‌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)