1. No category

MAS114 exam solutions
Dr James Cranch
2013–2014
Question 1
(i) The Chinese Remainder Theorem says that, if m and n are coprime positive integers, and a and b are any integers, then the simultaneous congruences
x≡a
(mod m)
x ≡ b (mod n)
have a solution modulo mn. (2 marks)
(ii) We can write
x = 11 + 14a,
x = 9 + 19b.
This gives us 11 + 14a = 9 + 19b, which rearranges to 19b − 14a = 2. (1
mark) We have
19 = 1 × 14 + 5,
14 = 2 × 5 + 4,
5 = 1 × 4 + 1.
(1 mark) This means that
1 = 5−4 = 5−(14−2×5) = 3×5−14 = 3×(19−14)−14 = 3×19−4×14,
and hence a = 8, b = 6 is a solution to 19b − 14a = 2. (1 mark)
To find a general solution, we can subtract 19 × 6 − 14 × 8 = 2 from
19b − 14a = 2, obtaining
19(b − 6) + 14(a − 8) = 0.
(1 mark) Since 19 | 19(b − 6), we have 19 | 14(a − 8), and since 14 and 19
are coprime we have 19 | (a − 8), or a = 19k + 8.(1 mark)
Substituting, we get b = 6 − 14k, and it’s quick to check that these do in
fact work. Hence the solution is a = 19k + 8, b = 6 − 14k, and substituting
in, we get that x = 11 + 14a = 11 + 14(19k + 8) = 123 + 266k, so x ≡ 123
(mod 266). (1 mark)
(iii) The former equation says that x is 9 more than a multiple of 22, and is
hence odd. The latter equation says that x is 14 more than a multiple of
20, and is hence even. (1 mark) Since x cannot be at once even and odd,
there are no solutions. (1 mark)
1
Question 2
(i) The function fn is injective exactly when n is odd. (1 mark)
Similarly, the function fn is surjective exactly when n is odd. (1 mark)
(ii) We take as our base case n = 0. Here we have n5 − n = 0, which is a
multiple of 5. (2 marks)
Now we do the induction step: we assume that k 5 − k is a multiple of 5
and prove that (k + 1)5 − (k + 1) is.
However,
(k + 1)5 − (k + 1) = (k 5 + 5k 4 + 10k 3 + 10k 2 + 5k + 1) − (k + 1)
= (k 5 − k) + 5(k 4 + 2k 3 + 2k 2 + k).
(2 marks) The former term is a multiple of 5 by the induction hypothesis,
and the latter term is visibly a multiple of 5. This means that (k + 1)5 −
(k +1) is a multiple of 5, and that completes the induction step. (2 marks)
(iii) Fermat’s Little Theorem says that if p is prime and n is not a multiple of
p, then np−1 ≡ 1 (mod p). (1 mark) In particular, choosing p to be the
prime 5 gives us n4 ≡ 1 (mod 5), which says that 5 | (n4 − 1) whenever n
is not a multiple of 5. (1 mark)
Question 3
(i) A sequence x0 , x1 , x2 is a Cauchy sequence if, for any real number > 0,
there is some N , such that for any m, n > N we have |xm − xn | < . (3
marks)
(ii) Replacing aa with 2a in the denominators makes the denominators smaller
and hence the fractions bigger (1 mark), so we have:
1
1
1
1
1
+ ··· +
≤ (a+1) + · · · + (a+k) < a
(a+1)
(a+k)
2
(a + 1)
(a + k)
2
2
(1 mark). The latter inequality is by summing the geometric progression
(2 marks).
(iii) For any > 0, if we choose N ≥ d− log2 e, then we have what we need.
(1 mark)
2
Indeed, for any n ≥ m > N , we get
1
1
+ ··· + n
(m+1)
n
(m + 1)
1
< m (using the inequality above)
2
1
≤ N
2
1
≤ − log 2
2
1
=
1/
= ,
xm − xn =
exactly as needed.(2 marks)
3