Mathematics and Logic - Skill and Concept Development

Operations with Fractions

The operations here are derived from physical meaning of what it means to tale a fraction of an object.

Multiplication of Fractions

Let m, n, p and q denote whole numbers - 5, 7, 11 and 13 if you like.

We assume one n-th of $m \times n = n \times m$ objects is m. We further assume p n-ths of IT is p times [one n-th of IT], provided the latter can be calculated. The latter can be calculated when IT is a multiple of n.

Now

\begin{eqnarray*} \\ \frac pn \times \frac mq & =& p \times \left[ \frac 1n \mbox{ of } \frac mq \right] \\ & =& p \times \left[ \frac 1n \mbox{ of } \frac {n \times m}{n \times q} \right] \\ & =& p \times \left[ \frac { m}{n \times q} \right] \\ & = &\frac {p \times m}{n \times q} \end{eqnarray*}

Observe how raising terms gives a multiple of n. The result gives multiply the numerators, multiply the denominators rule for fraction multiplication.

The rule can be made more efficient, applied indirectly, by showing how to find common factors to cancel before products in the numerator and denominator of the result are calculated.

Observe the multiplication of fractions

\[ \frac pn \times \frac mq = \frac mq \times \frac pn \]

commutes because the multiplication of whole numbers commutes.

Reciprocal of Fractions

When p and q are nonzero whole 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.

Now

\[ A \times \frac pq = 1\]

implies

\[\frac pq \times A = 1\]

Therefore answer to the question of what fractional multiple of $\frac pq$ gives 1 is given by the reciprocal $A = \frac qp$

In measuring with units and subunits, the question of how times $\frac pq $ units goes into 1 unit has the answer: $A = \frac qp$

Division of Whole Numbers - Forwards and Backwards

Let N =17, d= 5 and r = 2 below if you like.

Long division of a whole number N by a divisor d, another whole numbers, implies there is a natural number $q \ge 0$ and a remainder r $ 0 \le r \lt d$ such that

\[ N =q \times d + r = q \times d + r\times 1 \]

Thus d goes into N, q whole times with a remainder of r. That is only answer that can be given before multiplication by improper fractions or mixed numbers is understood. The case q = 0 arises when $N \lt d$.

Now

\[ 1= \frac 1d \times d \]

gives

\begin{eqnarray*} N &=& q \times d + r \frac 1d \times d \\ & =& q \times d + \frac rd \times d \\ & = & \left[q +\frac rd\right] \times d \end{eqnarray*} Thus with a knowledge of fractions and mixed numbers, we may say, d objects goes into N objects, $q +\frac rd$ times exactly. That result appears in some but not all primary mathematics work booklets.

By raising terms, we see that mixed number

\[ q +\frac rd = \frac {q \times d}d + \frac rd = \frac {qd+r}d = \frac Nd \]

is equivalent to the fraction $\frac Nd$ - improper when $N \ge d$. The foregoing leads us to write

\[ N \mbox{ objects} \div d \mbox{ objects} = \frac Nd \]

Thus the improper[?] fraction $\frac Nd$ gives the exact number of times d objects go into N objects. Long division or inspection allows the latter to be expressed as a mixed number $q +\frac rd$

Division with Like Denominators

Take m = 4 on first reading if you like.

Now suppose the object in question is an m-th of another object. That would imply

\[ N \mbox{ m-ths} \div d \mbox{ m-ths} = \frac Nd \]

Or, in fraction notation

\[ \frac Nm \div \frac dm = \frac Nd \]

Division with Unlike Denominators

Take A =3, B =5, C = 11 and D = 13 on on first reading if you like.

In the case of unlike denominators, we raise terms to transform the division question into a like denominator case:

\begin{eqnarray*} \frac AB \div \frac CD &=& \frac {A \times D}{B \times D} \div \frac{B \times C}{B \times D} \\ & =& \frac {A \times D} {B \times C} \end{eqnarray*}

Showing how to divide defines the operation. But

\[ \frac {A \times D} {B \times C} = \frac AB \times \frac DC \]

Hence division by a fraction $\frac CD$

\[\frac AB \div \frac CD = \frac AB \times \frac DC = \]

as well. So it has the same result as multiplication by the reciprocal $\frac DC$ - the latter serves as constant of proportionality $K$ for division by $\frac CD$. That is,

\begin{eqnarray*} \frac AB \div \frac CD &=& \frac AB \times \frac DC \\ &=& \frac DC \times \frac AB \\ &=& K \times \frac AB\end{eqnarray*}

with

\[K = \frac DC = 1 \div \frac CD \]

Addition, Comparision and Subtraction with Like Denominators

Take a =5, b =11 and m = 5 on on first reading if you like.

Addition

Given a pair of whole a and b the distributive law for counting says

\[ a \mbox{ objects } + b \mbox{ objects} = [a+b] \mbox{ objects} \]

Now if the object is given by an m-th of another, we likewise have

\[a \mbox{ m-ths } + b \mbox{m-ths} = [a+b] \mbox{ m-ths} \]

In fraction notation that gives the like denominator fraction addition rule:

\[ \frac am + \frac bm = \frac{a+b}m \]

Comparison

If in the pair, the number a is more than b, we may write

a objects is more than b objects

Now if the object is given by an m-th of another, we have

a m-ths is more than b m-ths

\[ \frac am > \frac bm \]

where we read > as more than.

Subtraction

Now if in the pair, the number a is more than b, then by a distributive law for counting

\[ a \mbox{ objects } - b \mbox{ objects} = [a-b] \mbox{ objects} \]

Now if the object is given by an m-th of another, we have

\[ a \mbox{ m-ths } - b \mbox{ m-ths} = [a-b] \mbox{ m-ths} \]

In fraction notation that gives the like denominator fraction addition rule:

\[ \frac am - \frac bm = \frac{a-b}m \]

Addition, Comparision and Subtraction with unlike Denominators

Take A =4, B =6, C = 11 and D = 13 on on first reading if you like.

The sum, comparision and difference of two fractions $\frac AB$ and $\frac CD$ may done by raising terms [if need-be] to apply like denominators. We will study the raising term parts.

One common multiple of the denominators B and D is their product

\[M = B \times D = D \times B\]

Let M denote a common multiple of the denominators B and D. Then

\[ M = b \times B = d \times D\]

where \[b = M \div B = \frac MB\] is the number of times B goes into M, and \[d = M \div D = \frac MD\] is 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 - see site lesson on inequalities^{providing it is site to do - 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 $C \times d$ /

\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. Then 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*}

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