Proof that e is irrational

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.

ummary 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 "b"th partial sum, which approximates the limiting value "e". By choosing the magnifying factor to be "b"!, the fraction "a"/"b" and the "b"th 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

: $x = b!,iggl(e - sum_{n = 0}^{b} frac{1}{n!}iggr)!$

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

: $x = b!,iggl(frac{a}{b} - sum_{n = 0}^{b} frac{1}{n!}iggr)= a(b - 1)! - sum_{n = 0}^{b} frac{b!}{n!},.$

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: $x =sum_{n = b+1}^{infty} frac{b!}{n!}< sum_{k=1}^inftyfrac1{(b+1)^k}=frac{1}{b+1}iggl(frac1{1-frac1{b+1iggr)= frac{1}{b}le 1.$

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

References

ee also

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

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