Homework questions – Week 1

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University of Toronto – MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

Homework questions – Week 1

Jean-Baptiste Campesato

January 18^th, 2021 to January 22^nd, 2021

Exercise 1.

Using only the definition of the multiplication, the properties of the addition and Peano axioms, prove that:

1. ∀𝑎 ∈ ℕ, 0 × 𝑎 = 𝑎 × 0 = 0 2. ∀𝑎 ∈ ℕ, 𝑎 × 1 = 𝑎

Exercise 2.

Given𝑚 ∈ ℕ, we define inductively the function𝑚^• ∶ ℕ → ℕ

𝑛 ↦ 𝑚^𝑛 by𝑚⁰ = 1and∀𝑛 ∈ ℕ, 𝑚^𝑠(𝑛)= 𝑚^𝑛× 𝑚.

Prove that:

1. ∀𝑚 ∈ ℕ, 𝑚¹ = 𝑚

2. ∀𝑎, 𝑏, 𝑛 ∈ ℕ, (𝑎 × 𝑏)^𝑛 = 𝑎^𝑛× 𝑏^𝑛 3. ∀𝑎, 𝑚, 𝑛 ∈ ℕ, 𝑎^𝑚+𝑛= 𝑎^𝑚× 𝑎^𝑛 4. ∀𝑛 ∈ ℕ ⧵ {0}, 0^𝑛 = 0

5. ∀𝑛 ∈ ℕ, 1^𝑛 = 1 Exercise 3.

For each of the followings, is the binary relationℛan order on𝐸? If so, is it total?

1. 𝐸 = ℤand∀𝑥, 𝑦 ∈ ℤ, 𝑥ℛ𝑦 ⇔ 𝑥 = −𝑦

2. 𝐸 = ℝand∀𝑥, 𝑦 ∈ ℝ, 𝑥ℛ𝑦 ⇔cos²𝑥 +sin²𝑦 = 1

3. 𝐸 = 𝒫(𝑆)is the set of subsets of a fixed set𝑆and∀𝐴, 𝐵 ∈ 𝒫(𝑆), 𝐴ℛ𝐵 ⇔ 𝐴 ⊂ 𝐵 Exercise 4.

We define a binary relationℛonℕby∀𝑥, 𝑦 ∈ ℕ, 𝑥ℛ𝑦 ⇔ ∃𝑝, 𝑞 ∈ ℕ ⧵ {0}, 𝑦 = 𝑝𝑥^𝑞. 1. Prove thatℛis an order.

2. Is it a total order?

Exercise 5.

We define a binary relation≺onℕ²by(𝑥₁, 𝑦₁) ≺ (𝑥₂, 𝑦₂) ⇔ (𝑥1≤ 𝑥₂and𝑦₁ ≤ 𝑦₂). 1. Prove that≺is an order.

2. Is it a total order?

Exercise 6.

Prove that

1. ∀𝑎, 𝑏, 𝑐, 𝑑 ∈ ℕ, (𝑎 ≤ 𝑏and𝑐 ≤ 𝑑) ⇒ 𝑎 + 𝑐 ≤ 𝑏 + 𝑑 2. ∀𝑎, 𝑏, 𝑐, 𝑑 ∈ ℕ, (𝑎 ≤ 𝑏and𝑐 ≤ 𝑑) ⇒ 𝑎𝑐 ≤ 𝑏𝑑 Exercise 7.

For which𝑐 ∈ ℕ, do we have∀𝑎, 𝑏 ∈ ℕ, 𝑎𝑐 ≤ 𝑏𝑐 ⟹ 𝑎 ≤ 𝑏?

Exercise 8.

Using the well-ordering principle, find an alternative proof of: there is no natural number𝑛between0and1.

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2 Homework Questions – Week 1

Sample solutions to Exercise 1.

1. Given𝑎 ∈ ℕ, we already know that𝑎 × 0 = 0by definition of the multiplication. So we only need to prove that∀𝑎 ∈ ℕ, 0 × 𝑎 = 0.

Set𝐴 = {𝑎 ∈ ℕ ∶ 0 × 𝑎 = 0}, then

• 𝐴 ⊂ ℕ

• 0 ∈ 𝐴since0 × 0 = 0by definition of the multiplication.

• 𝑠(𝐴) ⊂ 𝐴. Indeed, let𝑚 ∈ 𝑠(𝐴), then𝑚 = 𝑠(𝑎)for some𝑎 ∈ 𝐴. Then 0 × 𝑚 = 0 × 𝑠(𝑎)

= 0 × 𝑎 + 0by definition of the multiplication

= 0 + 0since𝑎 ∈ 𝐴

= 0 Thus𝑚 ∈ 𝐴.

Therefore, by the induction principle,𝐴 = ℕ. So∀𝑎 ∈ ℕ, 0 × 𝑎 = 0.

2. Let𝑎 ∈ ℕ. Then

𝑎 × 1 = 𝑎 × 𝑠(0)since1 = 𝑠(0)

= 𝑎 × 0 + 𝑎by definition of the multiplication

= 0 + 𝑎by definition of the multiplication

= 𝑎

1. Let𝑚 ∈ ℕthen𝑚¹= 𝑚^𝑠(0)= 𝑚⁰× 𝑚 = 1 × 𝑚 = 𝑚.

2. Let𝑎, 𝑏 ∈ ℕ. Set𝐴 = {𝑛 ∈ ℕ ∶ (𝑎𝑏)^𝑛= 𝑎^𝑛𝑏^𝑛}.

• 𝐴 ⊂ ℕ

• 0 ∈ 𝐴: indeed,(𝑎𝑏)⁰ = 1and𝑎⁰𝑏⁰ = 1 × 1 = 1.

• 𝑠(𝐴) ⊂ 𝐴: let𝑚 ∈ 𝑠(𝐴)then𝑚 = 𝑠(𝑛)for some𝑛 ∈ 𝐴. Next (𝑎𝑏)^𝑚= (𝑎𝑏)^𝑠(𝑛)since𝑚 = 𝑠(𝑛)

= (𝑎𝑏)^𝑛(𝑎𝑏)by definition of(𝑎𝑏)^•

= 𝑎^𝑛𝑏^𝑛𝑎𝑏since𝑛 ∈ 𝐴

= (𝑎^𝑛𝑎)(𝑏^𝑛𝑏)by properties of the product

= 𝑎^𝑠(𝑛)𝑏^𝑠(𝑛)by definition of𝑎^•and𝑏^•

= 𝑎^𝑚𝑏^𝑚since𝑚 = 𝑠(𝑛) Hence𝑚 ∈ 𝐴.

Therefore, by the induction principle,𝐴 = ℕ. So for all𝑛 ∈ ℕ,(𝑎𝑏)^𝑛= 𝑎^𝑛𝑏^𝑛. 3. Let𝑎, 𝑚 ∈ ℕ. Set𝐴 = {𝑛 ∈ ℕ ∶ 𝑎^𝑚+𝑛 = 𝑎^𝑚𝑎^𝑛}. Then

• 𝐴 ⊂ ℕ

• 0 ∈ 𝐴: indeed,𝑎^𝑚+0= 𝑎^𝑚= 𝑎^𝑚× 1 = 𝑎^𝑚× 𝑎⁰

• 𝑠(𝐴) ⊂ 𝐴: let𝑘 ∈ 𝑠(𝐴)then𝑘 = 𝑠(𝑛)for some𝑛 ∈ 𝐴. Next 𝑎^𝑚+𝑘= 𝑎^{𝑚+𝑠(𝑛)}since𝑘 = 𝑠(𝑛)

= 𝑎^{𝑠(𝑚+𝑛)}by definition of the addition

= 𝑎^𝑚+𝑛× 𝑎by definition of𝑎^•

= 𝑎^𝑚𝑎^𝑛𝑎since𝑛 ∈ 𝐴

= 𝑎^𝑚𝑎^𝑠(𝑛)by definition of𝑎^•

= 𝑎^𝑚𝑎^𝑘since𝑘 = 𝑠(𝑛)

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

Hence𝑘 ∈ 𝐴.

Therefore, by the induction principle,𝐴 = ℕ. So for all𝑛 ∈ ℕ,𝑎^𝑚+𝑛 = 𝑎^𝑚𝑎^𝑛.

4. Let𝑛 ∈ ℕ ⧵ {0}. Then there exists𝑚 ∈ ℕsuch that𝑛 = 𝑠(𝑚). Thus0^𝑛= 0^𝑠(𝑚) = 0^𝑚× 0 = 0.

5. Set𝐴 = {𝑛 ∈ ℕ ∶ 1^𝑛= 1}. Then

• 𝐴 ⊂ ℕ

• 0 ∈ 𝐴:1⁰= 1by definition of1^•.

• 𝑠(𝐴) ⊂ 𝐴: let𝑚 ∈ 𝑠(𝐴)then𝑚 = 𝑠(𝑛)for some𝑛 ∈ 𝐴. Next 1^𝑚= 1^𝑠(𝑛)since𝑚 = 𝑠(𝑛)

= 1^𝑛× 1by definition of1^•

= 1 × 1since𝑛 ∈ 𝐴

= 1

Hence𝑚 ∈ 𝐴.

Therefore, by the induction principle,𝐴 = ℕ. So for all𝑛 ∈ ℕ,1^𝑛 = 1.

1. This binary relation is not an order since it is not reflexive.

Indeed,1ℛ1is false since1 ≠ −1.

2. This binary relation is not an order since it is not antisymmetric.

Indeed,0ℛ(2𝜋)and(2𝜋)ℛ0are true but0 ≠ 2𝜋.

3. The inclusion is an order on𝒫(𝑆). Indeed

• ∀𝐴 ∈ 𝒫(𝑆), 𝐴 ⊂ 𝐴(reflexivity).

• ∀𝐴, 𝐵 ∈ 𝒫(𝑆), (𝐴 ⊂ 𝐵and𝐵 ⊂ 𝐴) ⟹ 𝐴 = 𝐵(antisymmetry).

• ∀𝐴, 𝐵, 𝐶 ∈ 𝒫(𝑆), (𝐴 ⊂ 𝐵and𝐵 ⊂ 𝐶) ⟹ 𝐴 ⊂ 𝐶 (transitivity).

If𝑆 = ∅then𝒫(𝑆) = {∅}: the order is obviously total.

If𝑆 = {∗}has only one element then𝒫(𝑆) = {∅, {∗}}: the order is obviously total.

If𝑆contains at least two elements𝑎, 𝑏then the order is not total.

Indeed, set𝐴 = 𝑆 ⧵ {𝑎}and𝐵 = 𝑆 ⧵ {𝑏}.

Then𝐴 ⊄ 𝐵since𝑏 ∈ 𝐴but𝑏 ∉ 𝐵, and,𝐵 ⊄ 𝐴since𝑎 ∈ 𝐵but𝑎 ∉ 𝐴.

Thus, if𝑆contains at least two elements, then⊂is not a total order on𝒫(𝑆).

1. • Reflexivity. Let𝑥 ∈ ℕ. Then𝑥 = 1 × 𝑥¹. Hence𝑥ℛ𝑥.

• Antisymmetry. Let𝑥, 𝑦 ∈ ℕbe such that𝑥ℛ𝑦and𝑦ℛ𝑥. Then𝑥 ≤ 𝑦and𝑦 ≤ 𝑥. Thus𝑥 = 𝑦.

• Transitivity. Let 𝑥, 𝑦, 𝑧 ∈ ℕbe such that 𝑥ℛ𝑦and 𝑦ℛ𝑧. Then 𝑦 = 𝑝𝑥^𝑞 and𝑧 = 𝑟𝑦^𝑠 for some 𝑝, 𝑞, 𝑟, 𝑠 ∈ ℕ ⧵ {0}. Hence𝑧 = 𝑟𝑦^𝑠= 𝑟𝑝^𝑠𝑥^𝑞𝑠with𝑟𝑝^𝑠, 𝑞𝑠 ∈ ℕ ⧵ {0}. Thus𝑥ℛ𝑧.

2. This order is not total since0ℛ1and1ℛ0are both false.

1. • Reflexivity.Let(𝑥, 𝑦) ∈ ℕ², then𝑥 ≤ 𝑥and𝑦 ≤ 𝑦hence(𝑥, 𝑦) ≺ (𝑥, 𝑦).

• Antisymmetry.Assume that(𝑥₁, 𝑦₁) ≺ (𝑥₂, 𝑦₂)and that(𝑥₂, 𝑦₂) ≺ (𝑥₁, 𝑦₁).

Then𝑥₁≤ 𝑥₂,𝑦₁≤ 𝑦₂,𝑥₂ ≤ 𝑥₁and𝑦₂ ≤ 𝑦₁.

Since≤is an order onℕ, we get that𝑥₁= 𝑥₂and𝑦₁= 𝑦₂. Thus(𝑥₁, 𝑦₁) = (𝑥₂, 𝑦₂).

• Transitivity. Assume that(𝑥₁, 𝑦₁) ≺ (𝑥₂, 𝑦₂)and that(𝑥₂, 𝑦₂) ≺ (𝑥₃, 𝑦₃).

Then𝑥₁≤ 𝑥₂,𝑦₁≤ 𝑦₂,𝑥₂ ≤ 𝑥₃and𝑦₂ ≤ 𝑦₃.

Since≤is an order onℕ, we get that𝑥₁≤ 𝑥₃and𝑦₁≤ 𝑦₃. Thus(𝑥₁, 𝑦₁) ≺ (𝑥₃, 𝑦₃).

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4 Homework Questions – Week 1

2. Note that(1, 0) ≺ (0, 1)and(0, 1) ≺ (1, 0)are both false. Hence≺is not a total order onℕ².

Method 1:using the definition.

1. Let𝑎, 𝑏, 𝑐, 𝑑 ∈ ℕ. Assume that𝑎 ≤ 𝑏and𝑐 ≤ 𝑑.

Then there exist𝑘, 𝑙 ∈ ℕsuch that𝑏 = 𝑎 + 𝑘and𝑑 = 𝑐 + 𝑙.

Hence𝑏 + 𝑑 = 𝑎 + 𝑘 + 𝑐 + 𝑙 = (𝑎 + 𝑐) + (𝑘 + 𝑙)with𝑘 + 𝑙 ∈ ℕ.

Thus𝑎 + 𝑐 ≤ 𝑏 + 𝑑.

2. Let𝑎, 𝑏, 𝑐, 𝑑 ∈ ℕ. Assume that𝑎 ≤ 𝑏and𝑐 ≤ 𝑑.

Then there exist𝑘, 𝑙 ∈ ℕsuch that𝑏 = 𝑎 + 𝑘and𝑑 = 𝑐 + 𝑙.

Hence𝑏𝑑 = (𝑎 + 𝑘)(𝑐 + 𝑙) = 𝑎𝑐 + (𝑎𝑙 + 𝑘𝑐 + 𝑘𝑙)with𝑎𝑙 + 𝑘𝑐 + 𝑘𝑙 ∈ ℕ.

Thus𝑎𝑐 ≤ 𝑏𝑑.

Method 2:using the properties proved in class.

1. Let𝑎, 𝑏, 𝑐, 𝑑 ∈ ℕbe such that𝑎 ≤ 𝑏and𝑐 ≤ 𝑑.

Then𝑎 ≤ 𝑏 ⟹ 𝑎 + 𝑐 ≤ 𝑏 + 𝑐and𝑐 ≤ 𝑑 ⟹ 𝑏 + 𝑐 ≤ 𝑏 + 𝑑. Finally

{

𝑎 + 𝑐 ≤ 𝑏 + 𝑐

𝑏 + 𝑐 ≤ 𝑏 + 𝑑 ⟹ 𝑎 + 𝑐 ≤ 𝑏 + 𝑑.

2. Let𝑎, 𝑏, 𝑐, 𝑑 ∈ ℕbe such that𝑎 ≤ 𝑏and𝑐 ≤ 𝑑.

Then𝑎 ≤ 𝑏 ⟹ 𝑎𝑐 ≤ 𝑏𝑐and𝑐 ≤ 𝑑 ⟹ 𝑏𝑐 ≤ 𝑏𝑑.

Finally {

𝑎𝑐 ≤ 𝑏𝑐

𝑏𝑐 ≤ 𝑏𝑑 ⟹ 𝑎𝑐 ≤ 𝑏𝑑.

• The statement is false for𝑐 = 0, indeed,2 × 0 ≤ 1 × 0but2 ≤ 1is false.

• The statement is true for𝑐 ≠ 0. We are going to prove the contrapositive,∀𝑎, 𝑏 ∈ ℕ, 𝑏 < 𝑎 ⟹ 𝑏𝑐 < 𝑎𝑐.

Let𝑎, 𝑏 ∈ ℕbe such that𝑏 < 𝑎. Then𝑏 ≤ 𝑎and hence𝑏𝑐 ≤ 𝑎𝑐.

Assume by contradiction that𝑏𝑐 = 𝑎𝑐then𝑏 = 𝑎since𝑐 ≠ 0. Hence𝑏𝑐 < 𝑎𝑐as expected.

Assume by contradiction that the set𝐸 = {𝑛 ∈ ℕ ∶ 0 < 𝑛 < 1}is not empty.

Then, by the well-ordering principle,𝐸admits a least element, i.e. there exists𝑙 ∈ 𝐸such that∀𝑛 ∈ ℕ, 𝑙 ≤ 𝑛.

Since𝑙 ∈ 𝐸, we get that𝑙 < 1. Note that0 ∉ 𝐸, so𝑙 ≠ 0. Hence𝑙 < 1 ⟹ 𝑙²< 𝑙.

We know that if0 = 𝑙²= 𝑙 × 𝑙then𝑙 = 0. Hence𝑙²is positive.

Finally0 < 𝑙² < 𝑙 < 1. So𝑙² ∈ 𝐸which contradicts the fact that𝑙is the least element of𝐸.