Can we add congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a+c≡b+d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
Can we subtract congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a−c≡b−d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
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Consider: $$a≡b \mod(m) \hspace{0.5cm} and \hspace{0.5cm} c≡d \mod(m)$$ Then, $$m \mid (a-b) \hspace{0.5cm} and \hspace{0.5cm} m \mid (c-d)$$ So, $$a-b=km \hspace{0.5cm} and \hspace{0.5cm} c-d=k'm$$ Adding, $$(a+c)-(b+d)=(k+k')m$$ Hence, $$ m \mid (a+c)-(b+d)$$
We get: $$(a+c)≡(b+d) \mod m$$
Similarly, proceed for: $$(a-c)≡(b-d) \mod m$$
You wanted numerical examples:
Take $$7≡5 \mod(2) \hspace{0.5cm} and \hspace{0.5cm} 10≡4 \mod(2)$$
$$7+10≡5+4 \mod(2)$$ or $$10≡4 \mod(2)$$ $$17≡9 \mod(2)$$
Similarly,
$$7-10≡5-4 \mod(2)$$ or $$-3≡1 \mod(2)$$
Usually we talk about adding congruence classes. For example, if $a=b$ mod(m), we usually say the congruence classes of a and b are equal, that is $[a]=[b]$, and we can add/subtract/multiply two congruence classes.
I'll show the first question and you can do the second one.
By definition, $a=b$ mod(m) $\iff$ $a-b=0$ mod(m) $\iff$ $a-b=tm$ for some $t\in\mathbb{Z}$.
Using this fact, we can see that $$a-b+c-d=tm+sm=(t+s)m$$ for some $t,s\in\mathbb{Z}$. If we add $b+d$ to both sides, we obtain $$a+c=b+d+(t+s)m.$$ Taking mod(m) of both sides, we obtain our result.