Proof that e is irrational

Part of a series of articles on The mathematical constant, e Natural logarithm Applications in Compound interest · Euler's identity & Euler's formula  · Half lives & Exponential growth/decay Defining e Proof that e is irrational  · Representations of e · Lindemann–Weierstrass theorem People John Napier  · Leonhard Euler Schanuel's conjecture

In mathematics, the series representation of Euler's number e

<math>e = \sum_{n = 0}^{\infty} \frac{1}{n!}\!</math>

can be used to prove that e is irrational. Of the many representations of e, this is the Taylor series for the exponential function e^y evaluated at y = 1.

Contents 1 Summary of the proof 2 Proof 3 eq is irrational 4 References 5 See also

Summary of the proof

This is a proof by contradiction. Initially e is assumed to be a rational number of the form a/b. We then analyze a blown-up difference x of the series representing e and its strictly smaller bth partial sum, which approximates the limiting value e. By choosing the magnifying factor to be b!, the fraction a/b and the bth partial sum are turned into integers, hence x must be a positive integer. However, the fast convergence of the series representation implies that the magnified approximation error x is still strictly smaller than 1. From this contradiction we deduce that e is irrational.

Proof

Suppose that e is a rational number. Then there exist positive integers a and b such that e = a/b. Define the number

<math>\ x = b!\,\biggl(e - \sum_{n = 0}^{b} \frac{1}{n!}\biggr)\!</math>

To see that x is an integer, substitute e = a/b into this definition to obtain

<math>

x = b!\,\biggl(\frac{a}{b} - \sum_{n = 0}^{b} \frac{1}{n!}\biggr) = a(b - 1)! - \sum_{n = 0}^{b} \frac{b!}{n!}\,. </math> The first term is an integer, and every fraction in the sum is an integer since n≤b for each term. Therefore x is an integer. We now prove that 0 < x < 1. First, insert the above series representation of e into the definition of x to obtain

<math>x = \sum_{n = b+1}^{\infty} \frac{b!}{n!}>0\,.\!</math>

For all terms with n ≥ b + 1 we have the upper estimate

<math>\frac{b!}{n!}

=\frac1{(b+1)(b+2)\cdots(b+(n-b))} \le\frac1{(b+1)^{n-b}}\,,\! </math> which is even strict for every n ≥ b + 2. Changing the index of summation to k = n – b and using the formula for the infinite geometric series, we obtain

<math>

x =\sum_{n = b+1}^{\infty} \frac{b!}{n!} < \sum_{k=1}^\infty\frac1{(b+1)^k} =\frac{1}{b+1}\biggl(\frac1{1-\frac1{b+1}}\biggr) = \frac{1}{b} \le 1. </math> Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so e must be irrational.

e^q is irrational

The above proof can be found in Proofs from THE BOOK. It is used as a stepping stone in Ivan Niven's 1947 proof that π² is irrational and also for the stronger result that e^q is irrational for any non-zero rational q.^[1]

References

1. ^ Aigner, Martin & Ziegler, Günter M. (1998), Proofs from THE BOOK, Berlin, New York: Springer-Verlag, pp. 27-36.

See also

• Characterizations of the exponential function
• Transcendental number
• Lindemann–Weierstrass theorem