- 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 theTaylor series for theexponential 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 irrationalThe above proof can be found in

Proofs from THE BOOK . It is used as a stepping stone inIvan Niven 's 1947 proof that π^{2}is irrational and also for the stronger result that "e"^{"q"}is irrational for any non-zero rational "q". [*Citation | last1=Aigner | first1=Martin | last2=Ziegler | first2=Günter M. | author2-link=Günter M. Ziegler | title=*]Proofs from THE BOOK | publisher=Springer-Verlag | location=Berlin, New York | year=1998|pages=27-36.**References****ee also***

Characterizations of the exponential function

*Transcendental number

*Lindemann–Weierstrass theorem

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