Let R be a relation on N×N defined by (a,b)R,(c,d)⇔a+d=b+c for all (a,b),(c,d)ϵN×N
Show that :
(i) (a, b) R (a, b) for all (a,b)ϵN×N
(ii) (a,b)R(c,d)⇒(c,d)R(a,b) for all (a,b),(c,d)ϵN×N
(iii) (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) ϵN×N.
We have,
(a,b)R,(c,d)⇔a+d=b+c for all (a,b),(c,d)ϵN×N
(i) We have,
a + b = b + a for all a,bϵN
∴(a,b)R(a,b) for all, a,bϵN
(ii) Now,
(a, b) R (c, d)
⇒a+d=b+c
⇒c+d=d+a
⇒(c,d)R(a,b)
(iii) Now,
(a, b) R (c, d) and (c, d) R (e, f)
⇒a+d=b+c and c + f = d + e
⇒a+d+c+f=b+c+d+e (Adding)
⇒a+f=b+e
⇒(a,b)R(e,f)