Algebraic Multiplication and Division

In the chapter Positive and Negative Quantities we explained algebraic addition and subtraction. In this chapter we shall cover simple algebraic multiplication and division.

Algebraic Multiplication

Definition: One number is said to be multiplied by another when we do to the former what is done to unity to obtain the latter.

Thus, since 4 = 1+1+1+1, we must have

$4 \times x \: or \: 4x = x+x+x+x$

Similarly, $4 \times 5 = 5+5+5+5 = 20$

$3 \times 6= 6+6+6 = 18$

$5 \times 3 = 3+3+3+3+3+3 = 15$

$3 \times (-5) = -5-5-5 =-15$

$4 \times (-3) = -3-3-3-3 = -12$

$5 \times (-4) = -4-4-4-4-4 = -20$

Again, since -4 = -1-1-1-1

$(-4) \times x = -x-x-x-x$

Similarly, $(-4) \times 5 = -5-5-5-5 = -20$

$(-3) \times 6 = -6-6-6 = -18$

$(-5) \times 3 = -3-3-3-3-3 = -15$

Also,

$(-3) \times (-5) = - (-5) - (-5) - (-5) = 15$

$(-4) \times (-3) = - (-3) - (-3) = (-3) - (-3) = 12$

$(-5) \times (-4) = - (-4) - (-4) - (-4) - (-4) -(-4) = 20$

The number multiplied is called the multiplicand and the number by which it is multiplied is called the multiplier, the result is called the product.

Fundamental Propositions of Multiplications

1. To prove that $a \times b$ and $b \times a$ , i.e., $a$ multiplied by $b$ and $b$ multiplied by $a$ gives the same result:

First let $a$ and $b$ be two positive integers

Place $b$ units in a horizontal row and write down $a$ rows in such a manner that units in $a$ similar positions in the different rows may be in the same vertical column

1 1 1 1 1 1 . . . . . . $b$ terms

……

To $a$ rows

This being done, evidently it may also be said that we have written down b columns each containing a unites,

Now let us count up the total number of units thus written down

Since we have got a rows each containing b units and the total number of units = (number in the 1^st row) + (number in the 2^nd row) + (number in the 3^rd row) …… + (number in the ath row) = b+b+b+b …… to a terms = a\timesb

Also since we have got b columns each containing a units the total number of units = (number in the 1^st column) + (number in the second column) …… = (number in the bth column) = a+a+a+a+a…. to b terms = b \times a

Hence from the above explanation, it is clear that a\timesb and b\timesa gives the same result.

2. To prove that $ab \times c = a \times bc \: or \: b \times ac$ , i.e., to multiply $c$ by the product of $a$ and $b$ is the same as to multiply $c$ first by either of them and then the result by the other:

Place b brackets in a horizontal row each containing c units and write down a rows in such a manner that the brackets in similar positions in the different rows may be in the same vertical column, thus

[c] [c] [c] [c]…. $b$ times

……

To $a$ rows

This being done, it may also be said that we have written down $b$ columns each containing $a$ in brackets

As we got together $a\times b$ brackets and as each bracket contains $c$ units, the total number of units $= ab \times c$

Again, since we got $b$ brackets in a row each containing $c$ units the number of units in a row $= bc$ and as there are $a$ rows altogether, therefore the total number of units $= a \times bc$

Again, since we have got $a$ brackets in a column each containing $c$ units, the number of units in a column $= ac$ and as there are $b$ columns altogether.

$\therefore$ the total number of units $= b \times ac$

Hence from the above explanation it is clear that,

$ab \times c = a \times bc = b \times ac$

Note:

Products of monomial expressions can be always found by the method illustrated in the last article. It is necessary, however, when dealing with more complicated cases of multiplication that such products should be found mentally. Hence the student must get thoroughly accustomed to this kind of mental work.

Algebraic Division

One quantity $a$ is said to be divided by another quantity $b$ , when a third quantity $c$ is found such that $c \times b =a$ . In other words $\dfrac{a}{b} = c$

Thus when, $x=y \times z$ , we have $\dfrac{x}{y} = z$

When one quantity is divided by another, the former is called the dividend and the latter is called the divisor, the result is called the quotient.

Fundamental Propositions of Divisions

1. To prove that $\dfrac{a}{b} \times b = a$ :

If we denote $\dfrac{a}{b} \: by \: x$ we must have, by definition,

$x \times b=a$

Hence, $\dfrac{a}{b} \times b = x \times b =a$

Note:

To divide any quantity successively by two others is the same as to divide it at once by their product.

2. To prove that $\dfrac{a}{b}= a \times \dfrac{1}{b}$ :

We have, $\dfrac{1}{b} \times b = 1$

Hence $a \times \dfrac{1}{b} \times b=a$

$\therefore$ by definition $\dfrac{a}{b}= a \times \dfrac{1}{b}$

Thus, to divide any quantity by another is the same as to multiply the former by the reciprocal of the latter

Note:

When the dividend and divisor have the same sign, the quotient is positive, and when they have different signs, the quotient is negative. In other words, like signs produce + and unlike signs produce -.

Law of Signs for Multiplications and Divisions

If ‘a’ and ‘b’ are two whole numbers, we have

$a \times b = ab$

$a \times (-b) = -ab$

$(-a) \times (-b) = ab$

$(-a) \times b= -ab$

Thus the product of two whole numbers is positive or negative according as the multiplicand and the multiplier have like or unlike signs

The same thing can be found when the numbers are fractional. For instance $-\dfrac{2}{3} = -\dfrac{1}{3} -\dfrac{1}{3}$ , since is it obtained by subtracting a third part of unity twice to multiply any number $x \: by \: -\dfrac{2}{3}$ we must subtract a third part of x twice.

Hence, $(-\dfrac{2}{3}) \times x = (-\dfrac{x}{3}) -(\dfrac{x}{3})= -\dfrac{2x}{3}$

Similarly, $(-\dfrac{2}{3}) \times \dfrac{4}{5} = -\dfrac{4}{15}-\dfrac{4}{15}=-\dfrac{8}{15}$

$(-\dfrac{2}{3}) \times (-\dfrac{4}{5}) = \dfrac{8}{15}$

Hence we can enunciate the law of signs in a more general way, thus: The sign of the product of any two quantities is a positive or negative according as the multiplicand and the multiplier have like or unlike signs. More briefly, like signs produce +, and unlike signs produce -.

Note:

$\because$ $(-x) \times (-x) = x^2$ and $x \times x = x^2$ , we have $\sqrt{x^2} = \pm x$

Thus, every algebraic quantity has got two square roots which are equal in absolute value but opposite in sign.

Examples

1. Divide $18m^3n^2p$ by $-6m^2n^2p$

The dividend $= 18m^3n^2p$

The divisor $= -6m^2n^2p$

Quotient $= -3m$

2. Divide $-24a^7b^3c$ by $-6a^4bc$

The dividend $=-24a^7b^3c$

The divisor $= -6a^4bc$

The quotient $= 4a^5b^2$

3. Show that $(-ab)^2 = a^2b^2$

$(-ab)^2 = (-ab) \times (-ab)$

$= ab \times ab$

$=a \times a \times b \times b$

$= a^2b^2$

4. Multiply $x^4-3x^3+5x^2-6x+4$ by $-6x^2$

$(-6x^2) \times (x^4-3x^3+5x^2-6x+4)$

$= -6x^6+a8x^5-30x^4+36x^4-24x^2$

Exercise

Show that $-a \times 6b = -6ab$
Write the product of $-8x^6y^2z^5$ and $-20x^9z^2y^5$
Divide $86a^8b^5$ by $43a^7b^4$
Find the value of: $\dfrac{12a^64b^4 \times 2}{24ab}$
Find the value of: $\dfrac{24 \times 12 \times 36}{2 \times 4 \times 12}$
I bought $x$ books, $y$ pens and $z$ pencils. The price of each book was the same as the number of books I bought. The price of each pen was the same as the number of pens I bought. The price of each pencil was the same as the number of pencils I bought. How much money did I spend in all?
There are $p$ boys and $q$ girls. Each boy contributes as many rupees as the number of girls and each girl contributes as many rupees as the number of boys. This entire money is divided equally among two men. How much money does each man get?