# Proof that *e* is irrational

Template:E (mathematical constant)

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 can not be expressed as the quotient of two integers.

## Contents

## 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

Since this continued fraction is infinite, *e* is irrational. A short proof of the previous equality is known.^{[4]} 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, *e*^{2} is irrational.

## Fourier's proof

The most well-known proof is Joseph Fourier's proof by contradiction,^{[5]} which is based upon the equality

Initially *e* is assumed to be a rational number of the form ^{a}⁄_{b}. Note that *b* couldn't be equal to one as *e* is not an integer. It can be shown using the above equality that *e* is strictly between 2 and 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

To see that if *e* is rational, then *x* is an integer, substitute *e* = ^{a}⁄_{b} into this definition to obtain

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

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

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

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^{[6]} can be obtained from the previous one by noting that

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^{[7]} can be obtained from the fact that

## Generalizations

In 1840, Liouville published a proof of the fact that *e*^{2} is irrational^{[8]} followed by a proof that *e*^{2} is not a root of a second degree polynomial with rational coefficients.^{[9]} This last fact implies that *e*^{4} 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.^{[10]} In particular, *e*^{3} is irrational.

More generally, *e*^{q} is irrational for any non-zero rational *q*.^{[11]}

## See also

- Characterizations of the exponential function
- Transcendental number, including a proof that
*e*is transcendental - Lindemann–Weierstrass theorem

## References

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