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

Polynomial factorization
ANSWERED Proof that if $$\displaystyle{a}\equiv{b}{\left({b}\mod{n}\right)}$$, then $$\displaystyle{a}+{c}\equiv{b}+{c}{\left({b}\mod{n}\right)}$$ and $$\displaystyle{a}{c}\equiv{b}{c}{\left({b}\mod{n}\right)}$$. 2021-02-28

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