Operations with Fractions
The operations here are derived from the axioms for real numbers.
Multiplicative Inverse
Let $a$ be a nonzero real number. According to the axioms, there is a
multiplicative inverse which we may denote by $c$
with the property that
\[ ac = ca = 1\]
There is only one multiplicative inverse. Exercise: Show if b and c
both denote a multiplicative inverse of a, then b =c.
We denote the inverse by $a^{-1}$ or, a site innovation to used with
caution, by $\div a$
Properties of Multiplicative Inverses
If a and b are nonzero numbers then
\[ (ab)^{-1} = b^{-1}a^{-1}
\]
because multiplication is associative, because the product
\begin{eqnarray*} [b^{-1}a^{-1}] [ab] & = & b^{-1} (a^{-1} [ab] ) \\
& = & b^{-1} [(a^{-1} a)b] \\
& = & b^{-1} [(1)b] \\
& = & b^{-1} [ b] \\
&=& 1
\end{eqnarray*}
and because multiplicative inverses are unique. The commutative axiom or property
implies
\[ (ab)^{-1} = b^{-1}a^{-1} = a^{-1}b^{-1}
\]
Fraction and Division
For real numbers a and b, with a nonzero, we let $b \div a$ and $\frac ba$
both equal the product $b \times a^{-1}$
So division and fractions are both given by the equalities
\[
$b \div a$ = b \times a^{-1} = \frac ba
\]
with the middle term being the formula for calculating the ends terms.
Multiplication of Fractions
Let m, n, p and q denote real numbers with n and q both nonzero.
Then
\begin{eqnarray*} \frac pn \times \frac mq & =& p(n^{-1}) \times m(q^{-1})
\\ & =& pm (n^{-1}q^{-1} )
\\ & =& pm (n q)^{-1} \\ & =
&\frac {p m}{n q}
\\ & =
&\frac {p \times m}{n \times q} \end{eqnarray*}
Reciprocal of Fractions
When p and q are nonzero real numbers, the reciprocal of their ratio
\[\frac pq \] is \[\frac qp \] Clearly, the reciprocal of a reciprocal is
the original fraction. Further the product of a fraction and its
reciprocal \[\frac pq \times \frac qp = \frac {p \times q}{q \times p}
= \frac {p \times q}{p \times q} = 1\] The question what fractional multiple of
$\frac pq $ gives 1 has the answer $\frac qp$ = the reciprocal of the
fraction. If \[\frac pq \times A = 1\] then one-qth of boths sides must
be equal. That gives \[ \frac 1q \times A = \frac 1p \] Now q times both
sides must be equal. Then gives \[ A =\frac qp \] Thus A is the
reciprocal. Therefore
\[ \left(\frac pq\right)^{-1} = \frac qp\]
\]
Raising and Lowering Terms
Suppose a, b and c are real numbers with the last two b and c, both nonzero.
Then
\begin{eqnarray*} \frac {ac}{bc} &=& (ac)(bc)^{-1} \\
&=& (ac) (c ^{-1}b^{-1}) \\
&=& ((ac) c ^{-1})b^{-1} \\
&=& ( a ( c c ^{-1}))b^{-1} \\
&=& ( a \times 1)b^{-1} \\
&=& \frac ab
\end{eqnarray*}
Division of Fractions
\begin{eqnarray*}
\frac AB \div \frac CD &=& \frac AB \times \left[\frac CD\right]^{-1} \\
&=& \frac AB \times \frac CD \\
\end{eqnarray*}
Thus division is provided by multiplication by a reciprocal.
Addition, Comparision and Subtraction with unlike Denominators
The sum, comparision and difference of two fractions $\frac AB$ and $\frac CD$
may be obtained by raising terms [if need-be] to apply like denominators. We will study
the raising term parts.
Common multiples of the denominators B and D are given their product \[M = B \times D = D \times B\]
When B and D are whole numbers, whole number theory implies the existence of a least common
multiple.
Let M denote a positive, real number. Then
\[ M = b \times B = d \times D\] where $b = M \div B = \frac MB$ = the number
of times B goes into M, and $d = M \div D = \frac MD$ = the number of times D goes into M.
are proportional to M. The least common multiple of B and D gives the smallest values for b and d.
The product common multiple $M = B \times D = D \times B$ gives $b=D$ and $d=B$
Raising terms gives two fractions
\begin{eqnarray*}
\frac AB = \frac {A \times b}{b \times D} = \frac {A \times b}M \\
\frac CD = \frac {C \times d}{d \times D} = \frac {C \times d}M
\end{eqnarray*}
with like denominators to add, subtract or compare. The foregoing
implies
$\frac AB$ is more than $\frac CD$ when and only when $ A \times b$
is more than $C \times d$
$\frac AB$ is equivalent to $\frac CD$ when and only when $ A \times b
= C \times d$
$\frac AB$ is less than $\frac CD$ when and only when $ A \times b$
is less than $C \times d$
The foregoing also implies the
addition formula
\begin{eqnarray*}
\frac AB + \frac CD &=& \frac {A \times b + C \times d} M \\
&=& \frac {A \times [M \div B] + C \times [M \div D]} M
\end{eqnarray*}
and the subtraction formula
addition formula
\begin{eqnarray*}
\frac AB -\frac CD &=& \frac {A \times b - C \times d} M \\
&=& \frac {A \times [M \div B] - C \times [M \div D]} M
\end{eqnarray*}
When $M = B \times D = D \times B$ is the product of the denominators, the
addition and subtraction formulas become
\begin{eqnarray*}
\frac AB + \frac CD &=& \frac {A \times D + C \times B} {B \times D} \\
\frac AB - \frac CD &=& \frac {A \times D -C \times B} {B \times D} \\
\end{eqnarray*}
The above addition and subtraction formulas represent the first step
of a multistep process in which the resulting fraction [by convention] needs
to be simplified. Due to that convention, taking the common multiple M to
be the least common multiple usually leads to the easiest and fewest
steps in the simplication part of the calculation. But in some cases, the decimal
representation of whole numbers sometimes leads to less work due to the arithmetic
properties of decimals.
When $M = B \times D = D \times B$ is the product of the denominators, the foregoing
implies the cross product rules
$\frac AB$ is more than $\frac CD$ when and only when $ A \times D$
is more than $C \times B$
$\frac AB$ is equivalent to $\frac CD$ when and only when $ A \times D
= C \times B$
$\frac AB$ is less than $\frac CD$ when and only when $ A \times D$
is less than $C \times B$
for fraction comparison. Fraction skills and comprehension may be better served
by skipping these formulas, and instead showing how to compare fractions
by raising terms.
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