Problem Set n°1

(1)

University of Toronto – MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

Problem Set n°1

Jean-Baptiste Campesato Due on February 5^th, 2021

Except otherwise stated, you can only use the material covered from Jan 12 to Jan 26 (i.e. Chapter 1 & Chapter 2 up to §3).

Exercise 1.

We define the binary relation≺onℕ²by(𝑥₁, 𝑦₁) ≺ (𝑥₂, 𝑦₂) ⇔ (𝑥1< 𝑥₂or(𝑥1 = 𝑥₂and𝑦₁ ≤ 𝑦₂)). Is it an order? If so, is it total?

Exercise 2.

Prove that given𝑛 ∈ ℕ ⧵ {0}there exist finitely many𝛼₁, … , 𝛼_𝑚∈ ℕpairwise distinct such that 𝑛 = 2^𝛼¹+ 2^𝛼²+ ⋯ + 2^𝛼^𝑚

Exercise 3.

Solve4𝑥(𝑥 + 1) = 𝑦(𝑦 + 1)for(𝑥, 𝑦) ∈ ℕ².

Your answer can only rely on the properties ofℕproved in Chapter 1.

Particularly, your proof should not involve negative integers, rationals, calculus…

Hint:compare2𝑥and𝑦.

Exercise 4.

Prove that for every𝑛 ≥ 3, there exist𝑥₁, … , 𝑥_𝑛 ∈ ℕ ⧵ {0}pairwise distinct such that 1 = 1

𝑥₁ + 1

𝑥₂ + ⋯ + 1 𝑥_𝑛 In this exercise, you may assume that you already knowℚorℝso that_𝑥¹

𝑖 is well-defined.

Hint:1 = ¹

2 +¹

2.

(2)

2 Problem Set 1

Sample solution to Exercise 1.

We are going to prove that≺is a total order onℕ². It is actually called thelexicographic order.

It is the one used in dictionaries: you compare the first letter, if it is the same, then you look at the next one…

• Reflexivity.Let(𝑥, 𝑦) ∈ ℕ². Then𝑥 = 𝑥and𝑦 ≤ 𝑦. Thus(𝑥, 𝑦) ≺ (𝑥, 𝑦).

• Antisymmetry.Let(𝑥₁, 𝑦₁), (𝑥₂, 𝑦₂) ∈ ℕ²satisfying(𝑥₁, 𝑦₁) ≺ (𝑥₂, 𝑦₂)and(𝑥₂, 𝑦₂) ≺ (𝑥₁, 𝑦₁).

Assume by contradiction that𝑥₁< 𝑥₂, then(𝑥₂, 𝑦₂) ⊀ (𝑥₁, 𝑦₁). Which is a contradiction.

Assume by contradiction that𝑥₂< 𝑥₁, then(𝑥₁, 𝑦₁) ⊀ (𝑥₂, 𝑦₂). Which is a contradiction.

Thus𝑥₁ = 𝑥₂.

Since(𝑥₁, 𝑦₁) ≺ (𝑥₂, 𝑦₂), we know that𝑦₁≤ 𝑦₂. Since(𝑥₂, 𝑦₂) ≺ (𝑥₁, 𝑦₁), we know that𝑦₂≤ 𝑦₁. Thus𝑦₁ = 𝑦₂.

We proved that(𝑥₁, 𝑦₁) = (𝑥₂, 𝑦₂).

• Transitivity. Let(𝑥₁, 𝑦₁), (𝑥₂, 𝑦₂), (𝑥₃, 𝑦₃) ∈ ℕ²satisfying(𝑥₁, 𝑦₁) ≺ (𝑥₂, 𝑦₂)and(𝑥₂, 𝑦₂) ≺ (𝑥₃, 𝑦₃).

– Case 1:𝑥₁= 𝑥₂and𝑥₂= 𝑥₃.

Then𝑥₁= 𝑥₃. Furthemore𝑦₁ ≤ 𝑦₂and𝑦₂ ≤ 𝑦₃, so𝑦₁≤ 𝑦₃. Hence(𝑥₁, 𝑦₁) ≺ (𝑥₃, 𝑦₃).

– Case 2:𝑥₁= 𝑥₂and𝑥₂< 𝑥₃.

Then𝑥₁< 𝑥₃. Hence(𝑥₁, 𝑦₁) ≺ (𝑥₃, 𝑦₃).

– Case 3:𝑥₁< 𝑥₂and𝑥₂= 𝑥₃.

– Case 4:𝑥₁< 𝑥₂and𝑥₂< 𝑥₃.

• ≺is a total order. Let(𝑥₁, 𝑦₁), (𝑥₂, 𝑦₂) ∈ ℕ². According to the lectures, exactly one of the follows occurs.

– Case 1:𝑥₁ < 𝑥₂. Then(𝑥₁, 𝑦₁) ≺ (𝑥₂, 𝑦₂).

– Case 2:𝑥₂ < 𝑥₁. Then(𝑥₂, 𝑦₂) ≺ (𝑥₁, 𝑦₁).

– Case 3:𝑥₁ = 𝑥₂. Since≤is a total order onℕthen

∗ either𝑦₁ ≤ 𝑦₂and then(𝑥₁, 𝑦₁) ≺ (𝑥₂, 𝑦₂)

∗ or𝑦₂≤ 𝑦₁and then(𝑥₂, 𝑦₂) ≺ (𝑥₁, 𝑦₁).

That’s the existence of the positional numeral system with base2(binary numeral system).

Method 1:

We are going to prove by strong induction that for every𝑛 ≥ 1, there exist finitely many𝛼₁, … , 𝛼_𝑚 ∈ ℕ pairwise distinct such that𝑛 = 2^𝛼¹+ 2^𝛼²+ ⋯ + 2^𝛼^𝑚.

• Base case at𝑛 = 1.1 = 2⁰.

• Induction step.Assume that the statement holds for1, 2, … , 𝑛where𝑛 ≥ 1.

By Euclidean division,𝑛 + 1 = 2𝑞 + 𝑟where𝑞 ∈ ℕand𝑟 ∈ {0, 1}.

Note that𝑞 ≠ 0since otherwise1 < 𝑛 + 1 = 𝑟 ≤ 1.

Hence1 ≤ 𝑞 < 2𝑞 + 𝑟 = 𝑛 + 1.

Thus, by the induction hypothesis, 𝑞 = 2^𝛼¹ + 2^𝛼² + ⋯ + 2^𝛼^𝑚 where 𝛼₁ > 𝛼₂ > ⋯ > 𝛼_𝑚 are natural numbers.

Therefore𝑛 + 1 = 2𝑞 + 𝑟 = 2^𝛼¹⁺¹+ 2^𝛼²⁺¹+ ⋯ + 2^𝛼^𝑚⁺¹+ 𝑟2⁰.

Note that𝛼₁+ 1 > 𝛼₂+ 1 > ⋯ > 𝛼_𝑚+ 1 > 0. Hence the exponents are pairwise distinct (it is possible for2⁰to not appear if𝑟 = 0).

Which ends the induction step.

(3)

MAT246H1-S – LEC0201/9201 – J.-B. Campesato 3

Method 2:

We are going to prove by strong induction that for every𝑛 ≥ 1, there exist finitely many𝛼₁, … , 𝛼_𝑚 ∈ ℕ pairwise distinct such that𝑛 = 2^𝛼¹+ 2^𝛼²+ ⋯ + 2^𝛼^𝑚.

• Base case at𝑛 = 1.1 = 2⁰.

• Induction step.Assume that the statement holds for1, 2, … , 𝑛where𝑛 ≥ 1.

(i) First case: 𝑛 + 1is even. Then𝑛 + 1 = 2𝑘for some𝑘 ∈ ℕ.

Note that𝑘 ≠ 0since otherwise1 ≤ 𝑛 + 1 = 2𝑘 = 0.

Since𝑘 ≠ 0, we get that𝑘 < 2𝑘 = 𝑛 + 1, i.e. 𝑘 ≤ 𝑛.

Thus, by the induction hypothesis,𝑘 = 2^𝛼¹+ 2^𝛼²+ ⋯ + 2^𝛼^𝑚 where the𝛼₁, … , 𝛼_𝑚∈ ℕare pairwise distinct.

So𝑛 + 1 = 2 × (2^𝛼¹+ 2^𝛼²+ ⋯ + 2^𝛼^𝑚) = 2^𝛼¹⁺¹+ 2^𝛼²⁺¹+ ⋯ + 2^𝛼^𝑚⁺¹.

Assume by contradiction that there exist𝑖 ≠ 𝑗such that𝛼_𝑖+ 1 = 𝛼_𝑗+ 1. Then, by the cancellation rule,𝛼_𝑖= 𝛼_𝑗. Which is a contradiction since the𝛼_𝑖are pairwise distinct.

Therefore the𝛼₁+ 1, 𝛼₂+ 1, … , 𝛼_𝑚+ 1are pairwise distinct as requested.

(ii) Second case: 𝑛 + 1is odd. Then𝑛 + 1 = 2𝑘 + 1for some𝑘 ∈ ℕ.

Note that𝑘 ≠ 0since otherwise𝑛 + 1 = 2 × 0 + 1 = 1 ⟹ 𝑛 = 0. Hence, as above,𝑘 < 2𝑘 = 𝑛.

Thus, by the induction hypothesis,𝑘 = 2^𝛼¹+ 2^𝛼²+ ⋯ + 2^𝛼^𝑚 where the𝛼₁, … , 𝛼_𝑚∈ ℕare pairwise distinct.

Hence𝑛 + 1 = 1 + 2𝑘 = 2⁰+ 2 × (2^𝛼¹+ 2^𝛼²+ ⋯ + 2^𝛼^𝑚) = 2⁰+ 2^𝛼¹⁺¹+ 2^𝛼²⁺¹+ ⋯ + 2^𝛼^𝑚⁺¹.

As above, the 𝛼_𝑖 + 1 are pairwise distinct. Moreover 𝛼_𝑖 + 1 > 0. Therefore the 0, 𝛼₁ + 1, 𝛼₂ + 1, … , 𝛼_𝑚+ 1are pairwise distinct, as requested.

Which ends the induction step.

Let(𝑥, 𝑦) ∈ ℕ²be such that4𝑥(𝑥 + 1) = 𝑦(𝑦 + 1).

1. First case: assume that𝑦 ≤ 2𝑥.Then

𝑦(𝑦 + 1) ≤ 2𝑥(2𝑥 + 1) ≤ 2𝑥(2𝑥 + 2) = 4𝑥(𝑥 + 1) = 𝑦(𝑦 + 1) Hence2𝑥(2𝑥 + 1) ≤ 2𝑥(2𝑥 + 2)and2𝑥(2𝑥 + 2) = 𝑦(𝑦 + 1) ≤ 2𝑥(2𝑥 + 1).

Thus2𝑥(2𝑥 + 1) = 2𝑥(2𝑥 + 2), from which we get that𝑥(2𝑥 + 1) = 𝑥(2𝑥 + 2).

• Either𝑥 = 0and then𝑦 ≤ 0so𝑦 = 0.

• Or𝑥 ≠ 0and then, by cancellation, we get2𝑥 + 1 = 2𝑥 + 2.

We derive from the previous equality that1 = 2, which is impossible.

Thus the only possible solution in this case is(𝑥, 𝑦) = (0, 0).

2. Second case: assume that2𝑥 < 𝑦, i.e.2𝑥 + 1 ≤ 𝑦.Then

𝑦(𝑦 + 1) ≥ (2𝑥 + 1)(2𝑥 + 2) ≥ 2𝑥(2𝑥 + 2) = 𝑦(𝑦 + 1) Hence, as above,(2𝑥 + 1)(2𝑥 + 2) = 2𝑥(2𝑥 + 2).

Note that2𝑥 + 2 ≠ 0since2𝑥 + 2 ≥ 2 > 0.

So, by cancellation,2𝑥 = 2𝑥 + 1and hence0 = 1, which is impossible.

Therefore there is no solution(𝑥, 𝑦) ∈ ℕ²satisfying2𝑥 < 𝑦.

We proved that the only possible solution is(𝑥, 𝑦) = (0, 0).

We have to check that conversely it is a solution, which is the case since then4𝑥(𝑥 + 1) = 0 = 𝑦(𝑦 + 1).

So the only solution is(𝑥, 𝑦) = (0, 0).

(4)

4 Problem Set 1

Method 1 (using my hint):

We are going to prove the statement by induction on𝑛.

• Base case at𝑛 = 3.Note that1 = 1 2+ 1

3+ 1 6

• Induction step.Assume that the statement holds for some𝑛 ≥ 3.

By the induction hypothesis, there exist𝑥₁< … < 𝑥_𝑛inℕ ⧵ {0}such that1 = 1 𝑥₁ + 1

𝑥₂ + ⋯ + 1 𝑥_𝑛. Note that𝑥₁ ≠ 1since otherwise 1

𝑥₁ + 1

𝑥₂ + ⋯ + 1

𝑥_𝑛 = 1 + 1

𝑥₂ + ⋯ + 1

𝑥_𝑛 > 1. Thus𝑥₁ > 1.

Hence1 = 1 2 +1

2 = 1 2 + 1

2 ( 1 𝑥₁ + 1

𝑥₂ + ⋯ + 1 𝑥_𝑛)= 1

2 + 1 2𝑥₁ + 1

2𝑥₂ + ⋯ + 1 2𝑥_𝑛. Besides, since1 < 𝑥₁< 𝑥₂ < ⋯ < 𝑥_𝑛, we get that2 < 2𝑥₁ < 2𝑥₂< ⋯ < 2𝑥_𝑛. So the𝑛 + 1denominators are pairwise distinct.

Method 2:

We are going to prove the following stronger statement by induction on𝑛: for𝑛 ≥ 3, there exist1 < 𝑥₁ <

𝑥₂< ⋯ < 𝑥_𝑛such that1 = ¹

𝑥₁ + ¹

𝑥₂ + ⋯ + ¹

𝑥_𝑛 and𝑥_𝑛is even.

3+ 1 6

By the induction hypothesis, there exist1 < 𝑥₁ < … < 𝑥_𝑛inℕsuch that1 = 1 𝑥₁ + 1

𝑥₂ + ⋯ + 1

𝑥_𝑛 and𝑥_𝑛 is even.

Hence𝑥_𝑛= 2𝑘for some𝑘 ∈ ℕ ⧵ {0}.

Note that 1 𝑥_𝑛 = 1

2𝑘 = 1 3𝑘+ 1

6𝑘. Hence

1 = 1 𝑥₁ + 1

𝑥₂ + ⋯ + 1 𝑥_𝑛−1+ 1

3𝑘 + 1 6𝑘 Besides6𝑘is even and1 < 𝑥₁< 𝑥₂ < ⋯ < 𝑥_𝑛= 2𝑘 < 3𝑘 < 6𝑘.

So the𝑛 + 1denominators are pairwise distinct.

Method 3:

We are going to prove the statement by induction on𝑛.

3+ 1 6

By the induction hypothesis, there exist𝑥₁< … < 𝑥_𝑛inℕ ⧵ {0}such that1 = 1 𝑥₁ + 1

𝑥₂ + ⋯ + 1 𝑥_𝑛. Note that𝑥₁ ≠ 1since otherwise 1

𝑥₁ + 1

𝑥₂ + ⋯ + 1

𝑥_𝑛 = 1 + 1

𝑥₂ + ⋯ + 1

𝑥_𝑛 > 1. Thus𝑥₁ > 1.

Note that for𝑥 ≠ 0, 1

𝑥(𝑥 + 1)+ 1

𝑥 + 1 = 𝑥 + 1 𝑥(𝑥 + 1) = 1

𝑥. Therefore

1 = 1 𝑥₁ + 1

𝑥₂ + ⋯ + 1 𝑥_𝑛−1 + 1

𝑥_𝑛 = 1 𝑥₁ + 1

𝑥₂ + ⋯ + 1

𝑥_𝑛−1+ 1

𝑥_𝑛+ 1+ 1 𝑥_𝑛(𝑥_𝑛+ 1) Since1 < 𝑥_𝑛and0 < 𝑥_𝑛+ 1, we get𝑥_𝑛+ 1 < 𝑥_𝑛(𝑥_𝑛+ 1).

Therefore1 < 𝑥₁ < 𝑥₂< ⋯ < 𝑥_𝑛−1< 𝑥_𝑛< 𝑥_𝑛+ 1 < 𝑥_𝑛(𝑥_𝑛+ 1).

So the𝑛 + 1denominators are pairwise distinct.