(a) An integer n ≥
2 is a prime if its only divisors are 1 and n itself (that is, if k |
n, then either k = 1 or
Take any integer n ≥
2. Prove by contradiction. Assume that n has no prime factors. Then n
is not a prime because n = n ×
1 and n is a factor of itself. Since n is not a prime, it has a
factor m1 such that
< n, n = m1
By applying the same reasoning to the number m1
we get its factor m2
, 1 < m2
× k2 etc.
> … cannot be continues indefinitely because there is only a
finite number of integers between n and 1. We arrived at a
contradiction, thus, n must have a prime factor.
Prove by contradiction. Suppose that there is a finite set of prime
… , pn.
Construct a number N = p1p2p3…pn
+ 1. Since (b) is
true, it must have a prime factor p, however, no prime number can be
a factor of N. Q.E.D.
Apr 18th, 2015
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