Exponential of Zero and One

Theorem

Let $\exp x$ be the exponential of $x$.

Then:

$\exp 0 = 1$

$\exp 1 = e$

where $e$ is Euler's number, i.e. $2.718281828\ldots$.

Proof 1

We have that the exponential function is the inverse of the natural logarithm function:

$\ln 1 = 0$

$\ln e = 1$

Hence the result.

$\blacksquare$

Proof 2

We use the definition of the exponential as a limit of a sequence:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \exp 0\)	\(=\)	\(\displaystyle \lim_{n \to \infty} \left({1 + \frac 0 n}\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

That $\exp 1 = e$ follows from the Equivalence of Definitions of Euler's Number.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 14.4$