Concepts in Abstract Mathematics

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Concepts in Abstract Mathematics

Homework questions – Week 2

Jean-Baptiste Campesato

January 25^th, 2021 to January 29^th, 2021

Exercise 1.

1. Prove that∀𝑛 ∈ ℕ, ∃𝑘 ∈ ℕ, 𝑛³+ 2𝑛 = 3𝑘.

2. Prove that∀𝑛 ∈ ℕ,

𝑛

∑𝑘=0

𝑘

2^𝑘 = 2 −𝑛 + 2 2^𝑛 . Exercise 2.

We define a sequence(𝑢_𝑛)_𝑛≥1by𝑢₁ = 3and∀𝑛 ∈ ℕ ⧵ {0}, 𝑢_𝑛+1= 2 𝑛

𝑛

∑𝑘=1

𝑢_𝑘. Prove that∀𝑛 ∈ ℕ ⧵ {0}, 𝑢_𝑛= 3𝑛.

Exercise 3. Bernoulli’s inequality.

Prove that∀𝑥 ∈ [−1, +∞), ∀𝑛 ∈ ℕ, (1 + 𝑥)^𝑛 ≥ 1 + 𝑛𝑥.

(here we consider the usual order≥onℝ) Exercise 4.

For𝑛 ∈ ℕ, we define the statement𝑃 (𝑛)by2^𝑛> 𝑛². 1. Prove that∀𝑛 ≥ 3, 𝑃 (𝑛) ⟹ 𝑃 (𝑛 + 1).

2. For which𝑛 ∈ ℕ, is𝑃 (𝑛)true?

Exercise 5.

What do you think about the following proof by induction?

We want to prove that for any𝑛 ≥ 2,𝑛distinct points of the plane are always on the same line.

Proof:

• Base case: when𝑛 = 2the property is known to be true.

• Induction step: we assume that the property is true for some𝑛 ≥ 2and we want to show that it also holds for𝑛 + 1.

Let𝐴₁, 𝐴₂, … , 𝐴_𝑛+1be𝑛 + 1distinct points of the plane. By the induction hypothesis, we have – 𝐴₁, 𝐴₂, … , 𝐴_𝑛are on the same line𝐿.

– 𝐴₂, 𝐴₃, … , 𝐴_𝑛+1are on the same line𝐿^′.

Then𝐴₂, 𝐴₃, … , 𝐴_𝑛are at the same time on𝐿and𝐿^′so that𝐿 = 𝐿^′. Thus𝐴₁, … , 𝐴_𝑛+1are on the same line. Which ends the induction step.

Exercise 6.

Given𝑛 ∈ ℕ ⧵ {0}, prove that there exists a unique couple(𝑎, 𝑏) ∈ ℕsuch that𝑛 = 2^𝑎(2𝑏 + 1).

Exercise 7.

Find all the increasing functions𝑓 ∶ ℕ → ℕsuch that𝑓 (2) = 2and∀𝑝, 𝑞 ∈ ℕ, 𝑓 (𝑝𝑞) = 𝑓 (𝑝)𝑓 (𝑞).

Recall that a function𝑓 ∶ ℕ → ℕis increasing if∀𝑥, 𝑦 ∈ ℕ, 𝑥 < 𝑦 ⟹ 𝑓 (𝑥) < 𝑓 (𝑦).

Exercise 8.

1. Prove that if𝑆 ⊂ ℤadmits a greatest element then it is unique.

2. Prove that a non-empty finite subset ofℤadmits a greatest element.

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Exercise 9.

Let𝑛 ∈ ℕ ⧵ {0}. Prove that if one square of a2^𝑛× 2^𝑛chessboard is removed, then the remaining squares can be covered by L-shaped trominoes.

Figure 1: A8 × 8chessboard with a removed square. Figure 2:An𝐿-shaped tromino on a chessboard.

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Sample solutions to Exercise 1.

1. We are going to prove by induction that∀𝑛 ∈ ℕ, ∃𝑘 ∈ ℕ, 𝑛³+ 2𝑛 = 3𝑘.

• Base case at𝑛 = 0:0³+ 2 × 0 = 3 × 0.

• Induction step:assume that for some𝑛 ∈ ℕthere exists𝑘 ∈ ℕsuch that𝑛³+ 2𝑛 = 3𝑘. Then (𝑛 + 1)³+ 2(𝑛 + 1) = 𝑛³+ 3𝑛²+ 3𝑛 + 1 + 2𝑛 + 2

= 3𝑘 + 3𝑛²+ 3𝑛 + 3 by the induction hypothesis

= 3(𝑘 + 𝑛²+ 𝑛 + 1)

The induction step is proved since𝑘 + 𝑛²+ 𝑛+ ∈ ℕ.

2. We are going to prove by induction that∀𝑛 ∈ ℕ,

𝑛

∑𝑘=0

𝑘

2^𝑘 = 2 −𝑛 + 2 2^𝑛 .

• Base case at𝑛 = 0:

0 𝑘=0∑

𝑘

2^𝑘 = 0and2 −⁰⁺²

2⁰ = 2 − 2 = 0.

• Induction step:assume that

𝑛

∑𝑘=0

𝑘

2^𝑘 = 2 −𝑛 + 2

2^𝑛 for some𝑛 ∈ ℕ.

𝑛+1

∑𝑘=0

𝑘 2^𝑘 =

𝑛

∑𝑘=0

𝑘

2^𝑘 +𝑛 + 1 2^𝑛+1

= 2 −𝑛 + 2

2^𝑛 +𝑛 + 1

2^𝑛+1 by the induction hypothesis

= 2 −2𝑛 + 4 − 𝑛 − 1

2^𝑛+1 = 2 −(𝑛 + 1) + 2 2^𝑛+1 which ends the induction step.

We are going to prove by (strong) induction that∀𝑛 ≥ 1, 𝑢_𝑛= 3𝑛.

• Base case at𝑛 = 1:𝑢₁ = 3 × 1.

• Induction step:assume that𝑢_𝑘= 3𝑘for𝑘 = 1, … , 𝑛where𝑛 ≥ 1. Then 𝑢_𝑛+1= 2

𝑛 + 1

𝑛

∑𝑘=1

𝑢_𝑘

= 2 𝑛

𝑛

∑𝑘=1

3𝑘 by the induction hypothesis

= 6 𝑛

𝑛

∑𝑘=1

𝑘 = 6 𝑛

𝑛(𝑛 + 1)

2 = 3(𝑛 + 1) which ends the induction step.

Let𝑥 ∈ [−1, +∞). We are going to prove by induction that∀𝑛 ∈ ℕ, (1 + 𝑥)^𝑛≥ 1 + 𝑛𝑥.

• Base case at𝑛 = 0:(1 + 𝑥)⁰= 1and1 + 0 × 𝑥 = 1.

• Induction step:assume that(1 + 𝑥)^𝑛 ≥ 1 + 𝑛𝑥for some𝑛 ∈ ℕ. Then (1 + 𝑥)^𝑛+1= (1 + 𝑥)^𝑛(1 + 𝑥)

≥ (1 + 𝑛𝑥)(1 + 𝑥) by the induction hypothesis since1 + 𝑥 ≥ 0

= 1 + 𝑥 + 𝑛𝑥 + 𝑛𝑥²

≥ 1 + 𝑥 + 𝑛𝑥 = 1 + (𝑛 + 1)𝑥

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which ends the induction step.

1. Let𝑛 ≥ 3. Assume that𝑃 (𝑛)is true, i.e. 2^𝑛> 𝑛², and let’s prove𝑃 (𝑛 + 1), i.e. 2^𝑛+1> (𝑛 + 1)².

From the assumption, we get that2^𝑛+1= 2 × 2^𝑛≥ 2𝑛². 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 +

Since𝑛 ≥ 3 > 1 + √2, we know that𝑛²− 2𝑛 − 1 > 0. Hence𝑃 (𝑛 + 1)holds.

2. 𝑃 (3)and𝑃 (4)are false, but𝑃 (5)is true. So by induction,∀𝑛 ≥ 5, 𝑃 (𝑛)is true.

Beware:even if the induction step is true for𝑛 ≥ 3, we can only start the induction proof at𝑛 = 5! The base case is very important in a proof by induction.

The induction step is false when 𝑛 = 2 (it only holds for 𝑛 ≥ 3). Indeed, for 𝑛 = 2, we only have that 𝐴₁, 𝐴₂∈ 𝐿and that𝐴₂, 𝐴₃∈ 𝐿^′. Which is not enough to get that𝐿 = 𝐿^′since we only know that they have one point in common (it works if they have at least two points in common).

Beware: if you start an induction proof with a base case at𝑛₀, you have to make sure that the induction step𝑃 (𝑛) ⟹ 𝑃 (𝑛 + 1)holds for every𝑛 ≥ 𝑛₀. Otherwise, you didn’t prove anything…

Existence.We are going to prove the existence of such a couple(𝑎, 𝑏)by a strong induction on𝑛.

• Base case at𝑛 = 1:1 = 2⁰(2 × 0 + 1).

• Induction step.Assume that for1, 2, … , 𝑛admit such an expression for some𝑛 ≥ 1.

– First case: 𝑛 + 1is even, i.e. 𝑛 + 1 = 2𝑘for some𝑘 ∈ ℕ.

Note that𝑘 ≠ 0since otherwise1 ≤ 𝑛 + 1 = 0.

Since1 < 2and𝑘 ≠ 0, we get that𝑘 < 2𝑘 = 𝑛 + 1, so that𝑘 ≤ 𝑛.

Hence, by the induction hypothesis,𝑘 = 2^𝑎(2𝑏 + 1)for some(𝑎, 𝑏) ∈ ℕ². Then𝑛 + 1 = 2𝑘 = 2^𝑎+1(2𝑏 + 1).

– Second case:𝑛 + 1is odd, i.e.𝑛 + 1 = 2𝑘 + 1for some𝑘 ∈ ℕ. But then𝑛 + 1 = 2⁰(2 × 𝑘 + 1).

Which ends the induction step.

Uniqueness.Assume that2^𝑎(2𝑏 + 1) = 2^𝛼(2𝛽 + 1)for𝑎, 𝑏, 𝛼, 𝛽 ∈ ℕ.

If𝑎 < 𝛼then, by cancellation, we obtain2𝑏 + 1 = 2^𝛼−𝑎(2𝛽 + 1). Which is impossible since the LHS is odd whereas the RHS is even.

If𝛼 < 𝑎then, by cancellation, we obtain2^𝑎−𝛼(2𝑏 + 1) = 2𝛽 + 1. Which is impossible since the RHS is odd whereas the LHS is even.

Therefore𝑎 = 𝛼, and by cancellation we obtain2𝑏 + 1 = 2𝛽 + 1, hence2𝑏 = 2𝛽and finally𝑏 = 𝛽.

We proved that(𝑎, 𝑏) = (𝛼, 𝛽).

The function𝑓 ∶ ℕ → ℕdefined by𝑓 (𝑛) = 𝑛satisfies the conditions of the question. Actually, as we are going to prove, it is the only one.

From now on, we assume that𝑓 ∶ ℕ → ℕsatisfigies𝑓 (2) = 2and∀𝑝, 𝑞 ∈ ℕ, 𝑓 (𝑝𝑞) = 𝑓 (𝑝)𝑓 (𝑞), and we want to prove that∀𝑛 ∈ ℕ, 𝑓 (𝑛) = 𝑛.

• We know that0 < 1 < 2hence0 ≤ 𝑓 (0) < 𝑓 (1) < 𝑓 (2) = 2.

Therefore, the only possibility is that𝑓 (0) = 0and𝑓 (1) = 1.

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• Let’s prove by strong induction that∀𝑛 ∈ ℕ, 𝑓 (𝑛) = 𝑛.

– Base case at𝑛 = 0:𝑓 (0) = 0.

– Induction step.Assume that𝑓 (0) = 0, 𝑓 (1) = 1, 𝑓 (2) = 2, 𝑓 (3) = 3, … , 𝑓 (𝑛) = 𝑛for some𝑛 ≥ 0.

∗ First case: 𝑛 + 1is even, i.e. there exists𝑘 ∈ ℕsuch that𝑛 + 1 = 2𝑘.

Note that𝑘 ≠ 0since otherwise1 ≤ 𝑛 + 1 = 0.

Since1 < 2and𝑘 ≠ 0, we get that𝑘 < 2𝑘 = 𝑛 + 1, so that𝑘 ≤ 𝑛.

Then, by the induction hypothesis,𝑓 (𝑛 + 1) = 𝑓 (2𝑘) = 𝑓 (2)𝑓 (𝑘) = 2𝑘 = 𝑛 + 1.

∗ Second case:𝑛 + 1is odd, i.e. there exists𝑘 ∈ ℕsuch that𝑛 + 1 = 2𝑘 + 1.

Either𝑘 = 0and then𝑓 (𝑛 + 1) = 𝑓 (1) = 1 = 𝑛 + 2.

Or𝑘 ≠ 0and then𝑘 + 1 < 2𝑘 + 1 = 𝑛 + 1, i.e.𝑘 ≤ 𝑛.

Then𝑓 (𝑛 + 2) = 𝑓 (2(𝑘 + 1)) = 𝑓 (2)𝑓 (𝑘 + 1) = 𝑛 + 2by the induction hypothesis.

Thus𝑛 = 𝑓 (𝑛) < 𝑓 (𝑛 + 1) ≤ 𝑓 (𝑛 + 2) = 𝑛 + 2.

The only possible value is that𝑓 (𝑛 + 1) = 𝑛 + 1.

1. Let𝑚, 𝑚^′∈ 𝑆 be two greatest elements of𝑆.

Since𝑚 ∈ 𝑆and𝑚^′is a greatest element, we have𝑚 ≤ 𝑚^′.

Similarly, since𝑚^′∈ 𝑆 and𝑚is a greatest element of𝑆, we have𝑚^′≤ 𝑚.

Hence𝑚 = 𝑚^′.

That’s why we saythegreatest element: if it exists, it is unique (whereas we sayanupper bound).

2. Let’s prove that a non-empty finite subset𝑆 ⊂ ℤhas a greatest element, by induction on𝑛 = #𝑆.

• Base case at𝑛 = 1:if𝑆is a singleton, then its unique element is its greatest element.

• Induction step.Assume that the statement holds for sets of cardinal𝑛, for some𝑛 ≥ 1.

Let𝑆 ⊂ ℤbe such that#𝑆 = 𝑛 + 1.

Particulalry𝑆 ≠ ∅, so there exists𝑎 ∈ 𝑆.

Set𝑇 = 𝑆 ⧵ {𝑎}. Then#𝑇 = 𝑛, so by the induction hypothesis𝑇 admits a greatest element𝑚 ∈ 𝑇. I claim that𝑀 =max(𝑚, 𝑎) ∈ 𝑇 ∪ {𝑎} = 𝑆is the greatest element of𝑆.

Indeed, let𝑛 ∈ 𝑆, either𝑛 = 𝑎and then𝑎 ≤ 𝑀, or𝑛 ∈ 𝑇 and then𝑛 ≤ 𝑚 ≤ 𝑀.

Which ends the inductive step.

We are going to prove by induction on𝑛 ≥ 1that if one square of a2^𝑛× 2^𝑛chessboard is removed, then the remaining squares can be covered with L-shaped trominoes.

• Base case at𝑛 = 1.There are only four possible cases and for each of them the remaining is exactly one 𝐿-shaped tromino:

• Assume that the statement holds for some𝑛 ≥ 1and consider a2^𝑛+1× 2^𝑛+1chessboard with a removed square.

We may split this chessboard into four2^𝑛× 2^𝑛chessboards as follows:

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We may place an𝐿-shaped tromino such that it covers the corner situated at the center for each2^𝑛× 2^𝑛 chessboard without a removed square, see below.

Now, each of the2^𝑛× 2^𝑛chessboards has a removed square: we may apply the induction hypothesis in order to cover the remaining squares with𝐿-shaped trominoes.