1.3: Rational Numbers
- Page ID
- 22638
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 \(p t=q s\).
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 \(a=b\).
Show that for any \(a \in \mathbb{Q},\) one and only one of the following must hold: (a) \(a<0,(b) a=0,(c) a>0\).
Show that if \(a, b \in \mathbb{Q}^{+},\) then \(a+b \in \mathbb{Q}^{+}\).
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 \(a b>0\).
Show that if \(a, b \in \mathbb{Q}\) with \(a>0\) and \(b<0,\) then \(a b<0\).
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 \(c<0\).
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 \(a\).
Show that for any \(a \in \mathbb{Q},-|a| \leq a \leq|a|\).
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
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 \(r=\inf A .\)
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.
\(\mathbb{Q}\) does not have an upper bound.
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.
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.
Show that there does not exist a rational number \(s\) with the property that \(s^{2}=3\).
Show that there does not exist a rational number \(s\) with the property that \(s^{2}=6\).
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
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 \(a_{i}=i^{2}, i=1,2,3, \ldots\).
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
\[\lim _{i \rightarrow \infty} a_{i}=L.\]
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 \(\frac{1}{\epsilon}\).
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
\[\left|a_{i}-a_{k}\right|<\epsilon \text { whenever both } i>N \text { and } k>N.\]
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.
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.