Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 5F (Numbers and Sets)

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.

Let $2^k\le n<2^{k+1}$. In the reduced common-denominator expression for $H_n$, the term $1/2^k$ is the unique term whose denominator contains the largest power $2^k$; after multiplication by the least common multiple it contributes an odd integer while all other terms contribute even integers. Thus the numerator is odd and the denominator remains even, so $H_n$ is not an integer.