To prove that if $a \equiv b \ (\text{mod} \ m)$ and $c \equiv d \ (\text{mod} \ m)$, then $a + c \equiv b + d \ (\text{mod} \ m)$, we proceed as follows:

Step 1: Understand the given assumptions

1. $a \equiv b \ (\text{mod} \ m)$ means that $m$ divides the difference $a - b$, i.e., $a - b = km \quad \text{for some integer} \ k.$
2. Similarly, $c \equiv d \ (\text{mod} \ m)$ means that $m$ divides $c - d$, i.e., $c - d = lm \quad \text{for some integer} \ l.$

Step 2: Add the two congruences

Now, add the two congruences: $a + c \equiv b + d \ (\text{mod} \ m)$ We can express this as: $(a - b) + (c - d) = (km) + (lm) = (k + l)m.$

Step 3: Conclusion

Since $(k + l)m$ is clearly divisible by $m$, we conclude that: $a + c - (b + d) = (k + l)m,$ which shows that $m$ divides $(a + c) - (b + d)$, or equivalently: $a + c \equiv b + d \ (\text{mod} \ m).$

Thus, the proof is complete!

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Would you like further clarification or details on any part of this proof?

Here are some related questions to help deepen the understanding:

1. How would this proof change if we had multiple terms, such as $a_1 + a_2 + \cdots + a_n \equiv b_1 + b_2 + \cdots + b_n \ (\text{mod} \ m)$?
2. What happens if $a \equiv b \ (\text{mod} \ m)$ but $c \not\equiv d \ (\text{mod} \ m)$?
3. Can this property of congruence addition be extended to multiplication? If so, how?
4. How do we handle negative integers in modular arithmetic, and does this proof still hold?
5. What is the difference between modular addition and ordinary addition in regular arithmetic?

Tip: Always remember that modular arithmetic is about equivalence classes, so operations like addition or multiplication maintain these equivalences if the individual terms are congruent.