Proof that e is irrational

Part of a series of articles on the
mathematical constant e
Properties
Natural logarithm Exponential function
Applications
compound interest Euler's identity Euler's formula half-lives exponential growth and decay
Defining e
proof that e is irrational representations of e Lindemann–Weierstrass theorem
People
John Napier Leonhard Euler
Related topics
Schanuel's conjecture
v t

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).^[1][2][3] He computed the representation of e as a simple continued fraction, which is

${\displaystyle e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,\ldots ,2n,1,1,\ldots ].}$

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.^[4][5] Since the simple continued fraction of e is not periodic, this also proves that e is not a root of a quadratic polynomial with rational coefficients; in particular, e² is irrational.

Fourier's proof[]

The most well-known proof is Joseph Fourier's proof by contradiction,^[6] which is based upon the equality

${\displaystyle e=\sum _{n=0}^{\infty }{\frac {1}{n!}}\cdot }$

Initially e is assumed to be a rational number of the form a/b. The idea is to then analyze the scaled-up difference (here denoted x) between the series representation of e and its strictly smaller b^th partial sum, which approximates the limiting value e. By choosing the scale factor to be the factorial of 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 x is still strictly smaller than 1. From this contradiction we deduce that e is irrational.

Now for the details. If e is a rational number there exist positive integers a and b such that e = a/b. Define the number

${\displaystyle x=b!\left(e-\sum _{n=0}^{b}{\frac {1}{n!}}\right).}$

Use the assumption that e = a/b to obtain

${\displaystyle x=b!\left({\frac {a}{b}}-\sum _{n=0}^{b}{\frac {1}{n!}}\right)=a(b-1)!-\sum _{n=0}^{b}{\frac {b!}{n!}}.}$

The first term is an integer, and every fraction in the sum is actually an integer because n ≤ b for each term. Therefore, under the assumption that e is rational, 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

${\displaystyle x=b!\left(\sum _{n=0}^{\infty }{\frac {1}{n!}}-\sum _{n=0}^{b}{\frac {1}{n!}}\right)=\sum _{n=b+1}^{\infty }{\frac {b!}{n!}}>0,}$

because all the terms are strictly positive.

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

${\displaystyle {\frac {b!}{n!}}={\frac {1}{(b+1)(b+2)\cdots (b+(n-b))}}\leq {\frac {1}{(b+1)^{n-b}}}.}$

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

${\displaystyle x=\sum _{n=b+1}^{\infty }{\frac {b!}{n!}}<\sum _{n=b+1}^{\infty }{\frac {1}{(b+1)^{n-b}}}=\sum _{k=1}^{\infty }{\frac {1}{(b+1)^{k}}}={\frac {1}{b+1}}\left({\frac {1}{1-{\frac {1}{b+1}}}}\right)={\frac {1}{b}}\leq 1.}$

Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so e is irrational, Q.E.D.

Alternate proofs[]

Another proof^[7] can be obtained from the previous one by noting that

${\displaystyle (b+1)x=1+{\frac {1}{b+2}}+{\frac {1}{(b+2)(b+3)}}+\cdots <1+{\frac {1}{b+1}}+{\frac {1}{(b+1)(b+2)}}+\cdots =1+x,}$

and this inequality is equivalent to the assertion that bx < 1. This is impossible, of course, since b and x are positive integers.

Still another proof ^[8][9] can be obtained from the fact that

${\displaystyle {\frac {1}{e}}=e^{-1}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\cdot }$

Define ${\displaystyle s_{n}}$ as follows:

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

Then:

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

which implies:

${\displaystyle 0<(2n-1)!\left(e^{-1}-s_{2n-1}\right)<{\frac {1}{2n}}\leq {\frac {1}{2}}}$

for any positive integer ${\displaystyle n}$.

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

Generalizations[]

In 1840, Liouville published a proof of the fact that e² is irrational^[10] followed by a proof that e² is not a root of a second degree polynomial with rational coefficients.^[11] 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.^[12] In particular, e³ is irrational.

More generally, e^q is irrational for any non-zero rational q.^[13]

See also[]

• Characterizations of the exponential function
• Transcendental number, including a proof that e is transcendental
• Lindemann–Weierstrass theorem

References[]