Let R be the relation on the set Z of integers given by R = {(a, b): ab < 0}. Show that R is symmetric but neither reflexive nor transitive.
How to lay out the answer. Follow the frame, then write it in your own words.
1 · Write what is given and what is to be found. Given: R = {(a, b): ab < 0} on Z, so a R b when a and b have opposite signs. To show: R is symmetric, not reflexive and not transitive.
2 · Write the rules.
Reflexive: (a, a) ∈ R for every a ∈ Z.
Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R.
Transitive: (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R.
3 · Symmetric: yes. Let (a, b) ∈ R, so ab < 0. Since ba = ab, ba < 0 too. Hence (b, a) ∈ R.
4 · Reflexive: no. Take a = 1: 1 × 1 = 1, which is not less than 0. So (1, 1) ∉ R. In fact a × a ≥ 0 for every integer a.
5 · Transitive: no. (1, −2) ∈ R, since 1 × (−2) = −2 < 0. (−2, 3) ∈ R, since (−2) × 3 = −6 < 0. But 1 × 3 = 3 > 0, so (1, 3) ∉ R.
6 · Check and box the answer. R is symmetric, but it is neither reflexive nor transitive.
Scoring tip: For each 'not' part, one clear counterexample is enough. Show the product and its sign in one line.
