Proof that e is irrational

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

e = sum_{n = 0}^{infty} frac{1}{n!}!

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.

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

x = b!,biggl(e - sum_{n = 0}^{b} frac{1}{n!}biggr)!

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

x = b!,biggl(frac{a}{b} - sum_{n = 0}^{b} frac{1}{n!}biggr)

a(b - 1)! - sum_{n

0}^{b} frac{b!}{n!},.

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

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

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

frac{b!}{n!}

=frac1{(b+1)(b+2)cdots(b+(n-b))} lefrac1{(b+1)^{n-b}},,! 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

sum_{n

b+1}^{infty} frac{b!}{n!} < sum_{k=1}^inftyfrac1{(b+1)^k} =frac{1}{b+1}biggl(frac1{1-frac1{b+1}}biggr) = frac{1}{b} le 1.

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.

Proof that e is irrational

Summary of the proof

Proof

a(b - 1)! - sum_{n

sum_{n

e^q is irrational

References

See also

Proof that e is irrational

Summary of the proof

Proof

a(b - 1)! - sum_{n

sum_{n

eq is irrational

References

See also

e^q is irrational