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Compsci 120 The University of Auckland

Assignment 4

Due: Friday, October 30th, by 23.59pm Semester 2, 2020

There are four problems listed below. To get full credit on this assignment, you should pick any

three of them to complete!

Note that you do not have to follow the suggested proof techniques. Any correct proof is worth the

max possible number of points.

The same rules from Assignments 1, 2, and 3 (show all working, late assignments cannot be accepted,

etc.) all still apply here. Best of luck, and enjoy the problems!

1. Direct proofs and proofs by cases.

(a) Let n be a positive integer number. Prove that if p is a prime factor of n

2

, then p is a

factor of n.

(b) Suppose n and x are two positive integers, n > 2. Prove that if n

x ≡ n − 2 (mod n − 1),

then n = 3.

(c) Consider a correct claim with its incorrect proof.

Claim: Prove that n

3 + 5n is divisible by 6 for every positive integer n.

Proof. Because every n can be written in the form 6q + n%6 consider the following 5

cases:

Case 1: n%6 = 1. Then n

3%6 = 13%6 = 1 and 5n%6 = 5. Hence (n

3 + 5n)%6 =

1 + 5 = 6, which is divisible by 6.

Case 2: n%6 = 2. Since n

3 + 5n = n(n

2 + 5) the only way n(n

2 + 5) can be divisible

by 6 is if n

2 + 5 is divisible by 6. Indeed, (n

2 + 5)%6 = n

2%6 + 5%6 = (n%6)2 + 5 =

4 + 5 = 9. The number is not divisible by 6, so this case is not possible.

Case 3: n%6 = 3. Then n

3%6 = 33%6 = 1 and 5n%6 = 3. Hence (n

3 + 5n)%6 =

3 + 3 = 6, which is divisible by 6.

Case 4: n%6 = 4. Then n

3%6 = 43%6 = 4 and 5n%6 = 2. Hence (n

3 + 5n)%6 =

4 + 2 = 6, which is divisible by 6.

Case 5: n%6 = 5. Then n

3%6 = 53%6 = 5 and 5n%6 = 1. Hence (n

3 + 5n)%6 =

5 + 1 = 6, which is divisible by 6.

Since in all the cases n

3 + 5n is divisible by 6 or a case is not possible n

3 + 5n is a multiple

of 6 for every positive integer n.

Find ALL logical mistakes in the proof. You do not need to prove this claim.

2. Proof by contradiction

(a) Given a graph G, define k(G) to be the greatest possible natural number n such that Kn

(the complete graph on n vertices) is a subgraph of G. Recall that the chromatic number

χ(G) of G is the smallest possible number of colours sufficient to color the vertices of G

so that no two adjacent vertices receive the same color (see Assignment 3 Question 3).

Prove that χ(G) ≥ k(G).

(b) Let x, y, z be three integer numbers. Prove that if x2 + y2 = z2, then x is even or y is

even.

1

(c) Consider a correct claim with its incorrect proof.

Claim: Consider the equation x

2+y2−3 = 0. A pair (x0, y0) is called a rational solution

of x2 + y2 − 3 = 0 if

• x0 and y0 are rational;

• x20 + y20 − 3 = 0

The equation x2 + y2 − 3 = 0 has no rational solutions.

Proof. Assume that it is not true that x2 + y2 − 3 = 0 has no rational solutions. That is,

every solution (x, y) of x2 + y2 − 3 = 0 is rational. Consider one of these solutions, say

(a, b), then

• a and b are rational;

• a2 + b2 − 3 = 0

Since a, b are rational there are integer numbers pa, Therefore qb is a multiple

of qa. So there is an integer t, such that qa = tqa.

Now qa = tqa and qb = kqa. It is not possible. Therefore our assumption was incorrect

and x

2 + y

2 − 3 = 0 has no rational solutions.

Find ALL logical mistakes in the proof. You do not need to prove this claim.

3. Proofs by construction

(a) Prove that for any given natural number n > 2 there is a graph G on n vertices with the

following properties:

i. G has exactly (n − 2)(n − 1)

2

many edges;

ii. At least one vertex of G is connected to exactly n − 2 other vertices of G by an edge.

(b) Design an algorithm which, on input a tree T represented in the form of the picture,

outputs a colouring of its vertices with the property that no two adjacent vertices have

the same colour.

Note that the following operations can be perform on a picture of a tree without any

explanation:

• choosing a vertex;

• checking if two vertices are adjacent;

• checking if a vertex is coloured;

• checking if all vertices are coloured;

2

• finding the colour of a vertex;

• finding all neighbours of a vertex.

• colouring a vertex

(c) Consider a correct claim with its incorrect proof.

Claim: For every integer n ≥ 2, there is a graph on 2n vertices which has exactly 4

vertices of degree 2 and (2n − 4) vertices of degree 3.

Proof. Define the graph G on 2n vertices in the following way. Take a cycle graph C2n

and connect vertex i to vertex 2n − i + 1 by an edge for every 2 ≤ i ≤ n − 1.

The resulting graph has all the needed properties. Consider a vertex i, 2 ≤ i ≤ 2n − 1. It

is adjacent to i − 1, i + 1 and 2n − i + 1. Hence the degree of i is 3. Therefore, G satisfies

all the properties.

Find ALL logical mistakes in the proof. You do not need to prove this claim.

4. Mathematical induction

(a) Let x be a non-negative real number. Prove that

(1 + x)

n ≥ 1 + nx

for every natural number n.

(b) Prove that 72n − 1 is a multiple of 24 for every natural n.

(c) Consider a correct claim with its incorrect proof.

Claim: The solution of the following recurrence relation

(a1 = 3, a2 = 5

an+1 = 3an − 2an−1

is an = 2n + 1 for every positive integer n.

Proof. The induction statement P(n) is

(a1 = 3, a2 = 5

an+1 = 3an − 2an−1

Base case: You need to check the formula for n = 1. Indeed, if n = 1, then a1 = 2n+1 =21 + 1 = 3 as needed.

Induction step: Assume that an+1 = 2n+1 + 1 satisfies the recurrence relation then you

need to prove that an = 2n + 1 also satisfies the recurrence relation.

Indeed,

an+1 = 3an −2an−1 = 3(2n + 1)−2(2n−1 + 1) = 3 · 2n + 3−2n −2 = 2 · 2n + 1 = 2n+1 + 1.

The induction step is proved.

By the principle of mathematical induction an = 2n + 1 is the solution of the recurrence

relation for every positive integer n.

Find ALL logical mistakes in the proof. You do not need to prove this claim.

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