Natural Logarithm of 2 is Greater than One Half
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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:
- $\forall x \in \R_{>0}: \map f x = \dfrac 1 x$
From Real Rational Function is Continuous, $\map f x$ is a continuous real function, in particular on the closed interval $\closedint a b$.
Hence the Mean Value Theorem for Integrals can be applied:
There exists some $k \in \closedint 1 2$ such that:
| \(\ds \int_1^2 \map f x \rd x\) | \(=\) | \(\ds \map f k \paren {2 - 1}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int_1^2 \frac 1 x \rd x\) | \(=\) | \(\ds \frac 1 k \paren {2 - 1}\) | substituting $\dfrac 1 x$ for $\map f x$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \ln 2 - \ln 1\) | \(=\) | \(\ds \frac 1 k \paren {2 - 1}\) | Definition 1 of Natural Logarithm | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \ln 2\) | \(=\) | \(\ds \frac 1 k\) | Logarithm of $1$ is $0$ |
Thus:
| \(\ds 1\) | \(\le\) | \(\, \ds k \, \) | \(\, \ds \le \, \) | \(\ds 2\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 1^{-1}\) | \(\ge\) | \(\, \ds k^{-1} \, \) | \(\, \ds \ge \, \) | \(\ds 2^{-1}\) | Ordering of Reciprocals | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds 1\) | \(\ge\) | \(\, \ds \ln 2 \, \) | \(\, \ds \ge \, \) | \(\ds \frac 1 2\) |
$\blacksquare$
Proof 2
| \(\ds 1 - \frac 1 x\) | \(\le\) | \(\ds \ln x\) | Lower Bound of Natural Logarithm | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 1 2\) | \(\le\) | \(\ds \ln 2\) | letting $x = 2$ |
$\blacksquare$
Also see
Sources
- 2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards: Calculus (8th ed.): Appendix $A$: Properties of the Natural Logarithmic Function