Problem Set n°5

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University of Toronto – MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

Problem Set n°5

Jean-Baptiste Campesato Due on April 12^th, 2021

Write your solutions concisely but without skipping important steps.

Make sure that your submission is readable on Crowdmark.

Exercise 1.

Prove that∀𝑥 ∈ ℝ ⧵ ℚ, ∀𝑎, 𝑏, 𝑐, 𝑑 ∈ ℚ, 𝑎𝑑 − 𝑏𝑐 ≠ 0 ⟹ ^{𝑎𝑥+𝑏}

𝑐𝑥+𝑑 ∉ ℚ.

Remark: note that𝑐𝑥 + 𝑑 ≠ 0under the given assumptions.

Either𝑐 = 0but then𝑐𝑥 + 𝑑 = 𝑑 ≠ 0since𝑎𝑑 − 𝑏𝑐 ≠ 0. Or𝑐 ≠ 0but then𝑐𝑥 ∈ ℝ ⧵ ℚand−𝑑 ∈ ℚthus𝑐𝑥 + 𝑑 ≠ 0.

Exercise 2.

Let𝐸be a finite set. Express

|{(𝐴, 𝐵) ∈ 𝒫(𝐸) × 𝒫(𝐸) ∶ 𝐴 ∪ 𝐵 = 𝐸}|

in terms of|𝐸|.

Hint: you may start studying the case where the cardinality of𝐴is fixed.

Exercise 3.

The following questions are independent.

1. Does it exist a set𝐸such that|𝒫(𝐸)| = ℵ₀? 2. Prove that|[0, 1]| = |(0, 1)|.

Exercise 4.

We set

𝑆 = {𝑥 ∈ ℝ ∶ ∃𝑛 ∈ ℕ, ∃𝑎0, 𝑎₁, … , 𝑎_𝑛∈ ℤ, 𝑎_𝑛≠ 0and𝑎_𝑛𝑥^𝑛+ 𝑎_𝑛−1𝑥^𝑛−1+ ⋯ + 𝑎₁𝑥 + 𝑎₀= 0}

What is|𝑆|?

Remark: you can use basic facts concerning polynomials and their roots.

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2 Problem Set 5

Sample solution to Exercise 1.

Let𝑥 ∈ ℝ ⧵ ℚ. Let𝑎, 𝑏, 𝑐, 𝑑 ∈ ℚbe such that𝑎𝑑 − 𝑏𝑐 ≠ 0.

Assume by contradiction that𝑞 ≔ ^{𝑎𝑥+𝑏}

𝑐𝑥+𝑑 ∈ ℚ, then 𝑎𝑥 + 𝑏

𝑐𝑥 + 𝑑 = 𝑞 ⇔ 𝑥(𝑎 − 𝑞𝑐) = 𝑞𝑑 − 𝑏

• First case: if𝑎 ≠ 𝑞𝑐then𝑥 = ^{𝑞𝑑−𝑏}

𝑎−𝑞𝑐 ∈ ℚ, so there is a contradiction.

• Second case: if𝑎 = 𝑞𝑐then𝑞𝑑 − 𝑏 = 𝑥(𝑎 − 𝑞𝑐) = 0, i.e.𝑏 = 𝑞𝑑.

Therefore𝑎𝑑 − 𝑏𝑐 = 𝑞𝑐𝑑 − 𝑞𝑑𝑐 = 0, so there is a contradiction.

Set𝑛 ≔ |𝐸|.

Let𝑖 = 0, … , 𝑛. SetΩ_𝑖≔ {(𝐴, 𝐵) ∈ 𝒫(𝐸) × 𝒫(𝐸) ∶ 𝐴 ∪ 𝐵 = 𝐸 and|𝐴| = 𝑖}.

There are(^𝑛𝑖)subsets𝐴 ∈ 𝒫(𝐸)such that|𝐴| = 𝑖(See Q10E09).

For a given𝐴as above, in order to have𝐴 ∪ 𝐵 = 𝐸,𝐵must be of the form𝐵 = 𝐴^𝑐⊔ 𝐶where𝐶 ⊂ 𝐴.

There are2^𝑖= |𝒫(𝐴)|choices for such a subset𝐶, and hence for𝐵(See Q10E10).

Therefore|Ω_𝑖| = (^𝑛𝑖)2^𝑖. Finally

|{(𝐴, 𝐵) ∈ 𝒫(𝐸) × 𝒫(𝐸) ∶ 𝐴 ∪ 𝐵 = 𝐸}| =

|

𝑛

⨆𝑖=0

Ω_𝑖

|=

𝑛

∑𝑖=0

|Ω_𝑖|

=

𝑛

∑𝑖=0(𝑛 𝑖)2^𝑖 =

𝑛

∑𝑖=0(𝑛

𝑖)2^𝑖1^𝑛−𝑖= (2 + 1)^𝑛= 3^𝑛 = 3^|𝐸|

1. Let𝐸be a set, then:

• Either𝐸is finite and then𝒫(𝐸)is finite too by Q10E10, so that|𝒫(𝐸)| < ℵ₀.

• Or𝐸is infinite and thenℵ₀≤ |𝐸| < |𝑃 (𝐸)|by Cantor’s theorem.

In both cases|𝒫(𝐸)| ≠ ℵ₀, so there is no set𝐸such that|𝒫(𝐸)| = ℵ₀. 2. Method 1.

Note that(0, 1) ⊂ [0, 1], therefore|(0, 1)| ≤ |[0, 1]|.

Define𝑓 ∶ [0, 1] → (0, 1)by𝑓 (𝑥) = ^𝑥+1

3 .

Note that𝑓 is well-defined since if0 ≤ 𝑥 ≤ 1then0 < ¹

3 ≤ ^𝑥+1

3 ≤ ²

3 < 1.

Besides𝑓 is injective since if𝑥, 𝑦 ∈ [0, 1]satisfy𝑓 (𝑥) = 𝑓 (𝑦), then^𝑥+1

3 = ^𝑦+1

3 which implies𝑥 = 𝑦.

Therefore|[0, 1]| ≤ |(0, 1)|.

By Cantor–Schröder–Bernstein theorem, we conclude that|[0, 1]| = |(0, 1)|.

Method 2.

We know that(0, 1) ⊂ [0, 1] ⊂ ℝand that|(0, 1)| = |ℝ|(see Q11E08).

Therefore|[0, 1]| = |(0, 1)|(see Q11E01).

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

Comment: a complex number that is the root of a non-zero polynomial with integers (or rational, it is equivalent) coeﬀicients is said to be analgebraic number. Complex numbers which are not roots of such polynomials are called transcendental numbers.

The field of algebraic real numbers is quite often denoted by

ℝ_alg≔ {𝑥 ∈ ℝ ∶ ∃𝑛 ∈ ℕ, ∃𝑎0, 𝑎₁, … , 𝑎_𝑛 ∈ ℤ, 𝑎_𝑛≠ 0and𝑎_𝑛𝑥^𝑛+ 𝑎_𝑛−1𝑥^𝑛−1+ ⋯ + 𝑎₁𝑥 + 𝑎₀ = 0}

The goal of this exercise was to prove that|ℝ_alg| = ℵ₀, i.e. there are infinitely countably many algebraic real numbers, so thatalmost allreal numbers are transcendental (but it is usually quite diﬀicult to prove that a number is tran- scendental: we still don’t know whether𝜋 + 𝑒or𝜋𝑒are transcendental or not). This was first proved by Cantor in is famous articleUeber eine Eigenschaft des Inbegriffes aller reellen algebraischen Zahlenpublished 1874.

Claim.∀𝑛 ∈ ℕ ⧵ {0}, |ℤ^𝑛| = ℵ₀. Proof by induction on𝑛 ≥ 1.

Base case at𝑛 = 1: |ℤ¹| = |ℤ| = ℵ₀(from the lecture notes).

Induction step.Assume that|ℤ^𝑛| = ℵ₀for some𝑛 ≥ 1.

Since|ℤ^𝑛| = |ℕ|and|ℤ| = |ℕ|, we have|ℤ^𝑛+1| = |ℤ^𝑛× ℤ| = |ℕ × ℕ| = ℵ₀.

Method 1.

For𝑛 ∈ ℕand𝑎₀, 𝑎₁, … , 𝑎_𝑛∈ ℤwith𝑎_𝑛≠ 0, the set

{𝑥 ∈ ℝ ∶ 𝑎_𝑛𝑥^𝑛+ 𝑎_𝑛−1𝑥^𝑛−1+ ⋯ + 𝑎₁𝑥 + 𝑎₀ = 0}

is finite since a polynomial of degree𝑛has at most𝑛roots.

For𝑛 ∈ ℕ, we set 𝐴_𝑛 =

(𝑎₀,𝑎₁,…,𝑎_𝑛⋃)∈ℤ^𝑛×(ℤ⧵{0})

{𝑥 ∈ ℝ ∶ 𝑎_𝑛𝑥^𝑛+ 𝑎_𝑛−1𝑥^𝑛−1+ ⋯ + 𝑎₁𝑥 + 𝑎₀ = 0}

Since|ℤ^𝑛| = |ℕ|and|ℤ ⧵ {0}| = |ℕ|, we have that|ℤ^𝑛× (ℤ ⧵ {0})| = |ℕ × ℕ| = ℵ₀. Therefore𝐴_𝑛is countable as a countable union of finite sets.

Hence𝑆 = ⋃_𝑛∈ℕ𝐴_𝑛is countable as a countable union of countable sets.

Note thatℤ ⊂ 𝑆, since𝑚 ∈ ℤis a root of𝑥 − 𝑚 = 0.

Therefore𝑆is countably infinite, i.e.|𝑆| = ℵ₀.

Method 2.

For𝑛 ∈ ℕ, we denote by𝑃_𝑛the set of polynomials of degree𝑛with integer coeﬀicients.

Note that|𝑃_𝑛| = |ℤ^𝑛× ℤ ⧵ {0}|since a polynomial𝑎_𝑛𝑥^𝑛+ 𝑎_𝑛−1𝑥^𝑛−1+ ⋯ + 𝑎₁𝑥 + 𝑎₀of degree𝑛is characterized by its coeﬀicients(𝑎₀, 𝑎₁, … , 𝑎_𝑛) ∈ ℤ^𝑛× (ℤ ⧵ {0}).

Since|ℤ^𝑛| = |ℕ|and|ℤ ⧵ {0}| = |ℕ|, we have that|𝑃_𝑛| = |ℤ^𝑛× (ℤ ⧵ {0})| = |ℕ × ℕ| = ℵ₀. Hence the set𝑃 =⋃

𝑛∈ℕ

𝑃_𝑛of non-zero polynomials with integer coeﬀicients is countable as a countable union of countable sets.

Given𝑓 ∈ 𝑃,𝑓⁻¹({0}) = {𝑥 ∈ ℝ ∶ 𝑓 (𝑥) = 0}is finite since a polynomial of degree𝑛has at most𝑛roots.

Therefore𝑆 = ⋃

𝑓 ∈𝑃

𝑓⁻¹(0)is countable as a countable union of finite sets.

Note thatℤ ⊂ 𝑆, since𝑚 ∈ ℤis a root of𝑥 − 𝑚 = 0.

Therefore𝑆is countably infinite, i.e.|𝑆| = ℵ₀.