Check whether the relation R = {(a, b): a < b²} on the set R of real numbers is reflexive, symmetric or 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): a < b²} on R. To find: whether R is reflexive, symmetric, transitive.
2 · Write the three rules.
Reflexive: (a, a) ∈ R for all a.
Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R.
Transitive: (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R.
3 · Reflexive: test a = 1/2. 1/2 < (1/2)² = 1/4 is false, so (1/2, 1/2) ∉ R. R is not reflexive.
4 · Symmetric: test (1, 2). 1 < 2² = 4, so (1, 2) ∈ R. But 2 < 1² = 1 is false, so (2, 1) ∉ R. R is not symmetric.
5 · Transitive: test the chain (3, 2), (2, 3/2). 3 < 4 and 2 < 9/4, so both pairs are in R. But 3 < 9/4 is false, so (3, 3/2) ∉ R. R is not transitive.
6 · Check and box the answer. R is neither reflexive, nor symmetric, nor transitive.
Scoring tip: Choose small, easy numbers for counterexamples and show the inequality failing in one line.
