I want to prove fundamental theorem of arithmetic by using strong induction and proof by smallest counterexample.
Can someone by any chance help me? (with great detail please)?
I would appreciate people's help.
if the theorem is not true, there exists a minimal counter-example, a.
by supposition, there exists two distinct prime factorizations of a:
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 q1 = p2, which implies p2 = q1 < qi = p1, contradicting our original factorization of a.
thus we must have p1 = q1, and since the factorization of IS unique,
pi = qi for all i > 1, and ki = mi, for all i > 1.
thus the only difference between:
is the power of p1.
but , so we must have k1 = m1, and a does not exist, which proves the theorem.