Algebra Know-It-ALL

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

denominator is positive, and no prime in the numerator is the same as any prime in the

denominator, then that fraction is in lowest terms.

With this knowledge, it’s not hard to think of a couple of fractions that are in lowest

terms individually, but that produce a fraction that is not in lowest terms when they are

multiplied by each other. Take a look at this:

3/5× 5/7 = (3 × 5) / (5 × 7)

= 15/35

Both of the original fractions are in lowest terms here, but the product is not. You can

factor 5 out of both the numerator and the denominator of 15/35, reducing it to 3/7.

6. When you have two fractions in lowest terms and divide one by the other, the quotient
   is sometimes in lowest terms, but not always. This can be seen by “mutating” the
   solution to Prob. 5. First, consider this:

(3/5) / (11/7) = 3/5 × 7/11

= (3 × 7) / (5 × 11)

= 21/55

This is in lowest terms, as you saw in the solution to Prob. 5. Now try this:

(3/5) / (7/5) = 3/5 × 5/7

= (3 × 5) / (5 × 7)

= 15/35

Again as you saw in the solution to Prob. 5, this quotient is not in lowest terms, even

though the dividend and divisor fractions are.

7. Table A-8 is an S/R derivation proving that

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

Table A-8. Solution to Prob. 7 in Chap. 6. The result is a formula

for repeated division of fractions, where the second pair of fractions is

divided first. 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

(a/b) / (cf / de) Apply the formula for division of c/d by e/f

(a/b) / (g/h) Temporarily let cf = g and de = h, and substitute

the new names in the previous expression

ah / bg Apply the formula for division of a/b by g/h

ade / bcf Substitute cf for g and de for h in the previous

expression