Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 5F / i

Whether $\log 2$ is irrational is not presently known; the given fact about $e$ does not decide it.

${\log }_{2}3$ is irrational: if it were $p/q$, then $2^p=3^q$, contradicting unique prime factorisation.

$ae+b/e$ is transcendental, hence irrational. If it were algebraic, $e$ would solve $a{x}^2-(ae+b/e)x+b=0$ over the algebraic numbers, contradicting transcendence of $e$.

The cubic is strictly increasing and has one real root. A rational root of the monic integer polynomial would be an integer divisor of $3$; testing $\pm 1,\pm 3$ gives none, so the root is irrational.