Using the Index Laws

xn Multiplying terms Simplify: x + x + x + x + x = 5x x to the power of 5 Simplify: x × x × x × x × x = x5 x5 as been written using index notation. The number n is called the index or power. Start by asking pupils to simplify x + x + x + x + x. This is 5 lots of x, which is written as 5x. Next ask pupils how we could simplify x × x × x × x × x. Make sure there is no confusion between this repeated multiplication and the previous example of repeated addition. If x is equal to 2, for example, x + x + x + x + x equals 10, while x × x × x × x × x equals 32. Some pupils may suggest writing xxxxx. While this is not incorrect, neither has it moved us on very far. Point out the problems of readability, especially with high powers. When we write a number or term to the power of another number it is called index notation. The power, or index (plural indices), is the superscript number, in this case 5. The number or letter that we are multiplying successive times, in this case, x, is called the base. Practice the relevant vocabulary: x2 is read as ‘x squared’ or ‘x to the power of 2’; x3 is read as ‘x cubed’ or ‘x to the power of 3’; x4 is read as ‘x to the power of 4’. xn The number x is called the base.

Multiplying terms involving indices We can use index notation to simplify expressions. For example, 3p × 2p = 3 × p × 2 × p = 6p2 q2 × q3 = q × q × q × q × q = q5 Discuss each example briefly. Remind pupils to multiply any numbers together first followed by letters in alphabetical order. In the last example 2t × 2t the use of brackets may need further clarification. We must put a bracket around the 2t since both the 2 and the t are squared. If we wrote 2t2, then only the t would be squared. Give a numerical example, if necessary. If t was 3 then 2t would be equal to 6. We would then have 62, 36. If we wrote 2t2, that would mean 2 × 32 or 2 × 9 which is 18. Remember the order of operations - BIDMAS. Brackets are worked out before indices, but indices are worked out before multiplication. 3r × r2 = 3 × r × r × r = 3r3 2t × 2t = (2t)2 or 4t2

Multiplying terms with the same base When we multiply two terms with the same base the indices are added. For example, a4 × a2 = (a × a × a × a) × (a × a) = a × a × a × a × a × a = a6 = a (4 + 2) In general, Stress that the indices can only be added when the base is the same. xm × xn = x(m + n)

Dividing terms Remember, in algebra we do not usually use the division sign, ÷. Instead, we write the number or term we are dividing by underneath like a fraction. For example, Point out that we do not need to write the brackets when we write a + b all over c. Since both letters are above the dividing line we know that it is the sum of a and b that is divided by c. The dividing line effectively acts as a bracket. (a + b) ÷ c is written as a + b c

Dividing terms Like a fraction, we can often simplify expressions by cancelling. For example, n3 n2 6p2 3p n3 ÷ n2 = 6p2 ÷ 3p = 2 n × n × n n × n 6 × p × p 3 × p = = In the first example, we can divide both the numerator and the denominator by n. n ÷ n is 1. We can divide the numerator and the denominator by n again to leave n. (n/1 is n). If necessary, demonstrate this by substitution. For example, 3 cubed, 27, divided by 3 squared, 9, is 3. Similarly 5 cubed, 125, divided by 5 squared, 25, is 5. In the second example, we can divide the numerator and the denominator by 3 and then by p to get 2p. Again, demonstrate the truth of this expression by substitution, if necessary. For example, if p was 5 we would have 6 × 5 squared, 6 × 25, which is 150, divided by 3 × 5, 15. 150 divided by 15 is 10, which is 2 times 5. This power of algebra is such that this will work for any number we choose for p. Pupils usually find multiplying easier than dividing. Encourage pupils to check their answers by multiplying (using inverse operations). For example, n × n2 = n3. And 2p × 3p = 6p2 = n = 2p

Dividing terms with the same base When we divide two terms with the same base the indices are subtracted. For example, a × a × a × a × a a × a = a5 ÷ a2 = a × a × a = a3 = a (5 – 2) 2 4 × p × p × p × p × p × p 2 × p × p × p × p 4p6 ÷ 2p4 = = 2 × p × p = 2p2 = 2p(6 – 4) Stress that the indices can only be subtracted when the base is the same. In general, xm ÷ xn = x(m – n)

Expressions of the form (xm)n Sometimes terms can be raised to a power and the result raised to another power. For example, (y3)2 = y3 × y3 (pq2)4 = pq2 × pq2 × pq2 × pq2 = (y × y × y) × (y × y × y) = p4 × q (2 + 2 + 2 + 2) = y6 = p4 × q8 = p4q8

Expressions of the form (xm)n When a term is raised to a power and the result raised to another power, the powers are multiplied. For example, (a5)3 = a5 × a5 × a5 = a(5 + 5 + 5) = a15 = a(3 × 5) In general, (xm)n = xmn

Expressions of the form (xm)n Rewrite the following without brackets. 1) (2a2)3 = 8a6 2) (m3n)4 = m12n4 3) (t–4)2 = t–8 4) (3g5)3 = 27g15 5) (ab–2)–2 = a–2b4 6) (p2q–5)–1 = p–2q5 You may wish to ask pupils to complete this exercise individually before talking through the answers. The zero index is introduced in the last question and discussed on the next slide. 7) (h½)2 = h 8) (7a4b–3)0 = 1

Any number or term divided by itself is equal to 1. The zero index Any number or term divided by itself is equal to 1. Look at the following division: y4 ÷ y4 = 1 But using the rule that xm ÷ xn = x(m – n) y4 ÷ y4 = y(4 – 4) = y0 That means that y0 = 1 Stress that this rule is only true for non-zero values of x. 00 is undefined. In general, for all , x0 = 1

Negative indices Look at the following division: b × b b × b × b × b = 1 b × b = 1 b2 b2 ÷ b4 = But using the rule that xm ÷ xn = x(m – n) b2 ÷ b4 = b(2 – 4) = b–2 That means that 1 b2 Discuss the general form of each result where x is any number and m and n are integers. b–2 = In general, x–n = 1 xn

This is the reciprocal of u. Negative indices Write the following using fraction notation: This is the reciprocal of u. 1 u u–1 = 2 b4 2b–4 = x2 y3 x2y–3 = A number or term raised to the power of –1 is the reciprocal of the number or term. See N2.4 Reciprocals 2a (3 – b)2 2a(3 – b)–2 =

Negative indices Write the following using negative indices: 2 a = x3 y4 = x3y–4 p2 q + 2 = p2(q + 2)–1 Use the last two examples to explain that when we have terms linked by + and – in the denominator, or bracketed expressions, the whole of that expression must be raised to the negative power and not the individual terms. 3m (n2 + 2)3 = 3m(n2 + 2)–3