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The second half of a university mathematics exam focusing on integer properties. It includes problems related to proving properties for integers, congruences, and permutations. The document also provides solutions to these problems.
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A3 1. State the Binomial Theorem.
r=
2 r
n r
= 3n,
b) ∑n
r=
(−1)r r! (n − r)!
A4) Show that the following relations on Z are transitive.
A5) 1. Prove that for a ∈ Z we must have one of
a^3 ≡ 0 , 1 or 6 mod 7.
B8) 1. Explain what is meant by a ≡ b mod m.
Assume a ≡ b mod m and c ≡ d mod m. Prove that
a + c ≡ b + d mod m, ac ≡ bd mod m.
33 u − v52 = 1
with the least possible non-negative value for u.
a) Solve x^33 ≡ 2 mod 53. (You may assume that 53 is prime.)
B10) 1. Suppose ∗ is a binary operation on a non-empty set S.
Explain what is meant by a) “∗ is associative”; b) “∗ is commutative”; and c) “e is an identity of S”
Prove that a) the identity in a group is unique, b) the inverse of an element in a group is unique.
What is the identity element? List the inverses of each element of {[3] 15 , [6] 15 , [9] 15 , [12] 15 }.
Please, never write out the questions in the exam. We, the examiners and markers, know what the questions are and you are simply wasting your time.
The Binomial Theorem. For a, b ∈ R and n ∈ N
(a + b)n^ =
∑^ n
r=
arbn−r
n r
[2 marks]
a) Choose a = 2 and b = 1 then
∑^ n
r=
2 r
n r
∑^ n
r=
2 r 1 n−r
n r
= (2 + 1)n^ = 3n.
[2 marks] b) (See Question 6 on Problem Sheet 1 ) Choose a = 1 and b = −1 then ∑^ n
r=
(−1)r r! (n − r)!
n!
∑^ n
r=
(−1)r^ n! r! (n − r)!
n!
∑^ n
r=
(−1)r
n r
(1 − 1)n n!
[2 marks]
n − 1 r − 1
n − 1 r
n r
n! r! (n − r)!
n r
= |Pr (A)| or 2 n^ =
∑^ n
r=
n r
a mod 7 a^3 mod 7 0 0 1 1 2 1 3 6 4 1 5 6 6 6
Hence a^3 ≡ 0 , 1 or 6 mod 7. [3 marks]
a^3 mod 7 b^3 mod 7 2 a^3 − 5 b^3 mod 7 0 0 0 0 1 − 5 ≡ 2 0 6 − 30 ≡ 5 1 0 2 1 1 − 3 ≡ 4 1 6 − 28 ≡ 0 6 0 12 ≡ 5 6 1 7 ≡ 0 6 6 − 2 − 5 (−1) ≡ 3
In no row do we see a final result of 1, hence 2a^3 − 5 b^3 is never ≡ 1 mod 7, hence no n ≡ 1 mod 7 can be written as 2a^3 − 5 b^3 for integers a and b. [3 marks]
a^3 mod 7 b^3 mod 7 2 (a^3 + b^3 ) mod 7 0 0 0 0 1 2 0 − 1 − 2 ≡ 5 1 0 2 1 1 4 1 − 1 0 − 1 0 − 2 ≡ 5 − 1 1 0 − 1 − 1 − 4 ≡ 3
Working backwards
1 = 5 − 4 = 5 − (14 − 2 × 5) = 3 × 5 − 14 = 3 × (19 − 1 × 14) − 14 = 3 × 19 − 4 × 14 = 3 × 19 − 4 × (33 − 1 × 19) = 7 × 19 − 4 × 33 = 7 × (52 − 1 × 33) − 4 × 33 = 7 × 52 − 11 × 33.
Thus (u, v) = (− 11 , −7) is a solution. We require a solution with smallest positive value of u. This solution given by
1 = (7 − 33) × 52 + (−11 + 52) × 33 ,
i.e. (u, v) = (41, 26). [4 marks]
ap−^1 ≡ 1 mod p.
[1 mark] (This is similar to an example in Chapter 7 II of my notes where we solved x^7 ≡ 3 mod 11) To solve x^33 ≡ 2 mod 53, we note from above that 33 × 41 = 1 + 26 × 52. So raising both sides of the equation to 41 gives
241 ≡ x^33 ×^41 = x1+26×^52 = xx^26 ×^52 = x
x^52
≡ x mod 53
by Fermat’s Little Theorem. Thus a solution is given by x ≡ 241 ≡ 39 mod 53 by the result above. [2 marks]
Notes to B
When writing a = b + tm, c = d + sm, the symbols t and s must be different. I often saw a = b + tm, c = d + tm and this was wrong and lost marks.
In the proofs above we see the justifications “Since t + s ∈ Z...” and “since bs + td + tsm ∈ Z...”. Such justifying was forgotten by almost all the students.
Students found other methods and other answers to x^33 ≡ 2 mod 53. For instance, squaring both sides gives (x^33 )^2 ≡ 22 mod 53, i.e. 4 ≡ x^66 = x^52 x^14 ≡ x^14 mod 53 by Fermat’s Little Theorem. Thinking that 4 × 14 is close to 52 we raise both sides of this congruence to the 4th power to get 44 ≡ (x^14 )^4 = x^56 = x^52 x^4 ≡ x^4 mod 53 again by Fermat’s Little Theorem. So we have x^4 ≡ 44 mod 53. Obviously x ≡ 4 mod 53 is a solution of this, (as is x ≡ −4 mod 53).
[2 marks]
a) The order of π is the least positive d : πd^ = 1A, the identity permuta- tion on A. [1 mark]
b) (See pp. 7 and 8 of Chapter 6 of my notes) Proof that πe^ = 1 if, and only if, d|e.
(⇒) Assume πe^ = 1A. By the division Algorithm write e = qd + r for some integers q and 0 ≤ r ≤ d − 1. Then
(^1) A = πe^ = πqd+r^ =
πd
)q πr^ = (1A)q^ πr^ = πr.
But d is the least positive integer with πd^ = 1A and 0 ≤ r ≤ d − 1, thus r = 0. That is, e = qd and so d|e.
(⇐) Assume d|e. So e = dq for some q ∈ Z. But then
πe^ =
πd
)q = (1A)q^ = 1A.
[3 marks] c) The order of ρ = lcm (2, 4 , 4) = 4. The order of σ = lcm (3, 5 , 2) = 30. And
ρ◦σ =
so order of ρσ = lcm (5, 2 , 2) = 10. [3 marks]
Notes to B
Some students found the inverse of ρ ◦ σ by swapping rows in the permu- tation found in the first part of 1.c), reordering columns and obtaining the permutation calculated in the third part of 1.c).
2.a) Most students forgot that the order of π was the least positive d for which πd^ = 1A.
2.c) Far too many students attempted to calculate the orders by simply multiplying the orders of the cycles in the decompositions into disjoint cycles. This was wrong. You need to find the least common multiple of the orders of the cycles.
[3 marks]
The identity element is [6] 15. [1 mark] The inverses are [3]− 151 = [12] 15 , [6]− 151 = [6] 15 , [9]− 151 = [9] 15 and [12]− 151 = [3] 15. [2 marks]
Notes to B
1.a) Many students confused associative with commutative. 1.a), b) and c) Most students forgot that the conditions (a ∗ b) ∗ c = a ∗ (b ∗ c); a ∗ b = b ∗ a; and a ∗ e = e ∗ a = a have to hold for all a, b and c.
2.b) The use of associativity is fundamental to the proof of uniqueness of inverses. If you didn’t use associativity you didn’t have a valid proof. Note that this proof appeared in the section on semi-groups, which are sets with a binary operation that is associative.
