Let A be the set of all functions f : Z Z and R be a relation on A such that R = \(f, g) : f(0) = g(1) and f(1) = g(0)\. Then R is :
- Symmetric and transitive but not reflexive
- Symmetric but neither reflexive nor transitive
- Transitive but neither reflexive nor symmetric
- Reflexive but neither symmetric nor transitive
Answer
(2) Symmetric but neither reflexive nor transitive
Solution
Reflexive: (f, f) R needs f(0) = f(1), which fails for f(x) = x.
R is not reflexive.
Symmetric: (f, g) R f(0) = g(1), f(1) = g(0)
g(0) = f(1), g(1) = f(0) (g, f) R
R is symmetric.
Transitive: take f(x) = x + 1 and g(x) = 2 - x.
f(0) = 1 = g(1) and f(1) = 2 = g(0) (f, g) R; by symmetry (g, f) R.
But (f, f) R since f(0) = 1 2 = f(1).
R is not transitive.
Answer: (2)