Emeli Hagan

2020-11-12

For the following statement, either prove that they are true or provide a counterexample:
Let a, b, m, $n\in Z$ such that m, n > 1 and $n\mid m$. If $a\equiv b\left(\text{mod}m\right)$, then
$a\equiv b\left(\text{mod}n\right)$

aprovard

Given that,
Let a, b, m, $n\in Z$ such that m, n > 1 and $n\mid m$.
$a\equiv b\left(\text{mod}m\right)$
By using the definition of congruence relation,
$a\equiv b\left(\text{mod}m\right)$ means m divides b-a.
$m\mid \left(b-a\right)$
Also given that $n\mid m$,
By using,
If $a|b\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}b|cthena\mid c$.
Thus, $n|m\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}m|\left(b-a\right)$ implies that $n\mid \left(b-a\right)$.
By using reverse definition of congruence,
$a\equiv b\text{mod}\left(n\right)$
Therefore, if $a\equiv b\left(\text{mod}m\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}n\mid m$ then
$a\equiv b\left(\text{mod}n\right)$.

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