In Exercises, verify Theorem for the given sets.
Theorem
Let a, b, and c be integers.
(a) If a | b and a | c, then a | (b + c).
(b) If a | b and a | c, where b >c, then a | (b − c).
(c) If a | b or a | c, then a | bc.
(d) If a | b and b | c, then a | c.
Proof
(a)If a | b and a | c, then b = k1a and c = k2a for integers k1 and k2. So b + c = (k1 + k2)a and a | (b + c).
(b) This can be proved in exactly the same way as (a).
(c) As in (a), we have b = k1a or c = k2a. Then either bc = k1ac or bc = k2ab, so in either case bc is a multiple of a and a | bc.
(d) If a | b and b | c, we have b = k1a and c = k2b, so c = k2b = k2(k1a) = (k2k1)a and hence a | c.
Exercise
(a) A = {a, b, c, d, e, f}, B = {a, c, f, g, h, i, r}
(b) A = {a, b, c, d, e}, B = {f, g, r, s, t, u}
We need at least 10 more requests to produce the solution.
0 / 10 have requested this problem solution
The more requests, the faster the answer.