Alternative Proof that ln(2) is Irrational

Question

Is the following a correct proof that $\ln(2)$ is irrational without using the fact that $e$ is transcendental?

The power series $\ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k+1}x^k}{k}$ for $x=1$ gives the alternating harmonic series $\ln(2)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}$. Consider the sequence of partial sums $x_n = \sum_{k=1}^n \frac{(-1)^{k+1}}{k}$.

Odd terms are decreasing: $x_1>x_3>x_5...$ while even terms are increasing: $x_2<x_4<x_6<...$. All odd terms are greater than even terms: $x_{2n+1}>x_{2m}$ for all $m,n \in \Bbb N \cup \{0\}$. It can be proven that $ \{x_n\}$ is a Cauchy sequence (omitted since it is relatively long):

$$ \forall 0<\epsilon \in \Bbb Q \ \ \exists N \in \Bbb N: \lvert x_n - x_m \rvert < \epsilon \ \ \forall n,m \geq N$$

Let $q$ be the limit of $\{x_n\}$, which we know exists by the Alternating Series Test.

If $q \leq x_{2m}$ for some $m \in \Bbb N \cup \{0\}$ consider $\epsilon = x_{2m + 2} - x_{2m} > 0$ and any $N \in \Bbb N$. Choose $n \in \Bbb N$ such that $2n > 2m+2 + N$. Then

$$ \lvert x_{2n} - q \rvert = x_{2n} - q > x_{2n} - x_{2m} > x_{2m+2} - x_{2m} > \epsilon $$

So $q \nleq x_{2m}$ and similarly prove $q \ngeq x_{2n+1}$. So $x_{2m} < q < x_{2n+1}$ for all $m,n \in \Bbb N \cup \{0\}$.

I am concerned that this part is insufficient to show $q = \ln(2)$ is irrational:

Since sequence $\{x_n\}$ is Cauchy, we have: $$\forall 0 < \epsilon \in \Bbb Q \ \ \exists N \in \Bbb N: \lvert x_{2n+1} - x_{2m} \rvert < \epsilon \ \ \forall \ \ 2n+1,2m \geq N$$ Let $0 < \epsilon \in \Bbb Q$ be arbitrary. Then for $2n+1,2m \geq N$, $$ 0 < q- x_{2m} < x_{2n+1}-x_{2m} = \lvert x_{2n+1}-x_{2m} \rvert < \epsilon$$ That is, $\forall 0 < \epsilon \in \Bbb Q, \ \ 0 < q- x_{2m} < \epsilon$. So $q = \ln(2)$ cannot be rational, else $q-x_{2m}$ is rational.

Thanks for your help in advance.

Answer 1

Firstly, we should be careful with some terminologies. In the place of '$\forall 0<\epsilon \in \mathbb{Q}$', we should have used, 'given any $\epsilon, 0<\epsilon \in \mathbb{Q}'$. This is because the value of $N$ depends on the $\epsilon$ chosen.

Now coming to the confusion: Assuming that $\{x_n\}$ is a sequence of rational numbers and it converges to $q$ and that $x_n \leq q$ for all $n$. You are assuming that there exists an $N$ such that for all $n\geq N$, the following inequality holds good for all rational numbers $\epsilon$: $$0< q-x_n < \epsilon.$$

No, this is not the case. It cannot even happen. Fix an $n\geq N$. Then there must exist a positive rational number $\epsilon$ which is strictly less than $q-x_n$. Isn't it? Forget about rational or irrational, choose any positive real $\epsilon$. If there exists an $N$ such that for all $n\geq N$ we have $q-x_n < \epsilon$, then it simply means that $q-x_n< \epsilon$ and that there are possibilities that it can even be rational say $\epsilon^{\prime}$ for some $n$. Now, we cannot argue that $q-x_n$ should have been less than $\epsilon^{\prime}$, which is what you are doing.

Answer 2

When you say $$\forall 0 < \epsilon \in \Bbb Q, \ \ 0 < q- x_{2m} < \epsilon$$

you are treating $2m$ as if it was fixed as $\epsilon$ changes, and yet in deriving your inequality you required $2m\ge N$, and $N$ may change as $\epsilon$ changes.

And in fact (though we haven't proved it), it is almost certain that $N$ gets unboundedly larger as $\epsilon$ gets smaller, eventually exceeding any fixed value of $2m$.

I.e. for any fixed value of $2m$, as $\epsilon $ gets small, $N$ gets large, and eventually it gets larger than $2m$, and you can no longer say that $\lvert x_{2n+1}-x_{2m} \rvert < \epsilon$.