The addition – 1

MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

NATURAL NUMBERS – 2

January 14^th, 2021

From last Tuesday: Peano’s axioms

Theorem: Peano axioms

There exists a setℕtogether with an element0 ∈ ℕ”zero” and a function𝑠 ∶ ℕ → ℕ”successor” such that:

1 0is not the successor of any element ofℕ, i.e.0is not in the image of𝑠:

0 ∉ 𝑠(ℕ)

2 If the successor of𝑛equals the successor of𝑚then𝑛 = 𝑚, i.e. 𝑠is injective:

∀𝑛, 𝑚 ∈ ℕ, 𝑠(𝑛) = 𝑠(𝑚) ⟹ 𝑛 = 𝑚

3 The induction principle.If a subset ofℕcontains0and is closed under𝑠then it isℕ:

∀𝐴 ⊂ ℕ,{ 0 ∈ 𝐴

𝑠(𝐴) ⊂ 𝐴 ⟹ 𝐴 = ℕ

ℕ 0 𝑠 1 𝑠 2 𝑠 3 𝑠 4

The addition – 1

We are going to define ℕ → ℕ

𝑏 ↦ 𝑎 + 𝑏 for a given𝑎 ∈ ℕ.

How to do so? What would be a good definition to obtain what you intuitively know about+?

The idea is to define it inductively using the following properties we would like to have:

• 𝑎 + 0 = 𝑎

• For𝑏 ∈ ℕ,𝑎 + (𝑏 + 1) = (𝑎 + 𝑏) + 1

So if𝑎 + 𝑏is already defined, then we can define𝑎 + (𝑏 + 1).

Remember that intuitively+1is ”taking the successor”.

Formally, we prove:

Proposition

Let𝑎 ∈ ℕ. Then there exists a unique function(𝑎 + •) ∶ ℕ → ℕ

𝑏 ↦ 𝑎 + 𝑏 such that

1 𝑎 + 0 = 𝑎 2 ∀𝑏 ∈ ℕ, 𝑎 + 𝑠(𝑏) = 𝑠(𝑎 + 𝑏) The above result is a consequence of the induction principle.

The addition – 2

Remark

Set1 ≔ 𝑠(0). Then, as expected, for𝑛 ∈ ℕ, we have

𝑛 + 1 = 𝑛 + 𝑠(0) = 𝑠(𝑛 + 0) = 𝑠(𝑛) Hence, from now on, I will use indistinctively𝑛 + 1or𝑠(𝑛).

We can prove the following properties.

Proposition

• ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐(the addition is associative)

• ∀𝑎, 𝑏 ∈ ℕ, 𝑎 + 𝑏 = 𝑏 + 𝑎(the addition is commutative)

• ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 + 𝑏 = 𝑎 + 𝑐 ⟹ 𝑏 = 𝑐(cancellation)

• ∀𝑎, 𝑏 ∈ ℕ, 𝑎 + 𝑏 = 0 ⟹ 𝑎 = 𝑏 = 0

The addition – 3

Proof that∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐.

Let𝑎, 𝑏 ∈ ℕ. Set𝐴 = {𝑐 ∈ ℕ ∶ 𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐}. Then

• 𝐴 ⊂ ℕ

• 0 ∈ 𝐴. Indeed,𝑎 + (𝑏 + 0) = 𝑎 + 𝑏 = (𝑎 + 𝑏) + 0.

• 𝑠(𝐴) ⊂ 𝐴. Indeed, let𝑛 ∈ 𝑠(𝐴)then𝑛 = 𝑠(𝑐)for some𝑐 ∈ 𝐴. Therefore

𝑎 + (𝑏 + 𝑛) = 𝑎 + (𝑏 + 𝑠(𝑐)) = 𝑎 + 𝑠(𝑏 + 𝑐) = 𝑠(𝑎 + (𝑏 + 𝑐)) = 𝑠((𝑎 + 𝑏) + 𝑐) = (𝑎 + 𝑏) + 𝑠(𝑐) = (𝑎 + 𝑏) + 𝑛.

Hence𝑛 ∈ 𝐴.

Thus, by the induction principle,𝐴 = ℕand for any𝑐 ∈ ℕ,𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐. ■ Proof that∀𝑎, 𝑏 ∈ ℕ, 𝑎 + 𝑏 = 0 ⟹ 𝑎 = 𝑏 = 0.

Let𝑎, 𝑏 ∈ ℕbe such that𝑎 + 𝑏 = 0. Assume by contradiction that𝑎 ≠ 0or𝑏 ≠ 0.

Without lost of generality, we may assume that𝑏 ≠ 0(using commutativity).

Then𝑏 = 𝑠(𝑛)for some𝑛 ∈ ℕ.

So0 = 𝑎 + 𝑏 = 𝑎 + 𝑠(𝑛) = 𝑠(𝑎 + 𝑛).

Which is a contradiction since0 ∉ 𝑠(ℕ). ■

The multiplication – 1

We define inductively ℕ → ℕ

𝑏 ↦ 𝑎 × 𝑏 for a given𝑎 ∈ ℕusing the following desired properties:

• 𝑎 × 0 = 0

• For𝑏 ∈ ℕ,𝑎 × (𝑏 + 1) = (𝑎 × 𝑏) + 𝑎

So if𝑎 × 𝑏is already defined, then we can define𝑎 × (𝑏 + 1).

Formally, we prove:

Proposition

Let𝑎 ∈ ℕ. Then there exists a unique function(𝑎 × •) ∶ ℕ → ℕ

𝑏 ↦ 𝑎 × 𝑏 such that

1 𝑎 × 0 = 0 ² ∀𝑏 ∈ ℕ, 𝑎 × 𝑠(𝑏) = (𝑎 × 𝑏) + 𝑎

Remark

It is common to simply write𝑎𝑏for𝑎 × 𝑏when there is no possible confusion.

The multiplication – 2

Proposition

• ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 × (𝑏 × 𝑐) = (𝑎 × 𝑏) × 𝑐 (the multiplication is associative)

• ∀𝑎, 𝑏 ∈ ℕ, 𝑎 × 𝑏 = 𝑏 × 𝑎(the multiplication is commutative)

• ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 × (𝑏 + 𝑐) = 𝑎 × 𝑏 + 𝑎 × 𝑐and(𝑎 + 𝑏) × 𝑐 = 𝑎 × 𝑐 + 𝑏 × 𝑐(×is distributive over+)

• ∀𝑎 ∈ ℕ, 𝑎 × 1 = 𝑎

• ∀𝑎, 𝑏 ∈ ℕ, 𝑎 × 𝑏 = 0 ⟹ (𝑎 = 0or𝑏 = 0)

• ∀𝑎, 𝑏, 𝑐 ∈ ℕ,{

𝑎 × 𝑏 = 𝑎 × 𝑐

𝑎 ≠ 0 ⟹ 𝑏 = 𝑐(cancellation)

Order – 1

Definition: binary relation

Abinary relationℛon a set𝐸consists in associating a truth value to every couple(𝑥, 𝑦) ∈ 𝐸². We say that𝑥is related to𝑦byℛ, denoted𝑥ℛ𝑦, if the valuetrueis assigned to(𝑥, 𝑦).

Examples

1 Let𝐸 = {𝑎, 𝑏, 𝑐}. We can define a binary relationℛusing a truth table as below:

𝑦

𝑥 𝑎 𝑏 𝑐

𝑎 3 7 7

𝑏 7 7 3

𝑐 3 3 7

Here𝑎ℛ𝑎,𝑎ℛ𝑐,𝑏ℛ𝑐and𝑐ℛ𝑏.

2 For𝐸 = ℝ, we can define a binary relation as follows:∀(𝑥, 𝑦) ∈ ℝ², 𝑥ℛ𝑦 ⇔ 𝑥²− 𝑦²= 𝑥 − 𝑦.

Order – 2

Definition: order

We say that a binary relationℛon a set𝐸is anorderif

1 ∀𝑥 ∈ 𝐸, 𝑥ℛ𝑥 (reflexivity)

2 ∀𝑥, 𝑦 ∈ 𝐸, (𝑥ℛ𝑦and𝑦ℛ𝑥) ⟹ 𝑥 = 𝑦 (antisymmetry)

3 ∀𝑥, 𝑦, 𝑧 ∈ 𝐸, (𝑥ℛ𝑦and𝑦ℛ𝑧) ⟹ 𝑥ℛ𝑧 (transitivity)

We say that the orderℛistotalif additionaly∀𝑥, 𝑦 ∈ 𝐸, 𝑥ℛ𝑦or𝑦ℛ𝑥.

Definition: the usual order ≤ on ℕ

We define the binary relation≤onℕby ∀𝑎, 𝑏 ∈ ℕ, (𝑎 ≤ 𝑏 ⇔ ∃𝑘 ∈ ℕ, 𝑏 = 𝑎 + 𝑘).

The intuition behind this definition is that𝑎 ≤ 𝑏if we need to add some𝑘to𝑎in order to reach𝑏.

Notation

We write𝑎 < 𝑏for(𝑎 ≤ 𝑏and𝑎 ≠ 𝑏).

Order – 3

Theorem

The set of natural numbersℕis totally ordered for≤.

Proof.

1 Reflexivity.Let𝑎 ∈ ℕ, then𝑎 = 𝑎 + 0with0 ∈ ℕ, hence𝑎 ≤ 𝑎.

2 Antisymmetry.Let𝑎, 𝑏 ∈ ℕbe such that𝑎 ≤ 𝑏and𝑏 ≤ 𝑎.

Then there exists𝑘, 𝑙 ∈ ℕsuch that𝑏 = 𝑎 + 𝑘and𝑎 = 𝑏 + 𝑙.

Therefore𝑎 = 𝑏 + 𝑙 = 𝑎 + 𝑘 + 𝑙. Hence0 = 𝑘 + 𝑙and thus𝑙 = 𝑘 = 0so that𝑎 = 𝑏.

3 Transitivity. Let𝑎, 𝑏, 𝑐 ∈ ℕbe such that𝑎 ≤ 𝑏and𝑏 ≤ 𝑐.

Then there exists𝑘, 𝑙 ∈ ℕsuch that𝑏 = 𝑎 + 𝑘and𝑐 = 𝑏 + 𝑙.

Therefore𝑐 = 𝑏 + 𝑙 = 𝑎 + (𝑘 + 𝑙)with𝑘 + 𝑙 ∈ ℕ, i.e.𝑎 ≤ 𝑐.

4 ≤is total.Let𝑎 ∈ ℕ. Set𝐴 = {𝑏 ∈ ℕ ∶ 𝑎 ≤ 𝑏or𝑏 ≤ 𝑎}. Then

•𝐴 ⊂ ℕ •0 ∈ 𝐴, indeed𝑎 = 0 + 𝑎so that0 ≤ 𝑎. •𝑠(𝐴) ⊂ 𝐴 Indeed, let𝑛 ∈ 𝑠(𝐴). Then𝑛 = 𝑠(𝑏)for some𝑏 ∈ 𝐴, i.e. 𝑎 ≤ 𝑏or𝑏 ≤ 𝑎.

- If𝑎 ≤ 𝑏then𝑏 = 𝑎 + 𝑘for some𝑘 ∈ ℕ,𝑛 = 𝑠(𝑏) = 𝑏 + 1 = 𝑎 + 𝑘 + 1with𝑘 + 1 ∈ ℕ, so that𝑎 ≤ 𝑛.

- If𝑏 ≤ 𝑎then𝑎 = 𝑏 + 𝑙for some𝑙 ∈ ℕ. We may assume that𝑙 ≠ 0(since𝑎 = 𝑏was in the above case).

Then𝑙 = ̃𝑙 + 1for some ̃𝑙 ∈ ℕ. Hence𝑎 = 𝑏 + 𝑙 = 𝑏 + ̃𝑙 + 1 = 𝑏 + 1 + ̃𝑙 = 𝑛 + ̃𝑙, i.e. 𝑛 ≤ 𝑎.

In both cases𝑛 ∈ 𝐴.

Therefore, by the induction principle,𝐴 = ℕ. So, for all𝑏 ∈ ℕ, either𝑎 ≤ 𝑏or𝑏 ≤ 𝑎. ■

Order – 4

Proposition

1 ∀𝑎 ∈ ℕ, 𝑎 ≤ 0 ⟹ 𝑎 = 0

2 ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 + 𝑏 ≤ 𝑎 + 𝑐 ⟹ 𝑏 ≤ 𝑐

3 There is no𝑎 ∈ ℕsuch that0 < 𝑎 < 1.

4 There is no𝑎 ∈ ℕsuch that∀𝑏 ∈ ℕ, 𝑏 ≤ 𝑎.

5 ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 ≤ 𝑏 ⟹ 𝑎𝑐 ≤ 𝑏𝑐

Proof.

1 Let𝑎 ∈ ℕbe such that𝑎 ≤ 0. Then there exists𝑘 ∈ ℕsuch that0 = 𝑎 + 𝑘. Hence𝑎 = 𝑘 = 0.

2 Let𝑎, 𝑏, 𝑐 ∈ ℕ. Assume that𝑎 + 𝑏 ≤ 𝑎 + 𝑐. Then there exists𝑘 ∈ ℕsuch that𝑎 + 𝑐 = 𝑎 + 𝑏 + 𝑘.

Thus𝑐 = 𝑏 + 𝑘so that𝑏 ≤ 𝑐as expected.

3 Let𝑎 ∈ ℕ. Assume that𝑎 < 1, then there exists𝑙 ∈ ℕ ⧵ {0}such that1 = 𝑎 + 𝑙.

Since𝑙 ≠ 0,𝑙 = 𝑘 + 1for some𝑘 ∈ ℕ, and1 = 𝑎 + 𝑘 + 1so that0 = 𝑎 + 𝑘. Therefore𝑎 = 0.

4 Assume by contradiction that there exists𝑎 ∈ ℕsuch that∀𝑏 ∈ ℕ, 𝑏 ≤ 𝑎. Then𝑎 + 1 ≤ 𝑎hence1 ≤ 0, i.e.0 = 1 + 𝑘for some𝑘 ∈ ℕ. Therefore1 = 0which is a contradiction (otherwise0 = 𝑠(0)).

5 Let𝑎, 𝑏, 𝑐 ∈ ℕ. Assume that𝑎 ≤ 𝑏. Then𝑏 = 𝑎 + 𝑘for some𝑘 ∈ ℕ.

Thus𝑏𝑐 = (𝑎 + 𝑘)𝑐 = 𝑎𝑐 + 𝑘𝑐with𝑘𝑐 ∈ ℕ. Therefore𝑎𝑐 ≤ 𝑏𝑐. ■

The well-ordering principle

Theorem: the well-ordering principle

A nonempty subset𝐴ofℕhas a least element, i.e. there exists𝑛 ∈ 𝐴such that∀𝑎 ∈ 𝐴, 𝑛 ≤ 𝑎.

Proof.

Let’s prove the contrapositive, i.e. if a subset𝐴 ⊂ ℕdoesn’t have a least element then it is empty.

Let𝐵 = {𝑎 ∈ ℕ ∶ ∀𝑖 ≤ 𝑎, 𝑖 ∉ 𝐴}.

• 𝐵 ⊂ ℕ

• 0 ∈ 𝐵(otherwise0would be the least element of𝐴).

• 𝑠(𝐵) ⊂ 𝐵

Indeed, if𝑛 ∈ 𝑠(𝐵), then𝑛 = 𝑠(𝑎)for𝑎 ∈ 𝐵, i.e. ∀𝑖 ≤ 𝑎, 𝑖 ∉ 𝐴.

Note that𝑛 = 𝑎 + 1 ∉ 𝐴otherwise it would be the least element of𝐴.

Therefore∀𝑖 ≤ 𝑛, 𝑖 ∉ 𝐴, i.e. 𝑛 ∈ 𝐵.

Thus, by the induction principle,𝐵 = ℕso𝐴is empty. ■