Proof that e is irrational

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 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,[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. 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:

${\displaystyle 2=1+{\tfrac {1}{1!}}<e=1+{\tfrac {1}{1!}}+{\tfrac {1}{2!}}+{\tfrac {1}{3!}}+\cdots <1+\left(1+{\tfrac {1}{2}}+{\tfrac {1}{2^{2}}}+{\tfrac {1}{2^{3}}}+\cdots \right)=3.}$

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 ^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.

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

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

To see that if e is rational, then x is an integer, substitute e = ^a⁄_b into this definition 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, 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

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

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[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 natural numbers.

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 integer ${\displaystyle n\geq 2.}$

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

1. Euler, Leonhard (1744). "De fractionibus continuis dissertatio" [A dissertation on continued fractions] (PDF). Commentarii academiae scientiarum Petropolitanae. 9: 98–137.
2. Euler, Leonhard (1985). "An essay on continued fractions". Mathematical Systems Theory. 18: 295–398. doi:10.1007/bf01699475. hdl:1811/32133.
3. Sandifer, C. Edward (2007). "Chapter 32: Who proved e is irrational?". How Euler did it. Mathematical Association of America. pp. 185–190. ISBN 978-0-88385-563-8. LCCN 2007927658.
4. A Short Proof of the Simple Continued Fraction Expansion of e
5. Cohn, Henry (2006). "A short proof of the simple continued fraction expansion of e". American Mathematical Monthly. Mathematical Association of America. 113 (1): 57–62. arXiv:math/0601660. doi:10.2307/27641837. JSTOR 27641837.
6. de Stainville, Janot (1815). Mélanges d'Analyse Algébrique et de Géométrie [A mixture of Algebraic Analysis and Geometry]. Veuve Courcier. pp. 340–341.
7. MacDivitt, A. R. G.; Yanagisawa, Yukio (1987), "An elementary proof that e is irrational", The Mathematical Gazette, London: Mathematical Association, 71 (457): 217, doi:10.2307/3616765, JSTOR 3616765
8. Penesi, L. L. (1953). "Elementary proof that e is irrational". American Mathematical Monthly. Mathematical Association of America. 60 (7): 474. doi:10.2307/2308411. JSTOR 2308411.
9. Apostol, T. (1974). Mathematical analysis (2nd ed., Addison-Wesley series in mathematics). Reading, Mass.: Addison-Wesley.
10. Liouville, Joseph (1840). "Sur l'irrationalité du nombre e = 2,718…". Journal de Mathématiques Pures et Appliquées. 1 (in French). 5: 192.
11. Liouville, Joseph (1840). "Addition à la note sur l'irrationnalité du nombre e". Journal de Mathématiques Pures et Appliquées. 1 (in French). 5: 193–194.
12. Hurwitz, Adolf (1933) [1891]. "Über die Kettenbruchentwicklung der Zahl e". Mathematische Werke (in German). 2. Basel: Birkhäuser. pp. 129–133.
13. Aigner, Martin; Ziegler, Günter M. (1998), Proofs from THE BOOK (4th ed.), Berlin, New York: Springer-Verlag, pp. 27–36, doi:10.1007/978-3-642-00856-6, ISBN 978-3-642-00855-9.