Past exam of the mathematics course of the University of Cambridge 2015 ia Paper 4 5E ii Solution Created 2026-09-24 Updated 2026-10-06
Reflexivity follows by taking the two exponents equal to , and symmetry is built into the two divisibility conditions. For transitivity, suppose , , and . Then and , with positive exponents. Thus the relation is an equivalence relation.
Its equivalence classes have a useful description by prime factorization. Let be the finite set of prime factors of , with . If , every prime factor of divides ; the reverse divisibility gives . Conversely, if , write and , with all exponents positive. Choosing and gives the required divisibilities. The empty case is exactly .
Therefore the prime-support equivalence relation iswhere the radical of an integer is the product of its distinct prime factors. The class with empty support is . For every nonempty finite support , all positive exponent choices give one class, and varying just one exponent produces infinitely many different integers in it.
There are infinitely many classes because each prime number gives a different singleton support. For completeness, if there were only finitely many prime numbers , a prime factor of would differ from them all. There are infinitely many classes; the unique finite class is .
Prime-support equivalence relation 2026-10-06
Positive integers are equivalent when they have the same finite set of prime factors. Equivalently, each divides a positive power of the other. This equivalence relation is induced by the radical of an integer; its equivalence classes correspond to finite prime supports. The empty support gives the singleton , and every nonempty support gives an infinite class by varying exponents.
Three to the distinct-prime-factor count 2026-10-06
The multiplicative arithmetic function counts assignments of three labels to each distinct prime factor. It also counts pairs of squarefree divisors whose least common multiple is the radical of an integer .