When 60 divides 9

To discuss divisibility, we must consider a set—the numbers we are working with—equipped with two operations, addition and multiplication. This is not enough, however, if we wish to avoid certain pitfalls (see FOCUS). Consider a familiar example involving terminating decimals. These are the real numbers with a finite decimal expansion. Thus, 1/5 = 0.2 is a terminating decimal. By contrast, 1/3 = 0.333… is not a terminating decimal in this sense. They can all be written as fractions: if such a number u has n decimal places, multiplying it by 10ⁿ gives an integer a, and hence u = a / 10ⁿ; conversely, every number of this form is a terminating decimal. But not every fraction is a terminating decimal (think of 1/3).

Divisibility can be defined in the set D of terminating decimals just as it usually is for integers: the terminating decimal a divides the terminating decimal b if there exists a terminating decimal c such that b = a × c. Since integers are terminating decimals, it follows that for any two integers a and b, if a divides b in the usual sense, it also does so in the world of terminating decimals. The converse, however, is false: 2 divides 3 (since 3 = 2 × 1.5), and 60 divides 9 (since 9 = 60 × 0.15).

In the set $\mathbb{Z}\backslash mathbb\{Z\}$ of integers, only 1 and – 1 are invertible. For every other nonzero integer n, the fraction 1 / n is not an integer. By contrast, infinitely many elements of D have this property. For example, all numbers of the form ± 2ⁿ5^p, where n and p are integers, are invertible; indeed, these are the only ones as n and p range over $\mathbb{Z}\backslash mathbb\{Z\}$. Two terminating decimals a and b are called associates if there exists an invertible terminating decimal u such that a = b × u. The point of this definition is that two associates play exactly the same role with respect to divisibility; in particular, they have the same divisors.