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 N2 defined by: (m, n) R (j, k) iff mrk = mtj. equivalence relation. Prove that Ris an
    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.

    Expert Answer

     
    A relation is said to be

    Let R be a binary association on N2 defined by: (m, n) R (j, k) iff mrk = mtj. equivalence association. Prove that Ris an

    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