0

Given two sets a and b. When a and b only contain primes above 1, will the product of every number in a ever be equivalent to the product of every number in b?

champfish
  • 3
  • 1
    As those are sets if they are different then there is at least one element in one of them which is not in the other one. Now just apply the Fundamental Theorem of Arithmetic. – Timbuc Jun 06 '15 at 12:26
  • 2
    The Fundamental Theorem of Arithmetic assures us that if the products of two sets of primes are equal, the sets are identical. – Joffan Jun 06 '15 at 12:29
  • 1
    If the sets are different, then there is at least one prime which belongs in one set but not in the other. Hence one product will be divisible by that prime and one product will not. Hence the answer to your question is No. For multi-sets, it is slightly (but not extremely) more difficult to show a similar argument. – barak manos Jun 06 '15 at 12:34

1 Answers1

1

Well, let's see about that, let

\begin{align} N=\prod_{p \in a} p, \\ M=\prod_{r \in b} r. \end{align}

If $N=M$, then the fundamental theorem of arithmetic tells us that $M$ and $N$ both have the same prime factors, i.e., if $p \in a$ then also $p \in b$ and vice versa. Therefore $N=M$ implies $a=b$

Pol van Hoften
  • 1,594
  • 1
  • 10
  • 17