Here R is a relation on `NxxN,` defined by
`(a,b)R(c,d)iffa+d=b+c" for all "(a,b),(c,d)inNxxN.`
We shall show that R satisfies the following properties.
(i) Reflexivity:
We know that `a+b=b+a" for all "a,binN.`
`:." "(a,b)R(a,b)" for all "(a,bin(NxxN).`
So, R is reflexive.
(ii) Symmetry:
Let `(a,b)R(c,d).` Then,
`(a,b)R(c,d)impliesa+d=b+c`
`implies" "c+b=d+a`
`implies" "(c,d)R(a,b).`
`:." "(a,b)R(c,d)implies(c,d)R(a,b)" for all "(a,b),(c,d)inNxN.`
This shows that R is symmetric.
(iii) Transitivity:
Let `(a,b)R(c,d)" and "(c,d)R(e,f)`. Then,
`" "(a,b)R(c,d)" and "(c,d)R(e,f)`
`implies" "a+d=b+c" and "c+f=d+e`
`implies" "a+d+c+f=b+c+d+e`
`implies" "a+f=b+e`
`implies" "(a,b)R(e,f).`
Thus, (a,b) R (c,d) and (c,d) R (e,f)`implies" "`(a,b) R (e,f).
This shows that R is transitive.
`:." "R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on `NxxN.`
