# Imperial College London Discrete Math Questions

User Generated

nyrkb101

Mathematics

Imperial College London

## Description

support with abstract math introductory exem -sets and functions, prime numbers ,, functions and limits of functions, continuity, groups

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View attached explanation and answer. Let me know if you have any questions.And entire exam in word and pdf:)

SECTION A
1
(a)

(i) Proof. 𝑛 = 𝑚3 − 𝑚 = (𝑚 − 1)𝑚(𝑚 + 1), i.e. it is a product of three consecutive numbers.
Because of this, at least one of the factors 𝑚 − 1, 𝑚, 𝑚 + 1 is divisible by 3 and at least one
is divisible by 2. This means that 𝑛 is divisible by 2 ∙ 3 = 6.

(ii) For a statement “if p then q” the converse is “if q then p”, here it is “if 𝑛 is divisible by 6,
then 𝑛 = 𝑚3 − 𝑚 for some 𝑚”. This statement is NOT true.
Proof. Consider the difference between two consecutive terms of the sequence
𝑎𝑚 = 𝑚3 − 𝑚: ((𝑚 + 1)3 − (𝑚 + 1)) − (𝑚3 − 𝑚) =
= 𝑚3 + 3𝑚2 + 3𝑚 + 1 − 𝑚 − 1 − 𝑚3 + 𝑚 =
= 3𝑚2 + 3𝑚 = 3𝑚(𝑚 + 1) > 0,
so the sequence 𝑎𝑚 is strictly increasing.
Consider 𝑛 = 12 which is divisible by 6. It is strictly between 𝑎2 = 6 and 𝑎3 = 24. Because
of this 12 is not a member of {𝑎𝑛 }, ∎

This statement uses three independent logical variables (p, q, r), so its truth table require 8
rows. Construct it:
𝒑
T
T
T
T
F
F
F
F

𝒒
T
T
F
F
T
T
F
F

𝒓 𝒑 ⇒ 𝒒 𝒒 ⇒ 𝒓 (𝒑 ⇒ 𝒒) ∧ (𝒒 ⇒ 𝒓) 𝒑 ⇒ 𝒓 ((𝒑 ⇒ 𝒒) ∧ (𝒒 ⇒ 𝒓)) ⇒ (𝒑 ⇒ 𝒓)
T
T
T
T
T
T
F
T
F
F
F
T
T
F
T
F
T
T
F
F
T
F
F
T
T
T
T
T
T
T
F
T
F
F
T
T
T
T
T
T
T
T
F
T
T
T
T
T

The last column represents the entire statement and consists of T’s only, so the statement
is always true.

Observe that for any m, n we have (𝑚 − 𝑛) mod 10 = (𝑚 mod 10 − 𝑛 mod 10) mod 10.
Consider representatives of 𝑟𝑘 = (𝑡𝑘 mod 10) so that 0 ≤ 𝑟𝑘 < 10. Find 𝑡𝑘 , 𝑡𝑗 so that 𝑡𝑘 − 𝑡𝑗
is divisible by 10 is the same that 𝑟𝑘 = 𝑟𝑗 .
There are 11 𝑟𝑘 ’s and only 10 possible values for them, so by the pigeonhole principle at
least two of 𝑟𝑘 ’s are the same, so at least two 𝑡𝑘 ’s have the same residue while divided by
10.

Induction base, 𝑛 = 1: 2 ∙ 4 =

1∙48
6

which is true (both =8).
1

Induction step, 𝑛 → 𝑛 + 1: let ∑𝑛𝑟=1(𝑟 + 1)(𝑟 + 3) = 6 𝑛(2𝑛2 + 15𝑛 + 31). Then the next sum,
∑𝑛+1
𝑟=1 (𝑟 + 1)(𝑟 + 3), is equal to
𝑛

(𝑟 + 1)(𝑟 + 3) + ((𝑛 + 1 + 1)(𝑛 + 1 + 3)) = |
𝑟=1

𝑏𝑦 𝑡ℎ𝑒 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
|=
𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

1
= 𝑛(2𝑛2 + 15𝑛 + 31) + 6(𝑛 + 2)(𝑛 + 4) =
6
1
= (2𝑛3 + 15𝑛2 + 31𝑛 + 6𝑛2 + 36𝑛 + 48) =
6
1
= (2𝑛3 + 21𝑛2 + 67𝑛 + 48).
6
At the other side of the equation to prove we have
1
((𝑛 + 1)(2(𝑛 + 1)2 + 15(𝑛 + 1) + 31)) =
6
1
= (𝑛 + 1)(2𝑛2 + 19𝑛 + 48) =
6
1
= (2𝑛3 + 2𝑛2 + 19𝑛2 + 19𝑛 + 48𝑛 + 48) =
6
1
= (2𝑛3 + 21𝑛2 + 67𝑛 + 48),
6
which is equal to the left side, Q.E.D.

Proof. Induction base, 𝑛 = 0: 𝑓(0) = 0 = 21 − 0 − 2 is true.
Induction step, let 𝑓(𝑛) = 2𝑛+1 − 𝑛 − 2 for some 𝑛, then we need to prove that
𝑓(𝑛 + 1) = 2𝑛+2 − (𝑛 + 1) − 2 = 2𝑛+2 − 𝑛 − 3.
By definition of 𝑓, 𝑓(𝑛 + 1) = 2𝑓(𝑛) + 𝑛 + 1 = 2𝑛+2 − 2𝑛 − 4 + 𝑛 + 1 = 2𝑛+2 − 𝑛 − 3,
which is what we want.

(i) The sequence 𝑓(𝑛) is clearly strictly increasing:
𝑓(𝑛 + 1) − 𝑓(𝑛) = 2𝑛+2 − 𝑛 − 1 − 2 − (2𝑛+1 − 𝑛 − 2) =
= 2𝑛+1 − 1 > 0, 𝑛 ≥ 0.
Now consider 𝑓(2) = 4 and 𝑓(3) = 11. Because 𝑓(𝑛) is strictly increasing, 𝑓 cannot take
any values between 4 and 11, so it is not a surjection.
(ii) The same fact shows that no 𝑓(𝑛)’s are equal for different 𝑛’s, so 𝑓 is an injection.

(iii) Suppose 𝑛 ≥ 0 is even, then 𝑓(𝑛) = 2𝑛+1 − 𝑛 − 2 is an algebraic sum of three even
numbers, i.e. it is even. This shows that if n is NOT odd then 𝑓(𝑛) is NOT prime, which is
equivalent to what we want.
(iv) It is not true, a counterexample is 𝑛 = 5, 𝑓(𝑛) = 26 − 5 − 2 = 64 − 7 = 57 = 3 ∙ 19.

Note that for 𝑥 ∈ (−1,1) we have 1 − 𝑥 2 > 0. Because of this, 𝑓(𝑥) and 𝑥 have the same
signs. To prove that 𝑓 is injective, prove that it is strictly increasing:
1 − 𝑥 2 − 𝑥 ∙ (−2𝑥)
1 + 𝑥2
𝑓 ′ (𝑥) =
=
> 0.
(1 − 𝑥 2 )2
(1 − 𝑥 2 )2
To prove that 𝑓 is surjective, solve for 𝑦 the equation 𝑥 = 𝑓(𝑦):
𝑦
𝑥=
, 𝑥 − 𝑥𝑦 2 = 𝑦, 𝑥𝑦 2 + 𝑦 − 𝑥 = 0.
1 − 𝑦2
The discriminant is 1 + 4𝑥 2 > 0, so there are two solutions, 𝑦1,2 =

−1±√1+4𝑥 2
2𝑥

.

Because the solution must be positive for positive x and negative for negative x,
choose “+” for any x: 𝑓 −1 (𝑥) =

−1+√1+4𝑥 2
2𝑥

.

Well, this is true only for 𝑥 ≠ 0, for 𝑥 = 0 the only solution is 𝑦 = 0, so finally
−𝟏 + √𝟏 + 𝟒𝒙𝟐
, 𝒙 ≠ 𝟎.
𝒇−𝟏 (𝒙) = {
𝟐𝒙
𝟎, 𝒙 = 𝟎

̅̅̅̅̅, so
(a) Let the n...

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