Let n > 1. If n has no prime factor less than or equal to cubed root of n, prove that n is either prime or the product of 2 primes.
2006-09-12 10:06:56 · 3 answers · asked by jjodom1010 3 in Science & Mathematics ➔ Mathematics
Every (natural) number is the product of one or more primes.
We must prove that under the given condition, n has at most 2 non-zero dividers.
Suppose that n is the product of 3 prime factors (and possible some more): n = a * b * c * D, with 2 <= a <= b <= c prime factors.
The assumption is that a > n^(1/3), that is, a^3 > n.
But then
n = a * b * c * D >= a * a * a = a^3 > n
which is impossible. Therefore the assumption that n has 3 prime factors cannot be true.
2006-09-12 10:20:43 · answer #1 · answered by dutch_prof 4 · 0⤊ 0⤋
Prove it by contradiction. Suppose it is the product of 3 primes. If they were all greater than the cube root, how would their product compare to the number?
2006-09-12 17:15:36 · answer #2 · answered by bh8153 7 · 2⤊ 0⤋
dude math proofs is the worst in all maths.
calculas is better or even statistics cause its challenging
2006-09-12 18:08:48 · answer #3 · answered by Anonymous · 0⤊ 1⤋