Congruence Modula - PRove - THANK YOU?

Question

Let n be a natural number and let a, b, c, and d be integers. Prove the following proposition using congruence modula n where n is a natural number.

1-If a≡b (mod n) and c≡d (mod 8) then (a+c)≡(b+d)(mod n)
2-If a≡b (mod n) and c≡d (mod n), then ac≡ bd (mod n).


If a and b are integers, we will use the notation a≡ b (mod n) to mean that (a) is not congruent to b Modula n.
1-write the contrapositive of the following contidional statement:
For all integers a and b, if a ≡ b (mod 6) and b ≡0 (mod 6) then ab ≡ 0 (mod 6)

2- Is this statement true or false? Please explain

Answer 1

1. If a≡b (mod n) and c≡d (mod n), then n | (a-b) and n | (c-d), so a - b = nk and c - d = nj for some integers k and j. Then (a-b) + (c-d) = nk + nj = n(k+j). Since k+j is again an integer and (a-b)+(c-d) = (a+c ) - (b+d), we have n | (a+c)-(b+d), so (a+c)≡(b+d)(mod n).

2. If a≡b (mod n) and c≡d (mod n), then ac - bd = ac - bc + bc - bd = (a-b)c + b(c-d) = (nk)c + b(nt) = n(kc+bt). Thus, n | (ac-bd), so ac≡ bd (mod n).

The contrapositive is: If ab is not congruent to 0 (mod 6), then either a is not congruent to b (mod 6) or b is not congruent to 0 (mod 6). Statement is true.