Hey jeremy5561.

Hint: Show that if 0 < |x| < b then 0 <|x-a| < b+a and if 0 < |x| < c then 0 < |dx| < cd.

To get you started consider that 0 < x < b and -b < x < 0 for x > 0 and x < 0 respectively. If I add a this means that:

0 < |x-a| < c which implies 0 < x-a < c and -c < a-x < 0. By shifting the a term out to the boundaries we get:

0 < x < a+c and -c-a < -x < 0. This proves that we have 0 < |x-a| < c+a for a > 0.

Try the same thing for the multiplication case and then combine those two results to show a proof.