# Proof that if a \equiv b (\bmod n), then a+c \equiv b+c(\bmod n) and ac \equiv bc (\bmod n).

Proof that if $a\equiv b\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$, then $a+c\equiv b+c\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$ and $ac\equiv bc\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$.

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Step 1
Apply the definition of congruence and divisibility, to express $a\equiv b\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$.
From the definition of congruence, $a\equiv b\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$ means that $n\mid \left(a-b\right)$.
By the definition of divisibility, $n\mid \left(a-b\right)$ implies that $\left(a-b\right)$ is a multiple of n.
That is, $a-b=nk$, where k is an integer.
Step 2
To prove $a+c\equiv b+c\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$, add and subtract c on the left hand side of the equation
$a-b=nk$ and use the definition of congruence.
$a-b=nk$
$a-b+c-c=nk$
$\left(a+c\right)-\left(b+c\right)=nk$
Since, $\left(a+c\right)-\left(b+c\right)$ is a multiple of n, implies that $n\mid \left[\left(a+c\right)-\left(b+c\right)\right]$.
Hence, by the definition of congruence, $a+c=\equiv b+c\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$.
Thus, if $a\equiv b\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$ then $a+c=\equiv b+c\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$.
Step 3
To prove $ac\equiv bc\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$, use the definition of congruence and the properties of divisibility.
$a\equiv b\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$ implies that $n\mid \left(a-b\right)$.
By the properties of divisibility, if n divides $\left(a-b\right)$ then n also divides the multiple of $\left(a-b\right)$.
That is, $n|\left(a-b\right)c=n|\left(ac-bc\right)$
Hence, by the definition of congruence, $ac\equiv bc\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$.
Therefore, if $a\equiv b\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$ then $ac\equiv bc\left(b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}n\right)$.