# THE FUNDAMENTAL THEOREM OF ARITHMETIC Suppose that an

advertisement ```THE FUNDAMENTAL THEOREM OF ARITHMETIC
Suppose that an integer n &gt; 1 has two factorizations into irreducibles,
n = p1 &middot; &middot; &middot; pk = q1 &middot; &middot; &middot; q` ,
all pi and all qj irreducible.
Thus
pk | q1 &middot; &middot; &middot; q` ,
and so since the irreducible pk is prime, in fact
pk | qj
for some j ∈ {1, &middot; &middot; &middot; , `}.
Since multiplication commutes we may freely re-index the q’s so that
pk | q` .
But q` is irreducible, so its only factors are 1 and itself. Since pk is a factor of q`
and pk 6= 1, we have
pk = q` .
So now we may cancel the rightmost terms of the equality p1 &middot; &middot; &middot; pk = q1 &middot; &middot; &middot; q` (where
now we know pk = q` ) to get a new equality
n0 = p1 &middot; &middot; &middot; pk−1 = q1 &middot; &middot; &middot; q`−1 ,
n0 &lt; n.
By Strong Induction, n0 has a unique factorization. That is, k − 1 = ` − 1 and
pi = qi for i ∈ {1, &middot; &middot; &middot; , k − 1}. Consequently the two factorizations of n agree as
well. That is, the factorization of n is unique.
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