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).

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

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

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

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

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

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

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

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

$\mathbf{a-b=(n\alpha)\lambda}$

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

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

$\implies \mathbf{n|(a-b)}$

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

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

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