Answers to problems
Page 294:
The first question simply uses the meaning of decimal notation, with its place-notation extended to fractions. The expression .111111...
simply means: one-tenth plus one-hundredth plus one-thousandth...
The second question gives another way of seeing why this recurring decimal is equal to 1/9. Use the symbol r for 1/10, then .111111... is written
r + r2 + r3 + r4...
Multiply this by (1 − r) and get
( r + r2 + r3 + r4...)
− (r2 + r3 + r4 ...) = r
Hence the recurring decimal is r/(1 − r) = (1/10)/(1 − 1/10) = 1/9
Page 298:
This problem shows another way of seeing why the recurring decimal for 1/7 is right. This time let r = 10−6, and then the same argument shows that
1/999999 = 10−6 + 10−12 + 10−18... = .000001000001000001000001000001...
Multiplying by 142857:
142857/999999 = 1/7 = .142857142857142857142857...
Page 301
1/11 = .09090909...
1/37 = .027027027...
1/101 = .009900990099...
1/41 = .0243902439...
These fit the rule because 2 divides into 10, 3 divides into 36, 4 divides into 100, and 5 divides into 40.
Page 302:
It is easy to see the every remainder has a unique successor. If the remainder is r, then its successor is just the remainder when r × 10 is divided by p.
To see that each remainder has a unique predecessor, suppose if possible that there are two remainders s and t which both have r as their successor. Then r is the remainder when s × 10 is divided by p, and also the remainder when t × 10 is divided by p. Then the number (s − t) × 10 must be divisible by p. But this is impossible, as p is a prime greater than 5.
This establishes that the remainders must form cycles, or carousels. But there will in general be several disjoint carousels. We want to show they all have the same length. This is most easily seen from the example of 1/13, which has the two cycles 1-10-9-12-3-4 and 2-7-5-11-6-8. The second cycle can be obtained from the first by multiplying each element by 2, modulo 13. Thus it must have the same length. The same argument applies in general.
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