The study of division of integers begins with the notion of divisibility, whether the division of two integers results in an integer. Since division does not always yield a result at all, let alone an integer result, we frame this question in terms of multiplication.
definition. Given two integers $ a, b $, we say that $ b $ divides $ a $ if there is some integer $ n $ for which $ a = b \cdot n $. If $ b $ divides $ a $, we write $ b \mid a $. If $ b $ does not divide $ a $, we write $b \nmid a $.
The following statements are equivalent ways of describing that $ b \mid a $.
- $ b $ divides $ a $
- $ b $ is a divisor of $ a $
- $ a $ is divisible by $ b $
- $ a $ is a multiple of $ b $
examples
- Does $ 15 $ divide $ 45 $?yes. $ 15 \mid 45 $ since $ 45 = 15 \cdot 3 $.
- $ 15 $ divides $ 45 $
- $ 15 $ is a divisor of $ 45 $
- $ 45 $ is divisible by $ 15 $
- $ 45 $ is a multiple of $ 15 $
- Does $ 10 $ divide $ 45 $?no. $ 10 \nmid 45 $ since $ 45 $ is strictly between $ 10 \cdot 4 $ and $ 10 \cdot 5 $.
- $ 10 $ does not divide $ 45 $
- $ 10 $ is not a divisor of $ 45 $
- $ 45 $ is not divisible by $ 10 $
- $ 45 $ is not a multiple of $ 10 $
careful!
This notation $ a \mid b $ is sometimes confused with notation for the division operation. If you see $ 15 / 45 $, it is the division operation that returns the rational number $ 1 / 3 $. If you see $ 15 \mid 45 $, it is the statement that $ 45 $ is divisible by $ 15 $, ie, that $ 45 / 15 $ is an integer.
exercises
- Write a naive
dividesfunction that takes two integer inputs and returnstrueif the first divides the second and returnsfalseotherwise. This naive function should only use addition, subtraction, multiplication, and comparison.Some test values:
divides(15, 45) ~> truedivides(-15, 45) ~> truedivides(15, -45) ~> truedivides(-15, -45) ~> truedivides(10, 45) ~> falsedivides(-10, 45) ~> falsedivides(10, -45) ~> falsedivides(-10, -45) ~> falsedivides(7, 0) ~> truedivides(7, 7) ~> truedivides(7, -7) ~> truedivides(1, 7) ~> truedivides(-1, 7) ~> truedivides(0, 7) ~> false
solutionmodule NumberTheory::Naive def divides(b, a) return false if b == 0 return divides(-b, a) if b < 0 return divides(b, -a) if a < 0 multiple = 0 while multiple < a multiple += b end multiple == a end end Prove that every nonzero integer divides 0.
solutionFor any nonzero integer $ b $, we can express $ 0 $ as $ 0 = b \cdot 0 $, which proves that $ b \mid 0 $.
Prove that every integer is divisible by both $ 1 $ and $ -1 $.
solutionLet $ a $ be any integer. We can express $ a $ as $ a = 1 \cdot a $, which shows that $ 1 \mid a $. We can also express $ a $ as $ a = -1 \cdot (-a) $, which shows that $ -1 \mid a $.
Prove that every nonzero integer divides itself.
solutionFor any nonzero integer $ a $, we can express $ a $ as $ a = a \cdot 1 $, which shows that $ a \mid a $.
If $ b \mid a $, prove that $ -b \mid a $ and $ b \mid -a $.
solutionSince $ b \mid a $, there is an integer $ n $ such that $ a = b \cdot n $. Then $$ a = (-b) \cdot (-n) \quad \implies \quad -b \mid a $$ and $$ -a = b \cdot (-n) \quad \implies \quad b \mid -a . $$
Prove that the "divides" relation is transitive. In other words, if $ c \mid b $ and $ b \mid a $, prove that $ c \mid a $.
solutionThe first assumption, that $ c \mid b $, means that we can write $ b = c \cdot m $ for some integer $ m $. The second assumption, that $ b \mid a $, means that we can write $ a = b \cdot n $ for some integer $ n $. Substituting the first equation in for $ b $ in the second equation yields \[ a = (c \cdot m) \cdot n \quad \implies \quad a = c \cdot (m \cdot n). \] Since we have expressed $ a $ as an integer multiple of $ c $, we conclude that $ c \mid a $.