Concepts in Abstract Mathematics

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Concepts in Abstract Mathematics

Homework questions – Week 8

Jean-Baptiste Campesato March 15^th, 2021 to March 19^th, 2021

Exercise 1.

Prove that√7 + 4√3 + √7 − 4√3 ∈ ℕ.

Exercise 2.

1. Prove that∀𝑎, 𝑏 ∈ ℝ, 𝑎𝑏 ≤ 𝑎²+ 𝑏² 2 .

2. Prove that∀𝑎, 𝑏, 𝑐 ∈ ℝ, 𝑎𝑏 + 𝑏𝑐 + 𝑎𝑐 ≤ 𝑎²+ 𝑏²+ 𝑐². 3. Prove that∀𝑎, 𝑏, 𝑐 ∈ ℝ, 3𝑎𝑏 + 3𝑏𝑐 + 3𝑎𝑐 ≤ (𝑎 + 𝑏 + 𝑐)². Exercise 3.

Prove that∀𝑥 ∈ ℝ, |𝑥 − 1| ≤ 𝑥²− 𝑥 + 1.

Exercise 4.

1. Prove that∀𝑥, 𝑦 ∈ ℝ, |𝑥| + |𝑦| ≤ |𝑥 + 𝑦| + |𝑥 − 𝑦|.

2. Prove that∀𝑥, 𝑦 ∈ ℝ, |𝑥 + 𝑦|

1 + |𝑥 + 𝑦| ≤ |𝑥|

1 + |𝑥| + |𝑦|

1 + |𝑦|. Exercise 5.

Let𝐴 ⊂ ℝbe non-empty and bounded. We set𝐵 = {|𝑥 − 𝑦| ∶ 𝑥, 𝑦 ∈ 𝐴}.

1. Prove that𝐵admits a supremum.

2. Prove that sup𝐵 =sup𝐴 −inf𝐴.

Exercise 6.

Prove that if𝑓 ∶ [0, 1] → [0, 1]is non-decreasing then𝑓 admits a fixed point, i.e.∃𝑎 ∈ [0, 1], 𝑓 (𝑎) = 𝑎.

Hint: study{𝑥 ∈ [0, 1] ∶ 𝑓 (𝑥) ≥ 𝑥}.

Exercise 7.

Let𝐴 ⊂ ℝbe non-empty and bounded from above. Set𝑀 =sup(𝐴).

Prove that if𝑀 ∉ 𝐴then for all𝜀 > 0the set(𝑀 − 𝜀, 𝑀) ∩ 𝐴contains infinitely many elements.

Exercise 8. Conway’s Soldiers, or why geometric series are useful

We consider an infinite checkerboard represented byℤ×ℤwith pieces on it. The pieces are allowed to move using the peg solitaire rules: a move consists of one piece jumping over another piece into an empty cell (either horizontally or vertically), the piece which was jumped over is then removed.

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The goal of this exercise is to show that there is no initial configuration with finitely many pieces located on ℤ × ℤ_≤0allowing to reach cells with𝑦-coordinate5.

1. Prove that there exists initial configurations allowing to reach cells with𝑦-coordinate1,2,3and4.

2. We denote by𝜎the positive root of𝑥²+ 𝑥 − 1 = 0and we fix a target cell onℤ × ℤ.

We label each cell ofℤ × ℤwith𝜎^𝑛where𝑛is the Manhattan distance from the target to the cell.

𝜎⁰ 𝜎¹

𝜎¹ 𝜎¹ 𝜎¹ 𝜎² 𝜎² 𝜎² 𝜎²

𝜎²

𝜎² 𝜎² 𝜎³

𝜎³

𝜎³ 𝜎³

𝜎⁴ 𝜎⁴

Given a finite configuration𝐶 (i.e. finitely pieces on the checkerboard), we define𝐹 (𝐶) = ∑_𝑖∈𝐶𝜎^𝑛^𝑖 where𝑛_𝑖is the Manhattan from the target cell to the cell𝑖.

Prove that if𝐶^′is a configuration obtained after one move then𝐹 (𝐶^′) − 𝐹 (𝐶) ≤ 0.

3. Compute

+∞

∑𝑛=2

𝜎^𝑛.

4. Assume that the target cell is(0, 5). Compute 𝐹 (𝐶)where𝐶 contains all the cells with non-positive 𝑦-coordinates (hence𝐶contains infinitely many cells).

5. Conclude that there is no finite initial configuration inℤ × ℤ_≤0allowing to reach(0, 5).

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Sample solutions to Exercise 1.

Set𝛼 = √7 + 4√3 + √7 − 4√3then

𝛼² = 14 + 2√(7 + 4√3) (7 − 4√3) = 14 + 2√7²− 4²× 3 = 14 + 2√1 = 16

So𝛼 = ±4, but since𝛼 > 0, we get𝛼 = 4.

1. Let𝑎, 𝑏 ∈ ℝ, then0 ≤ (𝑎 − 𝑏)²= 𝑎²+ 𝑏²− 2𝑎𝑏, so that𝑎𝑏 ≤ ^𝑎²^+𝑏²

2 . 2. Let𝑎, 𝑏, 𝑐 ∈ ℝ. We know from the previous question that𝑎𝑏 ≤ ^𝑎²^+𝑏²

2 ,𝑏𝑐 ≤ ^𝑏²^+𝑐²

2 and𝑎𝑐 ≤ ^𝑎²^+𝑐²

2 . By summing these three inequalities, we get𝑎𝑏 + 𝑏𝑐 + 𝑎𝑐 ≤ ^𝑎²^+𝑏²^+𝑏²^+𝑐²^+𝑎²^+𝑐²

2 = 𝑎²+ 𝑏²+ 𝑐². 3. Let𝑎, 𝑏, 𝑐 ∈ ℝ. Then

(𝑎 + 𝑏 + 𝑐)²= 𝑎²+ 𝑏²+ 𝑐²+ 2𝑎𝑏 + 2𝑏𝑐 + 2𝑎𝑐

≥ 𝑎𝑏 + 𝑏𝑐 + 𝑎𝑐 + 2𝑎𝑏 + 2𝑏𝑐 + 2𝑎𝑐 from the previous question.

= 3𝑎𝑐 + 3𝑏𝑐 + 3𝑎𝑐

Let𝑥 ∈ ℝ.

• First case:𝑥 ≤ 1then𝑥²− 𝑥 + 1 − |𝑥 − 1| = 𝑥²− 𝑥 + 1 + (𝑥 − 1) = 𝑥²≥ 0, therefore|𝑥 − 1| ≤ 𝑥²− 𝑥 + 1.

• Second case: 𝑥 > 1then𝑥²− 𝑥 + 1 − |𝑥 − 1| = 𝑥²− 𝑥 + 1 − (𝑥 − 1) = 𝑥² − 2𝑥 + 2 = (𝑥 − 1)²+ 1 > 0, therefore|𝑥 − 1| ≤ 𝑥²− 𝑥 + 1.

1. Let𝑥, 𝑦 ∈ ℝ. Then2|𝑥| = |2𝑥| = |(𝑥 + 𝑦) + (𝑥 − 𝑦)| ≤ |𝑥 + 𝑦| + |𝑥 − 𝑦|and similarly2|𝑦| ≤ |𝑥 + 𝑦| + |𝑥 − 𝑦|.

Summing these two inequalities, we obtain2(|𝑥| + |𝑦|) ≤ 2(|𝑥 + 𝑦| + |𝑥 − 𝑦|).

2. Define𝑓 ∶ [0, +∞) → ℝby𝑓 (𝑢) = ^𝑢

1+𝑢 then𝑓 is differentiable and𝑓^′(𝑢) = ¹

(1+𝑢)² > 0.

Therefore𝑓 is increasing.

Let𝑥, 𝑦 ∈ ℝ. Since|𝑥 + 𝑦| ≤ |𝑥| + |𝑦|, we obtain

|𝑥 + 𝑦|

1 + |𝑥 + 𝑦| ≤ |𝑥| + |𝑦|

1 + |𝑥| + |𝑦| = |𝑥|

1 + |𝑥| + |𝑦|+ |𝑦|

1 + |𝑥| + |𝑦| ≤ |𝑥|

1 + |𝑥| + |𝑦|

1 + |𝑦|

1. Since𝐴is non-empty, there exists𝑥 ∈ 𝐴and then0 = |𝑥 − 𝑥| ∈ 𝐵. Therefore𝐵 is non-empty.

Since𝐴is bounded, there exists𝑀 ∈ ℝsuch that∀𝑥 ∈ 𝐴, |𝑥| ≤ 𝑀.

Therefore, if𝑥, 𝑦 ∈ 𝐴, then|𝑥 − 𝑦| ≤ |𝑥| + |𝑦| ≤ 2𝑀. Thus2𝑀is an upper bound of𝐵.

Since𝐵is a non-empty subset ofℝwhich is bounded from above, it admits a supremum.

2. Since𝐴is non-empty and bounded from below, there exists𝑚 =inf(𝐴).

Similarly, since𝐴is non-empty and bounded from above, there exists𝑀 =sup(𝐴).

Let𝑥, 𝑦 ∈ 𝐴, then𝑚 ≤ 𝑥 ≤ 𝑀and−𝑀 ≤ −𝑦 ≤ −𝑚, thus−(𝑀 − 𝑚) ≤ 𝑥 − 𝑦 ≤ 𝑀 − 𝑚, i.e.|𝑥 − 𝑦| ≤ 𝑀 − 𝑚.

Thus𝑀 − 𝑚is an upper bound of𝐵. Let’s check it is the least one.

Let𝜀 > 0. Since𝑚 = inf(𝐴), there exists𝑦 ∈ 𝐴such that𝑦 ≤ 𝑚 + ^𝜀

2. Since𝑀 = sup(𝐴), there exists 𝑥 ∈ 𝐴such that𝑀 − ^𝜀

2 ≤ 𝑥. Therefore|𝑥 − 𝑦| ≥ 𝑥 − 𝑦 ≥ 𝑀 − 𝑚 + 𝜀.

We proved that for every𝜀 > 0, there exists|𝑥 − 𝑦| ∈ 𝐵such that|𝑥 − 𝑦| ≥ 𝑀 − 𝑚 + 𝜀.

Therefore sup(𝐵) = 𝑀 − 𝑚.

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Set𝐸 = {𝑥 ∈ [0, 1] ∶ 𝑓 (𝑥) ≥ 𝑥}. Since𝑓 (0) ∈ [0, 1], we have that𝑓 (0) ≥ 0, so0 ∈ 𝐸.

Besides𝐸is bounded from above by1.

Thence, by the least upper bound principle,𝐸admits a supremum𝑎 =sup(𝐸).

Assume by contradiction that𝑓 (𝑎) ≠ 𝑎, then

• Either𝑓 (𝑎) < 𝑎. Since𝑎is the least upper bound of𝐸,𝑓 (𝑎)is not an upper bound, so there exists𝑏 ∈ 𝐸 such that𝑓 (𝑎) < 𝑏 ≤ 𝑎.

But then𝑏 ≤ 𝑎and𝑓 (𝑎) < 𝑏 ≤ 𝑓 (𝑏)(since𝑏 ∈ 𝐸), which is impossible since𝑓 is non-decreasing.

• Or𝑓 (𝑎) > 𝑎. Then, since𝑓 is non-decreasing, we get𝑓 (𝑓 (𝑎)) ≥ 𝑓 (𝑎). So𝑓 (𝑎) ∈ 𝐸. Which is impossible since for every𝑥 ∈ [0, 1],𝑥 ≤ 𝑎 < 𝑓 (𝑎)(since𝑎is an upper bound).

Let𝜀 > 0. Assume by contradiction that(𝑀 − 𝜀, 𝑀) ∩ 𝐴contains finitely many elements, i.e.

(𝑀 − 𝜀, 𝑀) ∩ 𝐴 = {𝑎₁, 𝑎₂, … , 𝑎_𝑝}.

Note that𝑎 ≔max(𝑎₁, … , 𝑎_𝑝) < 𝑀.

Set𝛿 = 𝑀 − 𝑎. Since𝛿 > 0, there exists𝑏 ∈ 𝐴such that𝑀 − 𝛿 < 𝑏 ≤ 𝑀. Since𝑀 ∉ 𝐴, we have𝑏 < 𝑀. Besides𝑏 > 𝑀 − 𝛿 = 𝑎 ≥ 𝑀 − 𝜀.

Therefore𝑏 ∈ (𝑀 − 𝜀, 𝑀) ∩ 𝐴.

But,∀𝑖 = 1, … , 𝑝,𝑏 > 𝑎 > 𝑎_𝑖. Hence a contradiction

1. To reach the first row:

To reach the second row:

I let you continue for the third and fourth rows!

2. There are three cases to handle:

• The piece moves towards the target cell: then if the piece is initially located at a cell labeled𝜎^𝑛, then it jumps over piece in a cell labeled𝜎^𝑛−1to reach a cell labeled𝜎^𝑛−2.

Therefore𝐹 (𝐶^′) − 𝐹 (𝐶) = −𝜎^𝑛− 𝜎^𝑛−1+ 𝜎^𝑛−2= 𝜎^𝑛−2(−𝜎²− 𝜎 + 1) = 0.

• The piece remains at the same distance to the target cell: then if the piece is initially located at a cell labeled𝜎^𝑛, then it jumps over piece in a cell labeled𝜎^𝑛−1to reach a cell labeled𝜎^𝑛.

Therefore𝐹 (𝐶^′) − 𝐹 (𝐶) = −𝜎^𝑛− 𝜎^𝑛−1+ 𝜎^𝑛= −𝜎^𝑛−1.

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• The piece moves away from the target cell: then if the piece is initially located at a cell labeled𝜎^𝑛, then it jumps over piece in a cell labeled𝜎^𝑛+1to reach a cell labeled𝜎^𝑛+2.

Therefore𝐹 (𝐶^′) − 𝐹 (𝐶) = −𝜎^𝑛− 𝜎^𝑛+1+ 𝜎^𝑛+2= 𝜎^𝑛(−1 − 𝜎 + 𝜎²) = −2𝜎^𝑛+1. 3. For those who like geometric series, like Cherge: since0 < 𝜎 < 1, we have

+∞

∑𝑛=2

𝜎^𝑛= 𝜎² 1 − 𝜎 = 1

Otherwise, if you don’t like geometric series: since∀𝑛 ∈ ℕ, 𝜎^𝑛+2 = 𝜎^𝑛− 𝜎^𝑛+1, we have a telescoping series:

𝐾

∑𝑛=2

𝜎^𝑛 =

𝐾−2

∑𝑛=0

𝜎^𝑛+2=

𝐾−2

∑𝑛=0

(𝜎^𝑛− 𝜎^𝑛+1) = 𝜎⁰− 𝜎^𝐾−1−−−−−→

𝐾→+∞ 1 − 0 = 1 4.

𝜎¹⁰ 𝜎¹¹ 𝜎¹²

𝜎⁹ 𝜎¹⁰ 𝜎¹¹

𝜎⁸ 𝜎⁹ 𝜎¹⁰

𝜎⁷ 𝜎⁸ 𝜎⁹

𝜎⁶ 𝜎⁷ 𝜎⁸

𝜎⁵ 𝜎⁶ 𝜎⁷

𝜎⁶ 𝜎⁷ 𝜎⁸

𝜎⁷ 𝜎⁸ 𝜎⁹

𝜎⁸ 𝜎⁹ 𝜎¹⁰

𝜎⁹ 𝜎¹⁰ 𝜎¹¹

𝜎¹⁰ 𝜎¹¹ 𝜎¹² ⋯

⋯

The cells on𝑦 = 0give

𝜎⁵+ 2𝜎⁶

+∞

∑𝑘=0

𝜎^𝑘= 𝜎⁵+ 2𝜎⁶

1 − 𝜎 = 𝜎⁵+ 2𝜎⁴= 𝜎³(𝜎²+ 2𝜎) = 𝜎³(1 + 𝜎) = 𝜎²(𝜎 + 𝜎²) = 𝜎²

Therefore, the cells on𝑦 = 𝑛give𝜎^2+𝑛and

𝐹 (𝐶) =

+∞

∑𝑛=2

𝜎^𝑛= 1

5. Assume that we have a finite initial configuration 𝐶₀, then 𝐹 (𝐶₀) < 𝐹 (𝐶) = 1 from the previous question.

If𝐶_𝑛is a configuration obtained after𝑛moves then(𝐹 (𝐶_𝑛))_𝑛is decreasing by Question 2.

Assume that we reach5after𝑛moves then𝐹 (𝐶_𝑛) ≥ 𝜎⁰= 1(since it contains at least a piece located at (5, 0)). But𝐹 (𝐶_𝑛) ≤ 𝐹 (𝐶₀) < 1. Hence a contradiction.