have

limi1i=0,

since, for any rational number ϵ>0,

|1i0|=1i<ϵ

for any i>N, where N is any integer larger than 1ϵ.

Definition

Suppose {ai}iI is a sequence with values in Q. We call {ai}iI a Cauchy sequence if for every ϵQ+, there exists NZ such that

|aiak|<ϵ whenever both i>N and k>N.

Proposition 1.3.3

If {ai}iI converges, then {ai}iI is a Cauchy sequence.

Proof

Suppose limiai=L. Given ϵQ+, choose an integer N such that

|aiL|<ϵ2

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

|aiak|=|(aiL)+(Lak)||aiL|+|akL|<ϵ2+ϵ2=ϵ.

Hence {ai}iI is a Cauchy sequence. Q.E.D.

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

Example 1.3.1

Let

f(x)=x22

and consider the sequence constructed as follows: Begin by setting a1=1, b1=2, and x1=32. If f(a1)f(x1)<0, set

x2=a1+x12,

a2=a1, and b2=x1; otherwise, set

x2=x1+b12,

a2=x1, and b2=b1. In general, given an,xn, and bn, if f(an)f(xn)<0, set

xn+1=an+xn2,

an+1=an, and bn+1=xn; otherwise, set

xn+1=xn+bn2,

an+1=xn, and bn+1=bn. Note that for any positive integer N,f(aN)<0, f(bN)>0, and

aN<xi<bN

for all i>N. Moreover,

|bNaN|=12N1,

so

|xixk|<12N1

for all i,k>N. Hence given any ϵQ+, if we choose an integer N such that 2N1>1ϵ, then

|xixk|<12N1<ϵ

for all i,k>N, showing that {xi}i=1 is a Cauchy sequence. Now suppose {xi}i=1 converges to sQ. Note hat we must have

aisbi

for all iZ+. If f(s)<0, then, since the set {a:aQ+,a2<2} does not have a supremum, there exists tQ+ such that s<t and f(t)<0. If we choose N so that

12N1<ts,

then

|sbN||aNbN|=12N1<ts.

Hence bN<t, which implies that f(bN)<0, contradicting the construction of {bi}i=1. Hence we must have f(s)>0. But if f(s)>0, then there exists tQ+ such that t<s and f(t)>0. We can then choose N so that t<aN, implying that f(aN)>0, contradicting the construction of {ai}i=1. Hence we must have f(s)=0, which is not possible since sQ. Thus we must conclude that {xi}i=1 does not converge.


This page titled 1.3: Rational Numbers is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform.

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