Imperial College London Discrete Math Questions

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Mathematics

Imperial College London

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

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Explanation & Answer

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SECTION A
1
(a)

(i) Proof. ๐‘› = ๐‘š3 โˆ’ ๐‘š = (๐‘š โˆ’ 1)๐‘š(๐‘š + 1), i.e. it is a product of three consecutive numbers.
Because of this, at least one of the factors ๐‘š โˆ’ 1, ๐‘š, ๐‘š + 1 is divisible by 3 and at least one
is divisible by 2. This means that ๐‘› is divisible by 2 โˆ™ 3 = 6.

(ii) For a statement โ€œif p then qโ€ the converse is โ€œif q then pโ€, here it is โ€œif ๐‘› is divisible by 6,
then ๐‘› = ๐‘š3 โˆ’ ๐‘š for some ๐‘šโ€. This statement is NOT true.
Proof. Consider the difference between two consecutive terms of the sequence
๐‘Ž๐‘š = ๐‘š3 โˆ’ ๐‘š: ((๐‘š + 1)3 โˆ’ (๐‘š + 1)) โˆ’ (๐‘š3 โˆ’ ๐‘š) =
= ๐‘š3 + 3๐‘š2 + 3๐‘š + 1 โˆ’ ๐‘š โˆ’ 1 โˆ’ ๐‘š3 + ๐‘š =
= 3๐‘š2 + 3๐‘š = 3๐‘š(๐‘š + 1) > 0,
so the sequence ๐‘Ž๐‘š is strictly increasing.
Consider ๐‘› = 12 which is divisible by 6. It is strictly between ๐‘Ž2 = 6 and ๐‘Ž3 = 24. Because
of this 12 is not a member of {๐‘Ž๐‘› }, โˆŽ

This statement uses three independent logical variables (p, q, r), so its truth table require 8
rows. Construct it:
๐’‘
T
T
T
T
F
F
F
F

๐’’
T
T
F
F
T
T
F
F

๐’“ ๐’‘ โ‡’ ๐’’ ๐’’ โ‡’ ๐’“ (๐’‘ โ‡’ ๐’’) โˆง (๐’’ โ‡’ ๐’“) ๐’‘ โ‡’ ๐’“ ((๐’‘ โ‡’ ๐’’) โˆง (๐’’ โ‡’ ๐’“)) โ‡’ (๐’‘ โ‡’ ๐’“)
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 Tโ€™s only, so the statement
is always true.

Observe that for any m, n we have (๐‘š โˆ’ ๐‘›) mod 10 = (๐‘š mod 10 โˆ’ ๐‘› mod 10) mod 10.
Consider representatives of ๐‘Ÿ๐‘˜ = (๐‘ก๐‘˜ mod 10) so that 0 โ‰ค ๐‘Ÿ๐‘˜ < 10. Find ๐‘ก๐‘˜ , ๐‘ก๐‘— so that ๐‘ก๐‘˜ โˆ’ ๐‘ก๐‘—
is divisible by 10 is the same that ๐‘Ÿ๐‘˜ = ๐‘Ÿ๐‘— .
There are 11 ๐‘Ÿ๐‘˜ โ€™s and only 10 possible values for them, so by the pigeonhole principle at
least two of ๐‘Ÿ๐‘˜ โ€™s are the same, so at least two ๐‘ก๐‘˜ โ€™s have the same residue while divided by
10.

Induction base, ๐‘› = 1: 2 โˆ™ 4 =

1โˆ™48
6

which is true (both =8).
1

Induction step, ๐‘› โ†’ ๐‘› + 1: let โˆ‘๐‘›๐‘Ÿ=1(๐‘Ÿ + 1)(๐‘Ÿ + 3) = 6 ๐‘›(2๐‘›2 + 15๐‘› + 31). Then the next sum,
โˆ‘๐‘›+1
๐‘Ÿ=1 (๐‘Ÿ + 1)(๐‘Ÿ + 3), is equal to
๐‘›

โˆ‘

(๐‘Ÿ + 1)(๐‘Ÿ + 3) + ((๐‘› + 1 + 1)(๐‘› + 1 + 3)) = |
๐‘Ÿ=1

๐‘๐‘ฆ ๐‘กโ„Ž๐‘’ ๐‘–๐‘›๐‘‘๐‘ข๐‘๐‘ก๐‘–๐‘œ๐‘›
|=
๐‘Ž๐‘ ๐‘ ๐‘ข๐‘š๐‘๐‘ก๐‘–๐‘œ๐‘›

1
= ๐‘›(2๐‘›2 + 15๐‘› + 31) + 6(๐‘› + 2)(๐‘› + 4) =
6
1
= (2๐‘›3 + 15๐‘›2 + 31๐‘› + 6๐‘›2 + 36๐‘› + 48) =
6
1
= (2๐‘›3 + 21๐‘›2 + 67๐‘› + 48).
6
At the other side of the equation to prove we have
1
((๐‘› + 1)(2(๐‘› + 1)2 + 15(๐‘› + 1) + 31)) =
6
1
= (๐‘› + 1)(2๐‘›2 + 19๐‘› + 48) =
6
1
= (2๐‘›3 + 2๐‘›2 + 19๐‘›2 + 19๐‘› + 48๐‘› + 48) =
6
1
= (2๐‘›3 + 21๐‘›2 + 67๐‘› + 48),
6
which is equal to the left side, Q.E.D.

Proof. Induction base, ๐‘› = 0: ๐‘“(0) = 0 = 21 โˆ’ 0 โˆ’ 2 is true.
Induction step, let ๐‘“(๐‘›) = 2๐‘›+1 โˆ’ ๐‘› โˆ’ 2 for some ๐‘›, then we need to prove that
๐‘“(๐‘› + 1) = 2๐‘›+2 โˆ’ (๐‘› + 1) โˆ’ 2 = 2๐‘›+2 โˆ’ ๐‘› โˆ’ 3.
By definition of ๐‘“, ๐‘“(๐‘› + 1) = 2๐‘“(๐‘›) + ๐‘› + 1 = 2๐‘›+2 โˆ’ 2๐‘› โˆ’ 4 + ๐‘› + 1 = 2๐‘›+2 โˆ’ ๐‘› โˆ’ 3,
which is what we want.

(i) The sequence ๐‘“(๐‘›) is clearly strictly increasing:
๐‘“(๐‘› + 1) โˆ’ ๐‘“(๐‘›) = 2๐‘›+2 โˆ’ ๐‘› โˆ’ 1 โˆ’ 2 โˆ’ (2๐‘›+1 โˆ’ ๐‘› โˆ’ 2) =
= 2๐‘›+1 โˆ’ 1 > 0, ๐‘› โ‰ฅ 0.
Now consider ๐‘“(2) = 4 and ๐‘“(3) = 11. Because ๐‘“(๐‘›) is strictly increasing, ๐‘“ cannot take
any values between 4 and 11, so it is not a surjection.
(ii) The same fact shows that no ๐‘“(๐‘›)โ€™s are equal for different ๐‘›โ€™s, so ๐‘“ is an injection.

(iii) Suppose ๐‘› โ‰ฅ 0 is even, then ๐‘“(๐‘›) = 2๐‘›+1 โˆ’ ๐‘› โˆ’ 2 is an algebraic sum of three even
numbers, i.e. it is even. This shows that if n is NOT odd then ๐‘“(๐‘›) is NOT prime, which is
equivalent to what we want.
(iv) It is not true, a counterexample is ๐‘› = 5, ๐‘“(๐‘›) = 26 โˆ’ 5 โˆ’ 2 = 64 โˆ’ 7 = 57 = 3 โˆ™ 19.

Note that for ๐‘ฅ โˆˆ (โˆ’1,1) we have 1 โˆ’ ๐‘ฅ 2 > 0. Because of this, ๐‘“(๐‘ฅ) and ๐‘ฅ have the same
signs. To prove that ๐‘“ is injective, prove that it is strictly increasing:
1 โˆ’ ๐‘ฅ 2 โˆ’ ๐‘ฅ โˆ™ (โˆ’2๐‘ฅ)
1 + ๐‘ฅ2
๐‘“ โ€ฒ (๐‘ฅ) =
=
> 0.
(1 โˆ’ ๐‘ฅ 2 )2
(1 โˆ’ ๐‘ฅ 2 )2
To prove that ๐‘“ is surjective, solve for ๐‘ฆ the equation ๐‘ฅ = ๐‘“(๐‘ฆ):
๐‘ฆ
๐‘ฅ=
, ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฆ 2 = ๐‘ฆ, ๐‘ฅ๐‘ฆ 2 + ๐‘ฆ โˆ’ ๐‘ฅ = 0.
1 โˆ’ ๐‘ฆ2
The discriminant is 1 + 4๐‘ฅ 2 > 0, so there are two solutions, ๐‘ฆ1,2 =

โˆ’1ยฑโˆš1+4๐‘ฅ 2
2๐‘ฅ

.

Because the solution must be positive for positive x and negative for negative x,
choose โ€œ+โ€ for any x: ๐‘“ โˆ’1 (๐‘ฅ) =

โˆ’1+โˆš1+4๐‘ฅ 2
2๐‘ฅ

.

Well, this is true only for ๐‘ฅ โ‰  0, for ๐‘ฅ = 0 the only solution is ๐‘ฆ = 0, so finally
โˆ’๐Ÿ + โˆš๐Ÿ + ๐Ÿ’๐’™๐Ÿ
, ๐’™ โ‰  ๐ŸŽ.
๐’‡โˆ’๐Ÿ (๐’™) = {
๐Ÿ๐’™
๐ŸŽ, ๐’™ = ๐ŸŽ

ฬ…ฬ…ฬ…ฬ…ฬ…, so
(a) Let the n...


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