Claim: $\ln 2$ is a rational number. (Note that $\ln 2 = 1 - \frac12 + \frac13 - \frac14 + \cdots + \frac{(-1)^{n+1}}n = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n$). This is done by proving that $\ln 2 = 1 - \frac12 + \frac13 - \frac14 + \cdots + \frac{(-1)^{n+1}}n$ is rational.
Proof by induction on $n$:
Base case:
$n = 1$: 1 is obviously rational.
Inductive hypothesis:
Suppose that $1 - \frac12 + \frac13 - \frac14 + \cdots + \frac{(-1)^{k+1}}k$ is rational.
Inductive step:
We need to show that $1 - \frac12 + \frac13 - \frac14 + \cdots + \frac{(-1)^{k+2}}{k+1}$ is a rational number. Observe that $1 - \frac12 + \frac13 - \frac14 + \cdots + \frac{(-1)^{k+2}}{k+1} = (1 - \frac12 + \frac13 - \frac14 + \cdots + \frac{(-1)^{k+1}}k) + \frac{(-1)^{k+2}}{k+1}$. By the induction hypothesis, $1 - \frac12 + \frac13 - \frac14 + \cdots + \frac{(-1)^{k+1}}k$ is a rational number. Furthermore, $\frac{(-1)^{k+2}}{k+1}$ is a rational number, as it can be expressed as a fraction. Thus, summing two rational numbers will result in another rational number, and by induction we have proven that $\ln 2$ is rational.
My thoughts are
- First, in the inductive step, we should state that this holds for some $n$.
- Second, we know that $\ln(2)$ is irrational.
- This seems really sound, but I just can't find the mistake in the induction proof.
Any thoughts?
What the induction proves (somewhat verbosely, possibly to hide that not much is really happening) is that each of the partial sums is a rational number, which is true.
But $\log 2$ is the limit of the partial sums, and the limit of a sequence of rationals is not necessarily rational.
In fact, it is well known that every real number is the sum of a series of rationals. For example,
$$ \pi = 3 + \frac{1}{10} + \frac{4}{100} + \frac{1}{1000} + \frac{5}{10000} + \frac{9}{100000} + \cdots $$