7.1: Fundamental Theorem of Arithmetic
In this chapter we investigate divisibility. It may seem peculiar that we would investigate a topic that you have studied since elementary school, but don’t be fooled by the apparent simplicity of the subject. The study of divisibility of integers is part of number theory. Geometry and number theory are the oldest areas of mathematical study, and they are still active fields of mathematical research.
Fundamental Theorem of Arithmetic
DEFINITION. Divides, factor Let \(a, b \in \mathbb{Z}\) . We say that \(a\) divides \(b\) , or \(a\) is a factor of \(b\) , if \[(\exists c \in \mathbb{Z}) a \cdot c=b .\] We write this as \(a \mid b\) . If \(a\) does not divide \(b\) we write \(a \nmid b\) .
Divisibility is the central idea of number theory. It is precisely because one integer need not be a factor of another integer, or a pair of integers may fail to have non-trivial common factors, that divisibility provides insight into the structure of integers. Put another way, consider the definition of divisibility applied to rational numbers - you will find that it does not provide any insight at all since a nonzero rational number is a factor of any other rational number. Furthermore, many of the properties of integers with regard to divisibility generalize many of the properties of integers with regard to divisibility generalize of this in Section 7.5.
DEFINITION. Prime number Let \(p \in \mathbb{N}\) . We say that \(p\) is a prime number if \(p>1\) and the only positive factors of \(p\) are \(p\) and 1 . DEFINITION. Relatively prime Let \(a, b \in \mathbb{Z}\) . We say that \(a\) and \(b\) are relatively prime if they have no common factor greater than 1 .
DEFINITION. Integer combination Let \(a, b, c \in \mathbb{Z}\) . Then \(c\) is an integer combination of \(a\) and \(b\) if \[(\exists m, n \in \mathbb{Z}) c=m a+n b .\] PROPOSITION 7.1. Let \(a, b \in \mathbb{Z}\) . If \(a\) and \(b\) are relatively prime, then \(a-b\) and \(b\) are relatively prime.
Discussion. We shall prove the contrapositive by showing that any common factor of \(a-b\) and \(b\) is also a factor of \(a\) .
PROOF. Let \(c>1\) be a common factor of \(b\) and \(a-b\) . So \[(\exists m \in \mathbb{Z}) b=c m\] and \[(\exists n \in \mathbb{Z}) a-b=c n .\] Then \[c(m+n)=a\] and so \(c \mid a\) . Therefore if \(a\) and \(b\) are relatively prime, then \(a-b\) and \(b\) are relatively prime.
PROPOSITION 7.2. Let \(a\) and \(b\) be integers. If \(a\) and \(b\) are relatively prime, then \[(\exists m, n \in \mathbb{Z}) m a+n b=1 .\] Discussion. We shall argue for the case in which \(a\) and \(b\) are natural numbers. Given the proposition for all pairs of relatively prime natural numbers, we may easily extend it to arbitrary pairs of relatively prime integers by changing the sign of \(m\) or \(n\) in the integer combination. This assumption allows us to argue by induction on the sum of the integers. The base case for this argument by induction will be \(a+b=3\) . If \(a=0=b\) , then \(a\) and \(b\) are not relatively prime. If \(a+b=1\) , then \(a\) and \(b\) are relatively prime and the choice of \(m\) and \(n\) is obvious. If \(a=b=1\) then \(a\) and \(b\) are relatively prime and again the choice of \(m\) and \(n\) is obvious.
Proof. We may assume that \(a>b>0\) . We argue by induction on \(a+b\) .
Base case: \(a+b=3\) .
Then \(a=2\) and \(b=1\) . So \[a-b=1 .\] Induction step:
Assume that the result holds for all pairs of relatively prime natural numbers with sum less than \(a+b\) .
By Proposition 7.1, \(b\) and \(a-b\) are relatively prime. By the induction hypothesis, there are \(i, j \in \mathbb{Z}\) such that \[i(a-b)+j b=1 .\] DiscuSsiON. If \(a-b=b\) , we are not in the case where we have two distinct positive numbers. How do we handle this possibility?
Let \(m=i\) and \(n=j-i\) . Then \[m a+n b=1 .\] By the induction principle the result holds for all relatively prime pairs of natural numbers.
DEFINITION. Greatest common divisor, \(\operatorname{gcd}(a, b)\) Let \(a, b \in \mathbb{Z}\) . The greatest common divisor of \(a\) and \(b\) , written \(\operatorname{gcd}(a, b)\) , is the largest integer which divides both \(a\) and \(b\) .
So \(a\) and \(b\) are relatively prime iff \(\operatorname{gcd}(a, b)=1\) .
Proposition 7.3. Let \(a, b, c \in \mathbb{Z}\) , and assume that \(\operatorname{gcd}(a, b)=1\) . If \(a \mid c b\) , then \(a \mid c\) .
PROOF. By Proposition \(7.2\) there are \(m, n \in \mathbb{Z}\) such that \[m a+n b=1 .\] Therefore \[c m a+c n b=c .\] Clearly \(a \mid c n b(\) since \(a \mid c b)\) and \(a \mid c m a\) . So \[a \mid(c m a+c n b)\] and therefore \(a \mid c\) .
PROPOSITION 7.4. Let
\(a, b, c \in \mathbb{Z} .\)
If
\(\operatorname{gcd}(a, b)=1, a \mid c\)
and
\(b \mid c\)
, then
\[a b \mid c\]
PROOF. Let
\(m, n \in \mathbb{Z}\)
be such that
\(a m=c\)
and
\(b n=c\)
. Then
\[a \mid b n\]
By Proposition
\(7.3, a \mid n\)
. Hence there is
\(k \in \mathbb{Z}\)
such that
\[a k=n\]
Therefore
\[a k b=c\]
and
\[a b \mid c\]
LEMMA 7.5. Assume
(1)
\(p \in \mathbb{N}\)
is prime
(2)
\(N \geq 1\)
, and
\(a_{1}, \ldots, a_{N} \in \mathbb{Z}\)
(3)
\(p \mid\left(\prod_{n=1}^{N} a_{n}\right)\)
.
Then there is some \(n \leq N\) such that \(p \mid a_{n}\) .
Proof. Let \(p\) be a prime number. We argue by induction on \(N\) .
Base case: \(N=1\)
The base case is obvious.
Induction step:
Let \(N>1\) and assume that the result holds for all products of fewer than \(N\) factors.
Let \[a=\prod_{n=1}^{N-1} a_{n}\] and suppose that \[p \mid\left(\prod_{n=1}^{N} a_{n}\right) .\] Then \[p \mid a \cdot a_{N} .\] If \(p \mid a\) , then by the induction hypothesis, \[(\exists n<N) p \mid a_{n} .\] Assume that \(p\) is not a factor of \(a\) ; since \(p\) is prime, \(\operatorname{gcd}(p, a)=1\) . By Proposition 7.3, \(p \mid a_{N}\) .
THEOREM 7.6. Fundamental Theorem of Arithmetic Let \(N\) be a natural number greater than 1. Then \(N\) may be uniquely expressed as the product of prime numbers (up to the order of the factors).
DISCUSSION. We permit a "product" with only one factor. So any prime number is its own unique prime factoring.
PROOF. We argue by induction on the natural numbers greater than \(1 .\)
Base case: \((N=2)\)
By the discussion preceding the proof, 2 is its own prime factoring.
Induction step:
Assume that the result holds for all natural numbers greater than 1 and less than \(N\) . If \(N\) is prime, the result follows. If \(N\) is not prime, then there are \(a, b \in \mathbb{N}, a<N\) and \(b<N\) , such that \[a \cdot b=N \text {. }\] By the induction hypothesis, \(a\) and \(b\) have unique prime factorings. The product of the factorings will be a prime factoring of \(N\) . Is the factoring unique up to order? Suppose that \[N=\prod_{i=1}^{m} p_{i}=\prod_{j=1}^{n} q_{j}\] where \(p_{i}\) is prime for \(1 \leq i \leq m\) , and \(q_{j}\) is prime for \(1 \leq j \leq n\) . Then \[p_{1} \mid \prod_{j=1}^{n} q_{j} .\] By Lemma 7.5, \[(\exists j \leq n) p_{1} \mid q_{j} .\] We may reorder the factors \(q_{1}, \ldots, q_{n}\) so that \(p_{1} \mid q_{1}\) . Both \(p_{1}\) and \(q_{1}\) are prime, so \[p_{1}=q_{1} .\] Therefore \[\prod_{i=2}^{m} p_{i}=\prod_{j=2}^{n} q_{j}<N .\] By the induction hypothesis, \(p_{2}, \ldots, p_{m}\) is a unique prime factoring of \(\prod_{i=2}^{m} p_{i}\) , so \(m=n\) and \(q_{2}, \ldots, q_{n}\) is a reordering of \(p_{2}, \ldots, p_{m}\) . Therefore \(q_{1} \cdots q_{n}\) is a reordering of \(p_{1} \cdots p_{m}\) and the prime factoring of \(N\) is unique.
REMARK. Why is the number 1 not defined to be a prime? After all, it has no factors other than itself or 1 ! The reason is because it is very useful to have uniqueness in the Fundamental Theorem of Arithmetic. If 1 were considered prime, it could be included arbitrarily often in the factoring of \(N\) .