Fix m ∈ N and let a, b, c, d ∈ Z be such that a ≡ b (mod m) and c ≡ d (mod m ). How would I prove the below statement. I understand the properties of congruence mod m but not sure how to apply them.
a + c ≡ b + d (mod m )
What I have so far
a - b = k * m c - d = l * m
so a = km + b c = lm + d
therefore a + c = km + b + lm + d a + c = ( b + d) + m( k + l ) (a + c) - (b + d)= m (k + l)
But I don't know how to get rid of the m(k + l) part ?
Recall that $a \equiv b \mod m$ means, by definition, that $a = km + b$ for some integer $k$. Similarly, $c \equiv d \mod m$ means that $c = lm + d$ for some integer $l$. To show that $a + c = b + d \mod m$, you must show that $a + c = rm + (b + d)$ for some integer $r$ (since this is what it means to be congruent mod $m$).
We have $$a + c = km + b + lm + d = km + lm + (b + d) = (k + l)m + (b + d).$$ Since $k + l$ is an integer, we conclude that $a + c \equiv b + d \mod m$.