transfinite arithmetic, cardinal arithmetic, ordinal arithmetic
prime field, p-adic integer, p-adic rational number, p-adic complex number
arithmetic geometry, function field analogy
The irrational number conventionally denoted $e$ (a notation credited to Euler, hence also called the Euler number) is the base of the natural logarithm; it is approximately 2.71828182845 in decimal notation.
There are numerous ways of defining $e$. One is
This can be interpreted as the groupoid cardinality for $core(FinSet)$. Perhaps more important than the constant $e$ is the standard exponential function (defined for all complex numbers $x$)
for which $e = \exp(1)$. This exponential function is especially convenient because it is uniquely characterized as a function $f(x)$ equal to its own derivative such that $f(0) = 1$ (necessary in order that it satisfy the exponential law $f(x + y) = f(x)f(y)$).
Construct a polar coordinate grid (consisting of radial lines through a point called the origin, and concentric circles centered at the origin). Draw a curve starting at any point except the origin in such a way that at each of its points $p$, the tangent at $p$ meets the radial line at $p$ in a 45 degree angle. This curve is called a logarithmic spiral. Then, following the trajectory of the spiral inward (towards the origin, so to speak) through one radian from $p$ to a second point $q$, the distance from $p$ to the origin differs to the distance from $q$ to the origin by a factor of $e$.
Equivalently, imagine four ants situated at the corners of a square, and imagine that at some instant each begins crawling toward its neighbor looking clockwise from above, each at the same speed. The trajectory of each ant is a logarithmic spiral as described above, and the same description of $e$ applies.
It is a simple matter to show that $e$ is irrational. For if on the contrary we have $e = p/q$, then $e \cdot n!$ would be an integer for any $n \geq q$. However,
where the nonzero tail after the integer part is bounded above by $\sum_{k=1}^\infty 1/(n+1)^k = 1/n \lt 1$ for $n \gt 1$, giving a contradiction.
It is harder to show that $e$ is transcendental. An online proof (written up by David Richeson) may be found here.
Eli Maor, e: The Story of a Number, Princeton University Press (1994). ISBN 0-691-05854-7.
Last revised on September 2, 2019 at 23:20:54. See the history of this page for a list of all contributions to it.