Prime Factorization (Arithmetic)

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
cheiney
Messages
11
Reaction score
0

Homework Statement



Assume n = p_1*p_2*p_3*...*p_r = q_1*q_2*q_3*...*q_s, where the p's and q's are primes. We can assume that none of the p's are equal to any of the q's. Why?

Homework Equations





The Attempt at a Solution



I am completely stuck on this. My understanding of the Fundamental Theorem of Arithmetic is that each number n[itex]\geq[/itex]2 has a unique prime factorization. So how could we possibly assume that the p's aren't equal to the q's?
 
Physics news on Phys.org
Office_Shredder said:
Is this literally the statement of your homework question?

Yes. The only 2 approaches I could really think of is that, since they didn't specify that n is greater than or equal to 2, so if all of q_j are negative and if j is an even number, then it would hold up. The other approach would be to describe n as a member of a set with "primes" in the sense that they cannot be divisible by other numbers in the set other than 1 and itself.
 
I would be inclined to say that "prime" implies positive and so the "unique factorization property" says, to the contrary of what this purports, that if p_1*p_2*p_3*...*p_r = q_1*q_2*q_3*...*q_s, then the "p"s and "q"s must be equal.