Download (b) The quotient is 4 and the remainder is 4.

Exercises 4.1
1. (a) The quotient is 5 and the remainder is 0.
(b) The quotient is 4 and the remainder is 4.
(c) The quotient is 30 and the remainder is 4.
2. (a) 5.
(b) 9.
(c) 7.
3. (a) We have that 267 = 2 · 112 + 43, 112 = 2 · 43 + 26, 43 = 1 · 26 + 17,
26 = 1 · 17 + 9, 17 = 1 · 9 + 8, 9 = 1 · 8 + 1. Thus gcd(112, 267) = 1.
We now run these calculations backwards to get
1 = 31 · 112 − 13 · 267.
(b) We have that 1870 = 7·242+176, 242 = 1·176+66, 176 = 2·66+44,
66 = 1 · 44 + 22. Thus gcd(242, 1870) = 22. We now run these
calculations backwards to get
22 = 31 · 242 − 4 · 1870.
22·121
= 242.
11
= 144.
lcm(48, 72) = 48·72
24
25·116
lcm(25, 116) = 1 = 2, 900.
4. (a) gcd(22, 121) = 11. Thus lcm(22, 121) =
(b) gcd(48, 72) = 24. Thus
(c) gcd(25, 116) = 1. Thus
5. Let d be any natural number that divides a, b and c. Since d divides a
and b it must divide gcd(a, b). But since d divides gcd(a, b) and c it must
divide gcd(gcd(a, b), c). Thus the lefthand side divides the righthand
side. Now let h = gcd(gcd(a, b), c). Then h divides gcd(a, b) and c.
But if h divides gcd(a, b) then it must divide a and b. It follows that
the righthand side divides the lefthand side. Hence the two sides must
equal eachother being natural numbers that are mutually divisible.
The proof of the other claim follows by symmetry.
Define a binary operation on the set N \ {0} by a ∗ b = gcd(a, b). We
have proved that this operation is associative.
13
By generalized associativity, we have that
Thus
gcd(910, 780, 286, 195) = ((910 ∗ 780) ∗ 286) ∗ 195.
((910 ∗ 780) ∗ 286) ∗ 195 = (130 ∗ 286) ∗ 195 = 26 ∗ 195 = 13.
6. Let u and v be two non-zero natural numbers. First observe that if
u = v then u ◦ u = u and gcd(u, u) = u and there is nothing to prove.
Without loss of generality, assume that u ≥ v. But then
v ◦ u = v ◦ (v + (v − u)) = v ◦ (u − v)
where u − v < u. Observe that gcd(u, v) = gcd(v, u − v). This process
can be repeated using commutativity to swop sides if necessary. It will
terminate when the two numbers are equal and then we use our first
observation to deduce the result.
7. Draw up a table with three columns. The first row is labelled 0,1,2.
the second is labelled 3,4,5 and so forth. Circle those numbers that
can be written in the form 3x + 5y where x, y ∈ N. We say that a
row is complete if all numbers in that row are circled. Observe that
9 = 3 · 3 and 10 = 2 · 5 and 11 = 2 · 3 + 5. Since this row is complete all
subsequent row are complete. Now 8 = 3 + 5 but 7 cannot be written
in the given way. It follows that 7 is the largest value that cannot be
made.
8. The values are tabulated below.
n
1
2
3
4
5
6
7
8
9
10
11
12
14
φ(n)
1
1
2
2
4
2
6
4
6
4
10
4
9. (a) This follows from the fact that a = 1 × a.
(b) We are given that b = am and a = bn for some integers m and
n. Thus a = bn = amn. Cancelling a we get that mn = 1. Thus
either m = n = 1 or m = n = −1.
(c) We are given that b = am and c = bn. Thus c = bn = amn. But
this means that a | c, as required.
(d) We are given that b = am and c = an. Thus b + c = am + an =
a(m + n). It follows that a | (b + c) as required.
10. (a) If a = 0 then b ∈ X and is positive. If a > 0 then a − 0 · b ∈ X
and is positive. If a < 0 then a − 2a ∈ X and is positive. Thus in
all cases X contains positive elements.
(b) This is immediate from the question.
(c) Suppose that r ≥ b. Then we may write a = (q + 1)b + (r − b)
where r − b ≥ 0 and r − b < r. This contradicts the choice of r.
(d) Suppose that r 6= r0 . Without loss of generality, assume that
r0 − r ≥ 0. Then r0 − r = (q − q 0 )b. But r0 − r < b. Thus q − q 0 = 0
and so q = q 0 from which we get that r = r0 .
15
Exercises 4.2
1. The primes less than 100 are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
√
131 = 11 · . . .. We therefore try the primes 2, 3, 5, 7, and 11.
None of them works and so 131 is a prime.
√
(b) 689 = 26 · . . .. We therefore try the primes 2, 3, 5, 7, 11, 13, 17,
19, and 23. We find that 13 | 689 and 689 = 13 · 53. But 53 is a
prime and so this is the prime factorization of our number.
√
(c) 5491 = 74 · . . .. We therefore try the primes 2, 3, 5, 7, 11, 13,
17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, √
67, 71, and 73. We
find that 17 | 5491 and 5491 = 17 · 323. Now 323 = 17 · . . .. We
therefore try the primes 2, 3, 5, 7, 11, 13, and 17. We find that
323 = 17 · 19. But 19 is a prime. Thus 5491 = 172 · 19 is the prime
factorization of our number.
2. (a)
3. The greatest common divisor is therefore 22 · 54 · 112 and the least
common multiple is therefore 24 · 3 · 56 · 114 .
√
√ √
4. Since ab = a b it is enough to prove the result for powers of primes.
√
Let a = pn where n ≥ 1. Suppose that n = 2m then pn = pm .
√
√
Suppose that n = 2m + 1 then pn = pm p.
√ √
(a) 2 5.
√ √ √
(b) 2 3 7.
√ √
(c) 3 2 3.
5. Suppose that a does not divide b. Then gcd(a, b) = 1. Thus there
are integers x and y such that 1 = ax + by. Then c = acx + bcy.
But a divides the righthand side and so a divides the lefthand side, as
claimed.
16
Exercises 4.3
1. (a) 310145 .
(b) 11B512 .
(c) 7D916 .
2. (a) 3499.
(b) 19006.
(c) 386556.
3. (a) 0 · 5 finite.
(b) 0 · 3 periodic.
(c) 0 · 25 finite.
(d) 0 · 2 finite.
(e) 0 · 16 ultimately periodic.
(f) 0 · 142857 periodic.
529
534−5
= 990
. We check to see if this
4. (a) The required fraction is 1000−10
fraction is in its lowest terms by calculating gcd(529, 990). But
529
this is 1. Thus the solution is 990
.
2106−2
(b) The required fraction is 10,000−10
= 2104
. The gcd of the two
9990
1052
numbers is 2. Thus the solution is 4995 .
76923
. The gcd of the two numbers is
(c) The required fraction is 999,999
1
76923. Thus the solution is 13
.
17