Mathematics Grade 9 PDF

Numerical Value Arabic 1 2 3 5 10 20 21 100 Numeral Babylonian





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Numerical Value Arabic 1 2 3 5 10 20 21 100 Numeral Babylonian










< 0.  n b =  the negative nth root of b, if b < 0 and n is odd. 0, if b = 0.  i If b < 0 and n is even, there is no real nth root of b, because an even power of any real number is a non-negative number. ii The symbol n is called a radical sign, the expression n b is called a radical, n is called the index and b is called the radicand. When no index is written, the radical sign indicates square root. 26  Unit 1 The Number System Example 2 a 4 16 = 2 because 24 = 16 2 b 0.04 = 0.2 because (0. 2) = 0.04 − 1000 = − 10 because (– 10) = – 1000 3 3 c Numbers such as 23, 3 35 and 3 10 are irrational numbers and cannot be written as terminating or repeating decimals. However, it is possible to approximate irrational numbers as closely as desired using decimals. These rational approximations can be found through successive trials, using a scientific calculator. The method of successive trials uses the following property:  For any three positive real numbers a, b and c and a positive integer n if an < b < cn, then a < n b < c. Example 3 Find a rational approximation of 43 to the nearest hundredth. Solution: Use the above property and divide-and-average on a calculator. Since 62 = 36 < 43 < 49 = 72 6< 43 < 7 Estimate 43 to tenths, 43 ≈ 6.5 Divide 43 by 6.5 6.615 6.5 43.000 6.5 + 6.615 Average the divisor and the quotient = 6.558 2 Divide 43 by 6.558 6.557 6.558 43.000 Now you can check that (6.557)2 < 43 < (6.558)2. Therefore 43 is between 6.557 and 6.558. It is 6.56 to the nearest hundredth. 3 Example 4 Through successive trials on a calculator, compute 53 to the nearest tenth. 27 Mathematics Grade 9 Solution: 33 = 27 < 53 < 64 = 43. That is, 33 < 53 < 43. So 3 < 3 53 < 4 Try 3.5: 3.53 = 42.875 So 3.5 < 3 53 < 4 Try 3.7: 3.73 = 50.653 So 3.7 < 3 53 < 4 Try 3.8: 3.83 = 54.872 So 3.7 < 3 53 < 3.8 Try 3.75: 3.753 = 52.734375 So 3.75 < 3 53 < 3.8 3 Therefore, 53 is 3.8 to the nearest tenth. B Meaning of fractional exponents ACTIVITY 1.8 1 1 State another name for 2 4. 1 2 What meaning can you give to 2 or 20.5 ? 2 3 Show that there is at most one positive number whose fifth root is 2. 1 5 By considering a table of powers of 3 and using a calculator, you can define 35 as 3.   5  15  1   ×5 This choice would retain the property of exponents by which  3  = 3 5  = 3.   1 n n Similarly, you can define 5 , where n is a positive integer greater than 1, as 5. In 1 general, you can define b for any b ∈ ℝ and n a positive integer to be n n b whenever n b is a real number. Definition 1.10 The n th power If b ∈ R and n is a positive integer greater than 1, then 1 bn = n b Example 5 Write the following in exponential form: 1 a 7 b 3 10 Solution: 1 1 1 1 − a 7 = 72 b 3 = 1 = 10 3 10 3 10 28  Unit 1 The Number System Example 6 Simplify: 1 1 1 a 25 2 b ( −8) 3 c 64 6 Solution: 1 a 25 2 = 25 = 5 (Since 52 = 25) 1 b (−8) = 3 −8 = −2 (Since (−2)3 = −8) 3 1 64 6 = 6 64 = 2 (Since 2 = 64) 6 c Group Work 1.5 Simplify: 1 1 1 i a (8 × 27 ) 3 b 83 × 27 3 ii a 3 8 × 27 b 3 8 × 3 27 1 1 1 iii a ( 36 × 49 ) 2 b 36 × 49 2 2 iv a 36 × 49 b 36 × 49 What relationship do you observe between a and b in i, ii, iii and iv? 1 1 1 The observations from the above Group Work lead you to think that 5 3 × 33 = ( 5 × 3) 3. This particular case suggests the following general property (Theorem). Theorem 1.2 1 1 1 For any two real numbers a and b and for all integers n ≥ 2 , a b = ( ab ) n n n Example 7 Simplify each of the following. 1 1 a 9 ×33 3 b 5 16 × 5 2 Solution: 16 × 5 2 = 5 16 × 2 1 1 1 5 b a 9 × 3 = ( 9 × 3) (by Theorem 1.2) 3 3 3 1 = 5 32 = ( 27 ) 3 (multiplication) =2 =3 (33 = 27) 29 Mathematics Grade 9 ACTIVITY 1.9 Simplify: 1 1 64 5  64  5 i a 1 b   5  2  2 1 1 8 2  8 2 ii a 1 b   2 2 2 1 1 27 3  27  3 iii a 1 b   3  729  729 What relationship do you observe between a and b in i, ii and iii? The observations from the above Activity lead us to the following theorem: Theorem 1.3 For any two real numbers a and b where b ≠ 0 and for all integers n ≥ 2, 1 1 an  a n 1 =  n b b 1 6 16 3 128 Example 8 Simplify a 1 b 6 3 2 2 Solution: 1 1 =   (by Theorem 1.3) 3 16 16 3 a 1 3  2 2 1 = 8 = 2 (since 23 = 8) 3 6 128 6 128 b 6 = 2 2 = 6 64 = 2 (because 26 = 64) 30  Unit 1 The Number System ACTIVITY 1.10 1 Suggest, with reasons, a meaning for 7 9 1 a 22 b 2 2 in terms of 2 2 3 1. 2 Suggest a relation between 5 2 and 5 2 9 9 m n  1 mn Applying the property (a ) = a , you can write  710  as 710. In general, you can say   p  1q  p   q a = a , where p and q are positive integers and a ≥ 0. Thus, you have the following   definition: Definition 1.11 p  1 ( a) p p For a ≥ 0 and p and q any two positive integers, a =  a q  = q q     Exercise 1.5 1 1 1 1 Show that: a 64 = 4 b 256 = 2 3 c 125 = 5 8 3 2 Express each of the following without fractional exponents and without radical signs: 1 1 1 4 2 6  27  3 a 81 b 9 c 64 d    8  1 e ( 0.00032 ) 5 f 4 0.0016 g 6 729 3 Explain each step of the following: 1 1 1 ( 27 × 125 ) 3 = ( 3 × 3 × 3 ) × ( 5 × 5 × 5 )  3 = ( 3 × 5 ) × ( 3 × 5 ) × ( 3 × 5 )  3 = 3 × 5 = 15 4 In the same manner as in Question 3, simplify each of the following: 1 1 1 a ( 25 × 121) 2 b ( 625 × 16 ) 4 c (1024 × 243 ) 5 5 Express Theorem 1.2 using radical notation. 6 Show that: 1 1 5 × 3 = 5× 3 1 a 7 4 × 5 4 = ( 7 × 5) 4 b 1 1 7 × 3 9 = 3 7×9 1 117 × 6 7 = (11× 6 ) 7 3 c d 31 Mathematics Grade 9 7 Express in the simplest form: 1 1 1 1 1 1 32 × 2 9 ×3  1 6 a 6 6 b 3 3 c 128 ×   6 2 1 1 d 5 16 × 2 5 e 3 16 × 4 3 f 32 7 × 4 7 1 1 1 1 1 8 1 5 g 5 × 27 ×   × 9 5 8 5 h 3 5× 5 8× 3 × 4 5 5 8 Express Theorem 1.3 using radical notation. 9 Simplify: 1 1 1 1 4 128 5 9 3 16 4 32 a 1 b 1 c 1 d 1 5 3 81 4 162 4 4 243 3 5 6 3 16 64 512 625 e 3 f 5 g 6 h 3 2 2 8 5 p q 10 Rewrite each of the following in the form a : 9 11 5  15   15   16  a  13  b  12  c  11        p  1 11 Rewrite the following in the form  a q      7 6 5 2 5 3 6 3 a 3 b 5 c 64 d 729 12 Rewrite the expressions in Question 10 using radicals. 13 Rewrite the expressions in Question 11 using radicals. 14 Express the following without fractional exponents or radical sign: 5 5 1  13  a  27  b 27 3 c 83   15 Simplify each of the following: 1 3 3 a 64 6 b 812 c 6418 6 2 6 4 3 9 d 81 e 512 f 512 32  Unit 1 The Number System C Simplification of radicals ACTIVITY 1.11 1 Evaluate each of the following and discuss your result in groups. a 3 ( − 2)3 b ( − 3) 2 c 4 ( − 5) 4 d 5 45 e 22 f 7 ( − 1)7 2 Does the sign of your result depend on whether the index is odd or even? n Can you give a general rule for the result of an where a is a real number and i n is an odd integer? ii n is an even integer? To compute and simplify expressions involving radicals, it is often necessary to distinguish between roots with odd indices and those with even indices. For any real number a and a positive integer n, n a n = a , if n is odd. n an = a , if n is even. ( − 2) = − 2, ( − 2) = −2 =2 5 2 5 3 x3 = x, ( − 2) = − 2 = 2, 4 x2 = x , 4 4 x4 = x Example 9 Simplify each of the following: a y2 b 3 −27x 3 c 25x 4 d 6 x6 e 4 x3 Solution: a y2 = y b 3 −27 x 3 = 3 ( −3 x ) 3 = −3 x 1 3 25 x = 5 x = 5x x =x x = (x ) =x 4 2 2 6 6 4 3 3 4 c d e 4 n A radical a is in simplest form, if the radicand a contains no factor that can be expressed as an nth power. For example 3 54 is not in simplest form because 33 is a factor of 54. Using this fact and the radical notations of Theorem 1.2 and Theorem 1.3, you can simplify radicals. 33 Mathematics Grade 9 Example 10 Simplify each of the following: 32 a 48 b 3 9×3 3 c 4 81 Solution: a 48 = 16 × 3 = 16 × 3 = 4 3 b 3 9 × 3 3 = 3 9 × 3 = 3 27 = 3 32 4 16 × 2 4 16 4 4 16 2 c 4 = = × 2 = 4 ×4 2 = 4 2 81 81 81 81 3 Exercise 1.6 1 Simplify each of the following: a 8 b 5 32 c 3 8 x2 d 363 1 e 3 512 f 27 x3 y 2 g 4 405 3 2 Simplify each of the following if possible. State restrictions where necessary. 1 a 50 b 2 36 c 72 d 3 8x 2 e a3 3 180 f 0.27 g − 63 h i 3 16 j 3 −54 9 3 Identify the error and write the correct solution each of the following cases: a A student simplified 28 to 25 + 3 and then to 5 3 b A student simplified 72 to 4 18 and then to 4 3 c A student simplified 7x 9 and got x3 7 4 Simplify each of the following: a 8 250 b 3 16 × 3 5 c 4 5 × 4 125 3 2 81 12 96 d × 7 × 14 e 3 f 7 3 3 6 2 98 x3 y 2 g x > 0, y > 0. h 4 3 × 2 18 14 xy 34  Unit 1 The Number System 5 The number of units N produced by a company from the use of K units of capital and L units of labour is given by N = 12 LK. a What is the number of units produced, if there are 625 units of labour and 1024 units of capital? b Discuss the effect on the production, if the units of labour and capital are doubled. Addition and subtraction of radicals Which of the following do you think is correct? 1 2 + 8 = 10 2 19 − 3 = 4 3 5 2 + 7 2 = 12 2 The above problems involve addition and subtraction of radicals. You define below the concept of like radicals which is commonly used for this purpose. Definition 1.12 Radicals that have the same index and the same radicand are said to be like radicals. For example, 1 i 3 5, − 5 and 5 are like radicals. radicals. 2 ii 5 and 3 5 are not like radicals. iii 11 and 7 are not like radicals. By treating like radicals as like terms, you can add or subtract like radicals and express them as a single radical. On the other hand, the sum of unlike radicals cannot be expressed as a single radical unless they can be transformed into like radicals. Example 11 Simplify each of the following: 1 1 a 2+ 8 b 3 12 − 3 + 2 + 27 3 9 Solution: a 2 + 8 = 2 + 2× 4 = 2 + 4 2 = 2 + 2 2 = (1 + 2) 2 = 3 2 35 Mathematics Grade 9 1 1 1 3 1 b 3 12 − 3 + 2 + 27 = 3 4 × 3 − 3 + 2 × + 9 × 3 3 9 3 3 9 3 1 = 3 4× 3− 3+2 + 9× 3 9 9 2 1 = 6 3− 3+ 3+ 3 3 3  2 1 =  6 −1+ +  3 = 6 3  3 3 Exercise 1.7 Simplify each of the following if possible. State restrictions where necessary. 1 a 2× 5 b 3× 6 c 21 × 5 d 2 x × 8x 2 10 9 40 e f g 50 y 3 ÷ 2 y h 2 4 3 3 10 9 24 ÷15 75 i 4 3 16 ÷ 2 3 2 j 3 3 2 a 2 3 +5 3 b 9 2 −5 2 c 3 + 12 d 63 − 28 e 75 − 48 f 6 ( 12 − 3 ) 2 g 2 x 2 − 50 x 2 h 5 3 54 − 2 3 2 i 8 24 + 54 − 2 96 3 a + 2 ab + b j a+ b k (  1 a− b   a )+ 1   b 3 a Find the square of 7 − 2 10. b Simplify each of the following: 7 + 24 7 − 24 i 5+ 2 6 − 5−2 6 ii + 2 2 iii ( p2 + 1 − p2 −1 )( p2 −1 + p2 + 1 ) 4 Suppose the braking distance d for a given automobile when it is travelling v km/hr is approximated by d = 0.000213 v5 m. Approximate the braking distance when the car is travelling 64 km/hr. 36  Unit 1 The Number System 1.2.5 The Four Operations on Real Numbers The following activity is designed to help you revise the four operations on the set of rational numbers which you have done in your previous grades. ACTIVITY 1.12 1 Apply the properties of the four operations in the set of rational numbers to compute the following (mentally, if possible). 2 3 7 3  −11   −3   −11  a + +  b ×  +    9 5 9 7  21   7   21  3  5 −3   −9 23   −7  c + +  d  × ×  7 6 7   7 −27   9  2 State a property that justifies each of the following statements. −2  3 3   − 2 3  3 −7  3 −4  −7  −4 3  a  ×  =  × × b  + =  +  3  2 5  3 2 5 9 2 5  9  5 2  − 3   − 5   − 1   −5  −3 −1 c   +   <   +   , since 5 < 5  5   6   5   6  In this section, you will discuss operations on the set of real numbers. The properties you have studied so far will help you to investigate many other properties of the set of real numbers. Group Work 1.6 Work with a partner Required:- scientific calculator 1 Try this Copy and complete the following table. Then use a calculator to find each product and complete the table. Factors product product written as a power 3 2 2 × 2 101 × 101  −1   −1  3  ×   5   5  37 Mathematics Grade 9 2 Try this Copy the following table. Use a calculator to find each quotient and complete the table. Division Quotient Quotient written as a power 5 1 10 ÷ 10 35 ÷ 32 4 2 1 1   ÷    2 2 Discuss the two tables: i a Compare the exponents of the factors to the exponents in the product. What do you observe? b Write a rule for determining the exponent of the product when you multiply powers. Check your rule by multiplying 32 × 33 using a calculator. ii a Compare the exponents of the division expressions to the exponents in the quotients. What pattern do you observe? b Write a rule for determining the exponent in the quotient when you divide powers. Check your rule by dividing 75 by 73 on a calculator. 3 Indicate whether each statement is false or true. If false, explain: a Between any two rational numbers, there is always a rational number. b The set of real numbers is the union of the set of rational numbers and the set of irrational numbers. c The set of rational numbers is closed under addition, subtraction, multiplication and division excluding division by zero. d The set of irrational numbers is closed under addition, subtraction, multiplication and division. 4 Give examples to show each of the following: a The product of two irrational numbers may be rational or irrational. b The sum of two irrational numbers may be rational or irrational. c The difference of two irrational numbers may be rational or irrational. d The quotient of two irrational numbers may be rational or irrational. 5 Demonstrate with an example that the sum of an irrational number and a rational number is irrational. 6 Demonstrate with an example that the product of an irrational number and a non- zero rational number is irrational. 38  Unit 1 The Number System 7 Complete the following chart using the words ‘yes’ or ‘no’. Rational Irrational Real Number number number number 2 3 2 − 3 3 2 1.23 1.20220222… 2 − ×1.23 3 75 + 1.23 75 − 3 1.20220222… + 0.13113111… Questions 3, 4, 5 and in particular Question 7 of the above Group Work lead you to conclude that the set of real numbers is closed under addition, subtraction, multiplication and division, excluding division by zero. You recall that the set of rational numbers satisfy the commutative, associative and distributive laws for addition and multiplication. If you add, subtract, multiply or divide (except by 0) two rational numbers, you get a rational number, that is, the set of rational numbers is closed with respect to addition, subtraction, multiplication and division. From Group work 1.6 you may have realized that the set of irrational numbers is not closed under all the four operations, namely addition, subtraction, multiplication and division. Do the following activity and discuss your results. ACTIVITY 1.13 1 Find a + b, if a a = 3 + 2 and b = 3 − 2 b a = 3 + 3 and b = 2 + 3 39 Mathematics Grade 9 2 Find a – b, if a a= 3 and b = 3 b a= 5 and b = 2 3 Find ab, if a a= 3 − 1 and b = 3 +1 b a = 2 3 and b = 3 2 4 Find a ÷ b, if a a = 5 2 and b = 3 2 b a = 6 6 and b = 2 3 Let us see some examples of the four operations on real numbers. Example 1 Add a = 2 3 + 3 2 and 2 −3 3 Solution (2 ) ( 3 +3 2 + ) 2 −3 3 = 2 3 + 3 2 + 2 − 3 3 = 3 ( 2 − 3) + 2 ( 3 + 1) = − 3+4 2 Example 2 Subtract 3 2 + 5 from 3 5 − 2 2 Solution: (3 ) ( ) 5 −2 2 − 3 2 + 5 = 3 5 −2 2 −3 2 − 5 = 5 (3 − 1) + 2 ( − 2 − 3) = 2 5 −5 2 Example 3 Multiply a 2 3 by 3 2 b 2 5 by 3 5 Solution: ( 5) 2 a 2 3×3 2 =6 6 b 2 5 × 3 5 = 2 × 3× = 30 Example 4 Divide a 8 6 by 2 3 b 12 6 by ( 2× 3 ) Solution: 8 6 8 6 a 8 6 ÷2 3 = = × =4 2 2 3 2 3 b 12 6 ÷ ( ) 2× 3 = 12 6 2× 3 = 12 6 6 =12 40  Unit 1 The Number System Rules of exponents hold for real numbers. That is, if a and b are nonzero numbers and m and n are real numbers, then whenever the powers are defined, you have the following laws of exponents. 1 a m × a n = a m+n 2 (a )m n = a mn 3 am an = a m −n n an  a  a n × b n = ( ab ) =   , b ≠ 0. n 4 5 bn  b  ACTIVITY 1.14 1 Find the additive inverse of each of the following real numbers: 1 a 5 b − c 2 +1 2 d 2.45 e 2.1010010001... 2 Find the multiplicative inverse of each of the following real numbers: 1 a 3 b 5 c 1− 3 d 26 2 e 1.71 f g 1.3 3 3 Explain each of the following steps: ( 6 − 2 15 × ) 3 3 + 20 = 3 3 × ( 6 − 2 15 + 20 )  3 3  =  × 6− × 2 15  + 20  3 3   18 2 45  =  −  + 20  3 3   9× 2 2 9× 5  =  −  + 20  3 3   3× 2 2 × 3× 5  =  −  + 20  3 3  = ( 2 − 2 5 + 20 ) ( = 2 +  −2 5 + 2 5    ) = 2 41 Mathematics Grade 9 Let us now examine the basic properties that govern addition and multiplication of real numbers. You can list these basic properties as follows:
Closure property: The set R of real numbers is closed under addition and multiplication. This means that the sum and product of two real numbers is a real number; that is, for all a, b ∈ R, a + b ∈ R and ab ∈ R
Addition and multiplication are commutative in ℝ: That is, for all a, b ∈ R, i a+b=b+a ii ab = ba
Addition and multiplication are associative in ℝ: That is, for all, a, b, c ∈ R, i (a + b) + c = a + ( b + c) ii (ab)c = a(bc)
Existence of additive and multiplicative identities: There are real numbers 0 and 1 such that: i a + 0 = 0 + a = a, for all a ∈ R. ii a. 1 = 1. a = a, for all a ∈ R.
Existence of additive and multiplicative inverses: i For each a ∈ R there exists –a ∈ R such that a + (–a) = 0 = (–a) + a, and –a is called the additive inverse of a. ∈ℝ such that a ×   = 1 =   × a, 1 1 1 ii For each non-zero a ∈ R, there exists a a a 1 and is called the multiplicative inverse or reciprocal of a. a
Distributive property: Multiplication is distributive over addition; that is, if a, b, c, ∈ R then: i a ( b + c) = ab + ac ii (b + c) a = ba + ca 42  Unit 1 The Number System Exercise 1.8 1 Find the numerical value of each of the following: 3  1 (4 ) −1 4 × 2 ×   × ( 8−2 ) × ( 642 ) 5 3 a 5 b 176 − 2 275 + 1584 − 891  16  c 15 1.04 − 3 5 5 9 5 +6 1 18 − 5 0.02 − 300 ( ) d 4 0.0001 − 5 0.00032 e 2 3 0.125 + 4 0.0016 2 Simplify each of the following 3 5 1 2 3  12  74 a ( 216 ) 3 b 2 ×23 5 c 3  d 1   49 4 1 1 1 5 32 e 34 × 258 f 16 4 ÷ 2 g 4 3 7 h 5 243 3 What should be added to each of the following numbers to make it a rational number? (There are many possible answers. In each case, give two answers.) a 5− 3 b −2 − 5 c 4.383383338… d 6.123456… e 10.3030003… 1.2.6 Limits of Accuracy In this subsection, you shall discuss certain concepts such as approximation, accuracy in measurements, significant figures (s.f), decimal places (d.p) and rounding off numbers. In addition to this, you shall discuss how to give appropriate upper and lower bounds for data to a specified accuracy (for example measured lengths). ACTIVITY 1.15 1 Round off the number 28617 to the nearest a 10,000 b 1000 c 100 2 Write the number i 7.864 ii 6. 437 iii 4.56556555… a to one decimal place b to two decimal places 3 Write the number 43.25 to a two significant figures b three significant figures 4 The weight of an object is 5.4 kg. Give the lower and upper bounds within which the weight of the object can lie. 43 Mathematics Grade 9 1 Counting and measuring Counting and measuring are an integral part of our daily life. Most of us do so for various reasons and at various occasions. For example you can count the money you receive from someone, a tailor measures the length of the shirt he/she makes for us, and a carpenter counts the number of screws required to make a desk. Counting: The process of counting involves finding out the exact number of things. For example, you do counting to find out the number of students in a class. The answer is an exact number and is either correct or, if you have made a mistake, incorrect. On many occasions, just an estimate is sufficient and the exact number is not required or important. Measuring: If you are finding the length of a football field, the weight of a person or the time it takes to walk down to school, you are measuring. The answers are not exact numbers because there could be errors in measurements. 2 Estimation In many instances, exact numbers are not necessary or even desirable. In those conditions, approximations are given. The approximations can take several forms. Here you shall deal with the common types of approximations. A Rounding If 38,518 people attend a football game this figure can be reported to various levels of accuracy. To the nearest 10,000 this figure would be rounded up to 40,000. To the nearest 1000 this figure would be rounded up to 39,000. To the nearest 100 this figure would be rounded down to 38,500 In this type of situation, it is unlikely that the exact number would be reported. B Decimal places A number can also be approximated to a given number of decimal places (d.p). This refers to the number of figures written after a decimal point. Example 1 a Write 7.864 to 1 d.p. b Write 5.574 to 2 d.p. Solution: a The answer needs to be written with one number after the decimal point. However, to do this, the second number after the decimal point also needs to be considered. If it is 5 or more, then the first number is rounded up. That is 7.864 is written as 7.9 to 1 d.p 44  Unit 1 The Number System b The answer here is to be given with two numbers after the decimal point. In this case, the third number after the decimal point needs to be considered. As the third number after the decimal point is less than 5, the second number is not rounded up. That is 5.574 is written as 5.57 to 2 d.p. Note that to approximate a number to 1 d.p means to approximate the number to the nearest tenth. Similarly approximating a number to 2 decimal places means to approximate to the nearest hundredth. C Significant figures Numbers can also be approximated to a given number of significant figures (s.f). In the number 43.25 the 4 is the most significant figure as it has a value of 40. In contrast, the 5 is the least significant as it only has a value of 5 hundredths. When we desire to use significant figures to indicate the accuracy of approximation, we count the number of digits in the number from left to right, beginning at the first non-zero digit. This is known as the number of significant figures. Example 2 a Write 43.25 to 3 s.f. b Write 0.0043 to 1 s.f. Solution: a We want to write only the three most significant digits. However, the fourth digit needs to be considered to see whether the third digit is to be rounded up or not. That is, 43.25 is written as 43.3 to 3 s.f. b Notice that in this case 4 and 3 are the only significant digits. The number 4 is the most significant digit and is therefore the only one of the two to be written in the answer. That is 0.0043 is written as 0.004 to 1 s.f. 3 Accuracy In the previous lesson, you have studied that numbers can be approximated: a by rounding up b by writing to a given number of decimal place and c by expressing to a given number of significant figure. In this lesson, you will learn how to give appropriate upper and lower bounds for data to a specified accuracy (for example, numbers rounded off or numbers expressed to a given number of significant figures). 45 Mathematics Grade 9 Numbers can be written to different degrees of accuracy. For example, although 2.5, 2.50 and 2.500 may appear to represent the same number, they actually do not. This is because they are written to different degrees of accuracy. 2.5 is rounded to one decimal place (or to the nearest tenths) and therefore any number from 2.45 up to but not including 2.55 would be rounded to 2.5. On the number line this would be represented as 2.4 2.45 2.5 2.55 2.6 As an inequality, it would be expressed as 2.45 ≤ 2.5 < 2.55 2.45 is known as the lower bound of 2.5, while 2.55 is known as the upper bound. 2.50 on the other hand is written to two decimal places and therefore only numbers from 2.495 up to but not including 2.505 would be rounded to 2.50. This, therefore, represents a much smaller range of numbers than that being rounded to 2.5. Similarly, the range of numbers being rounded to 2.500 would be even smaller. Example 3 A girl’s height is given as 162 cm to the nearest centimetre. i Work out the lower and upper bounds within which her height can lie. ii Represent this range of numbers on a number line. iii If the girl’s height is h cm, express this range as an inequality. Solution: i 162 cm is rounded to the nearest centimetre and therefore any measurement of cm from 161.5 cm up to and not including 162.5 cm would be rounded to 162 cm. Thus, lower bound = 161.5 cm upper bound = 162.5 cm ii Range of numbers on the number line is represented as 161 161.5 162 162.5 163 iii When the girl’s height h cm is expressed as an inequality, it is given by 161.5 ≤ h < 162.5. 46  Unit 1 The Number System Effect of approximated numbers on calculations When approximated numbers are added, subtracted and multiplied, their sums, differences and products give a range of possible answers. Example 4 The length and width of a rectangle are 6.7 cm and 4.4 cm, respectively. Find their sum. Solution: If the length l = 6.7 cm and the width w = 4.4 cm Then 6.65 ≤ l < 6.75 and 4.35 ≤ w < 4.45 The lower bound of the sum is obtained by adding the two lower bounds. Therefore, the minimum sum is 6.65 + 4.35 that is 11.00. The upper bound of the sum is obtained by adding the two upper bounds. Therefore, the maximum sum is 6.75 + 4.45 that is 11.20. So, the sum lies between 11.00 cm and 11.20 cm. Example 5 Find the lower and upper bounds for the following product, given that each number is given to 1 decimal place. 3.4 × 7.6 Solution: If x = 3.4 and y = 7.6 then 3.35 ≤ x < 3.45 and 7.55 ≤ y < 7.65 The lower bound of the product is obtained by multiplying the two lower bounds. Therefore, the minimum product is 3.35 × 7.55 that is 25.2925 The upper bound of the product is obtained by multiplying the two upper bounds. Therefore, the maximum product is 3.45 × 7.65 that is 26.3925. So the product lies between 25.2925 and 26.3925. 54.5 Example 6 Calculate the upper and lower bounds to , given that each of the 36.0 numbers is accurate to 1 decimal place. Solution: 54.5 lies in the range 54.45 ≤ x < 54.55 36.0 lies in the range 35.95 ≤ x < 36.05 The lower bound of the calculation is obtained by dividing the lower bound of the numerator by the upper bound of the denominator. So, the minimum value is 54.45 ÷ 36.05. i.e., 1.51 (2 decimal places). The upper bound of the calculation is obtained by dividing the upper bound of the numerator by the lower bound of the denominator. So, the maximum value is 54.55 ÷ 35.95. i.e., 1.52 (2 decimal places). 47 Mathematics Grade 9 Exercise 1.9 1 Round the following numbers to the nearest 1000. a 6856 b 74245 c 89000 d 99500 2 Round the following numbers to the nearest 100. a 78540 b 950 c 14099 d 2984 3 Round the following numbers to the nearest 10. a 485 b 692 c 8847 d 4 e 83 4 i Give the following to 1 d.p. a 5.58 b 4.04 c 157.39 d 15.045 ii Round the following to the nearest tenth. a 157.39 b 12.049 c 0.98 d 2.95 iii Give the following to 2 d.p. a 6.473 b 9.587 c 0.014 d 99.996 iv Round the following to the nearest hundredth. a 16.476 b 3.0037 c 9.3048 d 12.049 5 Write each of the following to the number of significant figures indicated in brackets. a 48599 (1 s.f) b 48599 (3 s.f) c 2.5728 (3 s.f) d 2045 (2 s.f) e 0.08562 (1 s.f) f 0.954 (2 s.f) g 0.00305 (2 s.f) h 0.954 (1 s.f) 6 Each of the following numbers is expressed to the nearest whole number. i Give the upper and lower bounds of each. ii Using x as the number, express the range in which the number lies as an inequality. a 6 b 83 c 151 d 1000 7 Each of the following numbers is correct to one decimal place. i Give the upper and lower bounds of each. ii Using x as the number, express the range in which the number lies as an inequality. a 3.8 b 15.6 c 1.0 d 0.3 e –0.2 8 Each of the following numbers is correct to two significant figures. i Give the upper and lower bounds of each. ii Using x as the number, express the range in which the number lies as an inequality. a 4.2 b 0.84 c 420 d 5000 e 0.045 48  Unit 1 The Number System 9 Calculate the upper and lower bounds for the following calculations, if each of the numbers is given to 1 decimal place. 46.5 a 9.5 × 7.6 b 11.0 × 15.6 c 32.0 25.4 4.9 + 6.4 d e 8.2 2.6 10 The mass of a sack of vegetables is given as 5.4 kg. a Illustrate the lower and upper bounds of the mass on a number line. b Using M kg for the mass, express the range of values in which it must lie, as an inequality. 11 The masses to the nearest 0.5 kg of two parcels are 1.5 kg and 2.5 kg. Calculate the lower and upper bounds of their combined mass. 12 Calculate upper and lower bounds for the perimeter of a school football field shown, if its dimensions are correct to 1 decimal place. 109.7 m 48.8 m Figure 1.9 13 Calculate upper and lower bounds for the length marked x cm in the rectangle shown. The area and length are both given to 1 decimal place. x 8.4 cm Area = 223.2 cm2 Figure 1.10 1.2.7 Scientific Notation (Standard form) In science and technology, it is usual to see very large and very small numbers. For instance: The area of the African continent is about 30,000,000 km2. The diameter of a human cell is about 0.0000002 m. Very large numbers and very small numbers may sometimes be difficult to work with or write. Hence you often write very large or very small numbers in scientific notation, also called standard form. 49 Mathematics Grade 9 Example 1 1.86 × 10 - 6 is written in scientific notation. Number from 1 up to Times 10 to but not including 10. a power. 8.735 × 10 4 and 7.08 × 10-3 are written in scientific notation. 14.73 × 10-1 , 0.0863 × 104 and 3.864 are not written in standard form (scientific notation). ACTIVITY 1.16 1 By what powers of 10 must you multiply 1.3 to get: a 13? b 130? c 1300? Copy and complete this table. 13 = 1.3 × 101 130 = 1.3 × 102 1,300 = 1.3 × 13,000 = 1,300,000 = 2 Can you write numbers between 0 and 1 in scientific notation, for example 0.00013? Copy and complete the following table. 13.0 = 1.3 × 10 = 1.3 × 101 1.3 = 1.3 × 1 = 1.3 × 100 1 0.13 = 1.3 × = 1.3 × 10−1 10 1 0.013 = 1.3 × = 100 0.0013 = 0.00013 = 0.000013 = 0.0000013 = Note that if n is a positive integer, multiplying a number by 10n moves its decimal point n places to the right, and multiplying it by 10−n moves the decimal point n places to the left. 50  Unit 1 The Number System Definition 1.13 A number is said to be in scientific notation (or standard form), if it is written as a product of the form a × 10k where 1 ≤ a < 10 and k is an integer. Example 2 Express each of the following numbers in scientific notation: a 243, 900,000 b 0.000000595 Solution: a 243,900,000 = 2.439 × 108. The decimal point moves 8 places to the left. b 0.000000595 = 5.95 × 10−7. The decimal point moves 7 places to the right. Example 3 Express 2.483 × 105 in ordinary decimal notation. Solution: 2.483 × 105 = 2.483 × 100,000 = 248,300. Example 4 The diameter of a red blood cell is about 7.4 × 10–4 cm. Write this diameter in ordinary decimal notation. 1 1 Solution: 7.4 × 10–4 = 7.4 × 4 = 7.4 × = 7.4 × 0.0001 = 0.00074. 10 10, 000 So, the diameter of a red blood cell is about 0.00074 cm. Calculators and computers also use scientific notation to display large numbers and small numbers but sometimes only the exponent of 10 is shown. Calculators use a space before the exponent, while computers use the letter E.  The calculator display 5.23 06 means 5.23 × 106. (5,230,000). The following example shows how to enter a number with too many digits to fit on the display screen into a calculator. Example 5 Enter 0.00000000627 into a calculator. Solution: First, write the number in scientific notation. 0.00000000627 = 6.27 × 10−9 Then, enter the number. 6.27 exp 9 + / − giving 6.27 – 09 Decimal Scientific Calculator Computer notation notation display display 250,000 2.5 × 105 2.5 0.5 2.5 E + 5 0.00047 4.7 × 10−4 4.7 – 04 4.7 E – 4 51 Mathematics Grade 9 Exercise 1.10 1 Express each of the following numbers in scientific notation: a 0.00767 b 5,750,000,000 c 0.00083 d 400,400 e 0.054 2 Express each of the following numbers in ordinary decimal notation: a 4.882 × 105 b 1.19 × 10–5 c 2.021 × 102 3 Express the diameter of an electron which is about 0.0000000000004 cm in scientific notation. 1.2.8 Rationalization ACTIVITY 1.17 Find an approximate value, to two decimal places, for the following: 1 2 i ii 2 2 In calculating this, the first step is to find an approximation of 2 in a reference book 1 or other reference material. (It is 1.414214.) In the calculation of , 1 is divided by 2 2 1.414214 1.414214… which is a difficult task. However, evaluating as ≈ 0.707107 2 2 is easy. 1 2 Since is equivalent to (How?), you see that in order to evaluate an expression 2 2 with a radical in the denominator, first you should transform the expression into an equivalent expression with a rational number in the denominator. The technique of transferring the radical expression from the denominator to the numerator is called rationalizing the denominator (changing the denominator into a rational number). The number that can be used as a multiplier to rationalize the denominator is called the rationalizing factor. This is equivalent to 1. 52  Unit 1 The Number System 1 For instance, if n is an irrational number then can be rationalized by multiplying n n n it by = 1. So, is the rationalizing factor. n n Example 1 Rationalize the denominator in each of the following: 5 3 6 3 a b c 3 8 5 3 2 Solution: 5 a The rationalizing factor is. 5 5 3 5 3 5 5 15 5 15 5 15 15 So, = × = = = = 8 5 8 5 5 8 25 8 5 2 8× 5 8 3 b The rationalizing factor is 3 6 6 3 6 3 6 3 So, = × = = =2 3 3 3 3 32 3 3 22 c The rationalizing factor is because 3 2× 3 4 = 3 8 = 2 3 2 2 3 3 3 22 33 4 33 4 So, 3 = 3. = = 2 2 3 22 3 23 2  2 If a radicand itself is a fraction  for example  , then, it can be written in the  3  2 equivalent form so that the procedure described above can be applied to rationalize 3 the denominator. Therefore, 2 2 2 3 6 6 6 = = × = = = 3 3 3 3 9 3 2 3 In general, For any non-negative integers a, b (b ≠ 0) a a a b ab = = =. b b b b b 53 Mathematics Grade 9 Exercise 1.11 Simplify each of the following. State restrictions where necessary. In each case, state the rationalizing factor you use and express the final result with a rational denominator in its lowest term. 2 2 5 2 12 5 a b c d e 2 6 4 10 27 18 3 3 1 9 20 4 f g 3 h i j 3 2 3 4 a2 3 4 5 More on rationalizations of denominators ACTIVITY 1.18 Find the product of each of the following: 1 ( 2 + 3 )( 2 − 3 ) 2 (5 + 3 2 )(5 − 3 2 )  1  1  3  5 − 3  5 + 3  2  2  You might have observed that the results of all of the above products are rational numbers. This leads you to the following conclusion: Using the fact that (a – b) (a + b) = a2 – b2, you can rationalize the denominators of expressions such as 1 1 1 , , where a , b are irrational numbers as follows. a+ b a −b a − b 1 1 a− b  a− b a− b =   = = ( ) i a+ b  a − b  a2 − b ( ) a2 − b 2 a+ b 1 1  a +b a +b a +b ii =  = = a −b a − b  a + b  ( a) a − b2 2 − b2 1 1  a+ b a+ b a+ b =  = = ( ) iii a− b a − b  a + b  ( a) −( b) a−b 2 2 54  Unit 1 The Number System Example 2 Rationalize the denominator of each of the following: 5 3 a b 1− 2 6 +3 2 Solution: 1+ 2 a The rationalizing factor is 1+ 2 5 5(1 + 2) 5+5 2 So = = 1 − 2 (1 − 2)(1 + 2) 12 − 2 ( ) 2 5+5 2 = =−5−5 2 1−2 6 −3 2 b The rationalizing factor is 6 −3 2 3 = 3 6 − 3 2 = 3 ( 6 − 3 2 ) ( ) So 6 − 3 2 ( 6) (3 2 ) 2 2 6 +3 2 6 +3 2 − 3 ( 6 − 3 2 ) = 6 − 18 =− 1 4 ( 6 − 3 2 ) 3 2− 6 = 4 Exercise 1.12 Rationalize the denominator of each of the following: 1 18 2 a b c 3− 5 5 −3 5− 3 3+4 10 3 2+ 3 d e f 3−2 7− 2 3 2 − 2 3 1 g 2 + 3 −1 55 Mathematics Grade 9 1.2.9 Euclid's Division Algorithm A The division algorithm ACTIVITY 1.19 1 Is the set of non-negative integers (whole numbers) closed under division? 2 Consider any two non-negative integers a and b. a What does the statement "a is a multiple of b" mean? b Is it always possible to find a non-negative integer c such that a = bc? If a and b are any two non-negative integers, then a ÷ b (b ≠ 0) is some non-negative integer c (if it exists) such that a = bc. However, since the set of non-negative integers is not closed under division, it is clear that exact division is not possible for every pair of non-negative integers. For example, it is not possible to compute 17 ÷ 5 in the set of non-negative integers, as 17 ÷ 5 is not a non-negative integer. 15 = 3 × 5 and 20 = 4 × 5. Since there is no non-negative integer between 3 and 4, and since 17 lies between 15 and 20, you conclude that there is no non-negative integer c such that 17 = c × 5. You observe, however, that by adding 2 to each side of the equation 15 = 3 × 5 you can express it as 17 = (3 × 5) + 2. Furthermore, such an equation is useful. For instance it will provide a correct answer to a problem such as: If 5 girls have Birr 17 to share, how many Birr will each girl get? Examples of this sort lead to the following theorem called the Division Algorithm. Theorem 1.4 Division algorithm Let a and b be two non-negative integers and b ≠ 0, then there exist unique non-negative integers q and r, such that, a = (q × b) + r with 0 ≤ r < b. In the theorem, a is called the dividend, q is called the quotient, b is called the divisor, and r is called the remainder. Example 1 Write a in the form b × q + r where 0 ≤ r < b, a If a = 47 and b = 7 b If a = 111 and b = 3 c If a = 5 and b = 8 56  Unit 1 The Number System Solution: a 6 b 37 c 0 7 47 3 111 8 5 42 9 0 5 21 5 q = 6 and r = 5 21 q = 0 and r = 5 ∴ 47 = 7 (6) + 5 0 ∴ 5 = 8 (0) + 5. q = 37 and r = 0 ∴ 111 = 3 (37) + 0 Exercise 1.13 For each of the following pairs of numbers, let a be the first number of the pair and b the second number. Find q and r for each pair such that a = b × q + r, where 0 ≤ r < b: a 72, 11 b 16, 9 c 11, 18 d 106, 13 e 176, 21 f 25, 39 B The Euclidean algorithm ACTIVITY 1.20 Given two numbers 60 and 36 1 Find GCF (60, 36). 2 Divide 60 by 36 and find the GCF of 36 and the remainder. 3 Divide 36 by the remainder you got in Step 2. Then, find the GCF of the two remainders, that is, the remainder you got in Step 2 and the one you got in step 3. 4 Compare the three GCFs you got. 5 Generalize your results. The above Activity leads you to another method for finding the GCF of two numbers, which is called Euclidean algorithm. We state this algorithm as a theorem. Theorem 1.5 Euclidean algorithm If a, b, q and r are positive integers such that a = q × b + r, then, GCF (a, b) = GCF (b, r). 57 Mathematics Grade 9 Example 2 Find GCF (224, 84). Solution: To find GCF (224, 84), you first divide 224 by 84. The divisor and remainder of this division are then used as dividend and divisor, respectively, in a succeeding division. The process is repeated until a remainder 0 is obtained. The complete process to find GCF (224, 84) is shown below. Euclidean algorithm Computation Division algorithm Application of Euclidean form Algorithm 2 84 224 224 = (2 × 84) + 56 GCF (224, 84) = GCF (84, 56) 168 56 1 56 84 84 = (1 × 56) + 28 GCF (84, 56) = GCF (56, 28) 56 28 2 28 56 56 56 = (2 × 28) + 0 GCF (56, 28) = 28 (by inspection) 0 Conclusion GCF (224, 84) = 28. Exercise 1.14 1 For the above example, verify directly that GCF (224, 84) = GCF (84, 56) = GCF (56, 28). 2 Find the GCF of each of the following pairs of numbers by using the Euclidean Algorithm: a 18; 12 b 269; 88 c 143; 39 d 1295; 407 e 85; 68 f 7286; 1684 58  Unit 1 The Number System Key Terms bar notation principal nth root composite number principal square root divisible radical sign division algorithm radicand factor rational number fundamental theorem of arithmetic rationalization greatest common factor (GCF) real number irrational number repeating decimal least common multiple (LCM) repetend multiple scientific notation perfect square significant digits prime factorization significant figures prime number terminating decimal Summary Summary 1 The sets of Natural numbers, Whole numbers, Integers and Rational numbers denoted by ℕ, W, ℤ, and ℚ, respectively are described by ℕ = {1, 2, 3,…} W = {0, 1, 2,…} ℤ = {…,−3, −2, −1, 0, 1, 2, 3,…} a  ℚ =  : a ∈ ℤ, b ∈ ℤ, b ≠ 0  b  2 a A composite number is a natural number that has more than two factors. b A prime number is a natural number that has exactly two distinct factors, 1 and itself. c Prime numbers that differ by two are called twin primes. d When a natural number is expressed as a product of factors that are all prime, then the expression is called the prime factorization of the number. 59