MATH123 FOUNDATION MATHEMATICS. Lecture Notes

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1 MATH123 FOUNDATION MATHEMATICS Lecture Notes School of Science and Technology University of New England Update date: March 31, 2014

2 Contents 1 Basic Arithmetic Basic Rules for Elementary Operations Natural Numbers, Whole Numbers and Integers Elementary Operations for the Whole Numbers Rules of Elementary Operations Order of Operations with Parentheses Commutative Property Distributive Property Division, Fractions and Decimals Fractions and Decimals Adding and Subtracting Fractions Multiplication and Division of Fractions Remarks on the Use of Symbols in Mathematics Elementary Operations Involving Negative Numbers Absolute Value and Comparison of Decimals and Fractions Exponents and Scientific Notation Operations with Scientific Notations Rounding and Estimation Some Applications Conversion of Units Percentage i

3 ii CONTENTS 2 Introduction of Algebra Notations and Review of the Basic Rules in Arithmetic Some Commonly Used Terminologies and Notations Addition and Subtraction with Algebraic Expressions Multiplication and Division with Algebraic Expressions Solutions of Linear Equations Applications of Linear Equations Elementary Functions Linear Functions and Their Graphs Linear Functions in Applications Power Functions Exponential and Logarithmic Functions Polynomials and Quadratic Equations Probability and Statistics Sets Introduction of Probability Counting Methods and Probability Introduction of Statistics Elementary Geometry Lines and Angles Triangles Congruence and Similarity of Triangles Quadrilaterals and Polygons Some Applications and Examples Involving Triangles Circles and Arcs Trigonometric Functions and Triangles Definition and Basic Properties

4 CONTENTS iii Radian Measure of Angles and Trigonometric Functions Applications of Trigonometric Functions Solid Figures

5 Chapter 1 Basic Arithmetic 1.1 Basic Rules for Elementary Operations Natural Numbers, Whole Numbers and Integers The set of numbers {1, 2, 3, 4,...} is called the set of counting numbers or the set of natural numbers. If the number zero is included, then the set {0, 1, 2, 3, 4,...} is called the set of whole numbers. If the negative of the natural numbers are included, then the set {..., 4, 3, 2, 1, 0, 1, 2, 3, 4,...} is called the set of integers Elementary Operations for the Whole Numbers Addition, subtraction, multiplication, and division are called the elementary operations for the whole numbers, and it is assumed that you understand these operations. 1

6 2 CHAPTER 1. BASIC ARITHMETIC Rules of Elementary Operations The very basic rules are: 1. Multiplication and division first. 2. Left to right. These rules are best explained by concrete examples. (a) = (2 + 3) 1 = 5 1 = 4 (No multiplication or division occurs and we follow the left to right rule.) (b) = 2 + (3 4) + 5 = = (2 + 12) + 5 = = 19 (We follow the multiplication first rule to do 3 4 first, then follow the left to right rule.) (c) = 2 + (3 4) = = 2 + (12 6) + 5 = = (2 + 2) + 5 = = 9 (We follow the multiplication and division first rule to calculate first and use the left to right rule to do 3 4 first, 12 6 second, after that we follow the left to right rule to calculate ) Order of Operations with Parentheses When an operation involves parentheses, those operations inside the parentheses should be performed first. Examples: (12 8) 4 = 4 4 = (7 + 1) = = (3 1) (1 + 2) = = = (3 1) (8 4) + 5 = = = 7 (Draw parentheses to mark the priority of operations.)

7 1.1. BASIC RULES FOR ELEMENTARY OPERATIONS 3 To summarize, the order of operations obey the following rules: Step 1. Parentheses first Step 2. Multiplication and division, reading from left to right Step 3. Addition and subtraction, reading from left to right The mnemonic BOMDAS was drilled in primary school. It is a prompt to remember the order of arithmetic operations, (a) Brackets (b) Of (like multiplication) (c) Multiplication (d) Division (e) Addition (f) Subtraction Commutative Property The sum of two numbers does not change when the order of the two numbers are interchanged; for example, = = 3, = = 8, etc. Similarly, the product of two numbers does not change when the order of the two numbers are interchanged; for example, 1 2 = 2 1 = 2, 3 5 = 5 3 = 15, etc. This commutative property does not apply to subtraction and division; for example, 1 2 = 1, 2 1 = 1; so

8 4 CHAPTER 1. BASIC ARITHMETIC Distributive Property When calculating 2 ( ), one may follow the rules given above to obtain 2 ( ) = 2 (9 3) = 2 6 = 12. Another way of calculation is to use the distributive property : 2 ( ) = = = 12 In doing this, the number 2 is distributed to each number inside the parentheses which contain only additions and subtractions. We should note that 2 (5 4 3) (2 5) (2 4) 2 3. Due to the commutative property for multiplication, the earlier calculation may also be written as ( ) 2 = Interestingly, there is a similar property for division: ( ) 2 = = = 5 (The normal rules give ( ) 2 = 10 2 = 5.) However, you may use a calculator to confirm that 2 ( ) You may also calculate directly to obtain 12 (4 2) = 12 2 = = 3 6 = 3. Hence 12 (4 2)

9 1.2. DIVISION, FRACTIONS AND DECIMALS Division, Fractions and Decimals In the previous section, we denoted division by, such as 10 5, 6 3, etc. However, more often a division bar is used, as in 10 (to replace 10 5), 6 (to replace 6 3). 10, 6 are also called fractions. Division can be performed by a calculator, or by using long division manually. In the era of calculators, manual calculations may seem unnecessary. However, proficiency with manual calculations is the basis for being able to perform quick and reliable approximations on the spot. These situations occur in planning meetings, recognizing when presented figures are incorrect (and potentially misleading) and getting estimates in work situations where a calculator is impractical. Examples of the latter are field or factory work when hands are dirty or there is not time to stop the process. Long division is demonstrated with an example The setting out of the long division algorithm is critical in getting it working correctly. Numbers have to be aligned vertically under each other. Having learnt tables in primary school, we are proficient at doing division with single digits but our brains are wired so that it gets tougher when going beyond that. The technique of long division is an algorithm which decomposes the problem of a big division into a series of steps of small divisions which can be accomplished easily. The first step is to approximate the division by looking for the the first number of the quotient 1. We can see that = 1 with 91 remainder. That is equivalent to making the first approximation that This approximation needs to be improved because there is 915 still remaining. This part of the algorithm is accomplished by:- Step 1 Divide 305 into = 1 with 91 remaining. That would be done by 1 The number that results from the division.

10 6 CHAPTER 1. BASIC ARITHMETIC This has determined that the first number in the quotient is 1 and now the algorithm advances to find the second digit in the quotient by examining the remainder. Step 2 Bring down the 5. This gives the remainder from the first approximation which is Step 3 Divide 915 by 305, = 3 with remainder 0. This provides the second digit of the quotient by finding the multiple of 305 (the divisor) that is closest to the remainder Thus = 13. If at step 3 the remainder was not zero, the algorithm would continue with decimals. For example, =

11 1.2. DIVISION, FRACTIONS AND DECIMALS For this example, it is sufficient to stop at 2 decimal places. The division bar also acts like a grouping symbol. For example = 10 5 = 2. To use in the above calculation, we need to write instead (6 + 4) (2 + 3) = 10 5 = Fractions and Decimals When addition, subtraction and multiplication are applied to integers, such as 8 ( ) (3 + 5) 2, the calculation gives an integer as a result. This is not always so when division is involved. Firstly, division by 0 is not allowed. Secondly, not all division involving integers results with an integer. For example, 8 3 equals 2 with remainder 2; we write this as 8 3 = or 8 3 = is a fraction. In general, a fraction involves two numbers: one upstairs, called the numerator 3 and one downstairs, called the denominator. A fraction with both numerator and denominator

12 8 CHAPTER 1. BASIC ARITHMETIC integers is called a rational number, so 8 3, 2 3, 6 2 are all rational numbers. However, 6 2 simplified: 6 2 = 3, which is an integer. (Integers are special rational numbers.) Each fraction can be written in decimal form: 2 5 = 0.4, 1 10 = 0.1, = 0.01, 2 3 = In the first three examples above, the decimal form is finite, but 2 3 are infinitely many 6 in the decimal form. One usually denote this by = = = 3.45(= ). can be = is infinite; there = 0.6. Similarly, Generally speaking, to change a fraction into decimal form, one uses long division or uses a calculator; for example, for 2, we have ) Thus 2 5 = 0.4. For 5, we have ) repeats Thus 5 11 = = To change a finite decimal number to a fraction, it is useful to keep in mind that the number in front of the decimal point is the integer part of the decimal, and the digits following the decimal

13 1.2. DIVISION, FRACTIONS AND DECIMALS 9 point give the fractional part of the number; for example, = = = = = = 2 100, etc. As elementary operations (addition, subtraction, multiplication and division) of decimal numbers can be performed by the calculator, we will not go to more details about them here. Instead, we will concentrate on the elementary operations of fractions, which plays an important role in manipulating general algebraic expressions, which involve variables x, y etc. and hence cannot be calculated by a standard calculator Adding and Subtracting Fractions Let us recall that the distributive property for division gives ( ) 2 = Write this in the form of fractions, we have = or = In general, to add or subtract fractions with common denominators, add or subtract the numerators; the denominator of the sum or difference is the same as the common denominator. For example,

14 10 CHAPTER 1. BASIC ARITHMETIC l = = 4 5 = = = = = = or = = = = How to add and subtract fractions with different denominators? We need the notion common fractions. A common fraction may have different names or representations, but they have the same value, or have the same decimal representation. For example, 1 2, 0.5, 5 10, 2 4 all represent the same decimal number 0.5. We say a fraction is in reduced form if no positive integer other than 1 can divide both the numerator and the denominator. Thus 1 is in reduced form, but 2 5, 2 are not To add or subtract two fractions with different denominators, we need firstly rewrite the given fractions into equivalent forms which have the same denominators. For example, to calculate = = = 3 4.

15 1.2. DIVISION, FRACTIONS AND DECIMALS 11 Similarly, = = 5 6. In the above calculation, we used a general method: = = i.e. to multiply both the numerator and denominator of the first fraction by the denominator of the second fraction, and do the same to the second fraction by the denominator of the first fraction. If we use a 1 b 1 and a 2 b 2 follows: boxed to denote two general fractions, this method may be expressed as a 1 b 1 + a 2 b 2 = a 1 b 2 b 1 b 2 + b 1 a 2 b 1 b 2 = a 1 b 2 + b 1 a 2 b 1 b 2. Thus, we may calculate as follows = = = 6 8 = 3 4. Similarly, a 1 b 1 a 2 b 2 = a 1 b 2 b 1 b 2 b 1 a 2 b 1 b 2 = a 1 b 2 b 1 a 2 b 1 b 2

16 12 CHAPTER 1. BASIC ARITHMETIC Example = ( ) 1 2 = ( ) 1 2 = ( ) 1 2 = = = = = 5 12 For 11 1, one could also use the following way of calculation = = Multiplication and Division of Fractions As above, it is easy to explain the rules for multiplication and division by denoting two fractions in general from: a 1 b 1 and a 2 b 2. Then a 1 b 1 a 2 b 2 = a 1 a 2 b 1 b 2.

17 1.2. DIVISION, FRACTIONS AND DECIMALS 13 For example, = = 2 12 = = = = = = = = (Here we have followed the same rule of multiplication first.) = ( ) 6 11 = (From left to right) 11 = = = We could also calculate in the following manner: or = = = = = =

18 14 CHAPTER 1. BASIC ARITHMETIC In the last method of calculation, we canceled one common factor, 2, from both upstairs (6 becomes 3) and downstairs (2 becomes 1) in the second step. The division of fraction is calculated by the following formula: a 1 b 1 a 2 b 2 = a 1 b 1 b 2 a 2 = a 1 b 2 b 1 a 2. For example,

19 1.2. DIVISION, FRACTIONS AND DECIMALS = = 8 3 = = = ( ) = ( ) = = = = = = = 18 = or = = = = = = 12 =

20 16 CHAPTER 1. BASIC ARITHMETIC A common mistake in the calculations involving fractions is (These are wrong!) a 1 + a 2 = a 1 + a 2, b 1 b 2 b 1 + b 2 a 1 a 2 = a 1 a 2 b 1 b 2 b 1 b Remarks on the Use of Symbols in Mathematics As we have seen in the previous two sections, using a 1 b 1 and a 2 b 2 to represent two generic fractions makes it very easy to explain the rules of elementary operations for fractions. If we use similar notations, the rules that we have discussed earlier can also be better expressed: Commutative rule for addition and multiplication of two numbers a and b a + b = b + a, a b = b a. Distributive rule: a (b 1 + b b n ) = a b 1 + a b a b n (a 1 + a a n ) b = a 1 b + a 2 b + + a n b (a 1 + a a n ) b = a 1 b + a 2 b + + a n b Note however a (b 1 + b b n ) a b 1 + a b a b n in general. Additions and subtractions of fractions with a common denominator a c + b c a c b c = a + b, c = a b c

21 1.2. DIVISION, FRACTIONS AND DECIMALS Elementary Operations Involving Negative Numbers Again we will use symbols, a, b etc. to explain the general rules and use examples to illustrate them further. General rules for addition and subtraction with negative numbers The rules are stated using the letters a and b. In practice, we substitute numbers for these letters and I show that in the example. a + ( b) = a b (1.1) = (b a) using ( a) = +a ( a) + b = b + ( a) (1.2) = b a a ( b) = a + b (1.3) a ( b) = a + b (1.4) = b a a + ( b) = (a + b) (1.5) General rules for multiplication with negative numbers

22 18 CHAPTER 1. BASIC ARITHMETIC Example 1 Evaluate ( 2) ( 5) ( 1 2 ) a ( b) = (a b) (1.6) ( a) b = (a b) (1.7) ( a) ( b) = a b (1.8) a ( b) = (a b) or (1.9) a b = a b ( a) b = (a b) or (1.10) a b = a b ( a) ( b) = a b or (1.11) a b = a b We can separate it into 2 parts; the addition part and the multiplication part. This is done because we do the multiplication first. In this example, the order of operations is:- ( 2) ( 5) + 1 }{{} 3 ( 1 2 ) }{{} 2nd operation 1st operation because multiplication takes precedence over addition. For the first operation, use multiplication rule number (1.6) with a = 1 3 and ( b) = ( 1 2 ). 1 3 ( 1 2 ) = ( ) = 1 6 For the second operation, use addition rule (1.4) with a = 2 and ( b) = ( 5) ( 2) ( 5) = = 5 2 = 3

23 1.2. DIVISION, FRACTIONS AND DECIMALS 19 Now combine the 2 results so far }{{} nd }{{} 1st = = 17 6 This problem would be set out like this:- ( 2) ( 5) ( 2 1 = ) 2 = = = = 17 6 (1.12)

24 20 CHAPTER 1. BASIC ARITHMETIC Example 2 Evaluate 6 + ( 10) 2 5 ( 1 7 ). The operation done first is the division because it takes precedence over addition and subtraction (BOMDAS), i.e. Step 1 Step ( 10) 2 }{{} 5 ( 1 7 ). }{{} 2nd 1st 2 5 ( 1 7 ) = 2 ( 5 7 ) 1 ( ) 2 = 5 7 = 14 5 p9 of notes multiplication rule (1.7) 6 + ( 10) = 6 10 addition rule (1.1) = 4 Step ( 10) 2 }{{} 5 ( 1 7 ) }{{} 2nd 1st = }{{} 4 2nd ( = = = ) 5 }{{} 1st This is the setting out:- 6 + ( 10) 2 ( 5 1 ) 7 [ ( )] 2 = ( ) 2 = (10 6) = = 6 5

25 1.2. DIVISION, FRACTIONS AND DECIMALS Absolute Value and Comparison of Decimals and Fractions Let a stand for a decimal number (we regard integers as special decimal numbers). Its absolute value, denoted by a, equals a itself if a is positive, equals a if a is negative. If a is 0, its absolute value is 0, i.e. 0 = 0. Thus 1 = 1, 0.02 = 0.02, = For any fraction a, its absolute value can be calculated in the following way b Therefore a = a b b. 2 5 = 2 5, 2 5 = 2 5, 2 5 = 2 5. Given two decimal numbers a and b, we say a is smaller than b (or b is greater than a) if b a is a positive decimal number, and write this as a < b (or b > a). We say a is a positive number if a > 0, a negative number if a < 0. To compare two given fractions, one can reduce them to decimals first and then compare, but one can also compare them directly as follows: Let a 1 b 1, a 2 b 2 be two given fractions, then a 1 b 1 < a 2 b 2 if and only if a 2 b 2 a 1 b 1 equivalently, if and only if a 1 b 1 a 2 b 2 Hence a 2 b 2 a 1 b 1 is a negative number). Recall that a 2 b 2 a 1 b 1 = a 2 b 1 b 2 b 1 b 2 a 1 b 2 b 1 = a 2 b 1 b 2 a 1 b 2 b 1. is a positive number (or is a positive number if and only if a 2 b 1 b 2 a 1 and b 2 b 1 are of the same sign, i.e. both positive or both negative. Examples (a) Compare 2 5 and 3 7. Solution: = = = 1 35 positive number) Therefore 2 5 < 3 7. > 0 (Recall we use a > 0 to mean a is a

26 22 CHAPTER 1. BASIC ARITHMETIC (b) Compare and 2 3. Solution: 2 3 ( ) = result is a positive number) Thus < 2 3. (c) Compare and Solution: Therefore < > 0 (we do not need to find the exact value to know the ( ) ( ) = = = = = < 0 = Exponents and Scientific Notation Similar to the use of symbols, like a, b, a 1 b 1, a 2 b 2, etc. used in the previous sections, exponents and scientific notations are introduced to simplify mathematical expressions. So instead of writing we can write 10 5, where 10 is called the base and 5 is called the exponent is called the power or exponential notation for Similarly, 10 } {{ } = 10 25, = 2 3, 6 6 = 6 2. One may write 10 = 10 1, 2 = 2 1, 25 times 6 = 6 1, and generally a = a 1, but the exponent 1 is usually omitted. In general, It is easy to check that a } a {{ a } = a n. n times = = = = 10 2 In general, a m a n = am n

27 1.2. DIVISION, FRACTIONS AND DECIMALS 23 It is also easy to understand (by the definition) that a m a n = a m+n We understand that a 0 = 1, which is consistent with the formula am a n = a m n : One then easily sees that 1 = am a m = am m = a = = = = = = m = 0. 0 } {{ 0} 1 m 1 zeros However, it is better to think the following way: In general, = = = = = m = m = 1 10 m a 1 = 1 a, a m = 1 a m. Scientific notation is a standard way of expressing decimal numbers by making use of the power 10 m, to make the expression easy to write and read. For example, can be written as Similarly, = = = = or simply 10 3.

28 24 CHAPTER 1. BASIC ARITHMETIC The scientific notation of a number is that number written as a decimal number with value between 1 and 10 times a power of 10, or as a power of 10 only. The general method to work out the exponent on 10 is to count the number of positions needed to move the decimal point of the given number (in decimal form, so 283 = 283.0, etc) to obtain a number with the same digits but with value between 1 and 10 as required in the scientific notation. For example, to change to 1.2, the decimal point is moved to the right 4 positions and hence = In general, if the decimal point is moved to the right n positions, 10 n should be used in the scientific notation. Analogously, to change = to 1.2, the decimal point is moved to the left 6 positions, and hence = In general, if the decimal point is moved to the left m positions, 10 m should be used in the scientific notation. Examples: Write the following numbers in scientific notation: a b c Solutions: a = (Decimal point moved 5 positions to the left.) b = (Decimal point moved 3 positions to the right.) c = (Decimal point moved 3 positions to the left.)

29 1.2. DIVISION, FRACTIONS AND DECIMALS Operations with Scientific Notations We start with some examples. Consider first ( ) ( ) Using the commutative property for multiplications, we can rewrite the above product in a different form without changing its value: We may calculate ( ) ( ) = ( ) ( ). Thus the result is Similarly = , = = 10 3 ( ) ( ) =( ) ( ) = =( ) 10 5 = Here we have used the scientific notation for to reduce the final answer in scientific notation. Analogously, we have ( ) ( ) = = = ( 8) = =( ) =

30 26 CHAPTER 1. BASIC ARITHMETIC This example uses the result (in reverse order) and 10m 10 n = 10 m n. a 1 b 1 a 2 b 2 = a 1 a 2 b 1 b 2 One might be attempted to do the following: = ( ) ( ) Wrong! But a little thinking would convince you that this is not correct; no rules in mathematics supports such a calculation. It turns out that addition and subtraction in scientific notation are more difficult to do than multiplication and division! Let us first consider some simpler cases. (a) Examples ( ) + ( ) = [ ( ) 10 2] + ( ) = ( ) + ( ) = ( ) 10 3 = = Here we have used the distribution property and in order to be able to use this property, we first changed to the form We could do the calculation in a different way: ( ) + ( ) = ( ) + [ ( ) 10 3] = ( ) + ( ) = ( ) 10 2 = This second method gives the answer in scientific notation directly.

31 1.2. DIVISION, FRACTIONS AND DECIMALS 27 Similarly, (b) ( ) + ( ) = ( ) + ( ) = ( ) 10 8 = (c) ( ) ( ) = ( ) ( ) = ( ) 10 8 = (d) ( ) ( ) = ( ) ( ) = ( ) = (e) ( ) + ( ) = ( ) + [ ( ) 10 8] = ( ) + ( ) = ( ) 10 3 =

32 28 CHAPTER 1. BASIC ARITHMETIC Rounding and Estimation The number appearing in the last example is long but the last three digits 521 are perhaps not significant in most applications. In practice, people often round a decimal number to an extent that the insignificant digits are dropped to make a shorter but more meaningful decimal number. For example, one may round the above number to 2.31, and a budget of $252,892, for 99,200 students can be approximated as a $250 million budget for 100,000 students. Generally speaking, to round a given decimal number, one uses common sense unless the accuracy is specified. For example, we used common sense in the above two examples to do the rounding. If we are required to round to 5 decimal places, we would obtain There are two commonly used rules for rounding off decimals: (a) If the first of the digits to be dropped is less than 5, then the last retained digit is unchanged. Thus to round off to two decimal places, we obtain 3.13 (the first of the digits to be dropped is 2, and 2 < 5). (b) If the first of the digits be to dropped is 5 or bigger, then the last retained digit is increased by 1. Thus to round to three decimal places, we obtain (the first of the digit to be dropped is 5). To round to three decimal places, we obtain (the first of the dropped digits is 6, and when 1 is added to the last retained digit 9, a carry in occurs).

33 1.3. SOME APPLICATIONS Some Applications Mathematical thinking is required to apply basic arithmetic to real world problems. This requires problem solving skills, which are often more demanding than mastering the rules and calculation techniques in basic arithmetic. Let us start with a simple one. Example 1. Adam sold a $3 benefit ticket on Monday and six $4 tickets on Tuesday. What is the total amount collected? Solution: [Total sales]=[sales on Monday] + [sales on Tuesday] = 3 + (4 6) (dollars) = = 27 (dollars) Example 2. If the US federal budget is $1.5trillion (1,500,000,000,000), how much does it cost each individual, on average, if there are 240,000,000 people? Solution: We must divide the budget by the number of people. Since the involved numbers are big, we use scientific notation. The desired calculation is 1.5 trillion = , 240, 000, 000 = = = = , or 6250 dollars. Therefore, it costs each individual $6250. (Note that the scientific notation have no advantage over 6250 in this case; one uses common sense to represent the answer by You would not lose marks if your answer is given by $ , but $6250 is more natural.) In the following sections, we consider several typical types of applications.

34 30 CHAPTER 1. BASIC ARITHMETIC Conversion of Units Units are important in the real world, but they are not unified internationally at the moment. For example, the standard units of length in the U.S. system are: inch(in.), foot(ft; equals 12in), yard (yd; equals 36in), mile(mi; equals 63,360in); in the Metric system, they are: meter(m), centimeter(cm; equals 0.01m), kilometer(km; equals 1000m). is 15 3 Example If an object covers a distance of 15 meters in 3 seconds, then the average speed of the object = 5 meter per second. Written as 5 m/sec. So the unit is meter per second. The area of a square of 3cm by 3cm is 3 3 = 9 cm 2 (square centimeters). So the unit is square centimeter. The volume of a cube of 2m by 2m by 2m is = 8m 3 (cubic meters). Here one uses cubic meters for the volume. We know that 1cm=0.01m, 1km=1000m. It follows that 1m = 100 (0.01m) = 100 (1cm) = 100cm, 1m = 10 3 (1000m) = 10 3 (1km) = 10 3 km = 0.001km.

35 1.3. SOME APPLICATIONS 31 If we want to convert the speed 5m/sec into the unit km/sec, we use 5m = 5 (1m) = 5 (0.001km) = 0.005km. Therefore 5m/sec=0.005km/sec. How to convert 5m/sec into the unit km/h (kilometer per hour)? We know that 1 hour=60 minutes=60 60 seconds=3600 seconds. Hence 1 sec.= 1 hour. We may now use m/sec = 0.005km/sec = 0.005km 1sec = 0.005km 1/3600hour = /3600 km/h /3600 = = = = 18 Hence 5m/sec=18km/h.

36 32 CHAPTER 1. BASIC ARITHMETIC Let us now see how to convert 9cm 2 into the unit of m 2. We know that 1cm=0.01m. Hence 1cm 2 = (1cm) (1cm) = (0.01m) (0.01m) = m 2 = (0.01) 2 m 2 = m 2 9cm 2 = 9 (1cm 2 ) = = 9 (0.0001m 2 ) = m 2 Analogously, to convert 8m 3 into the unit cm 3, we use 1m=100cm, 1m 3 = (1m) (1m) (1m) = (100cm) (100cm) (100cm) = cm 3 (1.13) = cm 3 = 10 6 cm 3 and 8cm 3 = 8 (1cm 3 ) = cm 3

37 1.3. SOME APPLICATIONS 33 Example: Given that 1 meter=39.38 inch. Convert 15m/sec to in/hour. Solution: We have 1m=39.37 in, 1 sec= hour. Thus 15m/sec = 15 m 1sec in = 1/3600 hour = ( ) = in/hour in/hour = in/hour Percentage Percent is a commonly used word, which refers to the ration of a given number to 100, that is, a percent is the numerator of a fraction whose denominator is 100. For example, 5 percent means , 130 percent means. The symbol % is used to indicate percent, such as 5%, 130%, etc. 100 A percent may always be represented by a fraction and this fraction may in turn be expressed in decimal form. Thus 5% can be written as or 0.05, 130%= = 13 =

38 34 CHAPTER 1. BASIC ARITHMETIC Conversely, any fraction or decimal can be expressed in percentage form. For example, = 100 = = 125% 5 8 = = = = 62.5% We now look at several typical questions involving percent. 1. Calculate the percent of a given quantity. Example 1. What is 6% of 80? Solution: Since 6%= = 0.06, 6% of 80 is 6% 80 = = 4.8 Example 2. What is 12 1 % of 120? 2 Solution: % = 12.5% = = %of120 is 121 % 120 = = 15

39 1.3. SOME APPLICATIONS 35 Example 3. A copper compound contains 80% copper by weight. How much copper is there in a 1.95 gram sample of the compound? Solution: 80% = = % 1.95 = = 1.56 Therefore, the sample contains 1.56 gram of copper. Example 4. If the present length of a brass rod is 1.46m, what will its length be after 0.3% expansion caused by heating? Solution: Future length = present length + expansion Expansion = 0.3% 1.46 = = = Therefore the length after expansion is = m. 2. Determine what percent of one quantity is of another. Example 1. In a class of 36 students, 32 passed the exam. What is percent of the student passed the exam in the class? Solution: = 0.88 = 0. 8 using the 8 notation for repeating decimals Therefore 88.89% passed the exam. (Here we have rounded to )

40 36 CHAPTER 1. BASIC ARITHMETIC Example 2. Water is a compound consisting of oxygen and hydrogen. A sample of water is decomposed into 0.965g of oxygen and 0.120g of hydrogen. Calculate the mass percent of oxygen and hydrogen in the sample of water. Solution: We need to know that weight of the water sample first. Since it consists of 0.965g of oxygen and 0.120g of hydrogen, the weight of the water sample is =1.085g. The mass percent of oxygen is therefore = = 88.94% (after rounding) and the mass percent of hydrogen is = = 11.06% (after rounding. (This is also 100%-88.94%. Why?) 3. Find quantity from percentage and percent. Example is 15% of what number? Solution: [The desired number] 15%=20. Therefore [the desired number] = 20 15% = = = =

41 1.3. SOME APPLICATIONS 37 We could also represent the desired number in the form of a fraction, which is = = = = Example 2. Decomposition of limestone yields 56% quicklime. How much limestone is needed to yield 2000kg of quicklime? Solution: desired quicklime = limestone 56% so that 2000 = limestone 56% Therefore, limestone = % = = kg (after rounding according to common sense). Thus 3571kg of limestone is needed.

42 38 CHAPTER 1. BASIC ARITHMETIC 4. Percent error or uncertainty. The percent error or uncertainty in an experiment is calculated from the following formula: percent error = experimental value true value true value Recall that the notation x is the absolute value (or magnitude) of x, i.e. x = x = x. For example, if the experimental value is 120, the true value is 100, then = = 0.2 = 20% and the measurement error (or uncertainty) is 20%. Example 1. A flask is marked 250ml. The volume measured with a standard container is 250.5ml. What is the percent error in the flask calibration? Solution: 250.5ml is the true capacity of the flask = % (after rounding). Thus the percent error is 0.20%. Example 2. What is the percent reading uncertainty in the measurement 1.67±0.02g? Solution: The reading uncertainty is 0.02g when the true value is 1.67g. Hence, the percent uncertainty is = % (after rounding).

43 1.3. SOME APPLICATIONS Percent decrease and increase. Example 1. In , the Nasdaq experienced a decline from 5,000 points to 1,600 points. What is the percent of this decline? What is the percent increase necessary for the Nasdaq to climb back to its previous hight? Solution: From 5,000 to 1,600, the decline is = 3400, so we have 3400 is what percent of 5000? = 0.68 = 68%. The stock market decline is 68%. To increase from 1600 to 5000 (the previous high), the increase is = 3400, so we ask 3400 is what percent of 1600? = = 212.5%. The Nasdaq needs a 212.5% increase to restore the original amount. Example 2. After a rise of 5%, Sue s current salary is $65,880. What was Sue s salary before the rise? Solution: We have new salary = old salary + increase = old salary + old salary 5% = old salary (1 + 5%) old salary = (1.05) (after rounding). So Sue s salary before the rise was $62743.

44 40 CHAPTER 1. BASIC ARITHMETIC Example 3. Harry bought a new bike for $180, which includes 10% GST. How much money was spent on the GST? Solution: We work out the price without the GST first. We call this the original price and we should have price before GST + (price before GST 10%) = 180 or price before GST (1 + 10%) = 180. Thus price before GST = 180 ( ) (after rounding) Therefore = (dollars) was spent on the GST. Note: % is not the GST money spent.

45 Chapter 2 Introduction of Algebra Algebra is a generalised and extended type of arithmetic. In addition to the numbers used in arithmetic, algebra includes one or more letters or symbols. Apart from the reason that very often relationships between quantities are expressed most efficiently by algebra as in section 2.4, the use of letters or symbols helps with the logical thinking and analysis. To see this, let us reexamine some of the examples near the end of the previous chapter. (a) After a rise of 5%, Sue s current salary is $65,880. What was Sue s salary before the rise? In the solution, we used [old salary] in the logical thinking to obtain [old salary] + [old salary] 5%= If instead of [old salary], we use the letter x, we would obtain x + x 5% = x (1 + 5%) = x = (1 + 5%) = (after rounding) Clearly using x is more convenient than using [old salary]. 41

46 42 CHAPTER 2. INTRODUCTION OF ALGEBRA You may like to use this method, i.e., introducing a symbol, such as x, to stand for the unknown, in several other old examples to simplify the solutions. Let us now look at another question, where one can introduce symbols to help the logical thinking towards a solution to the problem. (b) In a room there are 33 tables, some of them are three-legged, the others are four-legged. If they have a total of 100 legs, how many are three-legged, and how many are four-legged? Solution: To help with the logical thinking, we use x to stand for the number of tables that are three-legged, and use y to stand for the number of tables that are four-legged. Then we should have the following relationship: x + y = 33 (There are 33 tables) (x 3) + (y 4) = 100 (They have a total of 100 legs). From the first relationship we find y = 33 x. Substituting this expression of y into the second relationship we deduce (x 3) + (33 x) 4 = 100 We treat x as a number (though its value is unknown at the moment) and can rewrite the above identity (by the distributive property) as (x 3) + (33 4) (x 4) = 100 (x 3) (x 4) = 100 x (3 4) = 100 x ( 1) = 100 x = 100 and we can figure out from the last relationship that x = 32, and then y = 33 x = = 1. So there are 32 tables with three legs, and only 1 table with four legs.

47 2.1. NOTATIONS AND REVIEW OF THE BASIC RULES IN ARITHMETIC 43 In the above, we have already applied algebraic calculations. These calculations will be discussed further in the following sections. 2.1 Notations and Review of the Basic Rules in Arithmetic In algebra, partly for simplification, partly for avoiding confusion of with x, the multiplication symbol is usually omitted from the algebraic expressions; so 2 x is written as 2x, a y is written as ay, etc. Sometimes, one also uses to replace, such as 2 x, a y, 5 6, 3 x or 3 ( x) etc. Also, a is often written as a/b. Thus 3 = 3/5, x = x/y, etc. b 5 y Let us recall the basic rules we have met in arithmetic: (a) Communicative rule: (b) Distributive rule: a + b = b + a, ab = ba (a a n )b = b(a a n ) = a 1 b + + a n b (a a n )/b = a 1 /b + + a n /b b/(a a n ) b/a b/a n! (c) Addition and subtraction of fractions: (i) Same denominator: (ii) Different denominator: a c + b c = a + b, c a c b c = a b c a 1 b 1 + a 2 b 2 = a 1b 2 b 1 b 2 + b 1a 2 b 1 b 2 = a 1b 2 + b 1 a 2 b 1 b 2 a 1 a 2 = a 1b 2 b 1a 2 b 1 b 2 b 1 b 2 b 1 b 2 = a 1b 2 b 1 a 2 b 1 b 2

48 44 CHAPTER 2. INTRODUCTION OF ALGEBRA (d) Multiplication and division of fractions: a 1 b 1 a2 b 2 = a 1a 2 b 1 b 2, a 1 b 1 a 2 b 2 = a 1/b 1 a 2 /b 2 = a 1 b 1 b2 a 2 = a 1b 2 b 1 a 2 (e) Powers a n = a }.{{.. a}, a n = 1/a n n times a n a m = a n+m, a n a m = a n /a m = a n m These rules are becoming increasingly important when we deal with calculations in this chapter, since we can not rely on the use of calculators as before. For example, to simplify the expression 3x8 x 2, we can only rely on these rules: 3x 8 x 2 = 3 x8 x 2 = 3 x8 2 = 3x Some Commonly Used Terminologies and Notations Algebra is used to refer to a generalized and extended type of arithmetic. In addition to the numbers used in arithmetic, algebra includes one or more letters or symbols. As we mentioned earlier, a symbol used in algebra, like x in x + x 5% = may represent an unknown, whose value can often be found through manipulations (or simplifications) of the identity it satisfied, like the one above. Introducing an unknown usually helps with the logical analysis to obtain a solution. A symbol may also used as a variable. For example, if an object moves at a constant speed 5 m/sec., then after 2 seconds, it travels 5 2 = 10 meters, after 4 seconds, it travels 5 4 = 20 meters,. In general, after t seconds, it travels 5t meters. Here t can be regarded as a variable. Expressions like 5t, x + x 5%, 2ab, 3x + 4y, 1 2 gt2, (ax 2 + 1)/(by)

49 2.2. SOME COMMONLY USED TERMINOLOGIES AND NOTATIONS 45 are called algebraic expressions, and in 3x + 4y, 3x and 4y are usually called the terms in the expression. Similarly, in (ax 2 + 1)/(by + 2), ax and by + 2 may be called terms, each of which in turn has two terms: ax 2 and 1 in ax 2 + 1; by and 2 in by + 2. These notions are not strictly defined and are used in loose terms in general. One should be careful to distinguish the differences between (ax 2 + 1)/(y 2 + 2), ax 2 + 1/(y 2 + 2), (ax 2 + 1)/y 2 + 2, and ax 2 + 1/y (ax 2 + 1)/(y 2 + 2) = ax2 + 1 y = (ax 2 + 1) (y 2 + 2) ax 2 + 1/(y 2 + 2) = ax y (ax 2 + 1)/y = ax2 + 1 y ax 2 + 1/y = ax y To avoid confusion, one should write a/(bc) instead of a/bc. The later may represent (a/b)c, and if that is the case, (a/b)c should be used. In general, in expressions like 5t, 3x + 4y, 1 2 gt2, the numbers 5, 3, 4 and 1 are called the 2 coefficients or numerical factors in the terms 5t, 3x + 4y, 1 2 gt2, respectively. Terms only differ by the coefficients (or numerical factors) are called like terms. So 3t and 105t are like terms, as are 2.81x 2 and 0.1x 2, 3ab and 8ab, 6x 2 y and 1 2 x2 y, etc. An algebraic expression with like terms can often be simplified by applying the distributive law. For example, 3t + 105t = ( )t = 108t 2.81x 2 0.1x 2 = ( )x 2 = 2.71x 2 6x 2 y 4( 1 2 x2 y 2) = 76x 2 y x2 y 4 ( 2) = 6x 2 y 2x 2 y + 8 = 4x 2 y + 8

50 46 CHAPTER 2. INTRODUCTION OF ALGEBRA 2x (6 + 2x) = 2x + [( 6) + ( 2x)] = 2x + ( 6) + ( 2x) = 2x + ( 2x) + ( 6) (commutative law) = 2x 2x 6 = (2 2)x 6 = (0 x) 6 = 6. One could simply write (once fluent with the manipulations) 2x (6 + 2x) = 2x 6 2x = 6 Similarly, t + 2t 2 (2 3t + 4t 2 ) = t + 2t 2 2 ( 3t) (4t 2 ) = t + 2t t 4t 2 = 2 + (t + 3t) + (2t 2 4t 2 ) = 2 + 4t 2t 2. In an expression like 1 + [2 (3 + 4)], we calculate (a) = 7 first, (b) then [2 (3 + 4)] = 2 7 = 5, (c) and finally 1 + [2 (3 + 4)] = 1 + ( 5) = 1 5 = 4 then [2 (3 + 4)] = 2 7 = 5.

51 2.2. SOME COMMONLY USED TERMINOLOGIES AND NOTATIONS 47 Similarly, 2 {1 + [2 (3 + 4)]} =2 {1 + [2 7]} =2 {1 5} =2 { 4} = 2 4 = 8. If symbols are involved, we have to apply the suitable arithmetic rules to simplify a similar expression, such as t {1 + [2 (t + t 2 )]} = t 1 + t [2 (t + t 2 )] = t + t 2 t (t + t 2 ) = t + 2t t t t t 2 = t + 2t t 2 t 3 = 3t t 2 t 3. We note that with numbers, the calculation starts from the inner most bracket, but with a general algebraic expression with symbols, the simplification starts from the out most bracket.

52 48 CHAPTER 2. INTRODUCTION OF ALGEBRA 2.3 Addition and Subtraction with Algebraic Expressions These are best illustrated by some examples. Example 1. Add the following (a) x 2 x + 2 and 6x 2 + 7x 8 (b) y 3 2y + 4 and 2y 3 + y 2 + 3y + 1 (c) 7ab 2 + 6ab + 1 and ab 2 7ab Solution: (a) x 2 x x 2 + 7x 8 =(x 2 + 6x 2 ) + ( x + 7x) + (2 8) =7x 2 + 6x 6 (group like terms) (b) y 3 2y y 3 + y 2 + 3y + 1 =(y 3 + 2y 3 ) + y 2 + ( 2y + 3y) + (4 + 1) =3y 3 + y 2 + y + 5 (group like terms) (c) 7ab 2 + 6ab ab 2 7ab =(7ab 2 + ab 2 ) + (6ab 7ab) + 1 =8ab 2 ab + 1 (group like terms) Example 2. Subtract the second expression from the first (a) 6x + y and 5x + y (b) 2a 5b + 6ab and a b + ab (c) 2p 3 + p 2 2p + 8 and 2p 2 + p + 4 Solution:

53 2.3. ADDITION AND SUBTRACTION WITH ALGEBRAIC EXPRESSIONS 49 (a) 6x + y (5x + y) (Note: 6x + y 5x + y is very different!) =6x + y 5x y =(6x 5x) + (y y) =x + 0 =x (b) 2a 5b + 6ab (a b + ab) =2a 5b + 6ab a ( b) ab =2a 5b + 6ab a + b ab =(2a a) + ( 5b + b) + (6ab ab) =a 4b + 5ab (c) 2p 3 + p 2 2p + 8 (2p 2 + p + 4) =2p 3 + p 2 2p + 8 2p 2 p 4 =2p 3 + (p 2 2p 2 ) + ( 2p p) + (8 4) =2p 3 p 2 3p + 4 In general, to add or subtract two or more algebraic expressions, one first group the like terms together (allowed by the commutative rule for additions, and one treats a b as a + ( b) and then add or subtract the like terms to obtain a simpler expression.

54 50 CHAPTER 2. INTRODUCTION OF ALGEBRA 2.4 Multiplication and Division with Algebraic Expressions Multiplication of algebraic expression is usually easier to calculate than division. In multiplication, one uses the distributive property repeatedly. For example, (a + b) 2 = (a + b)(a + b) = a(a + b) + b(a + b) = a a + a b + b a + b b = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2 (a b) 2 = (a b)(a b) = a(a b) b(a b) = a a a b b a b ( b) = a 2 ab ab + b 2 = a 2 2ab + b 2 (a + b)(a b) = a(a b) + b(a b) = a a a b + b a b b = a 2 ab + ab b 2 = a 2 b 2 Note that (a + b) 2 a 2 + b 2, (a b) 2 a 2 b 2. The formula (a + b) 2 = a 2 + 2ab + b 2, (a b) 2 = a 2 2ab + b 2, (a + b)(a b) = a 2 b 2 are very useful in algebraic calculations, and one should try to memorize them so that they can be used at ease.

55 2.4. MULTIPLICATION AND DIVISION WITH ALGEBRAIC EXPRESSIONS 51 Example 1. Expand the following expressions: (a) (3x + y)(x 6y) (b) (a 2b)(3a b) (c) (x + 2y) 2 (d) (x + 2y + 3z) 2 Solution: (a) (3x + y)(x 6y) =3x(x 6y) + y(x 6y) }{{}}{{} (i) (ii) =3x x 3x 6y + y x y 6y }{{}}{{} (i) (ii) (distributive law) (distributive law) =3x 2 18xy + xy 6y 2 (commutative law) =3x 2 17xy 6y 2 (b) (a 2b)(3a b) =a(3a b) 2b(3a b) }{{}}{{} (i) (ii) =a } 3a {{ a } b 2b 3a 2b ( b) }{{} (i) (ii) (distributive law) =3a 2 ab 6ab + 2b 2 (commutative law) =3a 2 7ab + 2b 2 (c) (x + 2y) 2 = (x + 2y)(x + 2y) (distributive law) = [x(x + 2y) + y(x + 2y)] (distributive law) = [x x + x 2y + 2y x + 2y 2y] (distributive law) = [x 2 + 2xy + 2xy + 4y 2 ] (commutative law) = [x 2 + 4xy + 4y 2 ] = x 2 4xy 4y 2 or, using the formula (a + b) 2 = a 2 + 2ab + b 2 with a = x, b = 2y to obtain

56 52 CHAPTER 2. INTRODUCTION OF ALGEBRA (x + 2y) 2 = [x 2 + 2x 2y + (2y) 2 ] = [x 2 + 4xy + 4y 2 ] = x 2 4xy 4y 2 (d) (x + 2y + 3z) 2 = (x + 2y + 3z)(x + 2y + 3z) x(x + 2y + 3z) = + 2y(x + 2y + 3z) + 3z(x + 2y + 3z) x 2 + 2xy + 3xz + 2xy + 4y 2 + 6yz = + 3xz + 6yz + 9z 2 x 2 + 4xy + 6xz + 4y yz + 9z 2 Example 2. Expand (x + y + z)(x y z) Solution: We could follow the steps as in Example 1 (d) above, but it is quicker to use the formula (a + b)(a b) = a 2 b 2 and (a + b) 2 = a 2 + 2ab + b 2 as follows: (x + y + z)(x y z) = [x + (y + z)][x (y + z)] = x 2 (y + z) 2 (using (a + b)(a b) = a 2 b 2 ) = x 2 (y 2 + 2yz + z 2 ) (using (a + b) 2 = a 2 + 2ab + b 2 ) = x 2 y 2 2yz z 2 Division of algebraic expressions can be written as a fraction, for example, (x 2 + x) (2x) = x2 + x 2x, (a 2 + 1) (a + 1) = a2 + 1 a + 1.

57 2.4. MULTIPLICATION AND DIVISION WITH ALGEBRAIC EXPRESSIONS 53 in Sometimes the fraction has a common factor in the numerator and the denominator, such as x 2 + x 2x = (x + 1) x, 2x where x is a common factor, which can be canceled, to obtain x 2 + x 2x = (x + 1) x 2 x = x Thus the fraction x2 +x 2x can be simplified to x+1 2. In the case of a2 +1, there is no common factor and no further simplification can be done. a+1 Therefore, generally speaking, to divide one algebraic expression by another algebraic expression, the main calculation is the cancelation process. So (x 2 + x) 2x = x2 + x 2x = (x + 1) x 2x = x (a 2 + 1) (a + 1) = a2 + 1 (no cancelation can be done) a + 1 (6x + 8)(9x + 3) 6 = 2(3x + 4) 3(3x + 1) 6 = 6 (3x + 4)(3x + 1) 6 = (3x + 4)(3x + 1) 1 = (3x + 4)(3x + 1) (expansion of this expression is not required)

58 54 CHAPTER 2. INTRODUCTION OF ALGEBRA 10a 2 b 2 2ab = 10a2 b 2 2ab 2ab 5ab = 2ab = 5ab 8xy 3 4x 2 y = 8xy3 4x 2 y 4xy 2y2 = 4xy x = 2y2 x. There are cases which do not look simplifiable at first sight, but can actually be simplified. Here are some examples, (a) a2 1 a + 1, (b) 4x2 y 2 2x y, (c) 4x 2 y 2 2x 2 xy y 2. For (a), we realize there is a common factor when using a 2 1 = (a + 1)(a 1) (a special case of a 2 b 2 = (a + b)(a b)). Hence a 2 1 a + 1 (a + 1)(a 1) = (a + 1) = a 1 For (b), we use 4x 2 y 2 = (2x) 2 y 2 = (2x + y)(2x y). So For (c), we use 4x 2 y 2 = (2x) 2 y 2 4x 2 y 2 2x y = (2x + y)(2x y) (2x + y)(2x y) = 2x y = 2x + y 2x 2 xy y 2 = 2x 2 + xy 2xy y 2 ( put xy = xy 2xy to allow factorization) = x(2x + y) y(2x + y) = (x y)(2x + y)

59 2.4. MULTIPLICATION AND DIVISION WITH ALGEBRAIC EXPRESSIONS 55 Hence In the above, the process of writing 4x 2 y 2 (2x + y)(2x y) = 2x 2 xy y 2 (x y)(2x + y) = 2x y x y a 2 1 = (a + 1)(a 1) 4x 2 y 2 = (2x + y)(2x y) 2x 2 xy y 2 = (x y)(2x + y) is called factorization. This is usually a difficult task, as we have seen for 2x 2 xy y 2. But in this unit, we will not encounter many difficult ones such as 2x 2 xy y 2 = (x y)(2x + y). However, we should be able to use a 2 b 2 = (a + b)(a b) fluently. Example 3. Simplify the following expressions: (a) (b) (c) 21rs 3 (2p + q) 4p2 q r 2 s V V S (1 + E V ) m r x m r + x Solution. (a) (b) 21rs 3 (2p + q) 4p2 q r 2 s 3 7 r s s2 = (2p + q) 2 = 7s2 2p + q 2p q 5r = 7s2 (2p q) 5r(2p + q). V V S (1 + E V ) = V V S (V V + E V ) (2p + q)(2p q) 3 5 r s r