Irrationality of Log Base n of n Plus One

Problem Statement

Prove that for any integer $n > 1$ , $\log_n(n+1)$ is irrational.

Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

Solutions

Proof by Contradiction

Step 1: Assume the negation

Assume, for contradiction, that $\log_n(n+1)$ is rational for some integer $n > 1$ .

Then we can write:

\log_n(n+1) = \frac{p}{q}

where $p, q$ are positive integers.

Step 2: Convert to exponential form

By definition of logarithm:

n^{p/q} = n+1

Raising both sides to the power $q$ :

n^p = (n+1)^q

Step 3: Analyze modulo $n$

Left side: $n^p \equiv 0 \pmod{n}$ (clearly divisible by $n$ )

Right side: By the Binomial Theorem:

(n+1)^q = \sum_{k=0}^{q} \binom{q}{k} n^k \cdot 1^{q-k} = n^q + \binom{q}{1}n^{q-1} + \ldots + \binom{q}{q-1}n + 1

All terms contain $n$ except the last term, so:

(n+1)^q \equiv 1 \pmod{n}

Step 4: Derive contradiction

From $n^p = (n+1)^q$ , we have modulo $n$ :

0 \equiv 1 \pmod{n}

This means $n \mid (0-1)$ , so $n \mid (-1)$ .

For $n > 1$ , this is impossible.

Conclusion

The assumption that $\log_n(n+1)$ is rational leads to a contradiction.

Therefore, $\log_n(n+1)$ is irrational for all integers $n > 1$ .

\hfill $\square$

Takeaways

Logarithm to Exponential: Converting $\log_n(n+1) = \frac{p}{q}$ to $n^p = (n+1)^q$ enables algebraic manipulation
Modular Analysis: Working mod $n$ reveals contradiction: LHS $\equiv 0$ but RHS $\equiv 1$
Binomial Expansion: $(n+1)^q \equiv 1 \pmod{n}$ since only constant term survives
Non-standard Irrationality: Unlike $\sqrt{p}$ proofs, this uses modular arithmetic rather than prime factorization

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Complex Numbers: https://vumaths.com/booklets/hsc-complex-numbers/
HSC Probability: https://vumaths.com/booklets/hsc-probability/

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