Editing Proof that e is irrational (section)

==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.