Proofs for multiplication and division of fractions

[Paul Tanner]

For all fellow teachers (and students) who are interested, I give
below some simple, rigorous proofs of the standard algorithms for
multiplying factions and for dividing fractions. The method for
dividing fractions is sometimes called the 'invert and multiply' or
the 'multiply by the reciprocal' method. If you are looking for some
definitive answers as to why these algorithms work, then I hope that
these proofs help you in your quest.

The first two theorems below make things so easy. They actually are a
version of two of the most fundamental theorems for groups. (Don't be
discouraged by the 'group' word!) These two group theorems have a
version in the real numbers with addition as the operation, because
the real numbers are an additive commutative group. These two also
have a version in the non-zero real numbers with multiplication as the
operation, because the non-zero real numbers are a multiplicative
commutative group. Since we see this last version below, we assume
that all the variables used here are non-zero real numbers.

Check out any good abstract algebra textbook to see these two group
theorems, so that you can infer the real number additive version of
all the theorems below. To teachers: After reading everything here,
you might begin to see why abstract algebra contains the answers to so
many common questions about arithmetic.

First, we assume that the usual notation for the multiplicative
inverse of x is 1 / x. And we assume also the definition of real
number division, which is x / y = x * (1 / y) = (1 / y) * x.

(In abstract algebra, there is no operation of subtraction or of
division. Subtraction is essentially adding an additive inverse.
Division is essentially multiplying by a multiplicative inverse. Try
teaching division, x / y, this way: Take a whole and divide it into y
equal parts. Take 1 of these y parts, and add it to itself x times. So
x / y = x * (1 / y) = (1 / y) * x is essentially repeated addition.
But notice that this explanation, and the NCTM's explanation, can't
work when x and y both are irrational numbers, because 'repeated'
can't make real sense with this type of number. Wouldn't it be nice if
some of the ancient Greeks were right, and all numbers were rational?)

Theorem 1. m = 1 / (1 / m).

Proof:

1 = m * (1 / m) (by the multiplicative inverse)
1 = (1 / m) * (1 / (1 / m)) (by the multiplicative inverse -
usual notation)
(1 / m) * m = (1 / m) * (1 / (1 / m)) (by substitution, and by
the commutative property)
m = 1 / (1 / m) (by the cancellation property)

Theorem 2. 1 / (m * n) = (1 / m) * (1 / n).

Proof:

1 = (m * n) * (1 / (m * n)) (by the multiplicative inverse)
1 = m * (1 / m) = m * 1 * (1 / m) = m * (n * (1 / n)) * (1 / m)
(by the multiplicative inverse and multiplicative identity)
(m * n) * (1 / (m * n)) = (m * n) * ((1 / m) * (1 / n)) (by
substitution, and by the associative and commutative properties)
1 / (m * n) = (1 / m) * (1 / n) (by the cancellation property)

Multiplication of Fractions Theorem. (a / b) * (c / d) = (a * c) / (b
* d).

Proof:

(a / b) * (c / d) = (a * (1 / b)) * (c * (1 / d)) (by the
definition of division)
= (a * c) * ((1 / b) * (1 / d)) (by the associative and
commutative properties)
= (a * c) * ((1 / (b * d))) (by Theorem 2)
= (a * c) / (b * d) (by the definition of division)

Division of Fractions Theorem. (a / b) * (1 / ( c / d)) = (a / b) * (d
/ c).

Proof: Notice that all we need to prove is 1 / ( c / d) = d / c.

1 / ( c / d) = 1 / (c * (1 / d)) (by the definition of
division)
= (1 / c) * (1 / (1 / d)) (by Theorem 2)
= (1 / c) * d (by Theorem 1)
= d / c (by the definition of division)

There it is. With the two group theorems, the rigorous proofs of the
multiplication and division methods are simple and elegant. Given the
two group theorems, I hope that every elementary school teacher is now
able to see the real reason why these multiplication and division
methods work for all real numbers.

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[Paul Tanner]

The last sentence was

"Given the two group theorems, I hope that every elementary school
teacher is now able to see the real reason why these multiplication
and division methods work for all real numbers."

The last phrase should of course be

"for all non-zero real numbers."

Sorry for the oversight.