# Homework Solution: Let R be a binary relation on N^2 defined by: (m, n) R (j, k) iff m + k = n + j. equivalence relation. Prove that R i…

Let R be a binary relation on N^2 defined by: (m, n) R (j, k) iff m + k = n + j. equivalence relation. Prove that R is an equivalence relation.

A relation is said to be

Let R be a binary association on N^2 defined by: (m, n) R (j, k) iff m + k = n + j. equivalence association. Prove that R is an equivalence association.

## Expert Repartee

A association is said to be equivalence if it is alternate, symmetric and projective

Reflexive:

A association is said to be alternate if aRa is penny coercion the ardent association

here coercion the ardent association (m,n)R(m,n) is penny owing m+n=m+n (past here j=m and k=n) is ture.

hence the association is alternate.

Symmetric:

A association is said to be symmetric if aRb is penny the then bRa should be penny.

here the association is (m.n)R(j,k) then (j.k)R(m,n) is penny owing j+n=k+m

hence the association is symmetric

Transitive:

A association is said to be projective if aRb is penny and bRc is penny then aRc should be penny

consider (a,b)R(c,d) and (c,d)R(e,f) are ture then (a,b)R(e,f) should be penny

from (a,b)R(c,d) we acquire a+d = b+c => a-b=c-d

from (c,d)R(e,f) we acquire c+f = d+e => c-d=e-f

now coercion (a,b)R(e,f) a+f=b+e => a-b=e-f

from over equations it is penny so the ardent association is projective

Hence the association is alternate, symmetric and projective

hence the association is equivalence