Discrete Math Proof Prove the statement in two ways by the c

Discrete Math Proof

Prove the statement in two ways, by the contrapositive and by contradictions.

For all integers a, b, and c, if a divides b and a does not divide c, then a does not divide b + c

Solution

By contrapositive

For the first show the contrapositive, namely if a divides b+c and a does not divide b then a divides c.

assume b=aq1, c=aq2 a divides b+c, and a does not divide b, then for any value of c a does not divide a

hence proved.For all integers a, b, and c, if a divides b and a does not divide c, then a does not divide b + c

that

By contradiction assume this does not hold,Now

Since a does not divides b, by definition b (not =to) aq1 for some integer q1, and a also does not divides c then we have c (not =to)aq2 for some integer q2. Therefore b + c (not =to)aq1 + aq2 (not =to) a(q1 + q2) (not =to) aq3 for some integer q3 (not =to) q2 + q1 . This states that a does not divides b + c . but by definition of divisiblity this is false . hence our contadiction becomes falseand our statement is thus true