Irrationality of Log Base n of n Plus One

Irrationality proofs often combine number theory with algebraic manipulation, and this problem gives you a chance to practise exactly that. It sits squarely in the Proof topic for HSC Mathematics Extension 1 (and also strengthens reasoning skills for Extension 2), showing how a simple-looking logarithm can be tackled with contradiction and modular arithmetic. By working through it, you’ll sharpen your ability to convert between logarithmic and exponential forms and to apply modular analysis to an otherwise opaque equation.

Problem Statement

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

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Hints

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

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Solutions

Proof by Contradiction

We’ll use proof by contradiction, assuming the opposite of what we want to prove and chasing the consequences until we reach an impossibility.

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

To unlock the relationship between $n$ and $n+1$, we rewrite the logarithmic statement in 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 $n$

Now we examine this equation modulo $n$ — a powerful way to exploit divisibility, because the left side is clearly a multiple of $n$.

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 $n > 1$.

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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

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Further Readings

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