# 1.4. divisors

We saw that a negative integer can be a divisor of another integer. However, when discussing the divisors of an integer we will often restrict ourselves to its positive divisors.

definition. For a nonzero integer $a$, the divisors of $a$ are of the positive integers that divide $a$, ie, $$\{ n \in \mathbb{Z} \mid n > 0 \text{ and } n \mid b \}.$$

Note that we do not define the set of divisors of $0$ because every integer divides $0$.

## examples

1. What are the divisors of $47$?
The divisors of $47$ are $$1, 47 .$$
2. What are the divisors of $48$?
The divisors of $48$ are $$1, 2, 3, 4, 6, 8, 12, 16, 24, 48 .$$
3. What are the divisors of $49$?
The divisors of $49$ are $$1, 7, 49 .$$
4. What are the divisors of $50$?
The divisors of $50$ are $$1, 2, 5, 10, 25, 50 .$$
5. What are the divisors of $51$?
The divisors of $51$ are $$1, 3, 17, 51 .$$
6. What are the divisors of ?

## exercises

1. If $a$ and $b$ are positive integers such that $b \mid a$, prove that $b \le a$.

solution

Since $b \mid a$, there is an integer $n$ such that $a = b \cdot n$. Since both $a$ and $b$ are positive, then $n$ must be as well, because if $n$ were zero or negative, then multiplying by $b$ would result in zero or a negative number, which is impossible. Therefore, we can be sure that $1 \le n$. Multiplying both sides of this inequality by the positive integer $b$ results in $b \le b \cdot n$, which can be rewritten as $b \le a$.

2. Write a divisors function that takes an integer input and returns the sorted array of its divisors. (Hint: use exercise #1)

Some test values:

• divisors(25) ~> [1, 5, 25]
• divisors(27) ~> [1, 3, 9, 27]
• divisors(29) ~> [1, 29]
• divisors(30) ~> [1, 2, 3, 5, 6, 10, 15, 30]
• divisors(32) ~> [1, 2, 4, 8, 16, 32]
• divisors(34) ~> [1, 2, 17, 34]
• divisors(36) ~> [1, 2, 3, 4, 6, 9, 12, 18, 36]
solution
module NumberTheory::Division
def divisors(a)
return if a == 0
(1..a.abs).select { |b| divides(b, a) }
end
end
3. If $a$ and $b$ are positive integers such that $a \mid b$ and $b \mid a$, prove that $a = b$. (Hint: use exercise #1)

solution

Since $b \mid a$, the previous exercise implies that $b \le a$. Similarly, since $a \mid b$, the previous exercise implies that $a \le b$. This is only possible if $a$ and $b$ are actually equal.