1. The problem statement, all variables and given/known data The problem and solution are attached as TheProblemAndSolution.jpg. I will also copy down the problem and solution here.: Problem: Consider the set ℤ of integers. Define aRb by b = a^r for some positive integer r. Show that R is a partial order on ℤ, that is, show that R is (a) reflexive; (b) antisymmetric; (c) transitive. Solution: (a) R is reflexive since a = a^1. (b) Suppose aRb and bRa, say b = a^r and a = b^s. Then a = (a^r)^s = a^(rs). There are three possibilities: (i) rs = 1, (ii) a = 1, and (iii) a = -1. If rs = 1 then r = 1 and s = 1 and so a = b. If a = 1 then b = 1^r = 1 = a, and, similarly, if b = 1 then a = 1. Lastly, if a = –1 then b = –1 (since b ≠ 1) and a = b. In all three cases, a = b. Thus R is antisymmetric. (c) Suppose aRb and bRc say b = a^r and c = b^s. Then c = (a^r)^s = a^(rs) and, therefore, aRc. Hence R is transitive. Accordingly, R is a partial order on ℤ. 2. Relevant equations Definitions of partial order, reflexive binary relations, antisymmetric binary relations, transitive binary relations. 3. The attempt at a solution I don't get why if a = –1, then b = –1. If I assume that b ≠ 1, then I get it, but why is b = 1 forbidden? Any input would be greatly appreciated!