4 ms·

What about prime integers? Aren't they a product of 1 and themselves?

• happyscrappy 13y ago
  FTA: What if we permitted 1 to be prime? In that case, 84 would also have the "prime" factorisation 1 x 1 x 1 x 2 x 2 x 3 x 7. That is, 84 could still be factorised, but it would no longer have a unique prime factorisation.
  • CatMtKing 13y ago
    I did read the article. I was just pointing out that the poster's original statement seemed to exclude prime integers -- well, unless the empty product is a product of primes.
• anaphor 13y ago
  If you think of every integer as being a product of some other integers, e.g. the product of [a,b,c] then you could say the algorithm for that is: take the first number of the list, multiply it with the product of the rest of the numbers, when you get to an empty list, just apply the empty product, which is 1 for obvious reasons (1a is identity). So then [p] = p1. Then by this definition [] has no factors, it's the empty product. Of course you don't have to think of it like this, but it might be useful to do so.
• thaumasiotes 13y ago
  This has been covered by anaphor, but to respond to the question directly: No, a prime number, for example 5, is not the product of 1 and itself (at least, no more than it's the product of 1/3 and 15). It is the product of itself only, so it can be represented as a set of factors like {5}. 1 is the product of nothing, and would be represented as the set of factors {}. Multiplication, obviously, just combines the factors of the multiplicands. 6 (= {2,3}) times 2 (= {2}) equals {2,2,3}, 12. Hopefully that makes it clear why 1 should be {}.