March9 ,2021 RATIONALNUMBERS ConceptsinAbstractMathematics

MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

RATIONAL NUMBERS

March 9^th, 2021

Introduction

We want to enlargeℤwith numbers of the form ^𝑝_𝑞, where𝑝 ∈ ℤand𝑞 ∈ ℤ ⧵ {0}satisfying the property that ^𝑝_𝑞 × 𝑞 = 𝑝.

Such numbers are called rational numbers and they form the setℚ.

We also want to extend our operations+, ×and the order≤fromℤtoℚ.

Set𝑥 =¹⁰₁₄ and𝑦 = ⁵₇. Then14𝑥 = 10.

So, assuming the multiplication onℚhas a cancellation rule, we get7𝑥 = 5, i.e. 𝑥 = ⁵₇= 𝑦.

Therefore, to have a coherent construction, we need to identify several fractions which are not equala priori.

The usual formal way to achieve this point relies onequivalence classes, which is the point of view adopted in the lecture notes. I will adopt a more naive/less formal approach in the slides.

Definition

Definition

We define

ℚ ={ 𝑝

𝑞 ∶ 𝑝 ∈ ℤ, 𝑞 ∈ ℤ ⧵ {0}

} with the relation

∀𝑝, 𝑟 ∈ ℤ, ∀𝑞, 𝑠 ∈ ℤ ⧵ {0}, 𝑝 𝑞 = 𝑟

𝑠 ⇔ 𝑝𝑠 = 𝑞𝑟

where the operation in the RHS takes place inℤ(there is no cycling definition).

Example

10

14 = ⁵₇ since10 × 7 = 5 × 14 = 70.

A few notations

Notation

Note that the function ℤ → ℚ

𝑛 ↦ ^𝑛₁ is injective.

Therefore we may seeℤas a subset ofℚby setting𝑛 ≔ 𝑛

1 ∈ ℚfor𝑛 ∈ ℤ.

Notation

Note that for(𝑎, 𝑏) ∈ ℤ × ℤ ⧵ {0}, we have −𝑎 𝑏 = 𝑎

−𝑏. Hence we set−𝑎 𝑏 ≔ −𝑎

𝑏 = 𝑎

−𝑏.

Notation

Note that 𝑎

𝑏 = 0 ⇔ 𝑎 = 0and that if𝑎 𝑏 = 𝑎^′

𝑏^′ ≠ 0then 𝑏 𝑎 = 𝑏^′

𝑎^′. Hence, if𝑥 = 𝑎

𝑏 ≠ 0, we set𝑥⁻¹≔ 𝑏

𝑎 which doesn’t depend on the representative of𝑥.

Lowest form representation

Proposition

Given𝑥 ∈ ℚ, there exists a unique couple(𝑎, 𝑏) ∈ ℤ × ℕ ⧵ {0}such that𝑥 = 𝑎

𝑏 andgcd(𝑎, 𝑏) = 1.

Then we say that𝑥 =𝑎

𝑏 is written inlowest form.

Proof.Let𝑥 ∈ ℚ.

Existence. There exist𝛼 ∈ ℤand𝛽 ∈ ℤ ⧵ {0}such that𝑥 = ^𝛼_𝛽.

Write𝑑 =gcd(𝛼, 𝛽), then there exist𝑎 ∈ ℤand𝑏 ∈ ℤ ⧵ {0}such that𝛼 = 𝑑𝑎and𝛽 = 𝑑𝑏.

Note thatgcd(𝑎, 𝑏) = 1(since𝑑 =gcd(𝛼, 𝛽) =gcd(𝑑𝑎, 𝑑𝑏) = 𝑑gcd(𝑎, 𝑏)) and that ^𝛼

𝛽 = ^{sign(𝑏)𝑎}_|𝑏|

sincesign(𝑏)𝑎𝛽 =sign(𝑏)𝑎𝑑𝑏 = |𝑏|𝑑𝑎 = |𝑏|𝛼.

Uniqueness. Assume that ^𝑎_𝑏 = ^𝑎_𝑏^′_′ where𝑎, 𝑎^′∈ ℤ,𝑏, 𝑏^′∈ ℕ ⧵ {0},gcd(𝑎, 𝑏) = 1,gcd(𝑎^′, 𝑏^′) = 1.

Then𝑎𝑏^′= 𝑎^′𝑏. By Gauss’ lemma, since𝑏|𝑎𝑏^′andgcd(𝑎, 𝑏) = 1, we get that𝑏|𝑏^′. Similarly𝑏^′|𝑏, so|𝑏| = |𝑏^′|, thus𝑏 = 𝑏^′.

Then, using the cancellation rule,𝑎𝑏^′= 𝑎^′𝑏gives𝑎 = 𝑎^′since𝑏 = 𝑏^′≠ 0. ■

Addition

Proposition

The addition+ ∶ ℚ × ℚ → ℚ

(^𝑎𝑏,_𝑑^𝑐) ↦ ^{𝑎𝑑+𝑏𝑐}_𝑏𝑑 is well-defined.

Proof.We need to prove that the addition doesn’t depend on the choice of the representatives, i.e. that if ^𝑎_𝑏 = ^𝑎_𝑏′^′ and ^𝑐_𝑑 = _𝑑^𝑐′_′ then ^𝑎_𝑏 +_𝑑^𝑐 = ^𝑎_𝑏′^′ +_𝑑^𝑐′_′.

Assume that ^𝑎

𝑏 = ^𝑎_𝑏′^′ and ^𝑐

𝑑 = _𝑑^𝑐′_′, i.e. 𝑎𝑏^′= 𝑏𝑎^′and𝑐𝑑^′= 𝑑𝑐^′.

Therefore(𝑎𝑑 + 𝑏𝑐)(𝑏^′𝑑^′) = 𝑎𝑑𝑏^′𝑑^′+ 𝑏𝑐𝑏^′𝑑^′= 𝑏𝑎^′𝑑𝑑^′+ 𝑑𝑐^′𝑏𝑏^′= (𝑎^′𝑑^′+ 𝑏^′𝑐^′)(𝑏𝑑), i.e. ^{𝑎𝑑+𝑏𝑐}

𝑏𝑑 = ^𝑎′𝑑_𝑏^′_′^+𝑏_𝑑′^′𝑐′. ■

Remark

Note that the addition defined onℚis compatible with the one onℤ.

Indeed, if𝑚, 𝑛 ∈ ℤthen ^𝑚

1 +^𝑛₁ = ^𝑚+𝑛₁ .

Multiplication

Proposition

The multiplication× ∶ ℚ × ℚ → ℚ

(^𝑎𝑏,^𝑐_𝑑) ↦ ^𝑎𝑐_𝑏𝑑 is well-defined.

Proof.We need to prove that the multiplication doesn’t depend on the choice of the representatives, i.e. that if ^𝑎_𝑏 = ^𝑎_𝑏^′_′ and_𝑑^𝑐 = _𝑑^𝑐′_′ then ^𝑎_𝑏 ×_𝑑^𝑐 = ^𝑎_𝑏^′_′ ×_𝑑^𝑐′_′.

Assume that ^𝑎

𝑏 = ^𝑎_𝑏′^′ and ^𝑐

𝑑 = _𝑑^𝑐′_′, i.e. 𝑎𝑏^′= 𝑏𝑎^′and𝑐𝑑^′= 𝑑𝑐^′.

Therefore(𝑎𝑐)(𝑏^′𝑑^′) = 𝑎𝑏^′𝑐𝑑^′= 𝑏𝑎^′𝑑𝑐^′= (𝑎^′𝑐^′)(𝑏𝑑), i.e. _𝑏𝑑^𝑎𝑐 = _𝑏^𝑎_′^′𝑐_𝑑^′_′ as desired. ■

Remark

Note that the multiplication defined onℚis compatible with the one onℤ.

Indeed, if𝑚, 𝑛 ∈ ℤthen ^𝑚₁ ×^𝑛₁ = ^𝑚×𝑛₁ .

Order

We want to define an order onℚ, satisfying the following natural properties:

• ^𝑎

𝑏 ≤ ^𝑐

𝑑 if and only if0 ≤ ^𝑐

𝑑−^𝑎

𝑏 = ^{𝑏𝑐−𝑎𝑑}

𝑏𝑑 , and,

• 0 ≤ ^𝑒

𝑓 if and only if0 ≤ 𝑒𝑓 (i.e. the sign rule).

Therefore the following definition is natural:

Definition

We define the binary relation≤onℚby 𝑎 𝑏 ≤ 𝑐

𝑑 ⇔ 0 ≤ (𝑏𝑐 − 𝑎𝑑)𝑏𝑑 where the order on the RHS of the equivalence is the order ofℤ.

We have to check that it doesn’t depend on the choice of the representative!

Proposition

The order≤defined onℚextends the usual order≤onℤ.

Proof.Let𝑚, 𝑛 ∈ ℤthen 𝑚 1 ≤ 𝑛

1⇔ 0 ≤ 𝑛 − 𝑚 ⇔ 𝑚 ≤ 𝑛. ■

Properties

(ℚ, +, ×, ≤)is a (totally) ordered field, meaning that

• +is associative:∀𝑥, 𝑦, 𝑧 ∈ ℚ, (𝑥 + 𝑦) + 𝑧 = 𝑥 + (𝑦 + 𝑧)

• 0is the unit of+:∀𝑥 ∈ ℚ, 𝑥 + 0 = 0 + 𝑥 = 𝑥

• −𝑥is the additive inverse of𝑥:∀𝑥 ∈ ℚ, 𝑥 + (−𝑥) = (−𝑥) + 𝑥 = 0

• +is commutative:∀𝑥, 𝑦 ∈ ℚ, 𝑥 + 𝑦 = 𝑦 + 𝑥

• ×is associative:∀𝑥, 𝑦, 𝑧 ∈ ℚ, (𝑥𝑦)𝑧 = 𝑥(𝑦𝑧)

• ×is distributive with respect to+: ∀𝑥, 𝑦, 𝑧 ∈ ℚ, 𝑥(𝑦 + 𝑧) = 𝑥𝑦 + 𝑥𝑧and(𝑥 + 𝑦)𝑧 = 𝑥𝑧 + 𝑦𝑧

• 1is the unit of×: ∀𝑥 ∈ ℚ, 1 × 𝑥 = 𝑥 × 1 = 𝑥

• If𝑥 ≠ 0then𝑥⁻¹ is the multiplicative inverse of𝑥: ∀𝑥 ∈ ℚ ⧵ {0}, 𝑥𝑥⁻¹= 𝑥⁻¹𝑥 = 1

• ×is commutative:∀𝑥, 𝑦 ∈ ℚ, 𝑥𝑦 = 𝑦𝑥

• ≤is reflexive:∀𝑥 ∈ ℚ, 𝑥 ≤ 𝑥

• ≤is antisymmetric:∀𝑥, 𝑦 ∈ ℚ, (𝑥 ≤ 𝑦and𝑦 ≤ 𝑥) ⟹ 𝑥 = 𝑦

• ≤is transitive:∀𝑥, 𝑦, 𝑧 ∈ ℚ, (𝑥 ≤ 𝑦and𝑦 ≤ 𝑧) ⟹ 𝑥 ≤ 𝑧

• ≤is total:∀𝑥, 𝑦 ∈ ℚ, 𝑥 ≤ 𝑦or𝑦 ≤ 𝑥

• ∀𝑥, 𝑦, 𝑟, 𝑠 ∈ ℚ, (𝑥 ≤ 𝑦and𝑟 ≤ 𝑠) ⇒ 𝑥 + 𝑟 ≤ 𝑦 + 𝑠

• ∀𝑥, 𝑦, 𝑧 ∈ ℚ, (𝑥 ≤ 𝑦and𝑧 > 0) ⇒ 𝑥𝑧 ≤ 𝑦𝑧

Between two rationals there is another rational

Proposition

∀𝑥, 𝑦 ∈ ℚ, 𝑥 < 𝑦 ⟹ (∃𝑧 ∈ ℚ, 𝑥 < 𝑧 < 𝑦)

Proof.Let𝑥, 𝑦 ∈ ℚbe such that𝑥 < 𝑦. Then𝑧 = ^𝑥+𝑦₂ is a suitable choice.

Indeed,𝑥 < 𝑦implies2𝑥 < 𝑥 + 𝑦and thus𝑥 < 𝑧. Similarly𝑥 + 𝑦 < 2𝑦, and thus𝑧 < 𝑦. ■

Remark

Note thatℚis not well-ordered.

Indeedℚ_>0= {𝑥 ∈ ℚ ∶ 𝑥 > 0}is non-empty (and even bounded from below) but it has no least element.

ℚ is archimedean

Theorem: ℚ is archimedean

∀𝜀 ∈ ℚ_>0, ∀𝐴 ∈ ℚ_>0, ∃𝑁 ∈ ℕ, 𝑁𝜀 > 𝐴.

Proof.

Since ^𝐴

𝜀 > 0, we may find a representative ^𝑎

𝑏 = ^𝐴_𝜀 where𝑎, 𝑏 ∈ ℕ ⧵ {0}.

Then𝑎 + 1 − ^𝐴_𝜀 = 𝑎 + 1 −^𝑎_𝑏 = ^{𝑎(𝑏−1)+𝑏}_𝑏 > 0, thus(𝑎 + 1)𝜀 > 𝐴.

So𝑁 = 𝑎 + 1is a suitable choice. ■

Remark

The above theorem means that lim

𝑛→+∞

1

𝑛 = 0, or equivalently thatℚdoesn’t contain infinitesimal elements (i.e. there is not infinitely large or infinitely small elements).

The rational root theorem

The rational root theorem

Let𝑓 (𝑥) = 𝑎_𝑛𝑥^𝑛+ 𝑎_𝑛−1𝑥^𝑛−1+ ⋯ + 𝑎₁𝑥 + 𝑎₀be a polynomial with integer coefficients𝑎_𝑘∈ ℤ.

If𝑥 =^𝑝_𝑞 is a rational root of𝑓 written in lowest terms (i.e.gcd(𝑝, 𝑞) = 1), then𝑝|𝑎₀and𝑞|𝑎_𝑛.

Proof.By assumption we have that

𝑎_𝑛( 𝑝 𝑞 )

𝑛

+ 𝑎_𝑛−1( 𝑝 𝑞 )

𝑛−1

+ ⋯ + 𝑎₁𝑝

𝑞+ 𝑎₀= 0 Therefore

𝑎_𝑛𝑝^𝑛+ 𝑎_𝑛−1𝑝^𝑛−1𝑞 + ⋯ + 𝑎₁𝑝𝑞^𝑛−1+ 𝑎₀𝑞^𝑛= 0 Thus𝑝|𝑎₀𝑞^𝑛. Sincegcd(𝑝, 𝑞) = 1, by Gauss’ lemma we obtain that𝑝|𝑎₀.

Similarly𝑞|𝑎_𝑛. ■