In this section we introduce the idea of divisibility for integers. It is important to understand that this concept involves only integers and is not the same thing as division. Divisibility is a property while division is an operation.
Warning:\(d\mid n\) is a relationship between two integers. It is a statement that is either true or false. It does not equal a number. It is not the same thing as a fraction. It is not an equation, but it can be translated to the equation \(n=dk\) for some integer \(k\text{.}\)
If \(d\) does not divide \(n\text{,}\) then for every \(k\in \mathbb{Z}, n\neq dk\text{.}\) This can be difficult to show in general, but if \(n\) and \(d\) are specific integers, you could show \(\frac{n}{d}\) is not an integer. This is the only time a fraction can show up in proofs about divibilty.
Let \(a, b, c\in \mathbb{Z}\text{.}\) Assume \(a\mid b\) and \(b\mid c\text{.}\) Then by definition, \(b=ak\) for some \(k\in \mathbb{Z}\) and \(c=bj\) for some \(j\in \mathbb{Z}\text{.}\) Substituting the first equation into the second, \(c=(ak)j=a(kj)\text{.}\) Since \(kj\in\mathbb{Z}\text{,}\)\(a\mid c\text{.}\)
Case 2: \(n\) is composite. Then \(n=r_0s_0\) where \(1< r_0 < n\text{,}\) and \(r_0, s_0\in \mathbb{Z}\text{.}\) If \(r_0\) is prime, then \(n\) is divisible by a prime.
If \(r_0\) is not prime, then \(r_0=r_1s_1\) where \(1< r_1 < r_0\text{,}\) and \(r_1, s_1\in \mathbb{Z}\text{.}\) If \(r_1\) is prime, then \(n=r_1s_1s_0\) and \(n\) is divisible by a prime.
If at any point \(r_k\) is prime, then we are done, as we have found a prime divisor of \(n\text{.}\) We know that there cannot be infinitely many non-prime \(r_k\) since there are only finitely many integers between 1 and \(n\text{.}\) Thus, the process must terminate with a prime divisor of \(n\text{.}\) Hence, every integer is divisible by a prime.
We now state a big theorem in discrete mathematics, but leave the proof for later. This theorem is more commonly called the Fundamental Theorem of Arthmetic.
Given any integer \(n>1\) there exists a positive integer \(k\text{,}\) distinct primes \(p_1, p_2, \ldots, p_k\text{,}\) and positive integers \(e_1, e_2, \ldots e_k\text{,}\) such that \(n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}\text{.}\) This factorization of \(n\) into powers of primes is unique except for the order of the factors.
TheoremΒ 4.3.6 is more powerful than it may seem. Although you have probably found prime factorizations before, you may not have thought much about the fact that they are unique. Consider the statement \(3\mid 2^{101}\text{.}\) Is this true or false? Well, since \(2^{101}\) is a prime factorization, the only prime divisor it has is 2. Thus, \(3\mid 2^{101}\) is false. If you have the prime factorization of an integer, then you know exactly which primes (and which powers of primes) divide the integer.
Prove, using the definition of divisibility, for all integers \(a\text{,}\)\(b\text{,}\) and \(c\text{,}\) if \(a\mid b\) and \(a\mid c\) then \(a\mid (b+c)\text{.}\)
Prove or disprove: For all integers \(a\text{,}\)\(b\text{,}\) and \(c\text{,}\) if \(a\) is a factor of \(c\) then \(ab\) is a factor of \(c\text{.}\)