Algebra Know-It-ALL

606 Worked-Out Solutions to Exercises: Chapters 1 to 9

9. Table A-10 shows that the associative property of multiplication holds for any three
   fractions. If we have fraction of the form a/b, another of the form c/d, and a third of
   the form e/f, where a,c, and e are integers, and b,d, and f are positive integers, then

[(a/b)(c/d)](e/f)= (a/b)[(c/d)(e/f)]

Do you notice that this proof has a certain symmetry? First, we mold the expression,

in two “steps up,” into a form where we can apply the associative law for multiplying

integers. Then we invoke that law. Finally we mold the expression, in two “steps back

down,” into the form we want.

10. Let’s explore this situation and see what sorts of “clues” we can find. We’ve been told
    that there are four integers a,b,c, and d, such that

(a/b) / (c/d)= (c/d) / (a/b)

If we use the rule for division of one fraction by another on both sides of this equation,

we get

ad / bc=cb / da

Invoking the commutative law for multiplication in the numerator and denominator on

the right-hand side of the equals sign, we see that

ad / bc=bc / ad

Table A-10. Solution to Prob. 9 in Chap. 6. This shows that the

associative law holds for the multiplication of fractions. As you

read down the left-hand column, each statement is equal to all the

statements above it.

Statements Reasons

[(a/b)(c/d)](e/f) Begin here

(ac / bd)(e/f) Formula for multiplication of a/b by c/d

(ac)e / (bd)f Formula for multiplication of (ac / bd) by e/f

a(ce) / b(df) Associative law for multiplication of integers

applied to numerator and denominator

(a/b)(ce / df) Formula for multiplication of fractions

applied “backward” to turn quotient of

products into product of quotients

(a/b)[(c/d)(e/f)] Formula for multiplication of fractions

applied “backward” (again!) to turn quotient

of products into product of quotients

Q.E.D. Mission accomplished