A student writes: '12 | 4, because 12 = 3 × 4.' What is wrong with this claim?
ANothing — if 12 = 3 × 4, then 12 divides 4 by definition
BThe notation is reversed: 'a | b' means b = k·a for some integer k, so 4 | 12 (since 12 = 3·4), not 12 | 4
CThe claim would be correct only if k were required to be a prime number
DDivisibility requires both numbers to be positive, so neither 4 nor 12 can appear in this notation
The definition is: a | b means b = k·a for some integer k. Here 12 = 3 × 4 means 4 | 12 (4 divides 12), not 12 | 4. The student has the relationship backwards. '12 | 4' would require an integer k such that 4 = k·12, which has no integer solution. This direction confusion — conflating 'a divides b' with 'a is divisible by b' — is the most persistent error in divisibility notation.
Question 2 Multiple Choice
If a | b and a | c, what can be concluded about a | (3b − 7c)?
ANothing can be concluded without knowing the specific values of a, b, and c
Ba | (3b − 7c), because divisibility is preserved under any integer linear combination
Ca | (3b − 7c) only if 3 and 7 are also divisible by a
Da | (3b − 7c) only if b > c
The linear combination property states: if a | b and a | c, then a | (xb + yc) for any integers x and y. Here x = 3 and y = −7, so a | (3b − 7c) follows directly. This is the seed from which the theory of gcd and Bézout's identity grow — the integers divisible by a fixed a are closed under linear combination, forming what will later be recognized as an ideal.
Question 3 True / False
Divisibility is defined primarily for positive integers; extending it to negative integers requires a separate definition.
TTrue
FFalse
Answer: False
Divisibility applies to all nonzero integers. The definition b = k·a holds when k is any integer — positive, negative, or zero (though a itself must be nonzero to avoid division by zero). For example, −4 | 12 because 12 = (−3)(−4), and 4 | (−12) because −12 = (−3)(4). This generality is essential when working in the integers as a ring, where negative numbers have equal status with positive ones.
Question 4 True / False
If a and b are positive integers with a | b and b | a, then a = b.
TTrue
FFalse
Answer: True
This is the antisymmetry of the divisibility partial order. If a | b, then b = k·a for some positive integer k (since both are positive). If b | a, then a = m·b for some positive integer m. Substituting: a = m·b = m·k·a, so mk = 1. Since m and k are positive integers, mk = 1 forces m = k = 1, so a = b. This property — along with reflexivity (a | a) and transitivity (a | b and b | c implies a | c) — makes divisibility a partial order.
Question 5 Short Answer
Why is divisibility called a 'partial order' on positive integers rather than a 'total order'? What property does it lack that would be required for a total order?
Think about your answer, then reveal below.
Model answer: A total order requires that any two elements be comparable — for any a and b, either a | b or b | a. Divisibility fails this: consider 4 and 6. Neither 4 | 6 (since 6 = k·4 has no integer solution) nor 6 | 4 (since 4 = k·6 has no integer solution). So 4 and 6 are incomparable under divisibility, making it only a partial order.
Divisibility satisfies the three axioms of a partial order (reflexivity, antisymmetry, transitivity) but not comparability. This is geometrically visible in the Hasse diagram of the divisibility lattice: elements that are not related by divisibility (like 4 and 6) appear side by side with no connecting path. The lattice structure — with gcd as meet (greatest lower bound) and lcm as join (least upper bound) — is a richer consequence of this partial order that does not exist for total orders.