Concepts in Abstract Mathematics
Homework questions – Week 10
Jean-Baptiste Campesato March 29th, 2021 to April 2nd, 2021 Exercise 1.
Let𝐸be a set.
1. Prove that∀𝐴, 𝐵, 𝐶 ∈ 𝒫(𝐸), 𝐴 ∪ 𝐵 = 𝐵 ∩ 𝐶 ⟹ 𝐴 ⊂ 𝐵 ⊂ 𝐶.
2. Prove that∀𝐴, 𝐵 ∈ 𝒫(𝐸), 𝐴 ∩ 𝐵 = 𝐴 ∪ 𝐵 ⟹ 𝐴 = 𝐵.
Exercise 2.
Let𝑓 ∶ 𝐴 → 𝐵,𝑔 ∶ 𝐵 → 𝐶 andℎ ∶ 𝐶 → 𝐷be three functions.
Prove that𝑔 ∘ 𝑓 andℎ ∘ 𝑔are bijective if and only if𝑓,𝑔andℎare bijective.
Exercise 3.
Let𝑓 ∶ 𝐸 → 𝐹.
1. Prove that∀𝐴 ∈ 𝒫(𝐸), 𝐴 ⊂ 𝑓−1(𝑓 (𝐴)).
2. Prove that∀𝐵 ∈ 𝒫(𝐹 ), 𝑓 (𝑓−1(𝐵)) ⊂ 𝐵.
3. Can these inclusions be strict?
Exercise 4.
Let𝑓 ∶ 𝐸 → 𝐹.
1. Prove that∀𝐴, 𝐵 ∈ 𝒫(𝐹 ), 𝐴 ⊂ 𝐵 ⟹ 𝑓−1(𝐴) ⊂ 𝑓−1(𝐵).
Does the converse hold?
2. Prove that∀𝐴, 𝐵 ∈ 𝒫(𝐹 ), 𝑓−1(𝐴 ∩ 𝐵) = 𝑓−1(𝐴) ∩ 𝑓−1(𝐵).
3. Prove that∀𝐴, 𝐵 ∈ 𝒫(𝐹 ), 𝑓−1(𝐴 ∪ 𝐵) = 𝑓−1(𝐴) ∪ 𝑓−1(𝐵).
Exercise 5.
Let𝑓 ∶ 𝐸 → 𝐹.
1. Prove that∀𝐴, 𝐵 ∈ 𝒫(𝐸), 𝐴 ⊂ 𝐵 ⟹ 𝑓 (𝐴) ⊂ 𝑓 (𝐵).
Does the converse hold?
2. Prove that∀𝐴, 𝐵 ∈ 𝒫(𝐸), 𝑓 (𝐴 ∩ 𝐵) ⊂ 𝑓 (𝐴) ∩ 𝑓 (𝐵).
Can the inclusion be strict?
3. Prove that∀𝐴, 𝐵 ∈ 𝒫(𝐸), 𝑓 (𝐴 ∪ 𝐵) = 𝑓 (𝐴) ∪ 𝑓 (𝐵).
Exercise 6.
Let𝑓 ∶ 𝐸 → 𝐹. Prove that𝑓 is injective if and only if∀𝐴, 𝐵 ∈ 𝒫(𝐸), 𝑓 (𝐴 ∩ 𝐵) = 𝑓 (𝐴) ∩ 𝑓 (𝐵).
Exercise 7.
Let𝐸be a finite set. For𝐴, 𝐵 ∈ 𝒫(𝐸)we define the symmetric difference of𝐴and𝐵by𝐴Δ𝐵 = (𝐴∪𝐵)⧵(𝐴∩𝐵).
Prove that∀𝐴, 𝐵 ∈ 𝒫(𝐸), |𝐴Δ𝐵| = |𝐴| + |𝐵| − 2|𝐴 ∩ 𝐵|.
Exercise 8.
Let𝐸and𝐹 be two finite sets.
1. Prove that𝐹𝐸 (the set of functions𝐸 → 𝐹) is finite and express|𝐹𝐸|in terms of|𝐸|and|𝐹 |.
2. Prove that the set{𝑓 ∈ 𝐸𝐹 ∶ 𝑓 is injective}is finite and express its cardinal in terms of|𝐸|and|𝐹 |.
3. Prove that the set{𝑓 ∈ 𝐸𝐸 ∶ 𝑓 is bijective}is finite and express its cardinal in terms of|𝐸|.
The case of surjective functions is more tricky.
Let𝐸be a finite set and𝑘 ∈ {0, 1, … , |𝐸|}. What is the cardinal of{𝐴 ∈ 𝒫(𝐸) ∶ |𝐴| = 𝑘}?
Exercise 10.
Prove that a set𝐸is finite if and only if𝒫(𝐸)is finite.
In this case, give an expression of|𝒫(𝐸)|in terms of|𝐸|.
Exercise 11. The pigeonhole principle or Dirichlet’s drawer principle I had no enough time to cover this topic in lectures, so here it is :-).
1. Let𝐸and𝐹 be two finite sets. Prove that|𝐸| ≤ |𝐹 |if and only if there exists an injection𝑓 ∶ 𝐸 → 𝐹. 2. Let𝐸and𝐹 be two finite sets. Prove that if|𝐸| > |𝐹 |then there is no injective function𝐸 → 𝐹.
This statement is pigeonhole principle or Dirichlet’s drawer principle: if you have𝑛elements put in𝑘 < 𝑛boxes, then at least one box contains two elements.
3. During a post-covid party with𝑛 > 1participants, we may always find two people who shook hands to the same number of people.
4. Let𝑛 ∈ ℕ ⧵ {0}. Let𝑎1, 𝑎2, … , 𝑎𝑛 ∈ ℤ. Prove that there exists distinct𝑖1, … , 𝑖𝑟 ∈ {1, … , 𝑛}, 𝑟 ≥ 1,so that𝑛| ∑𝑟𝑘=1𝑎𝑖
𝑘.
5. Prove that among 13 distinct real numbers, there always exist two𝑥, 𝑦satisfying0 < 𝑥 − 𝑦
1 + 𝑥𝑦 < 2 − √3.
Hint: it looks like a trigonometric formula you know!
Sample solutions to Exercise 1.
1. Let𝐴, 𝐵, 𝐶 ∈ 𝒫(𝐸). Assume that𝐴 ∪ 𝐵 = 𝐵 ∩ 𝐶.
Let𝑥 ∈ 𝐴then𝑥 ∈ 𝐴 ∪ 𝐵 = 𝐵 ∩ 𝐶. Therefore𝑥 ∈ 𝐵. So𝐴 ⊂ 𝐵. Let𝑥 ∈ 𝐵then𝑥 ∈ 𝐴 ∪ 𝐵 = 𝐵 ∩ 𝐶. Therefore𝑥 ∈ 𝐶. So𝐵 ⊂ 𝐶. 2. Using the previous question.
Let𝐴, 𝐵 ∈ 𝒫(𝐸). Assume that𝐴 ∩ 𝐵 = 𝐴 ∪ 𝐵.
From the previous question we get that𝐴 ⊂ 𝐵 ⊂ 𝐴. Hence𝐴 = 𝐵.
Direct proof.
Let𝐴, 𝐵 ∈ 𝒫(𝐸). Assume that𝐴 ∩ 𝐵 = 𝐴 ∪ 𝐵.
Let𝑥 ∈ 𝐴then𝑥 ∈ 𝐴 ∪ 𝐵 = 𝐴 ∩ 𝐵. Thus𝑥 ∈ 𝐵. Therefore𝐴 ⊂ 𝐵.
Let𝑥 ∈ 𝐵then𝑥 ∈ 𝐴 ∪ 𝐵 = 𝐴 ∩ 𝐵. Thus𝑥 ∈ 𝐴. Therefore𝐵 ⊂ 𝐴.
Hence𝐴 = 𝐵.
Proof by contrapositive.
Let𝐴, 𝐵 ∈ 𝒫(𝐸). Assume that𝐴 ≠ 𝐵. Then
• either𝐴 ⧵ 𝐵 ≠ ∅and then there exists𝑥 ∈ 𝐸 such that𝑥 ∈ 𝐴and𝑥 ∉ 𝐵. Thus𝑥 ∈ 𝐴 ∪ 𝐵 but 𝑥 ∉ 𝐴 ∩ 𝐵. Therefore𝐴 ∩ 𝐵 ≠ 𝐴 ∪ 𝐵.
• or𝐵 ⧵𝐴 ≠ ∅and then there exists𝑥 ∈ 𝐸such that𝑥 ∈ 𝐵and𝑥 ∉ 𝐴. Thus𝑥 ∈ 𝐴∪𝐵but𝑥 ∉ 𝐴∩𝐵.
Therefore𝐴 ∩ 𝐵 ≠ 𝐴 ∪ 𝐵.
Sample solutions to Exercise 2.
⇐Assume that𝑓,𝑔andℎare bijective then𝑔 ∘ 𝑓 andℎ ∘ 𝑔are too.
⇒Assume that𝑔 ∘ 𝑓andℎ ∘ 𝑔are bijective.
Since𝑔 ∘ 𝑓 is surjective,𝑔is too. Sinceℎ ∘ 𝑔is injective,𝑔is too.
Hence𝑔is bijective, so it admits an inverse𝑔−1∶ 𝐶 → 𝐵.
Then𝑓 = 𝑔−1∘ (𝑔 ∘ 𝑓 )andℎ = (ℎ ∘ 𝑔) ∘ 𝑔−1are bijective as composition of bijective functions.
Sample solutions to Exercise 3.
1. Let𝐴 ∈ 𝒫(𝐸). Let𝑥 ∈ 𝐴. Then𝑓 (𝑥) ∈ 𝑓 (𝐴). Therefore𝑥 ∈ 𝑓−1(𝑓 (𝐴)).
We proved that𝐴 ⊂ 𝑓−1(𝑓 (𝐴)).
2. Let 𝐵 ⊂ 𝒫(𝐹 ). Let𝑦 ∈ 𝑓 (𝑓−1(𝐵)). Then there exists 𝑥 ∈ 𝑓−1(𝐵) such that 𝑦 = 𝑓 (𝑥). But since 𝑥 ∈ 𝑓−1(𝐵),𝑦 = 𝑓 (𝑥) ∈ 𝐵.
We proved that𝑓 (𝑓−1(𝐵)) ⊂ 𝐵.
3. Define
𝑓 ∶⎧⎪
⎨⎪
⎩
{1, 2} → {1, 2}
1 ↦ 1
2 ↦ 1
Then𝑓 (𝑓−1({1, 2})) = 𝑓 ({1, 2}) = {1} ⊊ {1, 2}.
And𝑓−1(𝑓 ({1})) = 𝑓−1({1}) = {1, 2} ⊋ {1}.
Sample solutions to Exercise 4.
1. Let𝐴, 𝐵 ∈ 𝒫(𝐹 )be such that𝐴 ⊂ 𝐵. Take𝑥 ∈ 𝑓−1(𝐴). Then𝑓 (𝑥) ∈ 𝐴 ⊂ 𝐵. Thus𝑥 ∈ 𝑓−1(𝐵).
The converse doesn’t hold. Indeed, define 𝑓 ∶{
{1} → {1, 2}
1 ↦ 1
then𝑓−1({2}) = ∅ ⊂ 𝑓−1({1}). But{2} ⊄ {1}.
2. Let𝐴, 𝐵 ∈ 𝒫(𝐹 ).
Since𝐴 ∩ 𝐵 ⊂ 𝐴and𝐴 ∩ 𝐵 ⊂ 𝐵, we get that𝑓−1(𝐴 ∩ 𝐵) ⊂ 𝑓−1(𝐴)and𝑓−1(𝐴 ∩ 𝐵) ⊂ 𝑓−1(𝐵). Thus 𝑓−1(𝐴 ∩ 𝐵) ⊂ 𝑓−1(𝐴) ∩ 𝑓−1(𝐵).
For the other inclusion, let𝑥 ∈ 𝑓−1(𝐴) ∩ 𝑓−1(𝐵). Then𝑓 (𝑥) ∈ 𝐴and𝑓 (𝑥) ∈ 𝐵. Thus𝑓 (𝑥) ∈ 𝐴 ∩ 𝐵so that𝑥 ∈ 𝑓−1(𝐴 ∩ 𝐵). Thus𝑓−1(𝐴) ∩ 𝑓−1(𝐵) ⊂ 𝑓−1(𝐴 ∩ 𝐵).
3. Let𝐴, 𝐵 ∈ 𝒫(𝐹 ).
Since𝐴, 𝐵 ⊂ 𝐴 ∪ 𝐵, we get𝑓−1(𝐴), 𝑓−1(𝐵) ⊂ 𝑓−1(𝐴 ∪ 𝐵). Thus𝑓−1(𝐴) ∪ 𝑓−1(𝐵) ⊂ 𝑓−1(𝐴 ∪ 𝐵).
For the other inclusion, let𝑥 ∈ 𝑓−1(𝐴 ∪ 𝐵). Then𝑓 (𝑥) ∈ 𝐴 ∪ 𝐵. Either𝑓 (𝑥) ∈ 𝐴and then𝑥 ∈ 𝑓−1(𝐴) ⊂ 𝑓−1(𝐴) ∪ 𝑓−1(𝐵)or𝑓 (𝑥) ∈ 𝐵and then𝑥 ∈ 𝑓−1(𝐵) ⊂ 𝑓−1(𝐴) ∪ 𝑓−1(𝐵). Thus𝑥 ∈ 𝑓−1(𝐴) ∪ 𝑓−1(𝐵). We proved that𝑓−1(𝐴 ∪ 𝐵) ⊂ 𝑓−1(𝐴) ∪ 𝑓−1(𝐵).
Sample solutions to Exercise 5.
1. Let𝐴, 𝐵 ∈ 𝒫(𝐸)be such that𝐴 ⊂ 𝐵. Let𝑦 ∈ 𝑓 (𝐴). Then𝑦 = 𝑓 (𝑥)for some𝑥 ∈ 𝐴. But𝑥 ∈ 𝐴 ⊂ 𝐵.
Thus𝑦 = 𝑓 (𝑥) ∈ 𝑓 (𝐵). We proved that𝑓 (𝐴) ⊂ 𝑓 (𝐵).
The converse doesn’t hold. Indeed define
𝑓 ∶⎧⎪
⎨⎪
⎩
{1, 2} → {1}
1 ↦ 1
2 ↦ 1
then𝑓 ({1}) = 𝑓 ({2}) = {1}but{1} ⊄ {2}.
2. Let𝐴, 𝐵 ∈ 𝒫(𝐸). Since𝐴 ∩ 𝐵 ⊂ 𝐴, 𝐵we get𝑓 (𝐴 ∩ 𝐵) ⊂ 𝑓 (𝐴), 𝑓 (𝐵). Thus𝑓 (𝐴 ∩ 𝐵) ⊂ 𝑓 (𝐴) ∩ 𝑓 (𝐵).
The inclusion can be strict using the same example as above.
3. Let𝐴, 𝐵 ∈ 𝒫(𝐸). Since𝐴, 𝐵 ⊂ 𝐴 ∪ 𝐵, we get𝑓 (𝐴), 𝑓 (𝐴) ⊂ 𝑓 (𝐴 ∪ 𝐵). Thus𝑓 (𝐴) ∪ 𝑓 (𝐵) ⊂ 𝑓 (𝐴 ∪ 𝐵).
For the other inclusion, let𝑦 ∈ 𝑓 (𝐴 ∪ 𝐵). Then𝑦 = 𝑓 (𝑥)for some𝑥 ∈ 𝐴 ∪ 𝐵. So either𝑥 ∈ 𝐴and then 𝑦 = 𝑓 (𝑥) ∈ 𝑓 (𝐴) ⊂ 𝑓 (𝐴) ∪ 𝑓 (𝐵), or𝑥 ∈ 𝐵 and then𝑦 = 𝑓 (𝑥) ∈ 𝑓 (𝐵) ⊂ 𝑓 (𝐴) ∪ 𝑓 (𝐵). In both cases 𝑦 ∈ 𝑓 (𝐴) ∪ 𝑓 (𝐵). So we proved that𝑓 (𝐴 ∪ 𝐵) ⊂ 𝑓 (𝐴) ∪ 𝑓 (𝐵).
Sample solutions to Exercise 6.
⇒Assume that𝑓 is injective. Let𝐴, 𝐵 ∈ 𝒫(𝐸).
We already know that𝑓 (𝐴 ∩ 𝐵) ⊂ 𝑓 (𝐴) ∩ 𝑓 (𝐵)holds (see the previous exercise).
Let’s prove that𝑓 (𝐴) ∩ 𝑓 (𝐵) ⊂ 𝑓 (𝐴 ∩ 𝐵).
Let𝑦 ∈ 𝑓 (𝐴) ∩ 𝑓 (𝐵). Then𝑦 = 𝑓 (𝑥1)for some𝑥1 ∈ 𝐴and𝑦 = 𝑓 (𝑥2)for some𝑥2 ∈ 𝐵.
Since𝑓 (𝑥1) = 𝑓 (𝑥2)and𝑓 is injective, we obtain that𝑥1= 𝑥2∈ 𝐴 ∩ 𝐵. Therefore𝑦 = 𝑓 (𝑥1) ∈ 𝑓 (𝐴 ∩ 𝐵).
We proved that𝑓 (𝐴) ∩ 𝑓 (𝐵) ⊂ 𝑓 (𝐴 ∩ 𝐵). Thus𝑓 (𝐴) ∩ 𝑓 (𝐵) = 𝑓 (𝐴 ∩ 𝐵).
⇐Assume that∀𝐴, 𝐵 ∈ 𝒫(𝐸), 𝑓 (𝐴 ∩ 𝐵) = 𝑓 (𝐴) ∩ 𝑓 (𝐵).
Let𝑥1, 𝑥2∈ 𝐸 be such that𝑓 (𝑥1) = 𝑓 (𝑥2). Set𝑦 ≔ 𝑓 (𝑥1) = 𝑓 (𝑥2).
Then𝑓 ({𝑥1} ∩ {𝑥2}) = 𝑓 ({𝑥1}) ∩ 𝑓 ({𝑥2}) = {𝑦} ∩ {𝑦} = {𝑦}.
Particularly{𝑥1} ∩ {𝑥2} ≠ ∅, thus𝑥1= 𝑥2. Sample solutions to Exercise 7.
|𝐴Δ𝐵| = |(𝐴 ∪ 𝐵) ⧵ (𝐴 ∩ 𝐵)|
= |𝐴 ∪ 𝐵| − |𝐴 ∩ 𝐵|
= |𝐴| + |𝐵| − |𝐴 ∩ 𝐵| − |𝐴 ∩ 𝐵|
= |𝐴| + |𝐵| − 2|𝐴 ∩ 𝐵|
Sample solutions to Exercise 8.
1. Let𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝐸.
Then𝜓 ∶ 𝐹𝐸 → 𝐹|𝐸|defined by𝜓(𝑓 ) = (𝑓 (𝜑(0)), … , 𝑓 (𝜑(|𝐸| − 1)))is a bijection (prove it).
Therefore|𝐹𝐸| = |𝐹|𝐸|| = |𝐹 ||𝐸|.
2. According to the last exercise, there exists an injective function𝐸 → 𝐹 if and only if|𝐸| ≤ |𝐹 |.
Next, since𝐸is finite, there exists a bijection𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝐸.
For𝑓 (𝜑(0))we have|𝐹 |possible choices. For𝑓 (𝜑(1))we have|𝐹 ⧵ {𝑓 (𝜑(0))}| = |𝐹 | − 1choices. For 𝑓 (𝜑(2))we have|𝐹 ⧵ {𝑓 (𝜑(0)), 𝑓 (𝜑(1))}| = |𝐹 | − 2choices. And so on.
Therefore,|{𝑓 ∈ 𝐸𝐹 ∶ 𝑓 is injective}| = |𝐹 |(|𝐹 | − 1) ⋯ (|𝐹 | − |𝐸| + 1) = (|𝐹 |−|𝐸|)!|𝐹 |! . Thus|{𝑓 ∈ 𝐸𝐹 ∶ 𝑓is injective}| =
{
0 if|𝐸| > |𝐹 |
|𝐹 |!
(|𝐹 |−|𝐸|)! if|𝐸| ≤ |𝐹 | .
3. It a special case of the above question when|𝐸| = |𝐹 |:|{𝑓 ∈ 𝐸𝐸 ∶ 𝑓 is bijective}| = |𝐸|!
(|𝐸|−|𝐸|)! = |𝐸|!.
Sample solutions to Exercise 9.
The number of subsets with cardinality𝑘included in a set of cardinality𝑛is denoted(𝑛𝑘)read ”𝑛choose 𝑘”.
We are going to prove that(𝑛𝑘) = (𝑛−𝑘)!𝑘!𝑛! .
Let𝐸a finite set. Set𝑛 = |𝐸|. Fix𝑘 ∈ {0, 1, … , 𝑛}.
An ordered list of𝑘distinct elements is the same as fixing an injection{0, 1, … , 𝑘 − 1} → 𝐸. So, using the previous question there are 𝑛!
(𝑛−𝑘)! such ordered lists.
Two ordered lists of 𝑘 elements give the same subset if and only if one is obtained from the other one permuting its elements, which is the same as constructing a bijection {0, 1, … , 𝑘 − 1} → {0, 1, … , 𝑘 − 1}.
From the previous question there are𝑘!such bijections.
Therefore(𝑛𝑘) =
𝑛!
(𝑛−𝑘)!
𝑘! = 𝑛!
(𝑛−𝑘)!𝑘!. Sample solutions to Exercise 10.
⇒
Method 1 (by induction):
Let’s prove by induction on𝑛 = |𝐸|that𝒫(𝐸)is finite and that|𝒫(𝐸)| = 2|𝐸|.
• Base case at𝑛 = 0:if𝐸 = ∅then𝒫(𝐸) = {∅}is finite.
• Induction step:assume that the statement holds for some𝑛 ∈ ℕ, i.e. if𝐸is a set with|𝐸| = 𝑛then𝒫(𝐸) is finite and|𝒫(𝐸)| = 2𝑛.
Let𝐸be a set such that|𝐸| = 𝑛 + 1. Since|𝐸| > 0, there exists𝑥 ∈ 𝐸.
By the induction hypothesis, since|𝐸 ⧵ {𝑥}| = 𝑛, we get that𝒫(𝐸 ⧵ {𝑥})is finite and|𝒫(𝐸 ⧵ {𝑥})| = 2𝑛. Note that𝒫(𝐸 ⧵ {𝑥}) = {𝐴 ∈ 𝒫(𝐸) ∶ 𝑥 ∉ 𝐴}and that
{𝐴 ∈ 𝒫(𝐸) ∶ 𝑥 ∉ 𝐴} → {𝐴 ∈ 𝒫(𝐸) ∶ 𝑥 ∈ 𝐴}
𝐴 ↦ 𝐴 ∪ {𝑥}
is a bijection.
Therefore𝒫(𝐸) = {𝐴 ∈ 𝒫(𝐸) ∶ 𝑥 ∉ 𝐴} ⊔ {𝐴 ∈ 𝒫(𝐸) ∶ 𝑥 ∈ 𝐴}is finite and
|𝒫(𝐸)| = |{𝐴 ∈ 𝒫(𝐸) ∶ 𝑥 ∉ 𝐴}| + |{𝐴 ∈ 𝒫(𝐸) ∶ 𝑥 ∈ 𝐴}| = 2𝑛+ 2𝑛= 2𝑛+1.
Method 2 (using the previous exercise):
Let𝐸be a finite set. We know that for𝑘 = 0, … , |𝐸|, the number of subsets with𝑘elements is(𝑘𝑛). Therefore the number of subsets included in𝐸is
|𝒫(𝐸)| =
|𝐸|
∑𝑘=0(𝑛 𝑘)=
|𝐸|
∑𝑘=0(𝑛
𝑘)1𝑘1|𝐸|−𝑘 = (1 + 1)|𝐸| = 2|𝐸|
Method 3 (which generalizes to infinite sets):
Let𝐸be a finite set. We define𝜓 ∶ 𝒫(𝐸) → {0, 1}𝐸 by𝜓(𝐴)(𝑥) = {
1 if𝑥 ∈ 𝐴 0 otherwise .
Then𝜓is a bijection thus𝒫(𝐸)is finite since{0, 1}𝐸is and moreover|𝒫(𝐸)| = |{0, 1}𝐸| = 2|𝐸|.
⇐Let𝐸be a set. Assume that𝒫(𝐸)is finite.
Note thatΦ ∶ 𝐸 → 𝒫(𝐸)defined byΦ(𝑥) = {𝑥}is injective. Therefore𝐸is finite too.
Sample solutions to Exercise 11.
1. ⇒Assume that|𝐸| ≤ |𝐹 |.
There exist bijections𝜑 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐸|} → 𝐸and𝜓 ∶ {𝑘 ∈ ℕ ∶ 𝑘 < |𝐹 |} → 𝐹. Since|𝐸| ≤ |𝐹 |,𝑓 = 𝜓 ∘ 𝜑−1∶ 𝐸 → 𝐹 is well-defined and injective.
⇒Assume that there exists an injection𝑓 ∶ 𝐸 → 𝐹.
Then𝑓 induces a bijection𝑓 ∶ 𝐸 → 𝑓 (𝐸), so that|𝐸| = |𝑓 (𝐸)|.
And since𝑓 (𝐸) ⊂ 𝐹, we have|𝑓 (𝐸)| ≤ |𝐹 |.
2. It is a consequence of the previous question.
3. A participant shook either0, 1,… or𝑛 − 1hands. So we have𝑛”boxes”. Not that it is not possible to have at the same time the boxes0and𝑛 − 1non-empty. Therefore we have only𝑛 − 1boxes for𝑛 participants, so two participants must have shaken the same number of boxes.
Formally:
• First case: there is at least one participant who didn’t shake any hand. Then𝑓 ∶ {participants} → {0, 1, … , 𝑛 − 2}mapping each participant to the number of hands he shook is well-defined. Since
|{participants}| = 𝑛 > 𝑛 − 1 = |{0, 1, … , 𝑛 − 2}|, 𝑓 can’t be injective. Therefore at least two participants shooke the same number of hands.
• Second case: all participants shooke at least one hand. Then𝑓 ∶ {participants} → {1, … , 𝑛 − 1}
mapping each participant to the number of hands he shook is well-defined. Since|{participants}| = 𝑛 > 𝑛 − 1 = |{1, 2, … , 𝑛 − 1}|,𝑓 can’t be injective. Therefore at least two participants shooke the same number of hands.
4. For𝑟 = 1, 2, … , 𝑛, set𝑠𝑟 = ∑𝑟𝑘=1𝑎𝑘.
• First case: there exists𝑟such that𝑛|𝑠𝑟. Then we are done.
• Second case: otherwise, we have𝑛numbers𝑠1, … , 𝑠𝑛 whose remainders for the Euclidean divi- sion by𝑛are among1, … , 𝑛 − 1(i.e. 𝑛 − 1possible remainders). Hence at least two have the same remainders, let’s say𝑠𝑝and𝑠𝑞with𝑞 > 𝑝. Then𝑛|𝑠𝑞− 𝑠𝑝= ∑𝑞𝑘=𝑝+1𝑎𝑘.
5. First, note that 1+tantan𝑎−tan𝑎 𝑏
tan𝑏 =tan(𝑎 − 𝑏)and that
tan 𝜋
12 =tan(𝜋 3 − 𝜋
4 )= tan𝜋
3 −tan𝜋
4
1 +tan𝜋
3tan𝜋
4
= √3 − 1 1 + √3
= (√3 − 1)
2
2 = 2 − √3
Let𝑥1, … , 𝑥13 be 13 distinct real numbers. We set 𝛼𝑘 = arctan𝑥𝑘 ∈ (−𝜋2,𝜋
2). Note that the𝛼𝑘 are distinct since arctan is injective.
Note that(−𝜋2,𝜋
2) = 𝐼1⊔ 𝐼2⊔ 𝐼3⊔ ⋯ ⊔ 𝐼12where 𝐼1= (−𝜋
2, −𝜋 2 + 𝜋
12 ], 𝐼2 = (−𝜋 2 + 𝜋
12, −𝜋 2 + 2𝜋
12 ], … , 𝐼11 = (−𝜋
2 + 10 𝜋 12,𝜋
2 + 11𝜋
12 ], 𝐼12 = (−𝜋
2 + 11 𝜋 12,𝜋
2 ) We define𝑓 ∶ {𝛼1, … , 𝛼13} → {1, … , 12}by𝑓 (𝛼𝑘) = 𝑟where𝛼𝑘∈ 𝐼𝑟.
Since|{𝛼1, … , 𝛼13}| = 13 > 12 = |{1, … , 12}|, 𝑓 is not injective. So there exists 𝑎𝑙𝑝ℎ𝑎𝑘 < 𝛼𝑙 and 𝑟 = 1, … , 12such that𝛼𝑘, 𝛼𝑙∈ 𝐼𝑟. Then0 < 𝛼𝑙− 𝛼𝑘< 𝜋
12. Since tan is increasing on(−𝜋2,𝜋
2), we get that tan0 <tan(𝛼𝑙− 𝛼𝑘) <tan 𝜋
12 = 2 − √3.
Note that tan(𝛼𝑙− 𝛼𝑘) = tan𝛼𝑙−tan𝛼𝑘
1+tan𝛼𝑙tan𝛼𝑘 = 𝑥𝑙−𝑥𝑘
1+𝑥𝑙𝑥𝑘. Thus0 < 𝑥𝑙−𝑥𝑘
1+𝑥𝑙𝑥𝑘 <= 2 − √3as requested.
