1Peanoaxioms Contents 1-Naturalnumbers ConceptsinAbstractMathematics

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Concepts in Abstract Mathematics

1 - Natural numbers

Jean-Baptiste Campesato

In this chapter we introduce the set ℕ of natural numbers. We will start with a minimal axiomatic description of it from which we will derive the main properties ofℕ.

Intuitively, we describeℕstarting from0and repeatedly doing the operation+1(we say that we take the successor):1is the successor of0,2is the successor of1,3is the successor of2and so on…This operation is governed by a few rules in order to make sure that the set we obtain coincides with our intuitive expectation about what should beℕ.

The method ofproof by inductionis closely related to the nature ofℕ. Hence we will study it at the end of this chapter.

I use the convention thatℕis the set of non-negative integers, i.e. 0 ∈ ℕ.

1 Peano axioms

All the results concerning the natural numbers will derive from the next theorem, that we admit, and which states the existence ofℕ.

Theorem 1(Peano axioms). There exists a setℕ together with an element 0 ∈ ℕ read aszero and a function 𝑠 ∶ ℕ → ℕread assuccessorsuch that:

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

0 ∉ 𝑠(ℕ)

(ii) If the successor of𝑛equals the successor of𝑚then𝑛 = 𝑚, i.e. 𝑠is injective:

∀𝑛, 𝑚 ∈ ℕ, 𝑠(𝑛) = 𝑠(𝑚) ⟹ 𝑛 = 𝑚

(iii) The induction principle. If a subset ofℕcontains0and is closed under𝑠then it isℕ:

∀𝐴 ⊂ ℕ, {

0 ∈ 𝐴

𝑠(𝐴) ⊂ 𝐴 ⟹ 𝐴 = ℕ

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The setℕis the set ofnatural numbers. As we will see, all the results ofℕwill derive from the above basic properties. The last axiom basically means that all the elements ofℕcan be obtained from0by taking the successor iteratively. Intuitively, the successor of𝑛is𝑠(𝑛) = 𝑛 + 1(actually, it will become formal after we define the addition, see Remark6).

0 3

0 7 (i) doesn’t hold

0 7 (ii) doesn’t hold

0 7 𝑠is not a function

0 7 (iii) doesn’t hold

Below are some basic propositions relying only on Peano axioms.

Proposition 2. Any non-zero natural number is the successor of a natural number, i.e.

∀𝑛 ∈ ℕ ⧵ {0}, ∃𝑚 ∈ ℕ, 𝑛 = 𝑠(𝑚) Proof. Set𝐴 = 𝑠(ℕ) ∪ {0}. Then

• 𝐴 ⊂ ℕ

• 0 ∈ 𝐴

• 𝑠(𝐴) ⊂ 𝑠(ℕ) ⊂ 𝐴

Hence, by the induction principle,𝐴 = ℕ.

Let𝑛 ∈ ℕ ⧵ {0}, then𝑛 ∈ 𝐴but𝑛 ≠ 0, therefore𝑛 ∈ 𝑠(ℕ). So there exists𝑚 ∈ ℕsuch that𝑛 = 𝑠(𝑚). ■ Proposition 3. A natural number is never its own successor, i.e.

∀𝑛 ∈ ℕ, 𝑛 ≠ 𝑠(𝑛) Proof. Set𝐴 = {𝑛 ∈ ℕ ∶ 𝑛 ≠ 𝑠(𝑛)}. Then

• 𝐴 ⊂ ℕ

• 0 ∈ 𝐴since0 ∉ 𝑠(ℕ)(particularly0 ≠ 𝑠(0))

• 𝑠(𝐴) ⊂ 𝐴

Indeed, let𝑚 ∈ 𝑠(𝐴). Then𝑚 = 𝑠(𝑛)for some𝑛 ∈ 𝐴. So𝑠(𝑛) ≠ 𝑛.

Since𝑠is injective, we get that𝑠(𝑠(𝑛)) ≠ 𝑠(𝑛), i.e.𝑠(𝑚) ≠ 𝑚.

Hence𝑚 ∈ 𝐴.

So, by the induction principle,𝐴 = ℕ. Thus, for every𝑛 ∈ ℕwe have that𝑛 ≠ 𝑠(𝑛). ■ Remark 4(You can skip it). Up to a bijection,ℕis uniquely defined by the Peano axioms.

More precisely, if there exists a set𝑆, with an element𝑒 ∈ 𝑆 and a function𝜎 ∶ 𝑆 → 𝑆such that (i) 𝑒 ∉ 𝜎(𝑆)

(ii) ∀𝑥, 𝑦 ∈ 𝑆, 𝜎(𝑥) = 𝜎(𝑦) ⟹ 𝑥 = 𝑦 (iii) ∀𝐴 ⊂ 𝑆,

{ 𝑒 ∈ 𝐴

𝜎(𝐴) ⊂ 𝐴 ⟹ 𝐴 = 𝑆

then there exists a bijectionΦ ∶ 𝑆 → ℕsuch thatΦ(𝑒) = 0,𝑠(Φ(𝑥)) = Φ(𝜎(𝑥)).

𝑒

𝜎 𝜎 𝜎 𝜎

0

𝑠 𝑠 𝑠 𝑠

Φ

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2 Addition, multiplication and order

2.1 Addition

The following proposition defines inductively the functionaddition with𝑎.

Proposition 5. Let𝑎 ∈ ℕ. Then there exists a unique function(𝑎 + •) ∶ ℕ → ℕ

𝑏 ↦ 𝑎 + 𝑏 such that (i) 𝑎 + 0 = 𝑎

(ii) ∀𝑏 ∈ ℕ, 𝑎 + 𝑠(𝑏) = 𝑠(𝑎 + 𝑏)

Proof. This function is well defined by the induction principle (i.e. we didn’t miss any element of the domain ℕusing this iterative definition).

Let’s check that it is unique. Let𝜑 ∶ ℕ → ℕbe such that𝜑(0) = 𝑎and∀𝑏 ∈ ℕ, 𝜑(𝑠(𝑏)) = 𝑠(𝜑(𝑏)).

Set𝐴 = {𝑏 ∈ ℕ ∶ 𝜑(𝑏) = 𝑎 + 𝑏}. Then

• 𝐴 ⊂ ℕ

• 0 ∈ 𝐴since𝜑(0) = 𝑎 = 𝑎 + 0.

• 𝑠(𝐴) ⊂ 𝐴. Indeed, let𝑐 ∈ 𝑠(𝐴). Then𝑐 = 𝑠(𝑏)for some𝑏 ∈ 𝐴and 𝜑(𝑐) = 𝜑(𝑠(𝑏)) since𝑐 = 𝑠(𝑏)

= 𝑠(𝜑(𝑏)) by definition of𝜑

= 𝑠(𝑎 + 𝑏) since𝑏 ∈ 𝐴

= 𝑎 + 𝑠(𝑏) by definition of𝑎 + •

= 𝑎 + 𝑐 since𝑠(𝑏) = 𝑐 Hence𝑐 ∈ 𝐴.

Therefore, by the induction principle,𝐴 = ℕ. Thus, for every𝑏 ∈ ℕwe have that𝜑(𝑏) = 𝑎 + 𝑏. ■ Remark 6. We set1 ≔ 𝑠(0). Then, for𝑛 ∈ ℕ,𝑛 + 1 = 𝑛 + 𝑠(0) = 𝑠(𝑛 + 0) = 𝑠(𝑛). As expected…

Hence, from now on, I will use indistinctively𝑛 + 1or𝑠(𝑛)(depending on which notation seems to be the more convenient).

Similarly,2 ≔ 𝑠(1),3 ≔ 𝑠(2),4 ≔ 𝑠(3)and so on…

Proposition 7.

1. ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐(the addition is associative) 2. ∀𝑎, 𝑏 ∈ ℕ, 𝑎 + 𝑏 = 𝑏 + 𝑎(the addition is commutative)

3. ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 + 𝑏 = 𝑎 + 𝑐 ⟹ 𝑏 = 𝑐(cancellation) 4. ∀𝑎, 𝑏 ∈ ℕ, 𝑎 + 𝑏 = 0 ⟹ 𝑎 = 𝑏 = 0

Proof.

1. 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)) since𝑛 = 𝑠(𝑐)

= 𝑎 + 𝑠(𝑏 + 𝑐)

= 𝑠(𝑎 + (𝑏 + 𝑐))

= 𝑠((𝑎 + 𝑏) + 𝑐) since𝑐 ∈ 𝐴

= (𝑎 + 𝑏) + 𝑠(𝑐)

= (𝑎 + 𝑏) + 𝑛 Hence𝑛 ∈ 𝐴.

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Thus, by the induction principle,𝐴 = ℕand for any𝑐 ∈ ℕ,𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐.

2. Sketch of proof:

(a) Prove that∀𝑎 ∈ ℕ, 0 + 𝑎 = 𝑎 + 0using the induction principle.

Hint: 0 + 𝑠(𝑎) = 𝑠(0 + 𝑎) = 𝑠(𝑎 + 0) = 𝑠(𝑎) = 𝑠(𝑎) + 0

(b) Prove that∀𝑎 ∈ ℕ, 𝑠(𝑎) = 1 + 𝑎using the induction principle.

Hint: 𝑠(𝑠(𝑎)) = 𝑠(1 + 𝑎) = (1 + 𝑎) + 1 = 1 + (𝑎 + 1) = 1 + 𝑠(𝑎). (c) Let𝑎 ∈ ℕ. Prove that∀𝑏 ∈ ℕ, 𝑎 + 𝑏 = 𝑏 + 𝑎.

Hint: 𝑎 + 𝑠(𝑏) = 𝑠(𝑎 + 𝑏) = 𝑠(𝑏 + 𝑎) = 𝑏 + 𝑠(𝑎) = 𝑏 + (1 + 𝑎) = (𝑏 + 1) + 𝑎 = 𝑠(𝑏) + 𝑎 3. Set𝐴 = {𝑎 ∈ ℕ ∶ ∀𝑏, 𝑐 ∈ ℕ, 𝑎 + 𝑏 = 𝑎 + 𝑐 ⟹ 𝑏 = 𝑐}. Then

• 𝐴 ⊂ ℕ

• 0 ∈ 𝐴

• 𝑠(𝐴) ⊂ 𝐴

Indeed, let𝑛 ∈ 𝑠(𝐴). Let𝑏, 𝑐 ∈ ℕsuch that𝑛 + 𝑏 = 𝑛 + 𝑐. We want to prove that𝑏 = 𝑐.

There exists𝑎 ∈ 𝐴such that𝑛 = 𝑠(𝑎). Then 𝑛 + 𝑏 = 𝑛 + 𝑐

⇒𝑠(𝑎) + 𝑏 = 𝑠(𝑎) + 𝑐

⇒𝑏 + 𝑠(𝑎) = 𝑐 + 𝑠(𝑎) by commutativity

⇒𝑠(𝑏 + 𝑎) = 𝑠(𝑐 + 𝑎) by construction of the addition

⇒𝑏 + 𝑎 = 𝑐 + 𝑎 since𝑠is injective

⇒𝑎 + 𝑏 = 𝑎 + 𝑐 by commutativity

⇒𝑏 = 𝑐 since𝑎 ∈ 𝐴 Hence𝑛 ∈ 𝐴.

Thus, by the induction principle,𝐴 = ℕ.

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

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

Then, by Proposition2,𝑏 = 𝑠(𝑛)for some𝑛 ∈ ℕ. So0 = 𝑎 + 𝑏 = 𝑎 + 𝑠(𝑛) = 𝑠(𝑎 + 𝑛).

Which is a contradiction since0 ∉ 𝑠(ℕ).

■

2.2 Multiplication

The following proposition defines inductively the functionmultiplication with𝑎.

Proposition 8. Let𝑎 ∈ ℕ. Then there exists a unique function(𝑎 × •) ∶ ℕ → ℕ

𝑏 ↦ 𝑎 × 𝑏 such that (i) 𝑎 × 0 = 0

(ii) ∀𝑏 ∈ ℕ, 𝑎 × 𝑠(𝑏) = (𝑎 × 𝑏) + 𝑎 Proposition 9.

1. ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 × (𝑏 × 𝑐) = (𝑎 × 𝑏) × 𝑐(the multiplication is associative) 2. ∀𝑎, 𝑏 ∈ ℕ, 𝑎 × 𝑏 = 𝑏 × 𝑎(the multiplication is commutative)

3. ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 × (𝑏 + 𝑐) = 𝑎 × 𝑏 + 𝑎 × 𝑐and(𝑎 + 𝑏) × 𝑐 = 𝑎 × 𝑐 + 𝑏 × 𝑐(×is distributive over+) 4. ∀𝑎 ∈ ℕ, 𝑎 × 1 = 𝑎

5. ∀𝑎, 𝑏 ∈ ℕ, 𝑎 × 𝑏 = 0 ⟹ (𝑎 = 0or𝑏 = 0) 6. ∀𝑎, 𝑏, 𝑐 ∈ ℕ,

{

𝑎 × 𝑏 = 𝑎 × 𝑐

𝑎 ≠ 0 ⟹ 𝑏 = 𝑐(cancellation) We prove these properties similarly to the ones of the addition.

Remark 10. It is common to omit the symbol×when there is no possible confusion (i.e. to simply write𝑎𝑏 for𝑎 × 𝑏).

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2.3 Order

The following definition is a little bit informal, but it is enough for our purpose.

Definition 11. Abinary relationℛon a set𝐸consists in associating a truth value to every couple(𝑥, 𝑦) ∈ 𝐸² (beware, order matters here).

We say that𝑥is related to𝑦byℛ, denoted𝑥ℛ𝑦, if the valuetrueis assigned to(𝑥, 𝑦).

Examples 12.

1. Let𝐸 = {𝑎, 𝑏, 𝑐}. Since𝐸is finite, 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:

𝑥ℛ𝑦 ⇔ 𝑥²− 𝑦² = 𝑥 − 𝑦

The following definition highlights the important properties of the order≤that you intuitively know.

Definition 13. We say that a binary relationℛon a set𝐸is anorderif (i) ∀𝑥 ∈ 𝐸, 𝑥ℛ𝑥 (reflexivity)

(ii) ∀𝑥, 𝑦 ∈ 𝐸, (𝑥ℛ𝑦and𝑦ℛ𝑥) ⟹ 𝑥 = 𝑦 (antisymmetry) (iii) ∀𝑥, 𝑦, 𝑧 ∈ 𝐸, (𝑥ℛ𝑦and𝑦ℛ𝑧) ⟹ 𝑥ℛ𝑧 (transitivity) Definition 14. We say that an orderℛon a set𝐸istotalif

∀𝑥, 𝑦 ∈ 𝐸, 𝑥ℛ𝑦or𝑦ℛ𝑥 Definition 15. We define the binary relation≤onℕby

∀𝑎, 𝑏 ∈ ℕ, (𝑎 ≤ 𝑏 ⇔ ∃𝑘 ∈ ℕ, 𝑏 = 𝑎 + 𝑘)

We read”𝑎is less than or equal to𝑏”or”𝑏is greater than or equal to𝑎”when𝑎 ≤ 𝑏holds.

Proposition 16. The set of natural numbersℕis totally ordered for≤.

Proof.

(i) Reflexivity: let𝑎 ∈ ℕ, then𝑎 = 𝑎 + 0with0 ∈ ℕ, hence𝑎 ≤ 𝑎.

(ii) Antisymmetry: let𝑎, 𝑏 ∈ ℕbe such that𝑎 ≤ 𝑏and𝑏 ≤ 𝑎.

Then there exists𝑘 ∈ ℕsuch that𝑏 = 𝑎 + 𝑘and there exists𝑙 ∈ ℕsuch that𝑎 = 𝑏 + 𝑙.

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

(iii) Transitivity: let𝑎, 𝑏, 𝑐 ∈ ℕbe such that𝑎 ≤ 𝑏and𝑏 ≤ 𝑐.

Then there exists𝑘 ∈ ℕsuch that𝑏 = 𝑎 + 𝑘and there exists𝑙 ∈ ℕsuch that𝑐 = 𝑏 + 𝑙.

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

(iv) ≤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 ≤ 𝑎 then𝑎 = 𝑏 + 𝑙 for some𝑙 ∈ ℕ. The case𝑙 = 0is covered by the above case, so we may assume that𝑙 ≠ 0. 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.

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

■ Definition 17. Given𝑎, 𝑏 ∈ ℕ, we write𝑎 < 𝑏for(𝑎 ≤ 𝑏and𝑎 ≠ 𝑏).

Proposition 18.

1. ∀𝑎, 𝑏 ∈ ℕ, 𝑎 < 𝑏 ⇔ 𝑎 + 1 ≤ 𝑏.

2. Given𝑎, 𝑏 ∈ ℕ, exactly one of the followings occurs: either𝑎 < 𝑏, or𝑎 = 𝑏, or𝑏 < 𝑎. Particularly, the negation of𝑎 ≤ 𝑏is𝑏 < 𝑎.

Proof.

1. ⇒: Let𝑎, 𝑏 ∈ ℕbe such that𝑎 < 𝑏. Then𝑎 ≤ 𝑏so there exists𝑘 ∈ ℕsuch that𝑏 = 𝑎 + 𝑘.

Assume by contradiction that𝑘 = 0then𝑎 = 𝑏which is false. Hence𝑘 ≠ 0and there exists ̃𝑘 ∈ ℕsuch that𝑘 = ̃𝑘 + 1. Then𝑏 = (𝑎 + 1) + ̃𝑘. We proved that𝑎 + 1 ≤ 𝑏as expected.

⇐: Assume that𝑎 + 1 ≤ 𝑏then there exists𝑘 ∈ ℕsuch that𝑏 = 𝑎 + 1 + 𝑘.

• Then𝑏 = 𝑎 + (1 + 𝑘)with1 + 𝑘 ∈ ℕhence𝑎 ≤ 𝑏.

• Assume by contradiction that𝑎 = 𝑏then𝑎 = 𝑎 + 1 + 𝑘hence0 = 1 + 𝑘from which we get0 = 1, so0 = 𝑠(0). We get a contradiction with0 ∉ 𝑠(ℕ).

2. This property derives from the fact that the order≤is total.

■ Proposition 19.

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. Then 𝑐 = 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)but0 ∉ 𝑠(ℕ)).

5. Let𝑎, 𝑏, 𝑐 ∈ ℕ. Assume that𝑎 ≤ 𝑏. Then𝑏 = 𝑎 + 𝑘for some𝑘 ∈ ℕ. Thus𝑏𝑐 = (𝑎 + 𝑘)𝑐 = 𝑎𝑐 + 𝑘𝑐 with 𝑘𝑐 ∈ ℕ. Therefore𝑎𝑐 ≤ 𝑏𝑐.

■ Theorem 20(The well-ordering principle). The setℕiswell-orderedfor≤.

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. ■

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We proved that the induction principle implies the well-ordering principle, but they are actually equivalent. In the definition ofℕ, we could have replaced the induction principle by the well-ordering principle.

Proof that the well-ordering principle implies the induction principle.

Let𝐴 ⊂ ℕ. Assume that0 ∈ 𝐴and that𝑠(𝐴) ⊂ 𝐴. We want to prove that𝐴 = ℕ.

Assume by contradiction thatℕ ⧵ 𝐴 ≠ ∅. Then, by the well-ordering principle,ℕ ⧵ 𝐴admits a least element 𝑎 ∈ 𝐴. Obviously,𝑎 ≠ 0since0 ∈ 𝐴.

Since𝑎 ∈ ℕ ⧵ {0}, there exists𝑎 ∈ ℕ̃ such that𝑎 = 𝑠( ̃𝑎). Since𝑠(𝐴) ⊂ 𝐴,𝑎 ∉ 𝐴̃ (otherwise𝑎 = 𝑠( ̃𝑎) ∈ 𝐴).

But𝑎 < 𝑎. This contradicts the fact that̃ 𝑎is the least element of𝐴.

Henceℕ ⧵ 𝐴 = ∅and𝐴 = ℕ. ■

Proposition 21. ∀𝑎, 𝑏 ∈ ℕ, 𝑎𝑏 = 1 ⟹ 𝑎 = 𝑏 = 1.

Proof. Let𝑎, 𝑏 ∈ ℕbe such that𝑎𝑏 = 1. Since𝑎 = 0or𝑏 = 0implies that𝑎𝑏 = 0, we know that𝑎 ≠ 0and 𝑏 ≠ 0. We have0 ≤ 𝑎and𝑎 ≠ 0hence0 < 𝑎from which we get1 ≤ 𝑎. Similarly1 ≤ 𝑏.

Then𝑎 = 1 + 𝑘for some𝑘 ∈ ℕ. Then1 = 𝑎𝑏 = 𝑏 + 𝑏𝑘, i.e. 𝑏 ≤ 1. Hence𝑏 = 1and𝑎 = 𝑎 × 1 = 𝑎𝑏 = 1. ■ 2.4 Summary

The main properties of(ℕ, +, ×, ≤, 0, 1), where+, ×are two binary laws and≤is a binary relation, are:

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

• +is commutative:∀𝑎, 𝑏 ∈ ℕ, 𝑎 + 𝑏 = 𝑏 + 𝑎

• 0is the unit of+:∀𝑎 ∈ ℕ, 0 + 𝑎 = 𝑎 + 0 = 𝑎

• Cancellation rule: ∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 + 𝑏 = 𝑎 + 𝑐 ⇒ 𝑏 = 𝑐

• ∀𝑎, 𝑏 ∈ ℕ, 𝑎 + 𝑏 = 0 ⟹ 𝑎 = 𝑏 = 0.

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

• ×is commutative:∀𝑎, 𝑏 ∈ ℕ, 𝑎 × 𝑏 = 𝑏 × 𝑎

• 1is the unit of×: ∀𝑎 ∈ ℕ, 1 × 𝑎 = 𝑎 × 1 = 𝑎

• Cancellation rule: for∀𝑎, 𝑏, 𝑐 ∈ ℕ, {

𝑎 × 𝑏 = 𝑎 × 𝑐

𝑎 ≠ 0 ⇒ 𝑏 = 𝑐

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

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

• ∀𝑎, 𝑏 ∈ ℕ, 𝑎𝑏 = 1 ⟹ 𝑎 = 𝑏 = 1

• ≤is an order onℕ, i.e.

– Reflexivity:∀𝑎 ∈ ℕ, 𝑎 ≤ 𝑎

– Antisymmetry:∀𝑎, 𝑏 ∈ ℕ, (𝑎 ≤ 𝑏and𝑏 ≤ 𝑎) ⇒ 𝑎 = 𝑏 – Transitivity:∀𝑎, 𝑏, 𝑐 ∈ ℕ, (𝑎 ≤ 𝑏and𝑏 ≤ 𝑐) ⇒ 𝑎 ≤ 𝑐 Besides, this order is total:∀𝑎, 𝑏 ∈ ℕ, 𝑎 ≤ 𝑏or𝑏 ≤ 𝑎

• Well-ordering principle: a nonempty subset𝐴ofℕhas a least element.

• ≤is compatible with+:∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 ≤ 𝑏 ⇒ 𝑎 + 𝑐 ≤ 𝑏 + 𝑐

• ≤is compatible with×:∀𝑎, 𝑏, 𝑐 ∈ ℕ, 𝑎 ≤ 𝑏 ⇒ 𝑎𝑐 ≤ 𝑏𝑐

⋆ ∀𝑎, 𝑏, 𝑐, 𝑑 ∈ ℕ, (𝑎 ≤ 𝑏and𝑐 ≤ 𝑑) ⇒ 𝑎 + 𝑐 ≤ 𝑏 + 𝑑

⋆ ∀𝑎, 𝑏, 𝑐, 𝑑 ∈ ℕ, (𝑎 ≤ 𝑏and𝑐 ≤ 𝑑) ⇒ 𝑎𝑐 ≤ 𝑏𝑑

• ∀𝑎 ∈ ℕ, 𝑎 ≤ 0 ⟹ 𝑎 = 0

• There is no𝑎 ∈ ℕsuch that0 < 𝑎 < 1.

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

• ∀𝑎, 𝑏 ∈ ℕ, 𝑎 < 𝑏 ⇔ 𝑎 + 1 ≤ 𝑏.

• For𝑎, 𝑏 ∈ ℕwe have (exclusively) either𝑎 < 𝑏, or𝑎 = 𝑏, or𝑏 < 𝑎.

Particularly, the negation of𝑎 ≤ 𝑏is𝑏 < 𝑎.

The properties with a star were not proved in this chapter but will be proved as practice questions.

Except otherwise stated, you can directly use any of the above properties without proving them.

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3 Proof by induction

In this section we are going to highlight the connection between theprinciple of inductionas stated in Theorem 1and the notion ofproof by inductionthat you have already encountered.

3.1 Formal statement

Proof by induction is closely related to the fact that ℕ is defined by its initial term 0 and then by taking iteratively its successor. This fact is highlighted in the proof of the following theorem.

Theorem 22(Proof by induction). Let𝒫(𝑛)be a statement depending on𝑛 ∈ ℕ.

If𝒫(0)is true and if𝒫(𝑛) ⟹ 𝒫(𝑛 + 1)is true for all𝑛 ∈ ℕ, then𝒫(𝑛)is true for all𝑛 ∈ ℕ. Formally, {

𝒫(0)

∀𝑛 ∈ ℕ, (𝒫(𝑛) ⟹ 𝒫(𝑛 + 1)) ⟹ ∀𝑛 ∈ ℕ, 𝒫(𝑛)

The informal idea is that since𝒫(0)and𝒫(0) ⟹ 𝒫(1)are true then𝒫(1)is true. Then we can repeat the same process: since𝒫(1)and𝒫(1) ⟹ 𝒫(2)are true then𝒫(2)is true, and so on...

This way𝒫(0), 𝒫(1), 𝒫(2), 𝒫(3), …are all true.

Proof of Theorem22.

We define the set𝐴 = {𝑛 ∈ ℕ ∶ 𝒫(𝑛)is true}. Then:

• 𝐴 ⊂ ℕby definition of𝐴.

• 0 ∈ ℕsince𝒫(0)is true.

• 𝑠(𝐴) ⊂ 𝐴

Indeed, let𝑛 ∈ 𝑠(𝐴). Then𝑛 = 𝑠(𝑚) = 𝑚 + 1for some𝑚 ∈ 𝐴. By definition of𝐴,𝒫(𝑚)is true. But by assumption𝒫(𝑚) ⟹ 𝒫(𝑚 + 1)is also true. Hence𝒫(𝑚 + 1)is true, meaning that𝑛 = 𝑚 + 1 ∈ 𝐴.

Hence, by theinduction principleof Theorem1,𝐴 = ℕ. Finally, for every𝑛 ∈ ℕwe have that𝒫(𝑛)is true. ■ 3.2 In practice

How to write aproof by induction? There are several steps that you should make sure they appear clearly!

• What statement are you proving? What is your𝒫(𝑛)? Particularly, on which parameter are you doing the induction? You should make everything clear for the reader!

• Base case:prove that𝒫(0)is true.

• Induction step:prove that if𝒫(𝑛)is true for some𝑛 ∈ ℕthen𝒫(𝑛 + 1)is also true.

It is important to clearly write the induction hypothesis and what you want to prove in this step (the reader shouldn’t have to guess). Make sure that you used the induction hypothesis somewhere, otherwise there is something suspicious with your proof.

Below are two basic examples:

Proposition 23. For any𝑛 ∈ ℕ, the sum0 + 1 + 2 + ⋯ + 𝑛is equal to ^𝑛(𝑛+1)₂ . Proof. We are going to prove that∀𝑛 ∈ ℕ, 0 + 1 + 2 + 3 + ⋯ + 𝑛 = ^𝑛(𝑛+1)

2 by induction on𝑛.

• Base case: Let𝑛 = 0. Then the sum in the left hand side is equal to 0. And ^𝑛(𝑛+1)

2 = ^0⋅1

2 = 0. So the equality holds.

• Induction step: Assume that 0 + 1 + 2 + 3 + ⋯ + 𝑛 = ^𝑛(𝑛+1)

2 for some𝑛 ∈ ℕ and let’s prove that 0 + 1 + 2 + 3 + ⋯ + 𝑛 + (𝑛 + 1) = ^{(𝑛+1)(𝑛+2)}

2 .

0 + 1 + 2 + 3 + ⋯ + 𝑛 + (𝑛 + 1) = 𝑛(𝑛 + 1)

2 + (𝑛 + 1) by the induction hypothesis

= 𝑛(𝑛 + 1) + 2(𝑛 + 1)

2 = (𝑛 + 1)(𝑛 + 2) 2

Which proves the induction step. ■

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Proposition 24. For any𝑛 ∈ ℕ, the sum of the first𝑛odd numbers is equal to𝑛².

Proof. We are going to prove that∀𝑛 ∈ ℕ, 1 + 3 + ⋯ + (2𝑛 − 1) = 𝑛²by induction on𝑛.

• Base case:Let𝑛 = 0. Then the sum in the left hand side is empty, so it is equal to 0. And𝑛² = 0² = 0.

So the equality holds.

• Induction step: Assume that the sum of the first𝑛odd numbers is equal to𝑛² for some𝑛 ∈ ℕ, i.e.

1 + 3 + ⋯ + (2𝑛 − 1) = 𝑛².

Let’s prove that1 + 3 + ⋯ + (2𝑛 − 1) + (2𝑛 + 1) = (𝑛 + 1)².

1 + 3 + ⋯ + (2𝑛 − 1) + (2𝑛 + 1) = 𝑛²+ 2𝑛 + 1 by the induction hypothesis

= (𝑛 + 1)² by the binomial formula Which proves the induction step.

■

3.3 Variants of the induction 3.3.1 Strong induction

The strong induction is equivalent to the usual induction (i.e. one may prove that Theorem22holds assuming Theorem25, and that Theorem25holds assuming Theorem22). Nonetheless, in some cases, it may be easier to write a strong induction rather than a usual one.

Theorem 25(Strong induction). Let𝒫(𝑛)be a statement depending on𝑛 ∈ ℕ.

If𝒫(0)is true and if(𝑃 (0), 𝑃 (1), … , 𝒫(𝑛)) ⟹ 𝒫(𝑛 + 1)is true for all𝑛 ∈ ℕ, then𝒫(𝑛) is true for all𝑛 ∈ ℕ.

Formally,

{ 𝒫(0)

∀𝑛 ∈ ℕ, ((𝒫(0), 𝒫(1), … , 𝒫(𝑛)) ⟹ 𝒫(𝑛 + 1)) ⟹ ∀𝑛 ∈ ℕ, 𝒫(𝑛) Proof. For𝑛 ∈ ℕ, we defineℛ(𝑛)by

ℛ(𝑛)is true⇔ 𝒫(0), 𝒫(1), … , 𝒫(𝑛)are true

Assume that𝒫(0)is true and that(𝑃 (0), 𝑃 (1), … , 𝒫(𝑛)) ⟹ 𝒫(𝑛 + 1)is true for all𝑛 ∈ ℕ.

Thenℛ(0)is true since𝒫(0)is. And, for all𝑛 ∈ ℕ,ℛ(𝑛) ⟹ ℛ(𝑛 + 1)is true.

By the usual inductionℛ(𝑛)is true for any𝑛 ∈ ℕ. Particularly,𝒫(𝑛)is true for any𝑛 ∈ ℕas expected. ■ 3.3.2 Base case at𝑛₀

It may be easier to write a proof by induction starting at a base𝑛₀∈ ℕwhich is not necessarily0. Below is the corresponding statement for the usual induction, but it is possible to adapt the strong induction similarly.

Theorem 26. Let𝑛₀ ∈ ℕ. Let𝒫(𝑛)be a statement depending on a natural number𝑛 ≥ 𝑛₀.

If𝒫(𝑛₀)is true and if𝒫(𝑛) ⟹ 𝒫(𝑛 + 1)is true for every natural number𝑛 ≥ 𝑛₀, then𝒫(𝑛)is true for every natural number𝑛 ≥ 𝑛₀. Formally,

{ 𝒫(𝑛₀)

∀𝑛 ∈ ℕ_≥𝑛

0, (𝒫(𝑛) ⟹ 𝒫(𝑛 + 1)) ⟹ ∀𝑛 ∈ ℕ_≥𝑛

0, 𝒫(𝑛) Proof. For𝑛 ∈ ℕ, we defineℛ(𝑛)by

ℛ(𝑛)is true⇔ 𝒫(𝑛 + 𝑛₀)is true

Thenℛ(0)is true since𝒫(𝑛₀)is. And, for all𝑛 ∈ ℕ,ℛ(𝑛) ⟹ ℛ(𝑛 + 1)is true.

By the usual inductionℛ(𝑛)is true for any𝑛 ∈ ℕ, i.e.𝒫(𝑛)is true for any𝑛 ∈ ℕ_≥𝑛

0. ■

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Below is an example of induction starting at𝑛₀= 5.

Proposition 27. For any integer𝑛 ≥ 5,2^𝑛 > 𝑛².

Proof. We are going to prove that∀𝑛 ≥ 5, 2^𝑛 > 𝑛²by induction on𝑛.

• Base case at𝑛 = 5:2⁵ = 32 > 25 = 5².

• Induction step:Assume that2^𝑛> 𝑛²for some𝑛 ≥ 5and let’s prove that2^𝑛+1> (𝑛 + 1)².

Note that2^𝑛+1= 2×2^𝑛≥ 2𝑛²by the induction hypothesis. Hence it is enough to prove that2𝑛² > (𝑛+1)² which is equivalent to𝑛²− 2𝑛 − 1 > 0.

We study the sign of the polynomial𝑥²− 2𝑥 − 1. It is a polynomial of degree 2 with positive leading coeﬀicient and its discriminant is(−2)²− 4 × (−1) = 8 > 0. Therefore

𝑥 𝑥²−2𝑥−1

−∞ 1 − √2 1 + √2 +∞

+ 0 − 0 +

Since5 > 1 + √2, we know that𝑛²− 2𝑛 − 1 > 0for𝑛 ≥ 5. Which proves the induction step.

■ Remark 28. The above example is interesting because the induction step holds for𝑛 ≥ 3, but𝒫(3)and𝒫(4) are false: don’t forget the base case! It is crucial!