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MA 261 — Worksheet
Friday, March 7, 2014
1. Theorem 2.7 (Fundamental Theorem of Arithmetic - Existence Part)
Prove by strong induction on n > 1 that n can be written as a finite product of one or more
primes.
That is, for every natural number n greater than 1, there exists distinct primes p1 , p2 , . . . , pm
and natural numbers r1 , r2 , . . . , rm such that
n = pr11 pr22 · · · prmm .
Proof: Let us consider the base case. If n = 2 then there is nothing else to prove, as 2 is a
prime number. That is, n = 2 is the desired factorization.
For the inductive step, assume that any integer m with 1 < m < k can be written as
product of primes. We will prove the same for k. If k is prime, then k itself is the desired
expression. If k is a composite number, then k = ab for some integers a, b such that
1 < a, b < k. (Notice that neither a nor b can be 1 or k, as k is not prime.) By the (strong)
inductive hypothesis, both a and b can be expressed as product of one or more primes. Let
us say
a = q1a1 · · · qsas
b = qe1b1 · · · qetbt .
Thus we have that k = ab = q1a1 · · · qsas · qe1b1 · · · qetbt , as desired.
(Note that some of the qi ’s and qej ’s may be equal. Thus the expression in the last product
may be simplified.)
2. Lemma 2.8
Prove by induction on n ≥ 1 that if p and q1 , . . . , qn are primes and p|q1 · · · qn , then p = qi
for some i.
Proof: Let us consider the base case: n = 1. We need to show that if p and q1 are prime
numbers and p|q1 then p = q1 . Sincet p divides q1 , we have that q1 = pl for some integer
l. However p ̸= 1, as p is a prime number. As q1 is also a prime number, then l must be 1.
Thus p = q1 .
For the inductive step, assume that the lemma is true for an integer k ≥ 1. That is, whenever
p divides the product of k prime numbers, then p is equal to one of those prime numbers.
We want to prove the statement for the case in which p divides the product of k + 1 prime
numbers. That is, suppose p, q1 , . . . , qk , qk+1 are prime numbers and p|q1 · · · qk qk+1 . Either p = qk+1 , in which case we are done, or p ̸= qk+1 . In the latter case, notice that
gcd(p, qk+1 ) = 1. Theorem 1.41 says that if a, b, and c are integers such that a|bc and
gcd(a, c) = 1, then a|b. If we apply this result to our situation with
a=p
b = q1 · · · qk
c = qk+1 ,
we conclude that p|q1 · · · qk . By inductive assumption, since p divides the product of k
prime numbers, we conclude that p = qi for some 1 ≤ i ≤ k. Thus, by considering both
cases analyzed above, the lemma is true for n = k + 1.