Prove that log₂(3) is irrational

Question

Prove that $\log_2 3$ is irrational.

Solution — Step by Step

Suppose, for contradiction, that $\log_2 3$ is rational.

Then we can write $\log_2 3 = \dfrac{p}{q}$ where $p, q$ are integers, $q > 0$ , and the fraction is in lowest terms (gcd $(p, q) = 1$ ).

By the definition of logarithm:

\log_2 3 = \frac{p}{q} \implies 2^{p/q} = 3

Raising both sides to the power $q$ :

2^p = 3^q

The equation $2^p = 3^q$ says a power of 2 equals a power of 3.

The left side, $2^p$ , has only 2 as a prime factor. Its prime factorisation is $2^p$ .

The right side, $3^q$ , has only 3 as a prime factor. Its prime factorisation is $3^q$ .

For these to be equal, both sides must have the same prime factorisation. But $2^p$ contains only the prime 2, while $3^q$ contains only the prime 3. These cannot be equal unless both are 1 (i.e., $p = q = 0$ ).

But $p = q = 0$ would mean $\log_2 3 = 0$ , which implies $2^0 = 3$ , i.e., $1 = 3$ — a contradiction.

So $2^p = 3^q$ has no solution with positive integers $p, q$ . This contradicts our assumption.

Since assuming $\log_2 3$ is rational leads to a contradiction, $\log_2 3$ must be irrational.

Why This Works

The proof relies on the Fundamental Theorem of Arithmetic: every integer greater than 1 has a unique prime factorisation. Since 2 and 3 are different primes, no power of 2 can equal a power of 3 (except $2^0 = 3^0 = 1$ , but that would require $p = q = 0$ , which doesn’t work).

More generally: $\log_a b$ is irrational whenever $a$ and $b$ are “multiplicatively independent” — when there are no integers $p, q$ such that $a^p = b^q$ .

Alternative Method — Elegant One-Liner

By unique factorisation: $2^p = 3^q \implies$ the left side is even, the right side is odd. An even number cannot equal an odd number. Contradiction. ✓

This version is cleaner for exams — works because 2 is even and 3 is odd.

Common Mistake

A common error in proof by contradiction: students assume $\log_2 3 = p/q$ but then forget to check what constraints $p$ and $q$ must satisfy. The key is that $p, q$ are positive integers (since $\log_2 3 > 0$ ). If we allowed $p = q = 0$ , that’s a trivial case that’s easily ruled out separately.

The same proof technique works for many similar irrationality proofs: $\log_2 5$ , $\log_3 7$ , $\log_4 9$ (check: $4^p = 9^q \Rightarrow 2^{2p} = 3^{2q}$ — same argument). The method fails for $\log_4 8 = 3/2$ (rational!) because $4^3 = 8^2 = 64$ — here both have the same prime base (2).

Question

Solution — Step by Step

Why This Works

Alternative Method — Elegant One-Liner

Common Mistake

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