Answer to Question #136830 in Discrete Mathematics for Promise Omiponle

Question #136830
Show that if n|m, where n and m are integers greater than 1, and if a≡b(mod m), where a and b are integers, then a≡b(mod n).
1
Expert's answer
2020-10-20T18:25:45-0400

Given:  nm  where  n  and  m  are  integers  greater  than  1,and\mathbf{Given:\;n|m\;where\;n\;and \;m\;are\;integers\;greater\;than\;1,and}

ab(mod  m),  where  a  and  b  are  integers.\mathbf{a\equiv b(mod\;m),\;where\;a\;and\;b\;are\;integers.}


To  show:ab(mod  n)\mathbf{To\;show: a\equiv b(mod\;n)}


Proof:  Since  nm,i.e,  n  divides  m,we  have\mathbf{Proof:\;Since\;n|m,i.e,\;n\;divides\;m,we\;have-}

                            m=nα,where  α  is  an  integer.        ................(A)\mathbf{\;\;\;\;\;\;\;\;\;\;\;\;\;\;m=n\alpha,where\;\alpha\;is\;an\;integer.\;\;\;\;................(A)}


Also  ab(mod  m)    m  divides  (ab)    m(ab)\mathbf{Also\;a\equiv b(mod\;m) \implies m\;divides\;(a-b)\implies m|(a-b)}

          ab=mλ,where  λ  is  an  integer.      ................(B)\mathbf{\;\;\;\implies a-b=m\lambda,where\;\lambda\;is\;an\;integer.\;\;\;................(B)}


Substitute  m=nα  in  equation  (B),we  get\mathbf{Substitute\;m=n\alpha\;in\;equation\;(B),we\;get-}

ab=(nα)λ\mathbf{a-b=(n\alpha)\lambda}

    ab=n(αλ)          (Multiplication  is  associative.)\implies\mathbf{a-b=n(\alpha\lambda)\;\;\;\;\;(\because Multiplication\;is\;associative.)}

    n  divides  ab\implies \mathbf{n\;divides\;a-b}

    n(ab)\implies \mathbf{n|(a-b)}

    ab0(mod  n)\implies \mathbf{a-b\equiv0(mod\;n)}

    ab(mod  n)\implies \mathbf{a\equiv b(mod\;n)}

Hence  the  result.\mathbf{Hence\;the\;result.}






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