# Prove $a + b = c \implies a \space (\text{mod } n) + b \space (\text{mod } n) \equiv c \space (\text{mod } n)$

Prove $$a + b = c \implies a (\text{mod } N) + b (\text{mod } N) \equiv c (\text{mod } N)$$

My attempt:

$$a = k_1 \space \text{ (mod n)}$$ where $$k_1$$ is the remainder of $$a$$.

$$b = k_2 \space \text{ (mod n)}$$ where $$k_2$$ is the remainder of $$b$$.

We have $$n \mid a - k_1$$ and $$n \mid b - k_2$$. We can conclude that $$\tag {*} n \mid a - k_1 + b - k_2$$

Reordering $$(*)$$ gives

$$n \mid (a+b) - (k_1 + k_2)$$

Since $$a + b = c$$, we have

$$n \mid c - (k_1 + k_2)$$

We know that if $$n$$ divides $$x-y$$, then $$n$$ also divides $$y-x$$, hence

$$n \mid (k_1 + k_2) - c$$

And therefore

$$a \space (\text{mod } n) + b \space (\text{mod } n) \equiv c \space (\text{mod } n)$$

$$\Box$$

Is it correct?

Yes, your proof is correct. More simply, you can write $$a=q_1n+r_1,b=q_2n+r_2$$, then $$c=a+b=(q_1+q_2)n+(r_1+r_2)$$ and then the conclusion follows immediately.