6.1 Some Properties of Multiplication

6.1 Some Properties of Multiplication

In earlier classes we have explored how letter symbols can capture general statements, patterns and relations in a compact form. Algebra lets us justify and prove conjectures — not just verify them with one or two examples. In this chapter we examine the distributive property^? of multiplication over addition, use it to discover special algebraic identities^?, and see how these identities lead to quick shortcuts in computation.

Increments in Products

Consider the multiplication \(23 \times 27\).

1. By how much does the product increase if the first number (23) is increased by 1?
2. What if the second number (27) is increased by 1?
3. How much does the product increase if both are increased by 1?

Let us first consider a simpler problem — find the increase in the product when 27 is increased by 1. From the definition of multiplication (and the commutative property), it is clear that the product increases by 23. This is, 23 times the second number. In general, if \(a, b, c\) are three numbers, then:

This property can be visualised using a rectangular grid of dots: arrange a rows of \((b+c)\) dots each. The left block has \(a \times b\) dots, the right block has \(a \times c\) dots. Together \(a(b+c) = ab + ac\).

Using the identity with \(a=23,\, b=27, c=1\), we get \(23 \times (27+1) = 23 \times 27 + 23 \times 1\). So increasing the second number by 1 raises the product by 23.

Remember that \(a(b+c)\) and \(23(27+1)\) mean \(a \times (b+c)\) and \(23 \times (27+1)\) respectively. We usually drop the '\(\times\)' sign between brackets, just as in the case of expressions like \(5a, 4y\) etc. We can also similarly expand \(a + b \times c\) using the distributive property as follows:

\((a+b) \times c = (a+b) \cdot c = ac + bc\)   (commutativity of multiplication)
\(\quad = ca + cb\)   (commutativity of addition)

We can use the distributive property to find, in general, how much a product increases if one or both the numbers in the product are increased. Suppose the initial two numbers are \(a\) and \(b\). If one of the numbers, say \(b\), is increased by 1, then we have:

With \(a=23, b=27\): \((23+1)(27+1) = 23 \times 27 + 27 + 23 + 1\). Thus the product increases by \(a+b+1\) when both \(a\) and \(b\) are increased by 1.

When numbers in a product are increased by 1

What happens when one of the numbers in a product is increased by 1 and the other is decreased by 1? Will this property also be true if the number \((b-1)\) is used? Let us again use the distributive property, treating \((a+1)\) as a single term:

\((a+1)(b-1) = (a+1)\cdot b - (a+1)\cdot 1 = ab + b - a - 1\).

With \(a=23, b=27\): \(24 \times 26 = 23 \times 27 + 27 - 23 - 1 = 23 \times 27 + 3\). So the product increases by 3.

What happens when a product contains negative integers?

Check by substituting different values for \(a\) and \(b\) in each of the above cases. For example, \(a = -5, b = -8, c = -4, x = -5\) etc. We have seen that integers also satisfy the distributive property. That is, if \(x, y\) and \(z\) are any integers, then \((y + z) \times x = yx + zx\).

Thus, the expressions we have for increase of products hold when the letter-numbers take on negative integer values as well. Recall that two algebraic expressions are equal if they take on the same values when their letter-numbers are replaced by numbers. These numbers could be any integers.

If \(a\) and \(b\) are the initial numbers being multiplied, they become \(a+m\) and \(b+n\):

\((a+m)(b+n) = (a+m)\cdot b + (a+m)\cdot n = ab + bm + an + mn\).

The increase in the product is \(bm + an + mn\). Notice that this is the sum of the product of each term of \((a+m)\) with each term of \((b+n)\):

Product with Subtractions

This identity can be used to find how products change when the numbers being multiplied are increased or decreased by some amount. Can you now see how this identity can be used when one or both numbers are decreased?

For example, let us consider the case when one number is increased by 1 and the other is decreased by 1. Let us write \(b-1\) as \(b + (-1)\), taking \(n = -1\):

\((a+1)(b-1) = (a+1)(b + (-1)) = ab + 1\cdot b + a\cdot(-1) + 1\cdot(-1) = ab + b - a - 1\),

which is the same expression that we obtain earlier.

Verify the answers by finding the products without resorting to the subtractions in addition. Generalising this, we can find the product \((a+x)(b+y)\) as follows:

\((a-x)(b-y) = (a + (-x))(b + (-y)) = ab - bx - ay + xy\).

Check that this is the same as taking \(m=-x\) and \(n=-y\) in Identity 1. As in Identity 1, the product \((a+x)(b-y)\) is the sum of the product of each term of \((a+x)\) with each term of \(b\) and \((-y)\). Notice that the signs of the terms in the products can be determined using the rules of integer multiplication.

Example 1: Expand \(\frac{3}{2}(a - b + \frac{c}{3})\)

We get \((\frac{3}{2})a - (\frac{3}{2})b + (\frac{3}{2}) \cdot (\frac{c}{3}) = \frac{3}{2}a - \frac{3}{2}b + \frac{c}{2}\). The distributive property is not restricted to two terms within a bracket.

Example 2: Expand \((a+b)(a+b)\)

We have \((a+b)(a+b) = (a+b)\cdot a + (a+b)\cdot b = a^2 + ba + ab + b^2\). Since \(ba = ab\), we have two new terms having the same letter-numbers \(ab\) and they can be added. So we get \(ba + ab = 2ab\), and \((a+b)^2 = a^2 + 2ab + b^2\).

Can any two terms be added as a single term? For example, \(2a^2\) and \(3a^2\) have exactly the same letter-numbers — they can be simplified into a single term. A further simplification of the expression is not possible. Recall that we call terms having the same letter-numbers like terms.

Example 3: Expand \((a+b)^3\)

\((a+b)^3 = (a+b)(a+b)(a+b) = (a+b)(a^2 + 2ab + b^2) = a(a^2+2ab+b^2) + b(a^2+2ab+b^2) = a^3 + 2a^2 b + ab^2 + a^2 b + 2ab^2 + b^3 = a^3 + 3a^2 b + 3ab^2 + b^3\).

Square of the Sum / Difference of Two Numbers — Geometric View

The area of a square of side-length 60 units is 3600 sq. units (60²) and that of a square of side-length 5 units is 25 sq. units (5²). Can we use this to find the area of a square of side-length 65 units?

A square of side-length 65 can be split into 4 regions as shown in the figure — a square of side-length 60, a square of side-length 5, and two rectangles of side-lengths 60 and 5. The area of the square of side-length 65 is the sum of the areas of all its constituent parts. Can you find the area of the square of side-length 65 from the areas of the parts?

\(65^2 = 60^2 + 5^2 + 2 \times (60 \times 5) = 3600 + 25 + 600 = 4225\) sq. units.

Let us look at the general expression for the square of the sum of two numbers. Using the distributive property, \((a+b)^2\) can be expanded to \(a^2 + 2ab + b^2\) — giving us Identity 1A.

Use Identity 1A for \((m + 3)^2\) and \((6 + p)^2\)

\((m+3)^2 = m^2 + 6m + 9\). \((6+p)^2 = 36 + 12p + p^2\).

Expand \((6x + 5)^2\) — two methods

Expand \((3y + 2x)^2\) using the identity and by applying the distributive property

Using the identity: \((3y+2x)^2 = (3y)^2 + 2\cdot(3y)(2x) + (2x)^2 = 9y^2 + 12xy + 4x^2\).

Can we use \(60^2\) and \(5^2\) to find the value of \(55^2 = (60-5)^2\)?

Let us approach this through geometry by drawing a square of side-length 55 sitting inside a square of side-length 60. Area of a square of side-length 55 = 60² − (area of the two rectangles of side-lengths 60 and 5, each minus the corner square of side 5). We can subtract back the area of the small square of side 5 to this expression. In that way, we are only subtracting this area once.

So, \((60-5)^2 = 60^2 - (60\times 5) - (5 \times 60) + 5^2 = 3600 - 300 - 300 + 25 = 3025\).

Area of the square of side-length 55 is 3025 sq. units. We have seen what \(a + b\) squared gives when expanded. What do we get for \((a-b)^2\)? Using the distributive property:

\((a-b)^2 = (a-b)(a-b) = a(a-b) - b(a-b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2\).

Find the general expression of \((a-b)^2\) as we did for 55²

Use the identity \((a-b)^2 = a^2 - 2ab + b^2\) to find the values of (a) \(99^2\) and (b) \(58^2\).
(a) \(99^2 = (100-1)^2 = 10000 - 200 + 1 = 9801\).
(b) \(58^2 = (60-2)^2 = 3600 - 240 + 4 = 3364\).

Expand the following using Identity 1B and by applying the distributive property

(i) \((b-6)^2 = b^2 - 12b + 36\)
(ii) \((-2a + 3)^2 = (3-2a)^2 = 9 - 12a + 4a^2\)
(iii) \((7y - \frac{2}{3})^2 = 49y^2 - \frac{28y}{3} + \frac{4}{9}\).

Product of Sum and Difference — Identity 1C

Take a pair of natural numbers. Calculate the sum of their squares. Can you write twice this sum as a sum of two squares? Let us try this with other pairs of numbers. Have you figured out a pattern? Notice that \(2(5^2 + 6^2) = (6+5)^2 + (6-5)^2 = 121 + 1 = 122\). Do the identities below help in explaining the observed pattern?

\((a+b)^2 = a^2 + 2ab + b^2\) and \((a-b)^2 = a^2 - 2ab + b^2\). Adding the two: \((a+b)^2 + (a-b)^2 = 2a^2 + 2b^2\), so \(2(a^2+b^2) = (a+b)^2 + (a-b)^2\).

Pattern 2

Here is a related pattern. Try to describe the pattern using algebra.

\(9 \times 3 = 5 \times 10 - 23\)
\(8 \times 8 = 6 \times 6 + 14 \times 2\)
\(7 \times 7 = 2 \times 2 + 9 \times 5\)
\(10 \times 10 = 4 \times 4 + 14 \times 6\).

The pattern here appears to be \(a^2 - b^2 = (a+b)(a-b)\). This is a true identity! Using the distributive property, we get: \((a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2\).

Use Identity 1C to compute \(98 \times 102\) and \(45 \times 55\):
• \(98 \times 102 = (100-2)(100+2) = 10000 - 4 = 9996\).
• \(45 \times 55 = (50-5)(50+5) = 2500 - 25 = 2475\).

Show that \((a+b)(a-b) = a^2 - b^2\) geometrically (Try this). The rectangle of sides \(a+b\) and \(a-b\) can be rearranged into an L-shape equal in area to a large square of side \(a\) minus a small square of side \(b\).