Proof that e is irrational

Template:Short description Template:E (mathematical constant)

The [[e (mathematical constant)|number Template:Math]] 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 Template:Math 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 Template:Math is irrational in 1737 (but the text was only published seven years later).<ref>Template:Cite journal</ref><ref>Template:Cite journal</ref><ref>Template:Cite book</ref> He computed the representation of Template:Math 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, Template:Math is irrational. A short proof of the previous equality is known.<ref>A Short Proof of the Simple Continued Fraction Expansion of e</ref><ref>Template:Cite journal</ref> Since the simple continued fraction of Template:Math is not periodic, this also proves that Template:Math is not a root of a quadratic polynomial with rational coefficients; in particular, Template:Math is irrational.

Fourier's proof

The most well-known proof is Joseph Fourier's proof by contradiction,<ref>Template:Cite book</ref> which is based upon the equality

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

Initially Template:Math is assumed to be a rational number of the form Template:Math. The idea is to then analyze the scaled-up difference (here denoted Template:Math) between the series representation of Template:Math and its strictly smaller Template:Nowrap partial sum, which approximates the limiting value Template:Math. By choosing the scale factor to be the factorial of Template:Math, the fraction Template:Math and the Template:Nowrap partial sum are turned into integers, hence Template:Math must be a positive integer. However, the fast convergence of the series representation implies that Template:Math is still strictly smaller than 1. From this contradiction we deduce that Template:Math is irrational.

Now for the details. If Template:Math is a rational number, there exist positive integers Template:Math and Template:Math such that Template:Math. Define the number

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

Use the assumption that Template:Math 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 Template:Math for each term. Therefore, under the assumption that Template:Math is rational, Template:Math is an integer.

We now prove that Template:Math. First, to prove that Template:Math is strictly positive, we insert the above series representation of Template:Math into the definition of Template:Math 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 Template:Math. For all terms with Template:Math 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 Template:Math. Changing the index of summation to Template:Math and using the formula for the 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 Template:Math is irrational, Q.E.D.

Alternative proofs

Another proof<ref>Template:Cite journal</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 Template:Math. This is impossible, of course, since Template:Math and Template:Math are positive integers.

Still another proof<ref>Template:Cite journal</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:

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

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, Liouville published a proof of the fact that Template:Math is irrational<ref>Template:Cite journal</ref> followed by a proof that Template:Math is not a root of a second-degree polynomial with rational coefficients.<ref>Template:Cite journal</ref> This last fact implies that Template:Math is irrational. His proofs are similar to Fourier's proof of the irrationality of Template:Math. In 1891, Hurwitz explained how it is possible to prove along the same line of ideas that Template:Math is not a root of a third-degree polynomial with rational coefficients, which implies that Template:Math is irrational.<ref>Template:Cite book</ref> More generally, Template:Math is irrational for any non-zero rational Template:Math.<ref>Template:Cite book</ref>

Charles Hermite further proved that Template:Math is a transcendental number, in 1873, which means that is not a root of any polynomial with rational coefficients, as is Template:Math for any non-zero algebraic Template:Math.<ref>Template:Cite journal</ref>

References

Proof that e is irrational

Contents

Euler's proof

Fourier's proof

Alternative proofs

Generalizations

See also

References

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Proof that e is irrational

Euler's proof

Fourier's proof

Alternative proofs

Generalizations

See also

References

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