Next, we expand e back into its original form and substitute it into the above expression twice, getting
(a+ b)c+(a+ b)d
We can apply the right-hand distributive law twice, and rewrite this as
ac+bc+ad+bd
Now we employ the commutative law in a generalized way, obtaining
ac+ad+bc+bd
We know this is equal to the expression we began with, because we just got done “morphing” it using
known tactics, one step at a time. Therefore
(a+b)(c+d)=ac+ad+bc+bd
Q.E.D.
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems.
Don’t hurry! You’ll find worked-out answers in App. A. The solutions in the appendix may not
represent the only way a problem can be figured out. If you think you can solve a particular
problem in a quicker or better way than you see there, by all means try it!
- How does the absolute value change if you multiply an integer by −3?
- How does the absolute value change if you start with an integer and multiply by − 3
over and over without end?
- Evaluate the following expression:
4 + 32 / 8 × (−2)+ 20 / 5 / 2 − 8
- Do an S/R proof showing that for any four integers a, b, c, and d
abcd= dcba
Here’s a hint: you solved a problem like this for addition in Chap. 4. This proof
proceeds in the same way.
- Start with the integer −15, multiply by −45, then divide that result by −25, then
multiply that result by −9, then divide that result by −81, and finally multiply that
result by −5. What do you end up with?
- Show at least one situation where you can say that
a/(b/c)= (a/b)/c
where a, b, and c are integers. Do not use the trivial case a= 1, b= 1, and c= 1. Find
something more interesting!
Practice Exercises 81
