ProofWiki problem 25 Let \(a, b \in \mathbb{N}_{>0}\) such that there exists no \(m, n \in \mathbb{N}_{>0}\) such that \(a^m = b^n\). Prove that \(\log_b a\) is irrational.

Presented the problem statement above, the assistant gave a perfect answer with step-by-step calculations. To test if the assistant has a true mathematical understanding of the problem, we first asked for definitions of concepts used, and then varied the original problem by loosening some of the assumptions made, and asked the assistant for a proof in the new setting.

Asking for definitions We found that the assistant gave the correct definitions in the theorem statement as well as in its own proof. Concretely, it gave the right answers for: the definition of logarithm; the range of a logarithm's base; the meaning of the set \(\mathbb{N}_{>0}\); and whether \(\log_ba=\frac{p}{q}\) can be a negative number (\(p\) and \(q\) are variables arising from the assistant's own proof).

Loosening assumptions We started by asking the assistant whether the proof still holds if we instead have \(a, b \in \mathbb{R}_{>0}\) ? The assistant understood the meaning of \(\mathbb{R}_{>0}\) and confirmed the derivation still held, so the original lemma/proposition has been generalised (since one of its assumption has been relaxed). Later, we attempted to generalise the proposition further by dropping the assumption \(a \in \mathbb{R}_{>0}\) or \(b \in \mathbb{R}_{>0}\):

We continued by asking if dropping the assumption that \(b \in \mathbb{R}_{>0}\) or \(a \in \mathbb{R}_{>0}\) affects the original proof ? The assistant knew that these assumptions were necessary to make the log function well-defined, and pointed out that dropping either of the assumptions would invalidate our previous derivation.

These variations, though not impossible, are unlikely to appear together with the problem in the training data of the assistant. We think the assistant does have some understanding of the underlying mathematical concepts and its own proof, in the context of this problem.