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Chapter 7: Factorization and the Fundamental Theorem of Arithmetic
Practice HW p. 53 # 1-4, Additional Web Exercises
In this chapter, we began to examine some basic about primes and their importance.
Definition: A prime number is a positive integer p  2 (note that 1 is not considered to
be prime) whose only positive divisors are 1 and p (1 and itself). A number m  2 that
is not prime is said to be composite.
The following set represents the set of primes that are less than 100:
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97…}
Theorem: Let p be a prime number and suppose for two integers a and b that p | ab .
Then p | a or p | b .
Proof:
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For example, we know that 7 | 112 or 7 | 2  56 . Then the theorem says that 7 | 2 or
7 | 56 . It is indeed true that 7 | 56 .
The previous theorem can be extended to show the next result.
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Theorem: Prime Divisibility Property. Let p be a prime number and suppose that
p | (a1  a2 ar ) . Then p divides one of the ai ' s .
Proof:
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The Principle of Mathematical Induction
If S (1), S (2), , S (n  1), S (n) is a collection of n statements with the properties that
1. S (1) is true.
2. If S (i ) is true for all positive integers i  k , then S ( k  1) is true.
Then S (n ) is true.
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Example 1: Use induction to prove that
1  3  5    (2n  1)  n 2 .
Solution:
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The Fundamental Theorem of Arithmetic
We next discuss how primes represent the building blocks for all integers.
Theorem: The Fundamental Theorem of Arithmetic. Every integer n  2 can be factored
in a product of primes n  p1 p2  pr in exactly one way.
Proof: To prove the Fundamental Theorem, we will prove the following.
1. Any integer n  2, 3, 4, 5, 6, 7, then n can be factored into a product of primes in
some way.
To do this, we use induction.
i) If n  2 , then since 2 is prime, it can be factored as itself.
ii) Now, assume the assertion of true for all integers n where n  N . We will show
the assertion is true for the integer N  1 . If N  1 is prime, then we are done
since N  1 can be factored as itself. If N  1 is composite, we can factor N  1
as N  1  n1n2 , where n1 and n2 are integers satisfying the bound
2  n1, n2  N . Since both n1  N and n2  N , the induction hypothesis says
that both n1 and n2 can be factored into a product of primes, say n1  p1 p2  ps
and n2  q1q2  qt . Hence
N  1  n1n2  p1 p2  ps  q1q2  qt
is a product of primes. Hence, by induction, the result is true for all integers larger
than 2.
2. We next show the factorization can be done in only one way, disregarding order.
To show this, assume that n can be factored into a produce of primes in two ways, that is
say
n  p1 p2  pk and n  q1q2  ql
Then
p1 ( p2  pk )  q1q2  ql
By divisibility, this says that p1 | q1q2  ql . By the prime divisibility property, this says
that p1 divides one of the qi ' s . Reordering the qi ' s if necessary, say p1 | q1 . Since p1
and q1 are primes, this says that p1  q1 . Substituting p1 for q1 , we have
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p1 p2  pk  p1q2  ql
or
p2 ( p3  pk )  q2 q3  ql
By divisibility, this says that p2 | q2 q3  ql . By the prime divisibility property, this says
that p2 divides one of the qi ' s . Reordering the qi ' s if necessary, say p2 | q2 . Since p2
and q2 are primes, this says that p2  q2 . Substituting p2 for q2 , we have
p2 p3  pk  p2 q3  ql
or
p3 p4  pk  q3q4  ql
We continue in this fashion until all of the pi ' s or qi ' s cancel first. Suppose the qi ' s
cancel first. Then for the remaining pi ' s , we have
pl 1 pl  2  pk  1 ,
which says that all of the remainder pi ' s  1 for i  l  1,l  2 ,k . A similar argument
can be made with the qi ' s if the pi ' s cancel first. Thus, k  l and we can say that
n  p1 p2  pk  q1q2  qk
We have shown that p1  q1 , p2  q2 , …, pk  qk . Hence, the prime factorization is
unique.
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Example 2: Factor 5040 into a product of primes.
Solution:
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Example 3: Factor 58261 into a product of primes.
Solution:
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Notes
1. If d | a , then a  td . Hence, a must contain all of d ’s prime factors. That is, if
d  p1 p2  pk is d ’s prime factorization, then a  td  tp1 p2  pk .
2. gcd( a, b) is the product of the prime factors that a and b have in common.
Maple Command for Obtaining the Prime Factorization of a Number
> ifactor(5040);
> ifactor(58261);