Compare this with the result of the “challenge” problem, where you found out that
[(a/b) / (c/d)] / (e/f)=adf / bce
- Table A-9 is an S/R proof that if a, b,c, and d are nonzero integers, then
(a/b)(c/d)= (c/d)(a/b)
Therefore, the commutative property holds for the multiplication of fractions, and
indeed for the multiplication of any two rational numbers. Note the fourth step of this
proof, in which the rule for multiplication of fractions is applied “backwards.” We can
get away with this because equality works in both directions! This fact, which may seem
trivial to you but is really quite significant, is one of three aspects of equality known as the
reflexive,symmetric, and transitive properties. You should know what these terms mean.
The reflexive property tells us that for any quantity a,
a=a
The symmetric property tells us that for any two quantities a and b,
If a=b, then b=a
The transitive property tells us that for any three quantities a,b, and c,
If a=b and b=c, then a=c
Whenever any means of comparing things has all three of these properties, then it’s
called an equivalence relation. Equality is the most common example of an equivalence
relation. But there are others, such as the logical connector “if and only if ” or “iff,”
symbolized by a double-shafted, double-headed arrow, often with a little extra space
on either side (⇔).
Table A-9. Solution to Prob. 8 in Chap. 6. This shows that the
commutative 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) Begin here
ac / bd Formula for multiplication of fractions
ca / db Commutative law for multiplication of integers
applied to numerator and denominator
(c/d)(a/b) Formula for multiplication of fractions
applied “backwards”
Q.E.D. Mission accomplished
Chapter 6 605
