Can anybody please help me please?
if the theorem is not true, there exists a minimal counter-example, a.
by supposition, there exists two distinct prime factorizations of a:
with .
now so for some i = 1,2,...,s.
this, in turn, implies . since is prime, and , we conclude .
thus since ,
we conclude that are, in fact, the same factorization.
hence since we have r-1 factors on the left, and s-1 factors on the right:
r - 1 = s - 1, so r = s.
we now have two possibilities:
i = 1 (on the RHS),
i ≠ 1.
if i ≠ 1, then q_{1} = p_{2}, which implies p_{2} = q_{1} < q_{i} = p_{1}, contradicting our original factorization of a.
thus we must have p_{1} = q_{1}, and since the factorization of IS unique,
p_{i} = q_{i} for all i > 1, and k_{i} = m_{i}, for all i > 1.
thus the only difference between:
is the power of p_{1}.
but , so we must have k_{1} = m_{1}, and a does not exist, which proves the theorem.