Editing Proof that e is irrational

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{{E (mathematical constant)}}
{{DISPLAYTITLE:Proof that {{mvar|e}} is irrational}}
The [[e (mathematical constant)|number ''e'']] was introduced by [[Jacob Bernoulli]] in 1683. More than half a century later, [[Leonhard Euler|Euler]], who had been a student of Jacob's younger brother [[Johann Bernoulli|Johann]], proved that ''e'' is [[Irrational number|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).<ref>{{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}}</ref><ref>{{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 }}</ref><ref>{{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}}</ref> He computed the representation of ''e'' as a [[simple continued fraction]], which is

:<math>e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, \ldots, 2n, 1, 1, \ldots]. </math>

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.<ref>[https://arxiv.org/abs/math/0601660 A Short Proof of the Simple Continued Fraction Expansion of e]</ref><ref>{{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 }}</ref> Since the simple continued fraction of ''e'' is not [[Periodic continued fraction|periodic]], this also proves that ''e'' is not a root of a quadratic polynomial with rational coefficients; in particular, ''e''<sup>2</sup> is irrational.

==Fourier's proof==
The most well-known proof is [[Joseph Fourier]]'s [[proof by contradiction]],<ref>{{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}}</ref> which is based upon the equality

: <math>e = \sum_{n = 0}^\infty \frac{1}{n!}.</math>

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&nbsp;''b'', the fraction {{sfrac|''a''|''b''}} and the {{nowrap|''b''-th}} partial sum are turned into [[integer]]s, hence ''x'' must be a positive integer. However, the fast convergence of the series representation implies that ''x'' is still strictly smaller than&nbsp;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

: <math>x = b!\left(e - \sum_{n = 0}^{b} \frac{1}{n!}\right).</math>

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

: <math>x = b!\left (\frac{a}{b} - \sum_{n = 0}^{b} \frac{1}{n!}\right) = a(b - 1)! - \sum_{n = 0}^{b} \frac{b!}{n!}.</math>

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

: <math>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,</math>

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

: <math>\frac{b!}{n!} =\frac1{(b + 1)(b + 2) \cdots \big(b + (n - b)\big)} \le \frac1{(b + 1)^{n-b}}.</math>

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 [[Geometric series|infinite geometric series]], we obtain

:<math>x =\sum_{n = b + 1}^\infty \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.</math>

And therefore <math>x<1.</math>

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

==Alternative proofs==
Another proof<ref>{{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 }}</ref> can be obtained from the previous one by noting that

: <math>(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,</math>

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

Still another proof<ref>{{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 }}</ref><ref>Apostol, T. (1974). Mathematical analysis (2nd ed., Addison-Wesley series in mathematics). Reading, Mass.: Addison-Wesley.</ref> can be obtained from the fact that

: <math>\frac{1}{e} = e^{-1} = \sum_{n=0}^\infty \frac{(-1)^n}{n!}.</math>

Define <math>s_n</math> as follows:

: <math>s_n = \sum_{k=0}^n \frac{(-1)^{k}}{k!}.</math>

Then

: <math>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)!},</math>

which implies

: <math>0 < (2n - 1)! \left(e^{-1} - s_{2n-1}\right) < \frac{1}{2n} \le \frac{1}{2}</math> 

for any positive integer <math>n</math>.

Note that <math>(2n - 1)!s_{2n-1}</math> is always an integer. Assume that <math>e^{-1}</math> is rational, so <math>e^{-1} = p/q,</math> where <math>p, q</math> are co-prime, and <math>q \neq 0.</math> It is possible to appropriately choose <math>n</math> so that <math>(2n - 1)!e^{-1}</math> is an integer, i.e. <math>n \geq (q + 1)/2.</math> Hence, for this choice, the difference between <math>(2n - 1)!e^{-1}</math> and <math>(2n - 1)!s_{2n-1}</math> would be an integer. But from the above inequality, that is not possible. So, <math>e^{-1}</math> is irrational. This means that <math>e</math> is irrational.

==Generalizations==
In 1840, [[Joseph Liouville|Liouville]] published a proof of the fact that ''e''<sup>2</sup> is irrational<ref>{{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}}</ref> followed by a proof that ''e''<sup>2</sup> is not a root of a second-degree polynomial with rational coefficients.<ref>{{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}}</ref> This last fact implies that ''e''<sup>4</sup> is irrational. His proofs are similar to Fourier's proof of the irrationality of ''e''. In 1891, [[Adolf Hurwitz|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 ''e''<sup>3</sup> is irrational.<ref>{{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}}</ref> More generally, ''e''<sup>''q''</sup> is irrational for any non-zero rational ''q''.<ref>{{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}}</ref>

[[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''<sup>''&alpha;''</sup>}} for any non-zero [[algebraic number|algebraic]] ''&alpha;''.<ref>{{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}}</ref>

==See also==
* [[Characterizations of the exponential function]]
* [[Transcendental number]], including a [[Transcendental number#A proof that e is transcendental|proof that ''e'' is transcendental]]
* [[Lindemann–Weierstrass theorem]]
* [[Proof that π is irrational]]

==References==
<references/>

[[Category:Diophantine approximation]]
[[Category:Exponentials]]
[[Category:Article proofs]]
[[Category:E (mathematical constant)]]
[[Category:Irrational numbers]]