Prove "a-c = (b-d)(mod m)" Using Modular Arithmetic

[Cyborg31]

Homework Statement

Show that if a = (b mod m) and c = d(mod m) and m => 2, then a - c = (b - d)(mod m)

Homework Equations

c = d(mod m) <=> m|(c - d)
d = c + xm

The Attempt at a Solution

I don't know how any equivalences for a = (b mod m), is there a way to get b from a = (b mod m)?

I had a + c = (b mod m) + d(mod m) but I'm not sure where to go from there

 

[HallsofIvy]

a= b (mod m) means that a- b is divisible by m or that a- b= mp for some integer p.
Similarly, c= d (mod m) means that c- d= mq for some integer q.

Now what does a-c= b- d (mod m) mean?

 

[Cyborg31]

a - c = (b - d)(mod m) would be (a - c) - (b - d) = mx for some x

But its a = (b mod m)