(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

k =
n).

(b)
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

1 <
m_{1}
< n, n = m_{1}
× k_{1}.
By applying the same reasoning to the number m_{1}
we get its factor m_{2
, }1 < m_{2}
< m_{1},
m_{1
}=
m_{2}
× k_{2 }etc_{.
}However, the
sequence m_{1}
> m_{2}
> m_{3}
> … 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.

(c)
Prove by contradiction. Suppose that there is a finite set of prime
numbers: p_{1},
p_{2},
p_{3},
… , p_{n}.
Construct a number N = p_{1}p_{2}p_{3}…p_{n
}+ 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.