Exercise 3: Is "Divides" a Partial Order on the Positive Integers? — Possible Solution ==================================================================== REFLEXIVE ------------------------------ Does a divide a, for every positive integer a? Yes - a = 1 x a, so a is always a whole-number multiple of itself. Reflexive holds. ANTISYMMETRIC ------------------------------ Suppose a divides b AND b divides a. Does this force a = b? If a divides b, then b = k1 x a for some positive integer k1. If b divides a, then a = k2 x b for some positive integer k2. Substituting the first into the second: a = k2 x (k1 x a) = (k2 x k1) x a. Dividing both sides by a (valid since a is a positive integer, never zero): k2 x k1 = 1. Since k1 and k2 are both positive integers, the only way their product equals 1 is k1 = k2 = 1 - which means a = b. Antisymmetric holds, specifically because this relation is restricted to positive integers (it would fail for signed integers, where e.g. 1 divides -1 and -1 divides 1, but 1 != -1). TRANSITIVE ------------------------------ Suppose a divides b and b divides c. Does a then divide c? If a divides b, b = k1 x a. If b divides c, c = k2 x b = k2 x (k1 x a) = (k2 x k1) x a. Since k2 x k1 is itself a positive integer, c is a whole-number multiple of a, meaning a divides c. Transitive holds. CONCLUSION: THIS IS A PARTIAL ORDER ------------------------------ Since reflexive, antisymmetric, and transitive all hold, "divides" on the positive integers satisfies this chapter's own definition of a partial order exactly. AN INCOMPARABLE PAIR: 4 AND 6 ------------------------------ Does 4 divide 6? 6 / 4 = 1.5, not a whole number, so no. Does 6 divide 4? Since 6 > 4, 6 could only divide 4 if 4 were a whole-number multiple of 6, which it isn't (4 / 6 is less than 1) - so no. Neither 4 divides 6 nor 6 divides 4 - the two numbers are simply incomparable under this relation, exactly the kind of pair this chapter says a partial order (unlike a total order such as <=) is allowed to have. WHY THIS WORKS AS AN ANSWER ------------------------------ Each of the three required properties is proven algebraically rather than just asserted, with the antisymmetric proof specifically explaining why restricting to positive integers is what makes it work, and a concrete incomparable pair is verified explicitly in both directions to confirm the relation is genuinely partial, not total.