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The Euler Constant e is an Irrational Number 📂Analysis

The Euler Constant e is an Irrational Number

Theorem

Q\mathbb{Q} represents the set of rational numbers.

Proof

Using Maclaurin Expansion1

Strategy: Split e−1e^{-1} into two parts using the Maclaurin expansion and derive a contradiction. This proof cannot be conducted within the high school curriculum due to the necessity of the Maclaurin expansion.

N\mathbb{N} represents the set of natural numbers, Z\mathbb{Z} represents the set of integers.

  • Part 1. x1=x2x_{1} = x_{2}

    Assuming e∈Qe \in \mathbb{Q}, the Euler’s constant ee must be representable as e=abe = {{ a } \over {b}} for some a,b∈Na,b \in \mathbb{N}. By the Maclaurin expansion of the exponential function,

    e−1=ba=∑k=0∞(−1)kk! e^{-1} = {{ b } \over { a }} = \sum_{k=0}^{\infty} {{ (-1)^{k} } \over { k! }}

    Multiplying both sides by (−1)a+1a!(-1)^{a+1} a! and swapping their positions,

    ∑k=0∞(−1)k+a+1a!k!=b(−1)a+1(a−1)! \sum_{k=0}^{\infty} {{ (-1)^{k + a + 1} a! } \over { k! }} = b (-1)^{a+1} ( a - 1 )!

    If the sigma on the left side starts at k=a+1k=a+1, then ∑k=1a(−1)k+a+1a!k!\sum_{k=1}^{a} {{ (-1)^{k + a + 1} a! } \over { k! }} is missing from the right side, ∑k=a+1∞(−1)k+a+1a!k!=b(−1)a+1(a−1)!−∑k=1a(−1)k+a+1a!k! \sum_{k=a+1}^{\infty} {{ (-1)^{k + a + 1} a! } \over { k! }} = b (-1)^{a+1} ( a - 1 )! - \sum_{k=1}^{a} {{ (-1)^{k + a + 1} a! } \over { k! }} Now, if we let the left side be x1x_{1} and the right side x2x_{2}, obviously x1=x2x_{1} = x_{2} will hold. x1:=∑k=a+1∞(−1)k+a+1a!k! x_{1} := \sum_{k=a+1}^{\infty} {{ (-1)^{k + a + 1} a! } \over { k! }}

    x2:=b(−1)a+1(a−1)!−∑k=1a(−1)k+a+1a!k! x_{2} := b (-1)^{a+1} ( a - 1 )! - \sum_{k=1}^{a} {{ (-1)^{k + a + 1} a! } \over { k! }}

  • Part 2. x1∈(0,1)x_{1} \in (0,1)

    Directly expanding x1x_{1},

    x1=1a+1−1(a+1)(a+2)+1(a+1)(a+2)(a+3)−⋯ x_{1} = {{ 1 } \over { a+1 }} - {{ 1 } \over { (a+1)(a+2) }} + {{ 1 } \over { (a+1)(a+2)(a+3) }} - \cdots As kk increases, ∣(−1)k+a+1a!k!∣\left| (-1)^{k+a+1} {{ a! } \over { k! }} \right| decreases, hence x1x_{1} must be larger than 1a+1{{ 1 } \over { a+1 }} and smaller than 1a+1−1(a+1)(a+2){{ 1 } \over { a+1 }} - {{ 1 } \over { (a+1)(a+2) }}. Therefore, x1x_{1} must be some number between 00 and 11.

  • Part 3. x2∈Zx_{2} \in \mathbb{Z}

    If k≤ak \le a then a!k!{{ a! } \over { k! }} is a natural number, upon simplification,

    x1∈(0,1) x_{1} \in (0,1)

    x2∈Z x_{2} \in \mathbb{Z}

    However, since (0,1)∩Z=∅(0,1) \cap \mathbb{Z} = \emptyset, it means x1≠x2x_{1} \ne x_{2}, but it was clearly x1=x2x_{1} = x_{2} in Part 1, which is a contradiction.

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Using the Definition of ee2

Definition of Euler’s constant

e:=∑n=0∞1n! e: = \sum \limits_{n=0}^{\infty} \dfrac{1}{n!}

If we call sns_{n} the partial sum of ee, then by the definition of ee, it follows:

e−sn=1(n+1)!+1(n+2)!+1(n+3)!+⋯<1(n+1)!+1(n+1)!(n+1)+1(n+1)!(n+1)2+⋯=1(n+1)!(1+1(n+1)+1(n+1)2+⋯ )=1(n+1)!(n+1n)=1n!n \begin{align*} e - s_{n} =& \dfrac{1}{(n+1)!} + \dfrac{1}{(n+2)!} + \dfrac{1}{(n+3)!} + \cdots \\ &< \dfrac{1}{(n+1)!} + \dfrac{1}{(n+1)!(n+1)} + \dfrac{1}{(n+1)!(n+1)^{2}} + \cdots \\ =& \dfrac{1}{(n+1)!}\left( 1 + \dfrac{1}{(n+1)} + \dfrac{1}{(n+1)^{2}} + \cdots\right) \\ =& \dfrac{1}{(n+1)!}\left( \dfrac{n+1}{n} \right) \\ =& \dfrac{1}{n! n} \end{align*}

Now, let’s assume that ee is rational. Then there exists a positive integer p,qp, q that satisfies e=pqe=\dfrac{p}{q}, and accordingly:

0<q!(e−sq)<1q 0 < q!(e - s_{q}) < \dfrac{1}{q}

By assumption, q!e=(q−1)!pq!e=(q-1)!p is an integer. And,

q!sq=q!(1+1+12!+⋯+1q!) q! s_{q} = q! \left( 1 + 1 + \dfrac{1}{2!} + \cdots + \dfrac{1}{q!} \right)

so q!sqq!s_{q} is also an integer. Therefore, q!(e−sq)q!(e-s_{q}) is an integer, but since q≥1q\ge 1, q!(e−sq)q!(e-s_{q}) ends up being an integer between 00 and 11, which is a contradiction. Hence, the assumption is wrong, and by proof by contradiction, ee is irrational.

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See Also

  1. William R. Wade, An Introduction to Analysis (4th Edition, 2010), p263-264 ↩︎

  2. Walter Rudin, Principles of Mathematical Analysis (3rd Edition, 1976), p65 ↩︎