Show that if a equiv a mod m and b equiv b mod m then a b

Show that if a equiv a\' (mod m) and b equiv b\' (mod m), then a - b = a\' - b\' (mod m).

Solution

a=a\' mod m

Hence, a-a\'=km for some integer k

b=b\'   mod m

Hence, b-b\'=rm for some integer r

So, a-a\'-(b-b\')=(r-k)m=0 mod m

So,

a-a\'-(b-b\')=0 mod m

a-a\'=b-b\' mod m

a=a\'+b-b\' mod m

a-b=a\'-b\' mod m

Hence proved