Consider this proposition from Euclid’s Elements:
Let the number \(A\) be the least that is measured by the prime numbers \(B\), \(C\), and \(D\). I say that \(A\) is not measured by any other prime number except \(B\), \(C\), or \(D\). If possible, let it be measured by the prime number \(E\), and let \(E\) not be the same as any one of the numbers \(B\), \(C\), or \(D\).
Now, since \(E\) measures \(A\), let it measure it according to \(F\), therefore \(E\) multiplied by \(F\) makes \(A\). And \(A\) is measured by the prime numbers \(B\), \(C\), and \(D\). But, if two numbers multiplied by one another make some number, and any prime number measures the product, then it also measures one of the original numbers, therefore each of \(B\), \(C\), and \(D\) measures one of the numbers \(E\) or \(F\). Now they do not measure \(E\), for \(E\) is prime and not the same with any one of the numbers \(B\), \(C\), or \(D\). Therefore they measure \(F\), which is less than \(A\), which is impossible, for \(A\) is by hypothesis the least number measured by \(B\), \(C\), and \(D\). Therefore no prime number measures \(A\) except \(B\), \(C\), and \(D\). Therefore, if a number is the least that is measured by prime numbers, then it is not measured by any other prime number except those originally measuring it.
Can you explain what this proof is saying?
Now let’s consider a crazy set of numbers—all multiples of \(3\). Let’s use the symbol \(3\mathbb {Z}\) to denote the set consisting of all multiples of \(3\). As a gesture of friendship, I have written down the first \(100\) nonnegative integers in \(3\mathbb {Z}\):
As an example, I tell you that \(6\) is prome number in \(3\mathbb {Z}\). You may object because \(6 = 2\cdot 3\), but remember—\(2\) is not in \(3\mathbb {Z}\)!