REAL NUMBERS – 3

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MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

REAL NUMBERS – 3

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Theorem: ℚ is dense in ℝ

∀𝑥, 𝑦 ∈ ℝ, 𝑥 < 𝑦 ⇒ (∃𝑞 ∈ ℚ, 𝑥 < 𝑞 < 𝑦)

Proof.Let𝑥, 𝑦 ∈ ℝbe such that𝑥 < 𝑦. Set𝜀 = 𝑦 − 𝑥 > 0.

By the archimedean property, there exists𝑛 ∈ ℕ ⧵ {0}such that𝑛𝜀 > 1, i.e. ¹_𝑛 < 𝜀.

Set𝑚 = ⌊𝑛𝑥⌋ + 1. Then𝑛𝑥 < 𝑚 ≤ 𝑛𝑥 + 1, so𝑥 < ^𝑚_𝑛 ≤ 𝑥 +¹_𝑛 < 𝑥 + 𝜀 = 𝑦. ■

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 18, 2021 2 / 24

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ℚ is dense in ℝ – 2

Definition: interval

A subset𝐼 ⊂ ℝis anintervalif∀𝑥, 𝑦 ∈ 𝐼, ∀𝑧 ∈ ℝ, (𝑥 ≤ 𝑧 ≤ 𝑦 ⇒ 𝑧 ∈ 𝐼).

Crollary

If𝐼 ⊂ ℝis a non-empty interval not reduced to a singleton then𝐼 ∩ ℚ ≠ ∅.

Proof.Since𝐼 is non-empty and not reduced to a singleton, there exist𝑥, 𝑦 ∈ 𝐼with𝑥 < 𝑦.

Then, sinceℚis dense inℝ, there exists𝑞 ∈ ℚsuch that𝑥 < 𝑞 < 𝑦.

Since𝐼is an interval,𝑞 ∈ 𝐼. Hence𝑞 ∈ 𝐼 ∩ ℚ ≠ ∅. ■

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Theorem: ℝ ⧵ ℚ is dense in ℝ

∀𝑥, 𝑦 ∈ ℝ, 𝑥 < 𝑦 ⟹ (∃𝑠 ∈ ℝ ⧵ ℚ, 𝑥 < 𝑠 < 𝑦) Proof.Let𝑥, 𝑦 ∈ ℝsuch that𝑥 < 𝑦.

Since , there exists𝑞 ∈ ℚsuch that𝑥 < 𝑞 < 𝑦.

Similarly, there exists𝑝 ∈ ℚsuch that𝑥 < 𝑝 < 𝑞.

Hence we obtained𝑝, 𝑞 ∈ ℚsuch that𝑥 < 𝑝 < 𝑞 < 𝑦.

Set𝑠 = 𝑝 +^√2₂ (𝑞 − 𝑝). Then𝑠 ∈ ℝ ⧵ ℚ(otherwise, by contradiction,√2would be inℚ) and 𝑝 < 𝑠 < 𝑞(notice that0 <^√2₂ < 1so𝑠is a number between𝑝and𝑞).

We obtained𝑠 ∈ ℝ ⧵ ℚsuch that𝑥 < 𝑠 < 𝑦. ■

I wanted to give it as a question in PS4… before remembering it was in my lecture notes…

Corollary

If𝐼 ⊂ ℝis an interval which is non-empty and not reduced to a singleton then𝐼 ∩ (ℝ ⧵ ℚ) ≠ ∅.

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Decimal representation of real numbers – 1

Lemma

Let(𝑎_𝑘)_𝑘≥1 be a sequence such that∀𝑘 ∈ ℕ ⧵ {0}, 𝑎_𝑘∈ {0, 1, … , 9}. Then the series

𝑆 =

+∞

∑𝑘=1

𝑎_𝑘 10^𝑘 is convergent and𝑆 ≥ 0.

Proof.

Note that0 ≤ 𝑎_𝑘 10^𝑘 ≤ 9

10^𝑘 and that

+∞

∑𝑘=1

9

10^𝑘 is convergent (geometric series with ratio 1 10 < 1).

Therefore we may conclude using the BCT. ■

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Unfortunately the decimal representation may not be unique:0.9999 … =

+∞

∑𝑘=1

9 10^𝑘 = 9

10× 1

1 −₁₀¹ = 1.000 … In order to achieve uniqueness we are going to restrict to expansions which don’t end with infinitely many9.

Definition: proper decimal expansion

Let𝑥 ∈ ℝ. We say that⌊𝑥⌋.𝑎₁𝑎₂𝑎₃… ≔ ⌊𝑥⌋ +

+∞

∑𝑘=1

𝑎_𝑘

10^𝑘 is aproper decimal expansion of𝑥if 1 ∀𝑘 ∈ ℕ ⧵ {0}, 𝑎_𝑘∈ {0, 1, … , 9} 2 ∀𝑛 ∈ ℕ ⧵ {0},

𝑛

∑𝑘=1

𝑎_𝑘

10^𝑘 ≤ 𝑥 − ⌊𝑥⌋ <

𝑛

∑𝑘=1

𝑎_𝑘 10^𝑘+ 1

10^𝑛

Proposition

If⌊𝑥⌋ +

+∞

∑𝑘=1

𝑎_𝑘

10^𝑘is a proper decimal expansion of𝑥 ∈ ℝthen 1 𝑥 = ⌊𝑥⌋ +

+∞

∑𝑘=1

𝑎_𝑘

10^𝑘 2 ∀𝑁 ∈ ℕ ⧵ {0}, ∃𝑘 > 𝑁, 𝑎_𝑘≠ 9 Proof.

1 We already proved that𝑆 =

+∞

∑𝑘=1

𝑎_𝑘

10^𝑘 is convergent. Hence we get𝑆 ≤ 𝑥 − ⌊𝑥⌋ ≤ 𝑆. So𝑥 = ⌊𝑥⌋ + 𝑆.

2 Assume by contradiction that there exists𝑁 ∈ ℕ ⧵ {0}such that∀𝑘 > 𝑁, 𝑎_𝑘= 9.

Then𝑥 − ⌊𝑥⌋ =

+∞

∑𝑘=1

𝑎_𝑘 10^𝑘=

𝑁

∑𝑘=1

𝑎_𝑘 10^𝑘+

+∞

𝑘=𝑁+1∑ 9 10^𝑘 =

𝑁

∑𝑘=1

𝑎_𝑘 10^𝑘+ 1

10^𝑁. Which contradicts the definition of proper decimal expansion. ■

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Decimal representation of real numbers – 3

Theorem

A real number𝑥admits a unique proper decimal expansion.

Proof.Let𝑥 ∈ ℝ. Up to replacing𝑥with𝑥 − ⌊𝑥⌋, we may assume that⌊𝑥⌋ = 0.

Assume that

+∞

∑𝑘=1

𝑎_𝑘

10^𝑘is a proper decimal expansion of𝑥. Then𝑎_𝑛≤ 10^𝑛 (𝑥 −

𝑛−1

∑𝑘=1

𝑎_𝑘

10^𝑘)< 𝑎_𝑛+ 1.

So the only possible suitable sequence(𝑎_𝑛)is given by𝑎₁= ⌊10𝑥⌋and𝑎_𝑛+1=

⌊10^𝑛+1 (𝑥 −

𝑛

∑𝑘=1

𝑎_𝑘 10^𝑘)⌋. It proves the uniqueness, but we still need to check that it is valid.

1 Since⌊𝑥⌋ = 0, we have0 ≤ 𝑥 < 1. Thus0 ≤ 10𝑥 < 10. Therefore𝑎1= ⌊10𝑥⌋ ∈ {0, 1, … , 9}.

Let𝑛 ∈ ℕ ⧵ {0}, then0 ≤ 10^𝑛 (𝑥 −

𝑛−1

∑𝑘=1

𝑎_𝑘

10^𝑘)− 𝑎_𝑛< 1. Thus0 ≤ 10^𝑛+1 (𝑥 −

𝑛

∑𝑘=1

𝑎_𝑘 10^𝑘)< 10.

Therefore𝑎_𝑛+1∈ {0, 1, … , 9}.

2 We have∀𝑛 ∈ ℕ ⧵ {0},

𝑛

∑𝑘=1

𝑎_𝑘 10^𝑘≤ 𝑥 <

𝑛

∑𝑘=1

𝑎_𝑘 10^𝑘+ 1

10^𝑛 by construction.

■

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Theorem

A real number𝑥 ∈ ℝis rational if and only if its proper decimal expansion is eventually periodic, i.e. ∃𝑟 ∈ ℕ, ∃𝑠 ∈ ℕ ⧵ {0}, ∀𝑘 ∈ ℕ, 𝑎_{𝑟+𝑘+𝑠}= 𝑎_𝑟+𝑘, so that

𝑥 = ⌊𝑥⌋.𝑏₁𝑏₂… 𝑏_𝑟𝑐₁𝑐₂… 𝑐_𝑠≔ ⌊𝑥⌋.𝑏₁𝑏₂… 𝑏_𝑟𝑐₁𝑐₂… 𝑐_𝑠𝑐₁𝑐₂… 𝑐_𝑠𝑐₁…

Proof.

⇒Let𝑥 =^𝑎_𝑏where𝑎 ∈ ℤand𝑏 ∈ ℕ ⧵ {0}.

By Euclidean division,𝑎 = 𝑏𝑞₀+ 𝑟₀where0 ≤ 𝑟₀< 𝑏. Hence^𝑎

𝑏= 𝑞₀+^𝑟0

𝑏. Note that𝑞₀= ⌊^𝑎𝑏⌋. And we repeat:10𝑟𝑘= 𝑏𝑞𝑘+1+ 𝑟𝑘+1where0 ≤ 𝑟𝑘+1< 𝑏.

Since there are only𝑏possible remainders, the process will start looping after at most𝑏steps.

But note that the(𝑞𝑘)𝑘≥1defines exactly the decimal expansion of𝑥.

⇐Assume that the proper decimal expansion𝑥 = ⌊𝑥⌋ +

+∞

∑𝑘=1

𝑎_𝑘

10^𝑘 is eventually periodic, i.e.∃𝑟 ∈ ℕ, ∃𝑠 ∈ ℕ ⧵ {0}, ∀𝑘 ∈ ℕ, 𝑎_{𝑟+𝑘+𝑠}= 𝑎_𝑟+𝑘. Then𝑥 = ⌊𝑥⌋ +

𝑟

∑𝑘=1

𝑎_𝑘 10^𝑘+ 10^−𝑟

+∞

∑𝑘=1

𝑎_𝑟+𝑘

10^𝑘. Hence it is enough to prove that𝑦 =

+∞

∑𝑘=1

𝑎_𝑟+𝑘 10^𝑘 ∈ ℚ.

Note that10^𝑠𝑦 = 𝑁 + 𝑦where𝑁 = 𝑎_𝑟+1𝑎_𝑟+2… 𝑎_𝑟+𝑠¹⁰∈ ℕ. Hence𝑦 =_{10𝑠 −1}^𝑁 ∈ ℚ. ■

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Decimal representation of real numbers – 5

Example

We want to find the decimal expansion of ^1529327

24975 .

1 1529327 = 24975 × 61 + 5852

2 58520 = 24975 × 2 + 8570

3 85700 = 24975 × 3 + 10775

4 107750 = 24975 × 4 + 7850

5 78500 = 24975 × 3 + 3575

6 35750 = 24975 × 1 + 10775

And we start to loop. Therefore ^1529327₂₄₉₇₅ = 61.234314

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According to the above proof,

𝑎_𝑡𝑎_𝑡−1… 𝑎₀.𝑏₁𝑏₂… 𝑏_𝑟𝑐₁𝑐₂… 𝑐_𝑠= 𝑎_𝑡𝑎_𝑡−1⋯ 𝑎₀¹⁰+

𝑟

∑𝑘=1

𝑏_𝑘

10^𝑘 + 10^−𝑟𝑐₁𝑐₂… 𝑐_𝑠¹⁰ 10^𝑠− 1

= 𝑎_𝑡𝑎_𝑡−1… 𝑎₀¹⁰+ 𝑏₁𝑏₂… 𝑏_𝑟¹⁰

10^𝑟 + 𝑐₁𝑐₂… 𝑐_𝑠¹⁰ 10^𝑟+𝑠− 10^𝑟

= 𝑎_𝑡𝑎_𝑡−1… 𝑎₀𝑏₁𝑏₂… 𝑏_𝑟𝑐₁𝑐₂… 𝑐_𝑠¹⁰− 𝑎_𝑡𝑎_𝑡−1… 𝑎₀𝑏₁𝑏₂… 𝑏_𝑟¹⁰ 10^𝑟+𝑠− 10^𝑟

Examples

• 61.234314 = 61234314 − 61234

10⁶− 10³ = 61173080 999000

• 0.3 = 3 − 0 10 − 1 = 3

9

• 42.012 = 42012 − 42

10³− 1 = 41970 999

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√2 ∉ ℚ – 1

Proof 1 (Fundamental Theorem of Arithmetic).

Assume by contradiction that√2 = ^𝑎_𝑏 ∈ ℚ. Then2𝑏²= 𝑎².

The prime factorization of the LHS has an odd number of primes (counted with exponents) whereas the RHS has an even number of primes (counted with exponents).

Which is impossible since the prime factorization is unique up to order. ■

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√2 ∉ ℚ – 2

Proof 2 (Euclid’s lemma).

Assume by contradiction that√2 = ^𝑎_𝑏 ∈ ℚwritten in lowest form.

Then2𝑏²= 𝑎². Therefore2|𝑎². By Euclid’s lemma,2|𝑎, so𝑎 = 2𝑘.

Thus2𝑏²= 4𝑘², from which we get𝑏²= 2𝑘². By Euclid’s lemma,2|𝑏.

Hence2|gcd(𝑎, 𝑏) = 1, which is a contradiction. ■

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√2 ∉ ℚ – 3

Proof 3 (Gauss’ lemma).

Then2𝑏²= 𝑎². Therefore𝑏|𝑎².

Sincegcd(𝑎, 𝑏) = 1, by Gauss’ lemma (applied twice),𝑏|1and hence𝑏 = 1(since𝑏 ∈ ℕ ⧵ {0}).

Hence𝑎²= 2.

Which is impossible (2is not a perfect square: ∀𝑥 ∈ ℤ, 𝑥²≡ 0 mod3or𝑥²≡ 1 mod3). ■

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√2 ∉ ℚ – 4

Proof 4 (proof by infinite descent).

Assume by contradiction that√2 = ^𝑎_𝑏 ∈ ℚwhere𝑎 ∈ ℕand𝑏 ∈ ℕ ⧵ {0}.

Then2𝑏²= 𝑎². Then𝑎(𝑎 − 𝑏) = 𝑎²− 𝑎𝑏 = 2𝑏²− 𝑎𝑏 = 𝑏(2𝑏 − 𝑎). Hence√2 = ^𝑎_𝑏 = ^2𝑏−𝑎_𝑎−𝑏. Note that1 < √2 =^𝑎_𝑏, thus0 < 𝑎 − 𝑏. Therefore0 < 2𝑏 − 𝑎, so𝑎 − 𝑏 < 𝑏.

Therefore we obtained another expression of√2with a smaller positive denominator.

By repeating this process, we may construct an infinite sequence√2 =^𝑎_𝑏 = ^𝑎_𝑏¹

1 = ^𝑎_𝑏²

2 = ⋯such

that𝑎_𝑘> 0and0 < 𝑏_𝑘+1< 𝑏_𝑘.

Which is a contradiction since there is no decreasing infinite sequence of natural numbers. ■

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√2 ∉ ℚ – 5

Proof 5 (by congruences).

Assume by contradiction that√2 = ^𝑎_𝑏 ∈ ℚwritten in lowest form. Then2𝑏²= 𝑎².

We can’t have𝑎 ≡ 0 mod3and𝑏 ≡ 0 mod3simulatenously (otherwise3|gcd(𝑎, 𝑏) = 1).

• Either𝑎 ≡ ±1 mod3and𝑏 ≡ 0 mod3, then𝑎²− 2𝑏²≡ 1 mod3,

• or𝑎 ≡ 0 mod3and𝑏 ≡ ±1 mod3, then𝑎²− 2𝑏²≡ 1 mod3,

• or𝑎 ≡ ±1 mod3and𝑏 ≡ ±1 mod3, then𝑎²− 2𝑏²≡ 2 mod3.

Therefore𝑎²− 2𝑏²≢ 0 mod3and so𝑎²− 2𝑏²≠ 0. Which is a contradiction. ■

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√2 ∉ ℚ – 6

Proof 6 (by the well-ordering principle).

Assume by contradiction that√2 = ^𝑎_𝑏 ∈ ℚ. Then𝑎 = √2𝑏.

Therefore𝐸 = {𝑛 ∈ ℕ ∶ 𝑛√2 ∈ ℕ ⧵ {0}}is not empty since it contains|𝑏|as√2|𝑏| = |𝑎|.

By the well-ordering principle,𝐸admits a least element𝑝. Then𝑝√2 ∈ ℕ ⧵ {0}.

Set𝑞 = 𝑝√2 − 𝑝. Then𝑞 ∈ ℤ. Besides𝑞 = 𝑝(√2 − 1)so that0 < 𝑞 < 𝑝.

But𝑞√2 = 2𝑝 − 𝑝√2 = 𝑝 − 𝑞 ∈ ℕ ⧵ {0}. So𝑞 ∈ 𝐸.

Which is a contradiction since𝑝is the least element of𝐸and𝑞 < 𝑝. ■

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√2 ∉ ℚ – 7

Proof 7 (by the rational root theorem).

Since√2 =^𝑎_𝑏 is a root of𝑥²− 2 = 0, we deduce from the rational root theorem that𝑎|2and𝑏|1.

So either√2 = ±1or√2 = ±2.

We obtain a contradiction in both cases since(±1)²= 1 ≠ 2and(±2)²= 4 ≠ 2. ■

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√2 ∉ ℚ – 8

Proof 8 (by the archimedean property).

For𝑛 ∈ ℕ, set𝑢_𝑛= (√2 − 1)^𝑛. We may prove either by induction or using the binomial formula, that for every𝑛, there exist𝑎_𝑛, 𝑏_𝑛∈ ℤsuch that𝑢_𝑛= 𝑎_𝑛+ 𝑏_𝑛√2.

Since¹0 < √2 − 1 < ¹₂, we may also prove that0 < 𝑢_𝑛≤ ₂¹_𝑛. Assume by contradiction that√2 = ^𝑝_𝑞 ∈ ℚ, then

𝑢_𝑛= 𝑎_𝑛+ 𝑏_𝑛√2 = 𝑎_𝑛+ 𝑏_𝑛𝑝

𝑞 = 𝑞𝑎_𝑛+ 𝑝𝑏_𝑛 𝑞 Since𝑢_𝑛> 0we get that|𝑞𝑎_𝑛+ 𝑝𝑏_𝑛| ≥ 1and that𝑢_𝑛≥ _|𝑞|¹.

Therefore∀𝑛 ∈ ℕ, 0 < _|𝑞|¹ ≤ 𝑢_𝑛≤ ₂¹_𝑛. Which contradicts the archimedean property. ■

1Use the fact that(0, +∞) ∋ 𝑥 → 𝑥²∈ ℝis increasing and that2 ≤ (³2)

2

to conclude that√2 ≤³₂.

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√2 ∉ ℚ – 9

Proof 9 (geometric version of proof 4).

Assume by contradiction that√2 = ^𝑎

𝑏 ∈ ℚwhere𝑎 ∈ ℕand𝑏 ∈ ℕ ⧵ {0}. Then𝑎 = √2𝑏 > 𝑏.

𝑎 𝑏

𝑏 𝑎

𝑏 𝑏

(𝑎 − 𝑏)²

(𝑎 − 𝑏)² (2𝑏 − 𝑎)²

By a direct computation of the side length, the square at the center has an area of𝒜 = (2𝑏 − 𝑎)².

But, by inclusion-exclusion,𝒜also satisfies2(𝑎 − 𝑏)²+ 2𝑏²− 𝒜 = 𝑎². So𝒜 = 2(𝑎 − 𝑏)²+ 2𝑏²− 𝑎²= 2(𝑎 − 𝑏)²since𝑎²= 2𝑏².

Therefore2(𝑎 − 𝑏)²= 𝒜 = (2𝑏 − 𝑎)². Thus2 =^{(2𝑏−𝑎)}_{(𝑎−𝑏)}₂². Hence√2 = ^2𝑏−𝑎

𝑎−𝑏 with^a0 < 2𝑏 − 𝑎and0 < 𝑎 − 𝑏 < 𝑏.

aSee Proof 4 for0 < 2𝑏 − 𝑎.

By repeating this process, we may construct an infinite sequence√2 =^𝑎

𝑏 = ^𝑎¹

𝑏1 =^𝑎²

𝑏2 = ⋯such that𝑎_𝑘> 0 and0 < 𝑏_𝑘+1< 𝑏_𝑘.

Which is a contradiction since there is no decreasing infinite sequence of natural numbers. ■

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√2 ∉ ℚ – 10

Proof 10 (Pythagoras flavored).

Let𝐴𝐵𝐶be a isosceles right triangle in𝐴. By the Pythagorean theorem^𝐵𝐶

𝐴𝐵= √2.

Assuming that√2is rational means geometrically that𝐵𝐶and𝐴𝐵are commensurable, i.e. they are both integral multiple of a another length𝑑².

Put𝐷on[𝐵𝐶]such that𝐵𝐷 = 𝐴𝐵.

Define𝐸as the intersection of(𝐴𝐶)with the line through𝐷which is perpendicular to(𝐵𝐶).

Note that³𝐴𝐸 = 𝐸𝐷 = 𝐷𝐶.

Thus𝐶𝐷 = 𝐵𝐶 − 𝐴𝐵and𝐸𝐶 = 𝐴𝐶 − 𝐴𝐸 = 𝐴𝐵 − (𝐵𝐶 − 𝐴𝐵) = 2𝐴𝐵 − 𝐵𝐶.

Therefore𝐶𝐷and𝐸𝐶are integral multiple of𝑑.

Besides𝐷𝐸𝐶is a isosceles right triangle in𝐷, therefore we may repeat this construction on the triangle𝐷𝐸𝐶in order to construct an infinite sequence of segment lines (𝐴𝐶, 𝐸𝐶, 𝐹 𝐶, …, see below) which are all integral multiple of𝑑and with decreasing length.

Which is impossible.

𝐴 𝐵

𝐶

𝐴 𝐵

𝐶 𝐷 𝐸

𝐴 𝐵

𝐶 𝐷 𝐸

𝐹 𝐺

2That is the geometric version ofirrationalityused by ancient Greeks: ■

if√2 = 𝑎𝑏, set𝑑 = 𝐴𝐵𝑏 then𝐴𝐵 = 𝑏𝑑and𝐵𝐶 = √2 × 𝐴𝐵 = 𝑎𝑏𝑏𝑑 = 𝑎𝑑.

3Compare the triangles𝐵𝐴𝐸and𝐵𝐷𝐸which are respectively right in𝐴and𝐷with common hypotenuse and𝐴𝐵 = 𝐷𝐵, so, by the Pythagorean theorem,𝐴𝐸 = 𝐸𝐷.

Besides the triangle𝐶𝐷𝐸is isosceles right in𝐴by angle considerations, thus𝐸𝐷 = 𝐷𝐶.

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√2 ∉ ℚ – 11

Proof 11 (my favorite one).

The proof is left as an exercise to the reader. ■

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More about √2

Before leaving√2, I would like to show you a funny proof relying on thetertium non datur.

Proposition

There exist𝑎, 𝑏 > 0irrational numbers such that𝑎^𝑏∈ ℚ.

Proof.

• Assume that√2^√2∈ ℚ. Then we can take𝑎 = 𝑏 = √2.

• Assume that√2^√2∉ ℚ. Then we can take𝑎 = √2^√2and𝑏 = √2.

Indeed,

(√2^√2 )

√2

= √2²= 2.

■

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𝑒 ∉ ℚ

Recall from your first year calculus that𝑒 =

+∞

∑𝑛=0

1 𝑛!. Proof 1.

Assume by contradiction that𝑒 = ^𝑎_𝑏 where𝑎, 𝑏 ∈ ℕ ⧵ {0}. Note that𝑏 > 1since𝑒 ∉ ℕ. Besides

𝑏!

⎛⎜

⎜⎝ 𝑒 −

𝑏

∑𝑛=0

1 𝑛!

⎞⎟

⎟⎠

= 𝑏!( ∑_{𝑛≥𝑏+1} 1 𝑛! ) Note that the LHS is an integer.

We are going to derive a contradiction by proving that the RHS is not an integer. Indeed 0 < 𝑏!

( ∑_{𝑛≥𝑏+1} 1

𝑛! )≤∑

𝑛≥1

1 (𝑏 + 1)^𝑛 = 1

𝑏 < 1

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Proof 2.

For𝑛 ∈ ℕ, set𝑢_𝑛=

∫

1 0

𝑥^𝑛𝑒^𝑥d𝑥.

Using an induction and integration by part, we can prove that for𝑛 ∈ ℕ, there exist𝑎_𝑛, 𝑏_𝑛∈ ℤ such that𝑢_𝑛= 𝑎_𝑛+ 𝑒𝑏_𝑛.

Assume by contradiction that𝑒 = ^𝑝_𝑞 where𝑝, 𝑞 ∈ ℕ ⧵ {0}. Then0 < 𝑢_𝑛= 𝑎_𝑛+ 𝑏_𝑛^𝑝_𝑞 = ^𝑞𝑎^𝑛^+𝑝𝑏_𝑞 ^𝑛. Since𝑢_𝑛> 0we get that𝑞𝑎_𝑛+ 𝑝𝑏_𝑛≥ 1and that𝑢_𝑛≥ ¹_𝑞.

Therefore∀𝑛 ∈ ℕ, 0 < 1 𝑞 ≤ 𝑢_𝑛≤

∫

1 0

𝑥^𝑛𝑒d𝑥 = 𝑒 𝑛 + 1.

Which is impossible. ■