# Proof that e is irrational

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The number e was introduced by Jacob Bernoulli in 1683. More than half a century later, Euler, who had been a student of Jacob's younger brother Johann, proved that e is irrational; that is, that it cannot be expressed as the quotient of two integers.

## Euler's proof

Euler wrote the first proof of the fact that e is irrational in 1737 (but the text was only published seven years later). He computed the representation of e as a simple continued fraction, which is

$e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,\ldots ,2n,1,1,\ldots ].$ Since this continued fraction is infinite and every rational number has a terminating continued fraction, e is irrational. A short proof of the previous equality is known. Since the simple continued fraction of e is not periodic, this also proves that e is not a root of second degree polynomial with rational coefficients; in particular, e2 is irrational.

## Fourier's proof

The most well-known proof is Joseph Fourier's proof by contradiction, which is based upon the equality

$e=\sum _{n=0}^{\infty }{\frac {1}{n!}}\cdot$ Initially e is assumed to be a rational number of the form ab. Note that b could not be equal to 1 as e is not an integer. It can be shown using the above equality that e is strictly between 2 and 3:

$2=1+{\tfrac {1}{1!}} 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 the factorial of b, the fraction ab 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.

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

$x=b!\left(e-\sum _{n=0}^{b}{\frac {1}{n!}}\right).$ To see that if e is rational, then x is an integer, substitute e = ab into this definition to obtain

$x=b!\left({\frac {a}{b}}-\sum _{n=0}^{b}{\frac {1}{n!}}\right)=a(b-1)!-\sum _{n=0}^{b}{\frac {b!}{n!}}.$ The first term is an integer, and every fraction in the sum is actually an integer because n ≤ b for each term. Therefore, x is an integer.

We now prove that 0 < x < 1. First, to prove that x is strictly positive, we insert the above series representation of e into the definition of x and obtain

$x=b!\left(\sum _{n=0}^{\infty }{\frac {1}{n!}}-\sum _{n=0}^{b}{\frac {1}{n!}}\right)=\sum _{n=b+1}^{\infty }{\frac {b!}{n!}}>0,$ because all the terms are strictly positive.

We now prove that x < 1. For all terms with nb + 1 we have the upper estimate

${\frac {b!}{n!}}={\frac {1}{(b+1)(b+2)\cdots (b+(n-b))}}<{\frac {1}{(b+1)^{n-b}}}.$ This inequality is 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

$x=\sum _{n=b+1}^{\infty }{\frac {b!}{n!}}<\sum _{n=b+1}^{\infty }{\frac {1}{(b+1)^{n-b}}}=\sum _{k=1}^{\infty }{\frac {1}{(b+1)^{k}}}={\frac {1}{b+1}}\left({\frac {1}{1-{\frac {1}{b+1}}}}\right)={\frac {1}{b}}<1.$ Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so e must be irrational. Q.E.D.

## Alternate proofs

Another proof can be obtained from the previous one by noting that

$(b+1)x=1+{\frac {1}{b+2}}+{\frac {1}{(b+2)(b+3)}}+\cdots <1+{\frac {1}{b+1}}+{\frac {1}{(b+1)(b+2)}}+\cdots =1+x,$ and this inequality is equivalent to the assertion that bx < 1. This is impossible, of course, since b and x are natural numbers.

Still another proof  can be obtained from the fact that

${\frac {1}{e}}=e^{-1}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\cdot$ Define $s_{n}$ as follows:

$s_{n}=\sum _{k=0}^{n}{\frac {(-1)^{k}}{k!}}.$ Then:

$e^{-1}-s_{2n-1}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!}}-\sum _{k=0}^{2n-1}{\frac {(-1)^{k}}{k!}}<{\frac {1}{(2n)!}},$ which implies:

$0<(2n-1)!\left(e^{-1}-s_{2n-1}\right)<{\frac {1}{2n}}\leq {\frac {1}{2}}$ for any integer $n\geq 2.$ Note that $(2n-1)!s_{2n-1}$ is always an integer. Assume $e^{-1}$ is rational, so, $e^{-1}={\tfrac {p}{q}}$ where $p,q$ are co-prime and $q\neq 0.$ It's possible to appropriately choose $n$ so that $(2n-1)!e^{-1}$ is an integer i.e. $n\geq {\tfrac {q+1}{2}}.$ Hence, for this choice, the difference between $(2n-1)!e^{-1}$ and $(2n-1)!s_{2n-1}$ would be an integer. But from the above inequality, that's impossible.So, $e^{-1}$ is irrational. This means that $e$ is irrational.

## Generalizations

In 1840, Liouville published a proof of the fact that e2 is irrational followed by a proof that e2 is not a root of a second degree polynomial with rational coefficients. This last fact implies that e4 is irrational. His proofs are similar to Fourier's proof of the irrationality of e. In 1891, Hurwitz explained how it is possible to prove along the same line of ideas that e is not a root of a third degree polynomial with rational coefficients. In particular, e3 is irrational.

More generally, eq is irrational for any non-zero rational q.