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Lecture Notes on Engineering Mathematics

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Ch01-H8555.tex 1/8/2007 18: 7 page 3 Chapter 1 Revision of fractions, decimals and percentages Alternatively: 1.1 Fractions Step (2) Step (3) 2 When 2 is divided by 3, it may be written as or 2/3 or 3 ↓↓ 2 2/3. is called a fraction. The number above the line, (7 × 1) + (3 × 2) 3 1 2 + = i.e. 2, is called the numerator and the number below 3 7 21 the line, i.e. 3, is called the denominator. ↑ When the value of the numerator is less than the Step (1) value of the denominator, the fraction is called a proper 2 fraction; thus is a proper fraction. When the value 3 Step 1: the LCM of the two denominators; of the numerator is greater than the denominator, the 1 7 Step 2: for the fraction , 3 into 21 goes 7 times, fraction is called an improper fraction. Thus is 3 3 7 × the numerator is 7 × 1; an improper fraction and can also be expressed as a 2 mixed number, that is, an integer and a proper frac- Step 3: for the fraction , 7 into 21 goes 3 times, 7 7 tion. Thus the improper fraction is equal to the mixed 3 × the numerator is 3 × 2. 3 1 number 2 . 3 1 2 7 + 6 13 When a fraction is simplified by dividing the numer- Thus + = = as obtained previously. 3 7 21 21 ator and denominator by the same number, the pro- cess is called cancelling. Cancelling by 0 is not 2 1 permissible. Problem 2. Find the value of 3 − 2 3 6 1 2 Problem 1. Simplify + One method is to split the mixed numbers into integers 3 7 and their fractional parts. Then     The lowest common multiple (i.e. LCM) of the two 2 1 2 1 3 − 2 = 3 + − 2 + denominators is 3 × 7, i.e. 21. 3 6 3 6 Expressing each fraction so that their denominators 2 1 are 21, gives: = 3 + − 2 − 3 6 1 2 1 7 2 3 7 6 4 1 3 1 + = × + × = + = 1 + − = 1 = 1 3 7 3 7 7 3 21 21 6 6 6 2 7 + 6 13 Another method is to express the mixed numbers as = = 21 21 improper fractions.Ch01-H8555.tex 1/8/2007 18: 7 page 4 4 Engineering Mathematics 1 8 9 2 9 2 11  8 7 24 8 × 1 × 8   Since 3 = , then 3 = + = = × × = 3 3 3 3 3 5 3 7 5 × 1 × 1 1 1   1 12 1 13 64 4 Similarly, 2 = + = = = 12 6 6 6 6 5 5 2 1 11 13 22 13 9 1 Thus 3 − 2 = − = − = = 1 3 6 3 6 6 6 6 2 3 12 Problem 6. Simplify ÷ as obtained previously. 7 21 Problem 3. Determine the value of 3 3 12 7 5 1 2 ÷ = 4 − 3 + 1 12 7 21 8 4 5 21 Multiplying both numerator and denominator by the   reciprocal of the denominator gives: 5 1 2 5 1 2 4 − 3 + 1 = (4 − 3 + 1) + − + 8 4 5 8 4 5 1 3  3 21 3   3 × 5 × 5 − 10 × 1 + 8 × 2  3 7 12 7 1  4 4  = 2 + = = = 1 1 40 12 12  21  1 4   × 25 − 10 + 16 21 21  12   1  1 = 2 + 40 This method can be remembered by the rule: invert the 31 31 = 2 + = 2 second fraction and change the operation from division 40 40 to multiplication. Thus: 1 3 3 12 3 21  3   3 14 ÷ = × = as obtained previously. Problem 4. Find the value of × 7 21 12  4 7  1 4 7 15 3 1 Dividing numerator and denominator by 3 gives: Problem 7. Find the value of 5 ÷ 7 5 3 1 3 14 1 14 1 × 14  × = × =  7 15 7 5 7 × 5  5 The mixed numbers must be expressed as improper fractions. Thus, Dividing numerator and denominator by 7 gives: 2 1 × 14  1 × 2 2  14  3 1 28 22 28 3 42  = = 5 ÷ 7 = ÷ = × = 7 × 5 1 × 5 5 1  5 3 5 3 5 22 55  11 This process of dividing both the numerator and denom- inator of a fraction by the same factor(s) is called Problem 8. Simplify cancelling.     1 2 1 3 1 − + ÷ × 3 1 3 3 5 4 8 3 Problem 5. Evaluate 1 × 2 × 3 5 3 7 The order of precedence of operations for problems Mixed numbers must be expressed as improper frac- containing fractions is the same as that for integers, tions before multiplication can be performed. Thus, i.e. remembered by BODMAS (Brackets, Of, Division, Multiplication, Addition and Subtraction). Thus, 3 1 3 1 × 2 × 3 5 3 7           5 3 6 1 21 3 1 2 1 3 1 = + × + × + − + ÷ × 5 5 3 3 7 7 3 5 4 8 3 Section 1Ch01-H8555.tex 1/8/2007 18: 7 page 5 Revision of fractions, decimals and percentages 5 1 1 4 × 2 + 5 × 1 3  = − ÷ (B) 2 3 2 1 2 3 20 24  2. (a) + (b) − +  8 7 11 9 7 3 2   1 13 8  43 47 = − × (D) (a) (b) 3 20  1  5 77 63 1 26 3 2 1 4 5 = − (M) 3. (a) 10 − 8 (b) 3 − 4 + 1 3 5 7 3 4 5 6 (5 × 1) − (3 × 26)   = (S) 16 17 15 (a) 1 (b) 21 60 −73 13 = =−4 3 5 17 15 15 15 4. (a) × (b) × 4 9 35 119   Problem 9. Determine the value of 5 3 (a) (b)   12 49 7 1 1 1 3 1 of 3 − 2 + 5 ÷ − 3 7 2 13 7 4 6 2 4 8 16 2 5. (a) × × 1 (b) × 4 × 3 5 9 7 17 11 39     7 1 1 1 3 1 3 of 3 − 2 + 5 ÷ − (a) (b) 11 6 2 4 8 16 2 5 7 1 41 3 1 3 45 1 5 = of 1 + ÷ − (B) 6. (a) ÷ (b) 1 ÷ 2 6 4 8 16 2 8 64 3 9   8 12 7 5 41 3 1 (a) (b) = × + ÷ − (O) 15 23 6 4 8 16 2   2  1 3 8 1 7 7 5 41 16 1  7. + ÷ − 1 = × + × − (D) 2 5 15 3 24 6 4 8 3 2 1       35 82 1 7 5 3 15 4 = + − (M) 8. of 15 × + ÷ 5 24 3 2 15 7 4 16 5   35 + 656 1 1 2 1 3 2 13 = − (A) 9. × − ÷ + − 24 2 4 3 3 5 7 126       691 1 2 1 2 1 3 28 = − (A) 10. × 1 ÷ + + 1 2 24 2 3 4 3 4 5 55 691 − 12 = (S) 11. If a storage tank is holding 450 litres when it is 24 three-quarters full, how much will it contain 679 7 when it is two-thirds full? = = 28 24 24 400 litres 12. Three people, P, Q and R contribute to a fund. P provides 3/5 of the total, Q provides 2/3 of Now try the following exercise the remainder, and R provides £8. Determine (a) the total of the fund, (b) the contributions Exercise 1 Further problems on fractions of P and Q. (a) £60 (b) £36, £16 Evaluate the following: 1 2 7 1 1.2 Ratio and proportion 1. (a) + (b) − 2 5 16 4   9 3 The ratio of one quantity to another is a fraction, and (a) (b) 10 16 is the number of times one quantity is contained in another quantity of the same kind. If one quantity is Section 1Ch01-H8555.tex 1/8/2007 18: 7 page 6 6 Engineering Mathematics directly proportional to another, then as one quantity 1 person takes three times as long, i.e. doubles, the other quantity also doubles. When a quan- 4 × 3 = 12 hours, tity is inversely proportional to another, then as one 5 people can do it in one fifth of the time that one quantity doubles, the other quantity is halved. 12 person takes, that is hours or 2 hours 24 minutes. 5 Problem 10. A piece of timber 273 cm long is cut into three pieces in the ratio of 3 to 7 to 11. Determine the lengths of the three pieces Now try the following exercise The total number of parts is 3 + 7 + 11, that is, 21. Exercise 2 Further problems on ratio and Hence 21 parts correspond to 273 cm proportion 273 1 part corresponds to = 13 cm 1. Divide 621 cm in the ratio of 3 to 7 to 13. 21 81 cm to 189 cm to 351 cm 3 parts correspond to 3 × 13 = 39 cm 2. When mixing a quantity of paints, dyes of 7 parts correspond to 7 × 13 = 91 cm four different colours are used in the ratio of 11 parts correspond to 11 × 13 = 143 cm 7:3:19:5. If the mass of the first dye used 1 is 3 g, determine the total mass of the dyes 2 i.e. the lengths of the three pieces are 39 cm, 91 cm used. 17 g and 143 cm. 3. Determine how much copper and how much (Check: 39 + 91 + 143 = 273) zinc is needed to make a 99 kg brass ingot if they have to be in the proportions copper : Problem 11. A gear wheel having 80 teeth is in zinc: :8:3by mass. 72 kg : 27 kg mesh with a 25 tooth gear. What is the gear ratio? 4. It takes 21 hours for 12 men to resurface a stretch of road. Find how many men it takes to 80 16 Gear ratio = 80 : 25 = = = 3.2 resurface a similar stretch of road in 50 hours 25 5 24 minutes, assuming the work rate remains i.e. gear ratio =16:5 or 3.2:1 constant. 5 5. It takes 3 hours 15 minutes to fly from city A Problem 12. An alloy is made up of metals A and to city B at a constant speed. Find how long B in the ratio 2.5 : 1 by mass. How much of A has the journey takes if to be added to 6 kg of B to make the alloy? 1 (a) the speed is 1 times that of the original 2 speed and Ratio A : B: :2.5 : 1 (i.e. A is to B as 2.5 is to 1) or A 2.5 (b) if the speed is three-quarters of the orig- = = 2.5 B 1 inal speed. A (a) 2 h 10 min (b) 4 h 20 min When B = 6 kg, = 2.5 from which, 6 A = 6 × 2.5 = 15 kg 1.3 Decimals Problem 13. If 3 people can complete a task in 4 hours, how long will it take 5 people to complete The decimal system of numbers is based on the digits the same task, assuming the rate of work remains 0 to 9. A number such as 53.17 is called a decimal constant fraction, a decimal point separating the integer part, i.e. 53, from the fractional part, i.e. 0.17. The more the number of people, the more quickly the A number which can be expressed exactly as a deci- task is done, hence inverse proportion exists. mal fraction is called a terminating decimal and those 3 people complete the task in 4 hours. which cannot be expressed exactly as a decimal fraction Section 1Ch01-H8555.tex 1/8/2007 18: 7 page 7 Revision of fractions, decimals and percentages 7 3 are called non-terminating decimals. Thus, = 1.5 The sum of the positive decimal fractions is 2 4 is a terminating decimal, but = 1.33333... is a non- 3 23.4 + 32.68 = 56.08 terminating decimal. 1.33333... can be written as 1.3, called ‘one point-three recurring’. The sum of the negative decimal fractions is The answer to a non-terminating decimal may be expressed in two ways, depending on the accuracy 17.83 + 57.6 = 75.43 required: (i) correct to a number of significant figures, that is, Taking the sum of the negative decimal fractions from figures which signify something, and the sum of the positive decimal fractions gives: (ii) correct to a number of decimal places, that is, the 56.08 − 75.43 number of figures after the decimal point. i.e. −(75.43 − 56.08) = −19.35 The last digit in the answer is unaltered if the next digit on the right is in the group of numbers 0, 1, 2, 3 or 4, but is increased by 1 if the next digit on the right Problem 17. Determine the value of 74.3 × 3.8 is in the group of numbers 5, 6, 7, 8 or 9. Thus the non- terminating decimal 7.6183... becomes 7.62, correct to 3 significant figures, since the next digit on the right is When multiplying decimal fractions: (i) the numbers are 8, which is in the group of numbers 5, 6, 7, 8 or 9. Also multiplied as if they are integers, and (ii) the position of 7.6183... becomes 7.618, correct to 3 decimal places, the decimal point in the answer is such that there are as since the next digit on the right is 3, which is in the many digits to the right of it as the sum of the digits to group of numbers 0, 1, 2, 3 or 4. the right of the decimal points of the two numbers being multiplied together. Thus Problem 14. Evaluate 42.7 + 3.04 + 8.7 + 0.06 (i) 743 38 The numbers are written so that the decimal points are 5 944 under each other. Each column is added, starting from 22 290 the right. 28 234 42.7 3.04 (ii) As there are (1 + 1) = 2 digits to the right of 8.7 the decimal points of the two numbers being 0.06 multiplied together, (74.3 × 3.8), then 54.50 74.3 × 3.8 = 282.34 Thus 42.7 + 3.04 + 8.7 + 0.06 = 54.50 Problem 18. Evaluate 37.81÷ 1.7, correct to (i) 4 Problem 15. Take 81.70 from 87.23 significant figures and (ii) 4 decimal places The numbers are written with the decimal points under 37.81 each other. 37.81 ÷ 1.7 = 1.7 87.23 The denominator is changed into an integer by multi- −81.70 plying by 10. The numerator is also multiplied by 10 to 5.53 keep the fraction the same. Thus Thus 87.23 − 81.70 = 5.53 37.81 × 10 37.81 ÷ 1.7 = 1.7 × 10 Problem 16. Find the value of 378.1 = 23.4 − 17.83 − 57.6 + 32.68 17 Section 1Ch01-H8555.tex 1/8/2007 18: 7 page 8 8 Engineering Mathematics The long division is similar to the long division of (b) For mixed numbers, it is only necessary to convert integers and the first four steps are as shown: the proper fraction part of the mixed number to a 7 decimal fraction. Thus, dealing with the gives: 22.24117.. 8  378.100000 17 0.875  7 34 8 7.000 i.e. = 0.875 __ 8 38 7 Thus 5 = 5.875 34 __ 8 41 34 __ Now try the following exercise 70 68 __ Exercise 3 Further problems on decimals 20 In Problems 1 to 6, determine the values of the (i) 37.81 ÷ 1.7 = 22.24, correct to 4 significant expressions given: figures, and (ii) 37.81 ÷ 1.7 = 22.2412, correct to 4 decimal 1. 23.6 + 14.71 − 18.9 − 7.421 11.989 places. 2. 73.84 − 113.247 + 8.21 − 0.068 −31.265 Problem 19. Convert (a) 0.4375 to a proper 3. 3.8 × 4.1 × 0.7 10.906 fraction and (b) 4.285 to a mixed number 4. 374.1 × 0.006 2.2446 0.4375 × 10 000 5. 421.8 ÷ 17, (a) correct to 4 significant figures (a) 0.4375 can be written as without 10 000 and (b) correct to 3 decimal places. changing its value, (a) 24.81 (b) 24.812 4375 i.e. 0.4375 = 0.0147 10 000 6. , (a) correct to 5 decimal places and 2.3 By cancelling (b) correct to 2 significant figures. 4375 875 175 35 7 (a) 0.00639 (b) 0.0064 = = = = 10 000 2000 400 80 16 7. Convert to proper fractions: 7 (a) 0.65 (b) 0.84 (c) 0.0125 (d) 0.282 and i.e. 0.4375 = 16 (e) 0.024   285 57 13 21 1 141 3 (b) Similarly, 4.285 = 4 = 4 (a) (b) (c) (d) (e) 1000 200 20 25 80 500 125 8. Convert to mixed numbers: Problem 20. Express as decimal fractions: (a) 1.82 (b) 4.275 (c) 14.125 (d) 15.35 and 9 7 (e) 16.2125 (a) and (b) 5 ⎡ ⎤ 16 8 41 11 1 (a) 1 (b) 4 (c) 14 ⎢ 50 40 8 ⎥ ⎣ ⎦ (a) To convert a proper fraction to a decimal fraction, 7 17 (d) 15 (e) 16 the numerator is divided by the denominator. Divi- 20 80 sion by 16 can be done by the long division method, In Problems 9 to 12, express as decimal fractions or, more simply, by dividing by 2 and then 8: to the accuracy stated: 4.50 0.5625   4 2 9.00 8 4.5000 9. , correct to 5 significant figures. 9 9 0.44444 Thus = 0.5625 16 Section 1Ch01-H8555.tex 1/8/2007 18: 7 page 9 Revision of fractions, decimals and percentages 9 To convert fractions to percentages, they are (i) con- 17 verted to decimal fractions and (ii) multiplied by 100 10. , correct to 5 decimal places. 27 0.62963 5 5 (a) By division, = 0.3125, hence corresponds 16 16 9 11. 1 , correct to 4 significant figures. to 0.3125 × 100%, i.e. 31.25% 16 1.563 2 (b) Similarly, 1 = 1.4 when expressed as a decimal 31 5 12. 13 , correct to 2 decimal places. fraction. 37 13.84 2 Hence 1 = 1.4 × 100% = 140% 13. Determine the dimension marked x in 5 the length of shaft shown in Figure 1.1. The dimensions are in millimetres. Problem 23. It takes 50 minutes to machine a 12.52 mm certain part, Using a new type of tool, the time can be reduced by 15%. Calculate the new time taken 82.92 27.41 8.32 x 34.67 15 750 15% of 50 minutes = × 50 = 100 100 = 7.5 minutes. hence the new time taken is Figure 1.1 50 − 7.5 = 42.5 minutes. 14. A tank contains 1800 litres of oil. How many Alternatively, if the time is reduced by 15%, then tins containing 0.75 litres can be filled from it now takes 85% of the original time, i.e. 85% of this tank? 2400 85 4250 50 = × 50 = = 42.5 minutes,asabove. 100 100 1.4 Percentages Problem 24. Find 12.5% of £378 Percentages are used to give a common standard and 12.5 12.5% of £378 means × 378, since per cent means are fractions having the number 100 as their denomina- 100 25 1 ‘per hundred’. tors. For example, 25 per cent means i.e. and is 100 4 1   12.5 1 written 25%. Hence 12.5% of £378 = × 378 = × 378 =  100 8  8 378 Problem 21. Express as percentages: = £47.25 8 (a) 1.875 and (b) 0.0125 Problem 25. Express 25 minutes as a percentage A decimal fraction is converted to a percentage by of 2 hours, correct to the nearest 1% multiplying by 100. Thus, (a) 1.875 corresponds to 1.875 × 100%, i.e. 187.5% Working in minute units, 2 hours = 120 minutes. 25 (b) 0.0125 corresponds to 0.0125 × 100%, i.e. 1.25% Hence 25 minutes is ths of 2 hours. By cancelling, 120 25 5 = Problem 22. Express as percentages: 120 24 5 2 5 (a) and (b) 1 ˙ Expressing as a decimal fraction gives 0.2083 16 5 24 Section 1Ch01-H8555.tex 1/8/2007 18: 7 page 10 10 Engineering Mathematics Multiplying by 100 to convert the decimal fraction to a 4. When 1600 bolts are manufactured, 36 percentage gives: are unsatisfactory. Determine the percentage ˙ 0.2083 × 100 = 20.83% unsatisfactory. 2.25% Thus 25 minutes is 21% of 2 hours, correct to the 5. Express: (a) 140 kg as a percentage of 1 t nearest 1%. (b) 47 s as a percentage of 5 min (c) 13.4 cm as a percentage of 2.5 m (a) 14% (b) 15.67% (c) 5.36% Problem 26. A German silver alloy consists of 60% copper, 25% zinc and 15% nickel. Determine 6. A block of monel alloy consists of 70% nickel the masses of the copper, zinc and nickel in a 3.74 and 30% copper. If it contains 88.2 g of kilogram block of the alloy nickel, determine the mass of copper in the block. 37.8 g By direct proportion: 7. A drilling machine should be set to 100% corresponds to 3.74 kg 250 rev/min. The nearest speed available on 3.74 the machine is 268 rev/min. Calculate the 1% corresponds to = 0.0374 kg 100 percentage over speed. 7.2% 60% corresponds to 60 × 0.0374 = 2.244 kg 8. Two kilograms of a compound contains 30% 25% corresponds to 25 × 0.0374 = 0.935 kg of element A, 45% of element B and 25% of 15% corresponds to 15 × 0.0374 = 0.561 kg element C. Determine the masses of the three elements present. Thus, the masses of the copper, zinc and nickel are A 0.6 kg, B 0.9 kg, C 0.5 kg 2.244 kg, 0.935 kg and 0.561 kg, respectively. (Check: 2.244 + 0.935 + 0.561 = 3.74) 9. A concrete mixture contains seven parts by volume of ballast, four parts by volume of sand and two parts by volume of cement. Now try the following exercise Determine the percentage of each of these three constituents correct to the nearest 1% and the mass of cement in a two tonne dry Exercise 4 Further problems percentages mix, correct to 1 significant figure. 54%, 31%, 15%, 0.3 t 1. Convert to percentages: (a) 0.057 (b) 0.374 (c) 1.285 10. In a sample of iron ore, 18% is iron. How (a) 5.7% (b) 37.4% (c) 128.5% much ore is needed to produce 3600 kg of 2. Express as percentages, correct to 3 signifi- iron? 20 000 kg cant figures: 11. A screws’ dimension is 12.5 ± 8% mm. Cal- 7 19 11 culate the possible maximum and minimum (a) (b) (c) 1 33 24 16 length of the screw. (a) 21.2% (b) 79.2% (c) 169% 13.5 mm, 11.5 mm 3. Calculate correct to 4 significant figures: 12. The output power of an engine is 450 kW. If (a) 18% of 2758 tonnes (b) 47% of 18.42 the efficiency of the engine is 75%, determine grams (c) 147% of 14.1 seconds the power input. 600 kW (a) 496.4 t (b) 8.657 g (c) 20.73 s Section 1Ch02-H8555.tex 2/8/2007 9: 40 page 11 Chapter 2 Indices, standard form and engineering notation Laws of indices 2.1 Indices When simplifying calculations involving indices, cer- The lowest factors of 2000 are 2 × 2 × 2 × 2 × 5 × 5 × 5. tain basic rules or laws can be applied, called the laws 4 3 These factors are written as 2 × 5 , where 2 and 5 are of indices. These are given below. called bases and the numbers 4 and 5 are called indices. (i) When multiplying two or more numbers having When an index is an integer it is called a power. Thus, 4 the same base, the indices are added. Thus 2 is called ‘two to the power of four’, and has a base of 3 2 and an index of 4. Similarly, 5 is called ‘five to the 2 4 2+4 6 3 × 3 = 3 = 3 power of 3’ and has a base of 5 and an index of 3. Special names may be used when the indices are 2 and 3, these being called ‘squared’and ‘cubed’, respectively. (ii) When a number is divided by a number having the 2 3 Thus 7 is called ‘seven squared’ and 9 is called ‘nine same base, the indices are subtracted. Thus cubed’. When no index is shown, the power is 1, i.e. 2 1 5 means 2 . 3 5−2 3 = 3 = 3 2 3 Reciprocal (iii) When a number which is raised to a power is raised The reciprocal of a number is when the index is −1 to a further power, the indices are multiplied. Thus and its value is given by 1, divided by the base. Thus the 1 −1 5 2 5×2 10 reciprocal of 2 is 2 and its value is or 0.5. Similarly, (3 ) = 3 = 3 2 −1 1 the reciprocal of 5 is 5 which means or 0.2. 5 (iv) When a number has an index of 0, its value is 1. 0 Thus 3 = 1 Square root (v) A number raised to a negative power is the recip- 1 The square root of a number is when the index is , and rocal of that number raised to a positive power. 2 √ 1 1 −4 3 1/2 Thus 3 = Similarly, = 2 the square root of 2 is written as 2 or 2. The value 4 −3 3 2 (vi) When a number is raised to a fractional power of a square root is the value of the base which when the denominator of the fraction is the root of the multiplied by itself gives the number. Since 3 × 3 = 9, √ √ number and the numerator is the power. then 9 = 3. However, (−3) × (−3) = 9, so 9=−3. √ There are always two answers when finding the square 3 2/3 2 2 Thus 8 = 8 = (2) = 4 root of a number and this is shown by putting both √ √ 2 1/2 1 1 a + and a − sign in front of the answer to a square and 25 = 25 = 25 = ±5 √ √ 1/2 root problem. Thus 9=±3 and 4 = 4=±2, √ √ (Note that ≡ 2 ) and so on.Ch02-H8555.tex 2/8/2007 9: 40 page 12 12 Engineering Mathematics 2 3 2.2 Worked problems on indices (10 ) Problem 5. Evaluate: 4 2 10 × 10 2 3 2 4 Problem 1. Evaluate (a) 5 × 5 , (b) 3 × 3 × 3 2 5 and (c) 2 × 2 × 2 From the laws of indices: 2 3 (2×3) 6 From law (i): (10 ) 10 10 = = 4 2 (4+2) 6 10 × 10 10 10 2 3 (2+3) 5 (a) 5 × 5 = 5 = 5 = 5 × 5 × 5 × 5 × 5 = 3125 6−6 0 = 10 = 10 = 1 2 4 (2+4+1) 7 (b) 3 × 3 × 3 = 3 = 3 = 3 × 3× ··· to 7 terms = 2187 Problem 6. Find the value of 2 5 (1+2+5) 8 3 4 2 3 (c) 2 × 2 × 2 = 2 = 2 = 256 2 × 2 (3 ) (a) and (b) 7 5 9 2 × 2 3 × 3 Problem 2. Find the value of: From the laws of indices: 5 7 7 5 (a) and (b) 3 4 (3+4) 7 2 × 2 2 2 3 4 7 5 7−12 −5 (a) = = = 2 = 2 7 5 (7+5) 12 2 × 2 2 2 1 1 From law (ii): = = 5 2 32 5 2 3 2×3 6 7 (3 ) 3 3 (5−3) 2 6−10 −4 (a) = 7 = 7 = 49 (b) = = = 3 = 3 3 9 1+9 10 7 3 × 3 3 3 7 5 1 1 (7−4) 3 (b) = 5 = 5 = 125 = = 4 4 5 3 81 2 3 4 Problem 3. Evaluate: (a) 5 × 5 ÷ 5 and Now try the following exercise 5 2 3 (b) (3 × 3 ) ÷ (3 × 3 ) Exercise 5 Further problems on indices From laws (i) and (ii): In Problems 1 to 10, simplify the expressions 2 3 (2+3) 5 × 5 5 2 3 4 given, expressing the answers in index form and (a) 5 × 5 ÷ 5 = = 4 4 5 5 with positive indices: 5 5 (5−4) 1 = = 5 = 5 = 5 3 4 2 3 4 4 1. (a) 3 × 3 (b) 4 × 4 ×4 5 5 (1+5) 7 9 3 × 3 3 (a) 3 (b) 4 5 2 3 (b) (3 × 3 ) ÷ (3 × 3 ) = = 2 3 (2+3) 3 × 3 3 3 2 2 4 3 2. (a) 2 × 2 × 2 (b) 7 × 7 × 7 × 7 6 3 (6−5) 1 = = 3 = 3 = 3 6 10 (a) 2 (b) 7 5 3 4 7 2 3 3. (a) (b) 3 4 2 5 3 2 Problem 4. Simplify: (a) (2 ) (b) (3 ) , 2 3 5 (a) 2 (b) 3 expressing the answers in index form. 6 3 13 10 4. (a) 5 ÷ 5 (b) 7 /7 3 3 (a) 5 (b) 7 From law (iii): 3 4 3×4 12 2 5 2×5 10 2 3 3 2 6 6 (a) (2 ) = 2 = 2 (b) (3 ) = 3 = 3 5. (a) (7 ) (b) (3 ) (a) 7 (b) 3 Section 1Ch02-H8555.tex 2/8/2007 9: 40 page 13 Indices, standard form and engineering notation 13 Problem 9. Evaluate: 2 3 7 4 2 × 2 3 × 3 6. (a) (b) 1/2 3/4 2/3 −1/2 4 5 (a) 4 (b) 16 (c) 27 (d) 9 2 3 6 (a) 2 (b) 3 √ 7 5 5 13 1/2 (a) 4 = 4 = ±2 7. (a) (b) 2 3 2 √ 5 × 5 13 × 13 4 3/4 3 3 (b) 16 = 16 = ( ± 2) = ±8 2 2 (a) 5 (b) 13 (Note that it does not matter whether the 4th root 2 3 2 (9 × 3 ) (16 × 4) of 16 is found first or whether 16 cubed is found 8. (a) (b) 2 3 (3 × 27) (2 × 8) first — the same answer will result). 4 √ (a) 3 (b) 1 3 2/3 2 2 (c) 27 = 27 = (3) = 9 −2 2 −4 5 3 × 3 9. (a) (b) 1 1 1 1 −4 3 −1/2 5 3 (d) 9 = = √ = = ±   1/2 9 ±3 3 9 1 2 (a) 5 (b) 5 3 1.5 1/3 4 × 8 2 −3 3 −4 5 7 × 7 2 × 2 × 2 Problem 10. Evaluate: 2 −2/5 10. (a) (b) 2 × 32 −4 −2 6 7 × 7 2 × 2 × 2   1 √ 2 1.5 3/2 3 (a) 7 (b) 3 4 = 4 = 4 = 2 = 8 2 √ 3 1/3 2 8 = 8 = 2, 2 = 4 1 1 1 1 −2/5 and 32 = = √ = = 5 2/5 2 2 32 2 4 32 2.3 Further worked problems on 1.5 1/3 4 × 8 8 × 2 16 indices Hence = = = 16 2 −2/5 1 2 × 32 1 4 × 4 3 7 Alternatively, 3 × 5 Problem 7. Evaluate: 3 4 1.5 1/3 2 3/2 3 1/3 3 1 5 × 3 4 × 8 (2) × (2 ) 2 × 2 = = 2 −2/5 2 5 −2/5 2 −2 2 × 32 2 × (2 ) 2 × 2 The laws of indices only apply to terms having the 3+1−2−(−2) 4 = 2 = 2 = 16 same base. Grouping terms having the same base, and then applying the laws of indices to each of the groups 2 5 3 3 3 × 5 + 3 × 5 independently gives: Problem 11. Evaluate: 4 4 3 × 5 3 7 3 7 3 × 5 3 5 (3−4) (7−3) = = = 3 × 5 3 4 4 3 5 × 3 3 5 4 Dividing each term by the HCF (i.e. highest common 5 625 1 −1 4 2 3 = 3 × 5 = = = 208 factor) of the three terms, i.e. 3 × 5 , gives: 1 3 3 3 2 5 3 3 3 × 5 3 × 5 Problem 8. Find the value of + 2 5 3 3 2 3 2 3 3 × 5 + 3 × 5 3 × 5 3 × 5 3 5 2 2 = 2 × 3 × (7 ) 4 4 4 4 3 × 5 3 × 5 4 4 3 7 × 2 × 3 2 3 3 × 5 (2−2) (5−3) (3−2) 0 3 × 5 + 3 × 5 = 3 5 2 2 (4−2) (4−3) 2 × 3 × (7 ) 3 × 5 3−4 5−3 2×2−4 = 2 × 3 × 7 4 4 3 7 × 2 × 3 0 2 1 0 3 × 5 + 3 × 5 −1 2 0 = = 2 × 3 × 7 2 1 3 × 5 1 9 1 1 × 25 + 3 × 1 28 2 = × 3 × 1 = = 4 = = 2 2 2 9 × 5 45 Section 1Ch02-H8555.tex 2/8/2007 9: 40 page 14 14 Engineering Mathematics 3 2 3 Problem 12. Find the value of 4 5 2 = × × 2 5 3 2 3 3 × 5 3 3 5 4 4 3 3 3 × 5 + 3 × 5 2 3 3 (2 ) × 2 = (3+2) (3−2) 3 × 5 9 2 To simplify the arithmetic, each term is divided by the = 2 3 5 HCF of all the terms, i.e. 3 × 5 . Thus 3 × 5 2 5 3 × 5 4 4 3 3 Now try the following exercise 3 × 5 + 3 × 5 2 5 3 × 5 Exercise 6 Further problems on indices 2 3 3 × 5 = 4 4 3 3 3 × 5 3 × 5 In Problems 1 and 2, simplify the expressions + 2 3 2 3 given, expressing the answers in index form and 3 × 5 3 × 5 with positive indices: (2−2) (5−3) 3 × 5 = 3 2 −2 −2 (4−2) (4−3) (3−2) (3−3) 3 × 5 7 × 3 3 × 5 + 3 × 5 1. (a) (b) 4 4 5 4 −3 5 × 3 3 × 7 × 7 0 2 3 × 5 25 25   = = = 1 1 2 1 1 0 3 × 5 + 3 × 5 45 + 3 48 (a) (b) 2 3 7 3 × 5 7 × 3 2 3 −2 2 −4     3 −2 4 × 9 8 × 5 × 3 4 3 2. (a) (b) × 3 4 2 4 −2 8 × 3 25 × 2 × 9 3 5 Problem 13. Simplify:   −3   2 2 3 1 (a) (b) 5 5 10 2 2 2 × 5 giving the answer with positive indices   −1 1 0.25 3. Evaluate (a) (b) 81 2 3 A fraction raised to a power means that both the numer-   1/2 4 (−1/4) ator and the denominator of the fraction are raised to   (c) 16 (d) 3 3 4 4 9 that power, i.e. =   3 1 2 3 3 (a) 9 (b) ± 3 (c) ± (d) ± A fraction raised to a negative power has the same 2 3 value as the inverse of the fraction raised to a positive power. In Problems 4 to 8, evaluate the expressions given.   −2 2 2 3 1 1 5 5   Thus, = = = 1 × = 2 4   2 2 2 2 9 × 7 147 5 3 3 3 3 4. 4 4 3 2 3 × 7 + 3 × 7 148 2 5 5   4 2 −2 4 (2 ) − 3 × 4 1     −3 3 3 5. 2 5 5 2 3 9 2 × 16 Similarly, = = 3 5 2 2     3 −2 1 2     3 −2 −   3 2 4 3 4 5 2 3 65 × × 6. −5   3 5 3 2 2 3 3 72 3 Thus, =   −3 3 5 2 5 3 2 5 Section 1Ch02-H8555.tex 2/8/2007 9: 40 page 15 Indices, standard form and engineering notation 15 (ii) The laws of indices are used when multiplying   4 or dividing numbers given in standard form. For 4 example, 3 7. 64   2 2 3 2 (2.5 × 10 ) × (5 × 10 ) 9 3+2 = (2.5 × 5) × (10 )   2 3/2 1/3 2 5 6 (3 ) × (8 ) 1 = 12.5 × 10 or 1.25 × 10 8. 4 2 3 −1/2 1/2 2 (3) × (4 ) × (9) Similarly, 4 6 × 10 6 4−2 2 = × (10 ) = 4 × 10 2 2.4 Standard form 1.5 1.5 × 10 A number written with one digit to the left of the decimal 2.5 Worked problems on standard point and multiplied by 10 raised to some power is said to be written in standard form. Thus: 5837 is written form 3 as 5.837 × 10 in standard form, and 0.0415 is written −2 as 4.15 × 10 in standard form. Problem 14. Express in standard form: When a number is written in standard form, the first (a) 38.71 (b) 3746 (c) 0.0124 factor is called the mantissa and the second factor is 3 called the exponent. Thus the number 5.8 × 10 has a 3 mantissa of 5.8 and an exponent of 10 . For a number to be in standard form, it is expressed with (i) Numbers having the same exponent can be added only one digit to the left of the decimal point. Thus: or subtracted in standard form by adding or sub- (a) 38.71 must be divided by 10 to achieve one digit tracting the mantissae and keeping the exponent to the left of the decimal point and it must also be the same. Thus: multiplied by 10 to maintain the equality, i.e. 4 4 2.3 × 10 + 3.7 × 10 38.71 38.71 = ×10 = 3.871 × 10 in standard form 4 4 = (2.3 + 3.7) × 10 = 6.0 × 10 10 −2 −2 and 5.9 × 10 − 4.6 × 10 3746 3 (b) 3746 = × 1000 = 3.746 × 10 in standard −2 −2 1000 = (5.9 − 4.6) × 10 = 1.3 × 10 form 100 1.24 When the numbers have different exponents, (c) 0.0124 = 0.0124 × = one way of adding or subtracting the numbers is 100 100 to express one of the numbers in non-standard −2 = 1.24 × 10 in standard form form, so that both numbers have the same expo- nent. Thus: Problem 15. Express the following numbers, 4 3 2.3 × 10 + 3.7 × 10 which are in standard form, as decimal numbers: −2 4 0 4 4 (a) 1.725 × 10 (b) 5.491 × 10 (c) 9.84 × 10 = 2.3 × 10 + 0.37 × 10 4 4 = (2.3 + 0.37) × 10 = 2.67 × 10 1.725 −2 (a) 1.725 × 10 = = 0.01725 Alternatively, 100 4 3 4 2.3 × 10 + 3.7 × 10 (b) 5.491 × 10 = 5.491 × 10 000 = 54 910 = 23 000 + 3700 = 26 700 0 0 (c) 9.84 × 10 = 9.84 × 1 = 9.84 (since 10 = 1) 4 = 2.67 × 10 Section 1Ch02-H8555.tex 2/8/2007 9: 40 page 16 16 Engineering Mathematics Problem 16. Express in standard form, correct to 1 7 3 1 4. (a) (b) 11 (c) 130 (d) 3 significant figures: 2 8 5 32 3 2 9   −1 (a) (b) 19 (c) 741 (a) 5 × 10 (b) 1.1875 × 10 8 3 16 2 −2 (c) 1.306 × 10 (d) 3.125 × 10 3 In Problems 5 and 6, express the numbers given (a) = 0.375, and expressing it in standard form 8 as integers or decimal fractions: −1 gives: 0.375 = 3.75 × 10 3 2 5. (a) 1.01 × 10 (b) 9.327 × 10 2 ˙ (b) 19 = 19.6 = 1.97 × 10 in standard form, correct 4 0 3 (c) 5.41 × 10 (d) 7 × 10 to 3 significant figures (a) 1010 (b) 932.7 (c) 54 100 (d) 7 9 2 (c) 741 = 741.5625 = 7.42 × 10 in standard form, −2 −1 16 6. (a) 3.89 × 10 (b) 6.741 × 10 correct to 3 significant figures −3 (c) 8 × 10 (a) 0.0389 (b) 0.6741 (c) 0.008 Problem 17. Express the following numbers, given in standard form, as fractions or mixed −1 −2 numbers: (a) 2.5 × 10 (b) 6.25 × 10 (c) 2 1.354 × 10 2.6 Further worked problems on standard form 2.5 25 1 −1 (a) 2.5 × 10 = = = 10 100 4 Problem 18. Find the value of: 6.25 625 1 −2 (b) 6.25 × 10 = = = −2 −2 100 10 000 16 (a) 7.9 × 10 − 5.4 × 10 4 2 3 3 2 (b) 8.3 × 10 + 5.415 × 10 and (c) 1.354 × 10 = 135.4 = 135 = 135 10 5 2 3 (c) 9.293 × 10 + 1.3 × 10 expressing the answers in standard form. Now try the following exercise Numbers having the same exponent can be added or subtracted by adding or subtracting the mantissae and Exercise 7 Further problems on standard keeping the exponent the same. Thus: form −2 −2 (a) 7.9 × 10 − 5.4 × 10 In Problems 1 to 4, express in standard form: −2 −2 = (7.9 − 5.4) × 10 = 2.5 × 10 1. (a) 73.9 (b) 28.4 (c) 197.72   3 3 (a) 7.39 × 10 (b) 2.84 × 10 (b) 8.3 × 10 + 5.415 × 10 2 (c) 1.9772 × 10 3 3 = (8.3 + 5.415) × 10 = 13.715 × 10 4 2. (a) 2748 (b) 33 170 (c) 274 218 = 1.3715 × 10 in standard form   3 4 (c) Since only numbers having the same exponents can (a) 2.748 × 10 (b) 3.317 × 10 5 be added by straight addition of the mantissae, the (c) 2.74218 × 10 numbers are converted to this form before adding. Thus: 3. (a) 0.2401 (b) 0.0174 (c) 0.00923   2 3 −1 −2 9.293 × 10 + 1.3 × 10 (a) 2.401 × 10 (b) 1.74 × 10 −3 (c) 9.23 × 10 2 2 = 9.293 × 10 + 13 × 10 Section 1Ch02-H8555.tex 2/8/2007 9: 40 page 17 Indices, standard form and engineering notation 17 2 = (9.293 + 13) × 10 −3 3 −2 6 × 10 (2.4 × 10 )(3 × 10 ) 2 3 4. (a) (b) = 22.293 × 10 = 2.2293 × 10 −5 4 3 × 10 (4.8 × 10 ) 2 −3 in standard form. (a) 2 × 10 (b) 1.5 × 10 5. Write the following statements in standard Alternatively, the numbers can be expressed as form: decimal fractions, giving: −3 2 3 (a) The density of aluminium is 2710 kg m 9.293 × 10 + 1.3 × 10 3 −3 2.71 × 10 kg m = 929.3 + 1300 = 2229.3 (b) Poisson’s ratio for gold is 0.44 3 −1 = 2.2293 × 10 4.4 × 10 (c) The impedance of free space is 376.73  in standard form as obtained previously. This 2 3.7673 × 10  method is often the ‘safest’ way of doing this type (d) The electron rest energy is 0.511 MeV of problem. −1 5.11 × 10 MeV 3 4 (e) Proton charge-mass ratio is Problem 19. Evaluate (a) (3.75 × 10 )(6 × 10 ) 5 −1 9 5 789 700 C kg 3.5 × 10 and (b) 7 −1 2 9.57897 × 10 Ckg 7 × 10 expressing answers in standard form (f) The normal volume of a perfect gas is 3 −1 0.02241 m mol −2 3 −1 3 4 3+4 2.241 × 10 m mol (a) (3.75 × 10 )(6 × 10 ) = (3.75 × 6)(10 ) 7 = 22.50 × 10 8 = 2.25 × 10 5 3.5 × 10 3.5 5−2 (b) = × 10 2.7 Engineering notation and 2 7 × 10 7 common prefixes 3 2 = 0.5 × 10 = 5 × 10 Engineering notation is similar to scientific notation except that the power of ten is always a multiple of 3. Now try the following exercise −4 For example, 0.00035 = 3.5 × 10 in scientific Exercise 8 Further problems on standard notation, form −3 −6 but 0.00035 = 0.35 × 10 or 350 × 10 in In Problems 1 to 4, find values of the expressions engineering notation. given, stating the answers in standard form: Units used in engineering and science may be made 2 2 1. (a) 3.7 × 10 + 9.81 × 10 larger or smaller by using prefixes that denote multi- −1 −1 plication or division by a particular amount. The eight (b) 1.431 × 10 + 7.3 × 10 most common multiples, with their meaning, are listed 3 −1 (a) 1.351 × 10 (b) 8.731 × 10 in Table 2.1, where it is noticed that the prefixes involve 2 3 2. (a) 4.831 × 10 + 1.24 × 10 powers of ten which are all multiples of 3: −3 −4 For example, (b) 3.24 × 10 − 1.11 × 10 3 −3 (a) 1.7231 × 10 (b) 3.129 × 10 6 5MV means 5 × 1 000 000 = 5 × 10 −2 3 3. (a) (4.5 × 10 )(3 × 10 ) = 5 000 000 volts 4 (b) 2 × (5.5 × 10 ) 3 3.6 k means 3.6 × 1000 = 3.6 × 10 2 5 (a) 1.35 × 10 (b) 1.1 × 10 = 3600 ohms Section 1Ch02-H8555.tex 2/8/2007 9: 40 page 18 18 Engineering Mathematics Table 2.1 Prefix Name Meaning 12 T tera multiply by 1 000 000 000 000 (i.e. × 10 ) 9 G giga multiply by 1 000 000 000 (i.e. × 10 ) 6 M mega multiply by 1 000 000 (i.e. × 10 ) 3 k kilo multiply by 1000 (i.e. × 10 ) −3 m milli divide by 1000 (i.e. × 10 ) −6 μ micro divide by 1 000 000 (i.e. × 10 ) −9 n nano divide by 1 000 000 000 (i.e. × 10 ) −12 p pico divide by 1 000 000 000 000 (i.e. × 10 ) 7.5 7.5 μC means 7.5 ÷ 1 000 000 = or Now try the following exercise 6 10 −6 7.5 × 10 = 0.0000075 coulombs Exercise 9 Further problems on 4 −3 and 4mA means 4 × 10 or = engineering notation and 3 10 common prefixes 4 = = 0.004 amperes 1000 1. Express the following in engineering notation Similarly, and in prefix form: 0.00006 J = 0.06 mJ or 60 μJ (a) 100 000 W (b) 0.00054A 5 −4 (c) 15 × 10  (d) 225 × 10 V 5 620 000 N = 5620 kN or 5.62 MN −11 (e) 35 000 000 000 Hz (f) 1.5× 10 F 4 47 × 10  = 470 000  = 470 k or 0.47 M (g) 0.000017A (h) 46200  −5 (a) 100 kW (b) 0.54 mA or 540 μA and 12 × 10 A = 0.00012 A = 0.12 mA or 120 μA (c) 1.5 M (d) 22.5 mV (e) 35 GHz (f) 15 pF A calculator is needed for many engineering calcula- (g) 17 μA (h) 46.2 k tions, and having a calculator which has an ‘EXP’ and ‘ENG’ function is most helpful. 2. Rewrite the following as indicated: 4 −6 For example, to calculate: 3 × 10 × 0.5 × 10 (a) 0.025 mA= …….μA volts, input your calculator in the following order: (b) 1000 pF = …..nF x 4 (a) Enter ‘3’ (b) Press ‘EXP’ (or ×10 ) (c) (c) 62 × 10 V = …….kV Enter ‘4’(d) Press ‘×’(e) Enter ‘0.5’(f) Press ‘EXP’ (d) 1 250 000  = …..M x (or ×10 ) (g) Enter ‘−6’ (h) Press ‘=’ (a) 25 μA (b) 1 nF (c) 620 kV (d) 1.25 M   7 The answer is 0.015V or Now press the ‘ENG’ 200 3. Use a calculator to evaluate the following in −3 button, and the answer changes to 15 × 10 V. engineering notation: −7 4 The ‘ENG’ or ‘Engineering’ button ensures that the (a) 4.5 × 10 × 3 × 10 −5 3 (1.6 × 10 )(25 × 10 ) value is stated to a power of 10 that is a multiple of (b) 6 3, enabling you, in this example, to express the answer (100 × 10 ) −3 3 as 15 mV. (a) 13.5 × 10 (b) 4 × 10 Section 1Ch03-H8555.tex 31/7/2007 9: 50 page 19 Chapter 3 Computer numbering systems 4 3 2 From above: 11011 = 1 × 2 + 1 × 2 + 0 × 2 2 3.1 Binary numbers 1 0 + 1 × 2 + 1 × 2 The system of numbers in everyday use is the denary = 16 + 8 + 0 + 2 + 1 or decimal system of numbers, using the digits 0 to 9. = 27 10 It has ten different digits (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) and is said to have a radix or base of 10. Problem 2. Convert 0.1011 to a decimal fraction The binary system of numbers has a radix of 2 and 2 uses only the digits 0 and 1. −1 −2 −3 0.1011 = 1 × 2 + 0 × 2 + 1 × 2 2 −4 + 1 × 2 3.2 Conversion of binary to decimal 1 1 1 = 1 × + 0 × + 1 × 2 3 2 2 2 The decimal number 234.5 is equivalent to 1 + 1 × 4 2 1 0 −1 2 2 × 10 + 3 × 10 + 4 × 10 + 5 × 10 1 1 1 = + + i.e. is the sum of term comprising: (a digit) multiplied 2 8 16 by (the base raised to some power). = 0.5 + 0.125 + 0.0625 In the binary system of numbers, the base is 2, so = 0.6875 10 1101.1 is equivalent to: 3 2 1 0 −1 Problem 3. Convert 101.0101 to a decimal 1 × 2 + 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 2 number Thus the decimal number equivalent to the binary number 1101.1 is 2 1 0 101.0101 = 1 × 2 + 0 × 2 + 1 × 2 2 −1 −2 1 + 0 × 2 + 1 × 2 8 + 4 + 0 + 1 + , that is 13.5 −3 −4 2 + 0 × 2 + 1 × 2 i.e. 1101.1 = 13.5 , the suffixes 2 and 10 denoting = 4 + 0 + 1 + 0 + 0.25 2 10 binary and decimal systems of number respectively. + 0 + 0.0625 = 5.3125 10 Problem 1. Convert 11011 to a decimal number 2Ch03-H8555.tex 31/7/2007 9: 50 page 20 20 Engineering Mathematics Now try the following exercise Exercise 10 Further problems on conversion of binary to decimal numbers In Problems 1 to 4, convert the binary number given to decimal numbers. 1. (a) 110 (b) 1011 (c) 1110 (d) 1001 For fractions, the most significant bit of the result is the top bit obtained from the integer part of multiplication (a) 6 (b) 11 (c) 14 (d) 9 10 10 10 10 by 2. The least significant bit of the result is the bottom 2. (a) 10101 (b) 11001 (c) 101101 (d) 110011 bit obtained from the integer part of multiplication by 2. (a) 21 (b) 25 (c) 45 (d) 51 10 10 10 10 Thus 0.625 = 0.101 10 2 3. (a) 0.1101 (b) 0.11001 (c) 0.00111 (d) 0.01011   Problem 4. Convert 47 to a binary number 10 (a) 0.8125 (b) 0.78125 10 10 (c) 0.21875 (d) 0.34375 10 10 From above, repeatedly dividing by 2 and noting the 4. (a) 11010.11 (b) 10111.011 (c) 110101.0111 remainder gives: (d) 11010101.10111   (a) 26.75 (b) 23.375 10 10 (c) 53.4375 (d) 213.71875 10 10 3.3 Conversion of decimal to binary An integer decimal number can be converted to a cor- responding binary number by repeatedly dividing by 2 and noting the remainder at each stage, as shown below for 39 10 Thus 47 = 101111 10 2 Problem 5. Convert 0.40625 to a binary 10 number From above, repeatedly multiplying by 2 gives: The result is obtained by writing the top digit of the remainder as the least significant bit, (a bit is a binary digit and the least significant bit is the one on the right). The bottom bit of the remainder is the most significant bit, i.e. the bit on the left. Thus 39 = 100111 10 2 The fractional part of a decimal number can be converted i.e. 040625 = 0.01101 10 2 to a binary number by repeatedly multiplying by 2, as shown below for the fraction 0.625 Section 1Ch03-H8555.tex 31/7/2007 9: 50 page 21 Computer numbering systems 21 3.4 Conversion of decimal to binary Problem 6. Convert 58.3125 to a binary 10 number via octal For decimal integers containing several digits, repeat- The integer part is repeatedly divided by 2, giving: edly dividing by 2 can be a lengthy process. In this case, it is usually easier to convert a decimal number to a binary number via the octal system of numbers. This system has a radix of 8, using the digits 0, 1, 2, 3, 4, 5, 6 and 7. The denary number equivalent to the octal number 4317 is 8 3 2 1 0 4 × 8 + 3 × 8 + 1 × 8 + 7 × 8 i.e. 4 × 512 + 3 × 64 + 1 × 8 + 7 × 1 or 2255 10 The fractional part is repeatedly multiplied by 2 giving: An integer decimal number can be converted to a cor- responding octal number by repeatedly dividing by 8 and noting the remainder at each stage, as shown below for 493 10 Thus 58.3125 = 111010.0101 10 2 Now try the following exercise Exercise 11 Further problems on Thus 493 = 755 10 8 conversion of decimal The fractional part of a decimal number can be con- to binary numbers verted to an octal number by repeatedly multiplying by In Problem 1 to 4, convert the decimal numbers 8, as shown below for the fraction 0.4375 10 given to binary numbers. 1. (a) 5 (b) 15 (c) 19 (d) 29   (a) 101 (b) 1111 2 2 (c) 10011 (d) 11101 2 2 2. (a) 31 (b) 42 (c) 57 (d) 63   For fractions, the most significant bit is the top integer (a) 11111 (b) 101010 2 2 obtained by multiplication of the decimal fraction by 8, (c) 111001 (d) 111111 2 2 thus 3. (a) 0.25 (b) 0.21875 (c) 0.28125 0.4375 = 0.34 (d) 0.59375 10 8   (a) 0.01 (b) 0.00111 2 2 The natural binary code for digits 0 to 7 is shown (c) 0.01001 (d) 0.10011 2 2 in Table 3.1, and an octal number can be converted to a binary number by writing down the three bits 4. (a) 47.40625 (b) 30.8125 corresponding to the octal digit. (c) 53.90625 (d) 61.65625   (a) 101111.01101 (b) 11110.1101 2 2 Thus 437 = 100 011 111 8 2 (c) 110101.11101 (d) 111101.10101 2 2 and 26.35 = 010 110.011 101 8 2 Section 1