# Natural Logarithm of 2 is Greater than One Half

## Lemma

$\ln 2 \ge \dfrac 1 2$

where $\ln$ denotes the natural logarithm function.

## Proof 1

Let $f: \R_{>0} \to \R$ be the real function defined as:

$\forall x \in \R_{>0}: f \left({x}\right) = \dfrac 1 x$

From Real Rational Function is Continuous, $f \left({x}\right)$ is a continuous real function, in particular on the closed interval $\left[{a \,.\,.\, b}\right]$.

Hence the Mean Value Theorem for Integrals can be applied:

There exists some $k \in \left[{1 \,.\,.\, 2}\right]$ such that:

 $\displaystyle \int_1^2 f \left({x}\right) \ \mathrm d x$ $=$ $\displaystyle f \left({k}\right) \left({2 - 1}\right)$ $\displaystyle \implies \ \$ $\displaystyle \int_1^2 \frac 1 x \ \mathrm d x$ $=$ $\displaystyle \frac 1 k \left({2 - 1}\right)$ substituting $\dfrac 1 x$ for $f \left({x}\right)$ $\displaystyle \implies \ \$ $\displaystyle \ln 2 - \ln 1$ $=$ $\displaystyle \frac 1 k \left({2 - 1}\right)$ Definition of Natural Logarithm $\displaystyle \implies \ \$ $\displaystyle \ln 2$ $=$ $\displaystyle \frac 1 k$ Logarithm of $1$ is $0$

Thus:

 $\displaystyle 1$ $\le$ $\displaystyle k \le 2$ $\displaystyle \implies \ \$ $\displaystyle 1^{-1}$ $\ge$ $\displaystyle k^{-1} \ge 2^{-1}$ Ordering of Reciprocals $\displaystyle \implies \ \$ $\displaystyle 1$ $\ge$ $\displaystyle \ln 2 \ge \frac 1 2$

$\blacksquare$

## Proof 2

 $\displaystyle 1 - \frac 1 x$ $\le$ $\displaystyle \ln x$ Lower Bound of Natural Logarithm $\displaystyle \implies \ \$ $\displaystyle \frac 1 2$ $\le$ $\displaystyle \ln 2$ letting $x = 2$

$\blacksquare$