THE REAL NUMBERS
Section 2.5 Intervals
2.5 INTERVALS 49
Ifx2½0;1, we will use a repeated bisection procedure to associate a sequence (an) of 0s and 1s as follows. Ifx6¼12belongs to the left subinterval 0 ;12
we takea1:¼0, while if xbelongs to the right subinterval 12;1
we takea1¼1. Ifx¼12, then we may takea1to be either 0 or 1. In any case, we have
a1
2 xa1þ1 2 : We now bisect the interval 12a1;12ða1þ1Þ
. Ifxis not the bisection point and belongs to the left subinterval we takea2:¼0, and ifxbelongs to the right subinterval we takea2:¼1. If x¼14orx¼34, we can takea2to be either 0 or 1. In any case, we have
a1
2 þa2
22xa1
2 þa2þ1 22 :
We continue this bisection procedure, assigning at thenth stage the valuean:¼0 ifxis not the bisection point and lies in the left subinterval, and assigning the valuean:¼1 ifx lies in the right subinterval. In this way we obtain a sequence (an) of 0s or 1s that correspond to a nested sequence of intervals containing the pointx. For eachn, we have the inequality
ð2Þ a1
2 þa2
22þ þan
2nxa1
2 þa2
22þ þanþ1 2n :
Ifxis the bisection point at thenth stage, thenx¼m=2nwithmodd. In this case, we may choose either the left or the right subinterval; however, once this subinterval is chosen, then all subsequent subintervals in the bisection procedure are determined. [For instance, if we choose the left subinterval so thatan¼0, thenxis the right endpoint of all subsequent subintervals, and henceak¼1 for allknþ1. On the other hand, if we choose the right subinterval so thatan¼1, thenxis the left endpoint of all subsequent subintervals, and henceak¼0 for allknþ1. For example, ifx¼34, then the two possible sequences for x are 1, 0, 1, 1, 1,. . .and 1, 1, 0, 0, 0, . . . .]
To summarize:If x2½0;1,then there exists a sequence(an)of0s and1s such that inequality(2) holds for all n2N. In this case we write
ð3Þ x¼ð:a1a2 an Þ2;
and call (3) a binary representation of x. This representation is unique except when x¼m=2n formodd, in which casexhas the two representations
x¼ð:a1a2 an11000 Þ2¼ð:a1a2 an10111 Þ2; one ending in 0s and the other ending in 1s.
Conversely, each sequence of0s and1s is the binary representation of a unique real number in [0,1]. The inequality corresponding to (2) determines a closed interval with length 1=2nand the sequence of these intervals is nested. Therefore, Theorem 2.5.3 implies that there exists a unique real numberxsatisfying (2) for everyn2N. Consequently,xhas the binary representationð:a1a2 an Þ2.
Remark The concept of binary representation is extremely important in this era of digital computers. A number is entered in a digital computer on ‘‘bits,’’ and each bit can be put in one of two states—either it will pass current or it will not. These two states correspond to the values 1 and 0, respectively. Thus, the binary representation of a number can be stored in a digital computer on a string of bits. Of course, in actual practice, since only finitely many bits can be stored, the binary representations must be truncated. Ifnbinary digits are
used for a numberx2½0;1, then the accuracy is at most 1=2n. For example, to assure four- decimal accuracy, it is necessary to use at least 15 binary digits (or 15 bits).
Decimal Representations
Decimal representations of real numbers are similar to binary representations, except that we subdivide intervals intotenequal subintervals instead of two.
Thus, givenx2½0;1, if we subdivide [0,1] into ten equal subintervals, thenxbelongs to a subinterval½b1=10;ðb1þ1Þ=10 for some integerb1in 0f ;1;. . .;9g. Proceeding as in the binary case, we obtain a sequence (bn) of integers with 0bn9 for alln2Nsuch thatxsatisfies
ð4Þ b1
10þ b2
102þ þ bn
10nxb1
10þ b2
102þ þbnþ1 10n : In this case we say thatxhas adecimal representationgiven by
x¼:b1b2 bn :
Ifx1 and if B2N is such thatBx<Bþ1, then x¼B:b1b2 bn where the decimal representation of xB2½0;1 is as above. Negative numbers are treated similarly.
The fact that each decimal determines a unique real number follows from Theorem 2.5.3, since each decimal specifies a nested sequence of intervals with lengths 1=10n.
The decimal representation ofx2½0;1 is unique except whenxis a subdivision point at some stage, which can be seen to occur whenx¼m=10nfor somem;n2N; 1m10n. (We may also assume thatmis not divisible by 10.) Whenxis a subdivision point at thenth stage, one choice for bn corresponds to selecting the left subinterval, which causes all subsequent digits to be 9, and the other choice corresponds to selecting the right subinterval, which causes all subsequent digits to be 0. [For example, if x¼12 then x¼ :4999 ¼:5000 , and ify¼38=100 theny¼:37999 ¼:38000 .]
Periodic Decimals
A decimalB:b1b2 bn is said to beperiodic(or to berepeating), if there existk;n2N such that bn¼bnþm for all nk. In this case, the block of digits bkbkþ1 bkþm1 is repeated once thekth digit is reached. The smallest numbermwith this property is called the periodof the decimal. For example, 19=88¼:2159090 90 has periodm¼2 with repeating block 90 starting atk¼4. Aterminating decimalis a periodic decimal where the repeated block is simply the digit 0.
We will give an informal proof of the assertion:A positive real number is rational if and only if its decimal representation is periodic.
Suppose that x¼p=q where p;q2N have no common integer factors. For convenience we will also suppose that 0<p<q. We note that the process of ‘‘long division’’ ofqinto pgives the decimal representation of p=q. Each step in the division process produces a remainder that is an integer from 0 toq1. Therefore, after at mostq steps, some remainder will occur a second time and, at that point, the digits in the quotient will begin to repeat themselves in cycles. Hence, the decimal representation of such a rational number is periodic.
Conversely, if a decimal is periodic, then it represents a rational number. The idea of the proof is best illustrated by an example. Suppose that x¼7:31414 14 . We multiply by a power of 10 to move the decimal point to the first repeating block; here 2.5 INTERVALS 51
obtaining 10x¼73:1414 . We now multiply by a power of 10 to move one block to the left of the decimal point; here getting 1000x¼7314:1414 . We now subtract to obtain an integer; here getting 1000x10x¼731473¼7241, whence x¼7241=990, a rational number.
Cantor’s Second Proof
We will now give Cantor’s second proof of the uncountability of R. This is the elegant
‘‘diagonal’’ argument based on decimal representations of real numbers.
2.5.5 Theorem The unit interval½0;1 :¼fx2R:0x1g is not countable. Proof. The proof is by contradiction. We will use the fact that every real numberx2½0;1 has a decimal representationx¼0:b1b2b3 , wherebi¼0;1;. . .;9. Suppose that there is an enumerationx1;x2;x3 of all numbers in [0,1], which we display as:
x1 ¼ 0:b11b12b13 b1n ; x2 ¼ 0:b21b22b23 b2n ; x3 ¼ 0:b31b32b33 b3n ;
xn ¼ 0:bn1bn2bn3 bnn ;
We now define a real number y:¼0:y1y2y3 yn by setting y1:¼2 if b115 and y1:¼7 ifb114; in general, we let
yn:¼ 2 ifbnn5; 7 ifbnn4:
Theny2½0;1. Note that the numberyis not equal to any of the numbers with two decimal representations, since yn6¼0;9 for all n2N. Further, since y andxndiffer in the nth decimal place, theny6¼xnfor anyn2N. Therefore,yis not included in the enumeration of
[0,1], contradicting the hypothesis. Q.E.D.
Exercises for Section 2.5
1. IfI:¼½a;b andI0:¼½a0;b0 are closed intervals inR, show thatII0if and only ifa0aand bb0.
2. IfSRis nonempty, show thatSis bounded if and only if there exists a closed bounded interval Isuch thatSI.
3. IfSRis a nonempty bounded set, andIS:¼½infS;supS , show thatSIS. Moreover, ifJis any closed bounded interval containingS, show thatISJ.
4. In the proof of Case (ii) of Theorem 2.5.1, explain whyx,yexist inS.
5. Write out the details of the proof of Case (iv) in Theorem 2.5.1.
6. If I1I2 In is a nested sequence of intervals and if In¼½an;bn, show that a1a2 an andb1b2 bn .
7. LetIn:¼½0;1=n forn2N. Prove thatT1n¼1In¼f g0 . 8. LetJn:¼ð0;1=nÞforn2N. Prove thatT1n¼1Jn¼ ;.
9. LetKn:¼ðn;1Þforn2N. Prove thatT1n¼1Kn¼ ;.
10. With the notation in the proofs of Theorems 2.5.2 and 2.5.3, show that we haveh2T1n¼1In. Also show that½j;h ¼T1n¼1In.
11. Show that the intervals obtained from the inequalities in (2) form a nested sequence.
12. Give the two binary representations of38and167.
13. (a) Give the first four digits in the binary representation of13. (b) Give the complete binary representation of13.
14. Show that ifak;bk2f0;1;. . .;9gand if a1
10þ a2
102þ þ an
10n¼b1
10þ b2
102þ þ bm
10m6¼0; thenn¼mandak¼bk fork¼1;. . .;n.
15. Find the decimal representation of27. 16. Express17and192 as periodic decimals.
17. What rationals are represented by the periodic decimals 1:25137 137 and 35:14653 653 ?