Theorem 2.9. [Fundamental Theorem of Arithmetic, Uniqueness, as stated in our text].
Let n be a natural number. Let {p1 , p2 , . . . , pm } and {q1 , q2 , . . . , qs } be sets of primes with
pi 6= pj for i 6= j and qi 6= qj for i 6= j. Let {r1 , r2 , . . . , rm } and {t1 , t2 , . . . , ts } be sets of
natural numbers such that
n = pr11 pr22 · · · prmm = q1t1 q2t2 · · · qsts .
(1)
Then m = s and {p1 , p2 , . . . , pm } = {q1 , q2 , . . . , qs }. Moreover, if pi = qj then ri = tj . In
other words, if we express the same natural number as a product of powers of distinct primes,
then the expressions are identical except for the ordering of the factors.
Proof. Let everything be given as in the statement of the theorem. Since pr11 pr22 · · · prmm =
q1t1 q2t2 · · · qsts , we have p1 |q1t1 q2t2 · · · qsts by definition of divides. Rewrite this as
p1 | (q1 · · · q1 ) (q2 · · · q2 ) · · · (qs · · · qs ) .
| {z } | {z } | {z }
t1
t2
ts
Then by Lemma 2.8, it must be true that p1 |qj for some 1 ≤ j ≤ s. Since p1 6= pi for all
i 6= 1 and qj 6= qk for all k 6= j, then p1 is not equal to any other qi . As pr11 divides n and
q1t1 q2t2 · · · qsts , we may divide equation (1) by pr11 = qjr1 , giving
n
t −r
= pr22 · · · prmm = q1t1 · · · qjj 1 · · · qsts .
pr11
(2)
t −r
Following similar logic, p2 |q1t1 · · · qjj 1 · · · qsts and so by Lemma 2.8 p2 = qk where k 6= j.
t −r
As before, in fact pr22 divides n and q1t1 · · · qjj 1 · · · qsts , so
n
t −r1
pr11 pr22
= pr33 · · · prmm = q1t1 · · · qjj
· · · qktk −r2 · · · qsts .
(3)
Continuing for all pi , we see that each pi is contained in the set {q1 , q2 , . . . , qs }, so
{p1 , p2 , . . . , pm } ⊆ {q1 , q2 , . . . , qs }. Additionally, notice that we will have
pr11 pr22
n
= 1 = q1t1 −ru q2t2 −rv · · · qsts −rw .
· · · prmm
(4)
Since no prime would divide 1, the right side must also equal 1. This implies that all tj − ri
must equal zero, so ri = tj when pi = qj . It also implies that there is no qk such that there
does not exist a pi with pi = qk , meaning that m = s.
If we proceed in the same way with the qi dividing n, we will see that each qi is contained
in the set {p1 , p2 , . . . , pm }, implying {q1 , q2 , . . . , qs } ⊆ {p1 , p2 , . . . , pm }. Together with our
previous statement, we then have
{p1 , p2 , . . . , pm } = {q1 , q2 , . . . , qs }.
Then we have proven the theorem as the sets of primes are equal, m = s, and all
corresponding powers are equal.
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One way to clean this proof up is to renumber/rename the qi ’s. For example, we saide
p1 = qj for some 1 ≤ j ≤ s. We could have then said that we would reindex so that qj = q1 ,
so that after dividing off these factors we have
n
= pr22 · · · prmm = q1t1 −r1 q2t2 · · · qsts
pr11
which is perhaps easier to read.
This proof above was long and wordy. Let me give a slightly different setup that produces
a slightly shorter proof, though it is less explicit.
Theorem 2.9. [Fundamental Theorem of Arithmetic, Uniqueness, alternate]. Every natural number can be represented as a product of primes in exactly one way.
Proof. Suppose that n is a natural number and can be written as n = p1 p2 · · · px and n =
q1 q2 · · · qy where pi and qj are primes for all 1 ≤ i ≤ x and 1 ≤ j ≤ y, where the pi and qj
are not necessarily distinct. Since p1 p2 · · · px = q1 q2 · · · qy , by definition p1 |q1 q2 · · · qy . Then
by Lemma 2.8, p1 = qk for some 1 ≤ k ≤ y. Dividing by p1 yields
n
= p2 · · · px = q1 · · · qk−1 qk+1 · · · qy .
p1
Then p2 |q1 · · · qk−1 qk+1 · · · qy so by Lemma 2.8 we know that p2 = q` for some 1 ≤ ` ≤ y,
` 6= k. Divide this factor off as well. Proceeding in this way, we see that all pi = qj for some
i, j, and so {p1 , p2 , . . . , px } ⊆ {q1 , q2 , . . . , qy }.
Now proceed by considering q1 |p1 p2 · · · px . Then by Lemma 2.8 q1 = pv for some 1 ≤ v ≤
x. Divide off this factor so that
n
= p1 · · · pv−1 pv+1 · · · px = q2 · · · qy .
q1
Similarly, we show that all qi = pj for some i, j, and so {q1 , q2 , . . . , qy } ⊆ {p1 , p2 , . . . , px }.
Thus, {p1 , p2 , . . . , px } = {q1 , q2 , . . . , qy }, so the two factorizations n = p1 p2 · · · px and
n = q1 q2 · · · qy contain exactly the same primes, and the list of primes is unique.
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