R={(f,g):f(0)=g(1)and . f(1)=g(0)} Reflexive: (f,f)∈R =f(0)=f(1)and f(1)=f(0)⟶must hold ⇒ but this is not true for all function so not reflexive symmetricif f (f,g)∈R⇒(g,f)∈R Now, g(0)=f(1)and g(1)=f(0)⟶true ∴ symmetric Transitive: If(f,g)∈Rand (g,h)∈R ⇒(f,h)∈R Now (f,g)∈R⇒f(0)=g(1)and f(1)=g(0) (g,h)∈R⇒g(0)=h(1)and g(1)=h(0)
