Proof that e is irrational

Euler wrote the first proof of the fact that e is irrational in 1737 (but the text was only published seven years later).{{cite journal | last = Euler | first = Leonhard | year = 1744 | title = De fractionibus continuis dissertatio | url = http://www.math.dartmouth.edu/~euler/docs/originals/E071.pdf | journal = Commentarii Academiae Scientiarum Petropolitanae | volume = 9 | pages = 98–137 |trans-title=A dissertation on continued fractions}}{{cite journal | last = Euler | first = Leonhard | title = An essay on continued fractions | journal = Mathematical Systems Theory | year = 1985 | volume = 18 | pages = 295–398 | url = https://kb.osu.edu/dspace/handle/1811/32133 | publication-date = 1985 | doi=10.1007/bf01699475| hdl = 1811/32133 | s2cid = 126941824 | hdl-access = free }}{{cite book | last1 = Sandifer | first1 = C. Edward | title = How Euler did it | chapter = Chapter 32: Who proved e is irrational? | url=http://eulerarchive.maa.org/hedi/HEDI-2006-02.pdf |publisher = Mathematical Association of America | pages = 185–190 | year = 2007 | isbn = 978-0-88385-563-8 | lccn = 2007927658}} He computed the representation of e as a simple continued fraction, which is

:$e\; =\; [2;\; 1,\; 2,\; 1,\; 1,\; 4,\; 1,\; 1,\; 6,\; 1,\; 1,\; 8,\; 1,\; 1,\; \backslash ldots,\; 2n,\; 1,\; 1,\; \backslash ldots].$

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.[https://arxiv.org/abs/math/0601660 A Short Proof of the Simple Continued Fraction Expansion of e]{{cite journal | last = Cohn | first = Henry | journal = American Mathematical Monthly | volume = 113 | issue = 1 | pages = 57–62 | year = 2006 | title = A short proof of the simple continued fraction expansion of e | jstor = 27641837 | doi=10.2307/27641837| arxiv = math/0601660 | bibcode = 2006math......1660C }} 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.

The most well-known proof is Joseph Fourier's proof by contradiction,{{Cite book | last1 = de Stainville | first1 = Janot | year = 1815 | title = Mélanges d'Analyse Algébrique et de Géométrie |trans-title=A mixture of Algebraic Analysis and Geometry | publisher = Veuve Courcier | pages = 340–341}} which is based upon the equality

: $e\; =\; \backslash sum\_\{n\; =\; 0\}^\backslash infty\; \backslash frac\{1\}\{n!\}.$

Initially e is assumed to be a rational number of the form {{sfrac|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 {{nowrap|b-th}} partial sum, which approximates the limiting value e. By choosing the scale factor to be the factorial of b, the fraction {{sfrac|a|b}} and the {{nowrap|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 {{nowrap|1=e = {{sfrac|a|b}}}}. Define the number

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

Use the assumption that e = {{sfrac|a|b}} to obtain

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

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

We now prove that {{nowrap|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

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

because all the terms are strictly positive.

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

: $\backslash frac\{b!\}\{n!\}\; =\backslash frac1\{(b\; +\; 1)(b\; +\; 2)\; \backslash cdots\; \backslash big(b\; +\; (n\; -\; b)\backslash big)\}\; \backslash le\; \backslash frac1\{(b\; +\; 1)^\{n-b\}\}.$

This inequality is strict for every {{nowrap|n ≥ b + 2}}. Changing the index of summation to {{nowrap|1=k = n – b}} and using the formula for the infinite geometric series, we obtain

:$x\; =\backslash sum\_\{n\; =\; b\; +\; 1\}^\backslash infty\; \backslash frac\{b!\}\{n!\}$

< \sum_{n=b+1}^\infty \frac1{(b + 1)^{n-b}}

=\sum_{k=1}^\infty \frac1{(b + 1)^k}

=\frac{1}{b+1} \left (\frac1{1 - \frac1{b + 1}}\right)

= \frac{1}{b} \le 1.

And therefore

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

Another proof{{cite journal | last1 = MacDivitt | first1 = A. R. G. | last2 = Yanagisawa | first2 = Yukio | title = An elementary proof that e is irrational | journal = The Mathematical Gazette | volume = 71 | issue = 457 | pages = 217 | year = 1987 | publisher =Mathematical Association | place = London | jstor = 3616765 | doi=10.2307/3616765| s2cid = 125352483 }} can be obtained from the previous one by noting that

: $(b\; +\; 1)x\; =$

1 + \frac1{b + 2} + \frac1{(b + 2)(b + 3)} + \cdots <

1 + \frac1{b + 1} + \frac1{(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{{cite journal | last = Penesi | first = L. L. | year = 1953 | title = Elementary proof that e is irrational | journal = American Mathematical Monthly | publisher = Mathematical Association of America | volume = 60 | issue = 7 | pages = 474 | jstor = 2308411 | doi = 10.2307/2308411 }}Apostol, T. (1974). Mathematical analysis (2nd ed., Addison-Wesley series in mathematics). Reading, Mass.: Addison-Wesley. can be obtained from the fact that

: $\backslash frac\{1\}\{e\}\; =\; e^\{-1\}\; =\; \backslash sum\_\{n=0\}^\backslash infty\; \backslash frac\{(-1)^n\}\{n!\}.$

Define $s\_n$ as follows:

: $s\_n\; =\; \backslash sum\_\{k=0\}^n\; \backslash frac\{(-1)^\{k\}\}\{k!\}.$

Then

:

which implies

:

for any positive integer $n$.

Note that $(2n\; -\; 1)!s\_\{2n-1\}$ is always an integer. Assume that $e^\{-1\}$ is rational, so $e^\{-1\}\; =\; p/q,$ where $p,\; q$ are co-prime, and $q\; \backslash neq\; 0.$ It is possible to appropriately choose $n$ so that $(2n\; -\; 1)!e^\{-1\}$ is an integer, i.e. $n\; \backslash geq\; (q\; +\; 1)/2.$ 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 is not possible. So, $e^\{-1\}$ is irrational. This means that $e$ is irrational.

In 1840, Liouville published a proof of the fact that e² is irrational{{cite journal | last = Liouville | first = Joseph | journal = Journal de Mathématiques Pures et Appliquées | title = Sur l'irrationalité du nombre e = 2,718… | series = 1 | volume = 5 | pages = 192 | year = 1840 | language = fr}} followed by a proof that e² is not a root of a second-degree polynomial with rational coefficients.{{cite journal | last = Liouville | first = Joseph | journal = Journal de Mathématiques Pures et Appliquées | title = Addition à la note sur l'irrationnalité du nombre e | series = 1 | volume = 5 | pages = 193–194 | year = 1840 | language = fr}} 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, which implies that e³ is irrational.{{cite book | last1 = Hurwitz | first1 = Adolf | year = 1933 | orig-year = 1891 | title = Mathematische Werke | volume = 2 | language = de | chapter = Über die Kettenbruchentwicklung der Zahl e | publisher = Birkhäuser | location = Basel | pages = 129–133}} More generally, e^q is irrational for any non-zero rational q.{{cite book | last1=Aigner | first1=Martin | author1-link = Martin Aigner | 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 |isbn=978-3-642-00855-9 |doi=10.1007/978-3-642-00856-6 |edition=4th}}

Charles Hermite further proved that e is a transcendental number, in 1873, which means that is not a root of any polynomial with rational coefficients, as is {{math|e^α}} for any non-zero algebraic α.{{cite journal |last=Hermite |first=C. |author-link=Charles Hermite |year=1873 |title=Sur la fonction exponentielle |lang=fr |journal=Comptes rendus de l'Académie des Sciences de Paris |volume=77 |pages=18–24}}

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

Category:Diophantine approximation

Category:Exponentials

Category:Article proofs

Category:E (mathematical constant)

Category:Irrational numbers