Arithmetic for 5th grade, Summaries of Mathematics

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ARITHMETIC: A Textbook for Math 01 5th edition (2015) Anthony Weaver Department of Mathematics and Computer Science Bronx Community College Page 2 Contents 1 Whole numbers 9 1.0.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.1 Adding Whole Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.1.1 Commutativity, Associativity, Identity . . . . . . . . . . . . . . . . . . . . . . . 10 1.1.2 Multi-digit addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.2 Subtracting Whole Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.2.1 Commutativity, Associativity, Identity . . . . . . . . . . . . . . . . . . . . . . . 15 1.2.2 Multi-digit subtractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.2.3 Checking Subtractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.2.4 Borrowing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 1.3 Multiplying Whole Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.3.1 Commutativity, Associativity, Identity, the Zero Property . . . . . . . . . . . . . 21 1.3.2 Multi-digit multiplications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 1.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 1.4 Powers of Whole Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 1.4.1 Squares and Cubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1.4.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1.4.3 Square Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 1.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.5 Division of Whole Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.5.1 Quotient and Remainder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 1.5.2 Long Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 1.5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 1.6 Division and 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 1.7 Order of operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 1.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 1.8 Average . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 1.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 1.9 Perimeter, Area and the Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . 40 1.9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 2 Signed Numbers 49 2.1 Absolute value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 2.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 5 2.2 Inequalities and signed numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 2.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 2.3 Adding signed numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 2.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 2.3.2 Opposites, Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 2.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 2.3.4 Associativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 2.3.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 2.4 Subtracting signed numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 2.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 2.5 Multiplying Signed Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 2.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 2.6 Dividing Signed Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 2.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 2.7 Powers of Signed Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 2.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 2.8 Square Roots and Signed Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 2.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 3 Fractions and Mixed Numbers 71 3.1 What positive fractions mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 3.2 Proper and Improper Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 3.2.1 Zero as Numerator and Denominator . . . . . . . . . . . . . . . . . . . . . . . . 74 3.2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 3.3 Mixed Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 3.3.1 Converting an improper fraction into a mixed or whole number . . . . . . . . . . 75 3.3.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 3.3.3 Converting a mixed or whole number to an improper fraction . . . . . . . . . . . 77 3.3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 3.4 Multiplication of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 3.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 3.5 Equivalent Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 3.5.1 Cancellation and Lowest Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 3.5.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 3.6 Prime Factorization and the GCF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 3.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 3.6.2 Finding the GCF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 3.6.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 3.6.4 Cancelling the GCF for lowest terms . . . . . . . . . . . . . . . . . . . . . . . . 87 3.6.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 3.7 Pre-cancelling when Multiplying Fractions . . . . . . . . . . . . . . . . . . . . . . . . . 89 3.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.8 Adding and Subtracting Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 3.8.2 Adding and Subtracting Unlike Fractions . . . . . . . . . . . . . . . . . . . . . . 93 3.8.3 The LCM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 3.8.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Page 6 3.8.5 The LCD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 3.8.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 3.9 Comparison of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 3.9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 3.10 Division of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 3.10.1 Reciprocals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 3.10.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 3.10.3 Division is Multiplication by the Reciprocal of the Divisor . . . . . . . . . . . . . 101 3.10.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 3.11 Mixed Numbers and Mixed Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 3.11.1 Vertical Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . 103 3.11.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 3.11.3 Measurements in Mixed Units . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 3.11.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 3.12 Signed fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 3.12.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 3.13 Combined operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 3.13.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 4 Decimals and Percents 119 4.1 Decimal place values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 4.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 4.2 Significant and Insignificant 0’s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 4.3 Comparing Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 4.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 4.4 Rounding-off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 4.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 4.5 Adding and Subtracting Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 4.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 4.6 Multiplying and Dividing Decimals by Powers of 10 . . . . . . . . . . . . . . . . . . . . 129 4.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 4.7 Multiplication of general decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 4.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 4.8 Division of a decimal by a whole number . . . . . . . . . . . . . . . . . . . . . . . . . . 134 4.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 4.9 Division of a decimal by a decimal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 4.9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 4.10 Order of magnitude, Scientific notation . . . . . . . . . . . . . . . . . . . . . . . . . . 141 4.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 4.11 Percents, Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 4.11.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 4.12 Fractional parts of numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 4.12.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 Page 7 1.0.1 Exercises 1. 35 stands for 2. 209 stands for 3. 9532 stands for 4. 21045 stands for 1.1 Adding Whole Numbers When we add two or more integers, the result is called the sum. We assume you know the sums of single-digit numbers. For practice, do the following example. Example 2. Fill in the missing squares in the digit-addition table below. For example, the number in the row labelled 3 and the column labelled 4 is the sum 3 + 4 = 7. + 0 1 2 3 4 5 6 7 8 9 0 0 1 2 3 4 5 6 7 8 9 1 2 3 7 4 5 6 7 12 8 9 Table 1.1: The digit addition table 1. Do you notice any patterns or regularities in the digit-addition table? Can you explain them? 2. Why is the second line from the top identical to the top line? 3. What can you say about the left-most column and the second-from-left column? 1.1.1 Commutativity, Associativity, Identity The first question above leads us to an important property of addition, namely, that for any two numbers x and y , x + y = y + x . Page 10 In other words, the order in which two numbers are added does not effect the sum. This property of addition is called commutativity. The last two questions lead us to an important property of 0, namely, for any number x , x + 0 = x = 0 + x . In other words, when 0 is added to any number, x , you get the identical number, x , again. Because of this property, 0 is called the additive identity. One final property of addition is expressed in the following equation (x + y) + z = x + (y + z), which says that if three numbers are added, it doesn’t matter how you “associate” the additions: you can add the first two numbers first, and then add the third to that, or, you could add the second two numbers first, and then add the first to that. This property of addition is called associativity. Example 3. Find the sum of 2, 3, and 5 by associating in two different ways. Solution. Associating 2 and 3, we calculate (2 + 3) + 5 = 5 + 5 = 10. On the other hand, associating 3 and 5, we calculate 2 + (3 + 5) = 2 + 8 = 10. The sum is the same in both cases. 1.1.2 Multi-digit addition To add numbers with more than one digit, we line up the numbers vertically so that the ones places (right-most) are directly on top of each other, and all other places are similarly arranged. Then we add the digits in each place to obtain the sum. Example 4. Find the sum of 25 and 134. Solution. We line up the numbers vertically so that the 5 in the ones place of 25 is over the 4 in the ones place of 134. If we do this carefully, all the other places line up vertically, too. So there is a “ones” column, a “tens” column to the left of it, and a “hundreds” column to the left of that: 2 5 1 3 4 Then we draw a line and add the digits in each column to get the sum: 2 5 + 1 3 4 1 5 9 Page 11 Sometimes we need to “carry” a digit from one place to the next higher place. For example, when adding 38 + 47, we first add the ones places, 8 + 7 = 15. But 15 has two digits: it stands for 1 ten + 5 ones. We “put down” the 5 in the ones place, and “carry” the 1 (standing for a ten) to the tens column. So we have: Example 5. Find the sum 3 8 + 4 7 Solution. Put down the 5 in the ones place: 3 8 + 4 7 5 and carry the 1 to the top of the tens column: {1} 3 8 + 4 7 5 and finish the job by adding up the tens column, including the carried one: {1} 3 8 + 4 7 8 5 The sum is 85. Since carrying alters the sum in the column immediately to the left of the one we are working on, in multi-digit addition, we always start with the right-most column and proceed leftward. 1.1.3 Exercises Find the sums, carrying where necessary. 1. 2 6 + 5 5 Page 12 − 0 1 2 3 4 5 6 7 8 9 0 * * * * * * * * * 1 * * * * * * * * 2 * * * * * * * 3 0 * * * * * * 4 * * * * * 5 * * * * 6 * * * 7 5 * * 8 * 9 4 Table 1.2: The digit subtraction table 1.2.1 Commutativity, Associativity, Identity When we study negative numbers, we will see that subtraction is not commutative. We can see by a simple example that subtraction is also not associative. Example 7. Verify that (7− 4)− 2 is not equal to 7− (4− 2). Solution. Associating the 7 and 4, we get (7− 4)− 2 = 3− 2 = 1, but associating the 4 and the 2, we get 7− (4− 2) = 7− 2 = 5, a different answer. Until we establish an order of operations, we will avoid examples like this! It is true that x − 0 = x for any number x . However, 0 is not an identity for subtraction, since 0 − x is not equal to 0 (unless x = 0). To make sense of 0− x , we will need negative numbers. 1.2.2 Multi-digit subtractions To perform subtractions of multi-digit numbers, we need to distinguish the number “being diminished” from the number which is “doing the diminishing” (being taken away). The latter number is called the subtrahend, and the former, the minuend. For now, we take care that the subtrahend is no larger than the minuend. To set up the subtraction, we line the numbers up vertically, with the minuend over the subtrahend, and the ones places lined up on the right. Example 8. Find the difference of 196 and 43. Solution. The subtrahend is 43 (the smaller number), so it goes on the bottom. We line up the numbers vertically so that the 6 in the ones place of 196 is over the 3 in the ones place of 43. 1 9 6 4 3 Then we draw a line and subtract the digits in each column, starting with the ones column and working right to left, to get the difference: Page 15 1 9 6 − 4 3 1 5 3 1.2.3 Checking Subtractions Subtraction is the “opposite” of addition, so any subtraction problem can be restated in terms of addition. Using the previous example, and adding the difference to the subtrahend, we obtain 1 5 3 + 4 3 1 9 6 which is the original minuend. In general, if subtraction has been performed correctly, adding the difference to the subtrahend returns the minuend. This gives us a good way to check subtractions. Example 9. Check whether the following subtraction is correct: 9 4 − 5 1 3 3 Solution. Adding the (supposed) difference to the subtrahend, we get 3 3 + 5 1 8 4 which is not equal to the minuend (94). Thus the subtraction is incorrect. We leave it to you to fix it! 1.2.4 Borrowing Sometimes, when subtracting, we need to “borrow” a digit from a higher place and add its equivalent to a lower place. Example 10. Find the difference: 8 5 − 4 6 Solution. The digit subtraction in the ones column is not possible (we can’t take 6 from 5). Instead we remove or ”borrow” 1 ten from the tens place of the minuend, and convert it into 10 ones, which we add to the ones in the ones place of the minuend. The minuend is now represented as {7}{15}, Page 16 standing for 7 tens + 15 ones. It looks funny (as if 15 were a digit), but it doesn’t change the value of the minuend, which is still 7× 10 + 15 = 85. We represent the borrowing operation like this: {7}{15} 8 5 − 4 6 Ignoring the original minuend, we have 15− 6 = 9 for the ones place, and 7− 3 = 4 for the tens place, as follows: {7}{15} 8 5 − 4 6 3 9 The difference is 39. We can check this by verifying that the difference + the subtrahend = the original minuend: 3 9 + 4 6 8 5 Sometimes you have to go more than one place to the left to borrow successfully. This happens when the next higher place has a 0 digit – there is nothing to borrow from. Example 11. Find the difference: 2 0 7 − 6 9 Solution. In the ones column we can’t take 9 from 7, so we need to borrow from a higher place. We can’t borrow from the tens place, because it has 0 tens. But we can borrow from the hundreds place. We borrow 1 hundred, and convert it into 9 tens and 10 ones. The minuend is now represented as {1}{9}{17}, standing for 1 hundred + 9 tens + 17 ones, (as if 17 were a digit). We represent the borrowing as before: {1}{9}{17} 2 0 7 − 6 9 Ignoring the original minuend, we have 17− 9 = 8 for the ones place, 9− 6 = 3 for the tens place, and 1− 0 = 1 for the hundreds place: {1}{9}{17} 2 0 7 − 6 9 1 3 8 Page 17 1.3 Multiplying Whole Numbers Multiplication is really just repeated addition. When we say “4 times 3 equals 12,” we can think of it as starting at 0 and adding 3 four times over: 0 + 3 + 3 + 3 + 3 = 12. We can leave out the 0, since 0 is the additive identity ( 0+3 = 3). Using the symbol× for multiplication, we write 3 + 3 + 3 + 3 = 4× 3 = 12. The result of multiplying two or more numbers is called the product of the numbers. Instead of the × symbol, we often use a central dot (·) to indicate a product. Thus, for example, instead of 2× 4 = 8, we can write 2 · 4 = 8. Example 12. We assume you remember the products of small whole numbers, so it should be easy for you to reproduce the partial multiplication table below. For example, the number in the row labelled 7 and the column labelled 5 is the product 7 · 5 = 35. × 0 1 2 3 4 5 6 7 8 9 10 11 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 8 9 10 11 12 2 0 2 4 6 8 10 12 14 16 18 20 22 24 3 0 3 6 9 12 15 18 21 24 27 30 33 36 4 0 4 8 12 16 20 24 28 32 36 40 44 48 5 0 5 10 15 20 25 30 35 40 45 50 55 60 6 0 6 12 18 24 30 36 42 48 54 60 66 72 7 0 7 14 21 28 35 42 49 56 63 70 77 84 8 0 8 16 24 32 40 48 56 64 72 80 88 96 9 0 9 18 27 36 45 54 63 72 81 90 99 108 10 0 10 20 30 40 50 60 70 80 90 100 110 120 11 0 11 22 33 44 55 66 77 88 99 110 121 132 12 0 12 24 36 48 60 72 84 96 108 120 132 144 Table 1.3: Multiplication table 1. Do you notice any patterns or regularities in the multiplication table? Can you explain them? 2. Why does the second row from the top contain only 0’s? 3. Why is the third row from the top identical to the first row? 4. Is multiplication commutative? How can you tell from the table? Page 20 1.3.1 Commutativity, Associativity, Identity, the Zero Property An examination of the multiplication table leads us to an important property of 0, namely, when any number, N, is multiplied by 0, the product is 0: 0 · N = N · 0 = 0. It also shows us an important property of 1, namely, when any number, N, is multiplied by 1, the product is the identical number, x , again: 1 · N = N · 1 = N. For this reason, 1 is called the multiplicative identity. The following example should help you to see that multiplication is commutative. Example 13. The figure shows two ways of piling up twelve small squares. On the left, we have piled up 3 rows of 4 squares (3 × 4); on the right, we have piled up 4 rows of 3 squares (4 × 3). In both cases, of course, the total number of squares is the product 3× 4 = 4× 3 = 12. Figure 1.1: 3× 4 = 4× 3 = 12 Examples like the following help you to see that multiplication is associative. Example 14. We can find the product 3× 4× 5 in two different ways. We could first associate 3 and 4, getting (3× 4)× 5 = 12× 5 = 60, or we could first associate 4 and 5, getting 3× (4× 5) = 3× 20 = 60. The product is the same in both cases. 1.3.2 Multi-digit multiplications To multiply numbers when one of them has more than one digit, we need to distinguish the number “being multiplied” from the number which is “doing the multiplying.” The latter number is called the multiplier, and the former, the multiplicand. It really makes no difference which number is called the multiplier and which the multiplicand (because multiplication is commutative!). But it saves space if we choose the multiplier to be the number with the fewest digits. To set up the multiplication, we line the numbers up vertically, with the multiplicand over the multiplier, and the ones places lined up on the right. Page 21 Example 15. To multiply 232 by 3, we write 2 3 2 × 3 We multiply place-by-place, putting the products in the appropriate column. 3 × 2 ones is 6 ones, so we put 6 in the ones place 2 3 2 × 3 6 3 × 3 tens is 9 tens, so we put 9 in the tens place 2 3 2 × 3 9 6 Finally, 3 × 2 hundreds is 6 hundreds, so that we put down 6 in the hundreds place, and we are done: 2 3 2 × 3 6 9 6 Sometimes we need to carry a digit from one place to the next higher place, as in addition. For example, 4 × 5 = 20, a number with two digits. So we would “put down” the 0 in the current place, and “carry” the 2 to the column representing the next higher place, as in the next example. Example 16. Multiply 251 by 4. Solution. The steps are 2 5 1 × 4 For the ones place, 2 5 1 × 4 4 For the tens place, 4× 5 = 20, so we put down the 0 in the tens place and carry the 2 to the hundreds column: {2} 2 5 1 × 4 0 4 Page 22 1.3.3 Exercises Find the products. 1. 1 2 2 × 4 2. 8 3 × 5 3. 1 0 4 × 7 4. 3 0 0 8 × 9 5. 2 1 2 × 4 3 6. 8 3 × 5 6 Page 25 7. 1 3 6 × 2 7 8. 3 0 8 × 1 0 9 9. 2 1 0 3 × 4 4 10. 8 3 7 × 5 4 1.4 Powers of Whole Numbers If we start with 1 and repeatedly multiply by 3, 4 times over, we get a number that is called the 4th power of 3, written 34 = 1× 3× 3× 3× 3. The factor of 1 is understood and usually omitted. Instead we simply write 34 = 3× 3× 3× 3. In the expression 34, 3 is called the base, and 4 the exponent (or power). Example 19. The 5th power of 2, or 25, is the product 2× 2× 2× 2× 2 = 32. The 3rd power of 4, or 43, is the product 4× 4× 4 = 64. We make the following definition in the cases where the exponent is 0. Page 26 For any nonzero number N, N0 = 1. (00 is undefined.) This makes sense if you remember the harmless factor of 1 that is understood in every exponential expression. (You’ll see another justification when you study algebra.) Although 00 is undefined, expressions such as 02, 03, etc., with 0 base and nonzero exponent, make perfect sense (e.g., 03 = 0× 0× 0 = 0). Example 20. 170 = 1. 05 = 0. 01 = 0. 00 is undefined. 1.4.1 Squares and Cubes Certain powers are so familiar that they have special names. For example, the 2nd power is called the square and the 3rd power is called the cube. Thus 52 is read “5 squared,” and 73 is read as “7 cubed.” The source of these special names is geometric (see Section 1.9). The area of a square, x units on a side, is x2 square units. This means that the square contains x2 small squares, each one unit on a side. For example, the figure below shows a square 6 units on a side, with area 62 = 36 square units. = Similarly, the volume of a cube, y units on a side, is y3 cubic units. This means that the cube contains y3 little cubes, each one unit on a side. For example, the volume of an ice cube that measures 2 cm (centimeters) on a side is 23 = 8 cubic centimeters. 1.4.2 Exercises 1. Rewrite using an exponent: 8× 8× 8× 8 2. Rewrite using an exponent: 4 · 4 · 4 · 4 · 4 · 4 3. Evaluate 25 4. Evaluate 90 5. Evaluate 07 6. Evaluate 54 7. Evaluate 102 Page 27 easy to see that there is something “left over,” namely, 8. Here are the computations: 1 0 0 − 2 3 7 7 − 2 3 5 4 − 2 3 3 1 − 2 3 8 This operation (repeated subtraction) is called division. The number we start with (100 in the example) is called the dividend, and the number we repeatedly subtract (23 in the example) is called the divisor. We use the symbol ÷, and note that the dividend is written first: dividend÷ divisor. 1.5.1 Quotient and Remainder Unlike the other three operations (addition, subtraction, multiplication), the result of a division of whole numbers consists of not one but two whole numbers: the number of subtractions performed (4 in the example), and the number left over (8 in the example). These two numbers are called the quotient and the remainder, respectively. Whole number divisions with remainder 0 are called exact. For example, 48÷ 6 has quotient 8 and remainder 0, so the division is exact and we can write 48÷ 6 = 8, with the understanding that the remainder is 0. Exact divisions can be restated in terms of multiplication. Subtracting 6 (8 times) from 48 yields exactly 0. On the other hand, starting at 0 and adding 6 (8 times) returns 48. Recalling that multiplication is a shorthand for this kind of repeated addition, we see that the two statements 48÷ 6 = 8 and 48 = 6 · 8 say exactly the same thing. In general, a ÷ b = c and a = b · c are equivalent statements. Example 23. Express the statement 72 = 8 · 9 as an exact division in two ways. Solution. We can get to 0 by starting at 72 and repeatedly subtracting 8 (9 times), or by repeatedly subtracting 8 (9 times). So, using the division symbol, we can write 72÷ 8 = 9 or we can write 72÷ 9 = 8. Page 30 Example 24. The division 63÷ 7 is exact. Express this fact using an appropriate multiplication. Solution. Since the quotient is 9 and the remainder is 0, we can write 63÷ 7 = 9 or 63 = 9 · 7. 1.5.2 Long Division Divisions with multi-digit divisors and/or dividends can get complicated, so we remind you of a standard way (long division) of organizing the computations. Here is what it looks like: dividenddivisor quotient *** *** *** *** *** remainder The horizontal lines indicate subtractions of intermediate numbers; there is one subtraction for each digit in the quotient. For example, the fact that the division 100 ÷ 23 has quotient 4 and remainder 8 is expressed in the long division form as follows: 1 0 02 3 4 − 9 2 8 The 4 repeated subtractions of 23 are summarized as the single subtraction of 4 · 23 = 92. In long division, we try to estimate the number of repeated subtractions that will be needed, multiply this estimate by the divisor, and hope for a number that is close to, but not greater than, the dividend. It will be easy to see when our estimate is too large, and to adjust it downward. If it is too small, the result of the subtraction will be too big – there was actually “room” for further subtraction. Going back to our example, 100÷ 23, it is more or less clear that we will need more than 2 subtractions, since 23× 2 = 46, which leaves a big remainder of 54 (bigger than the divisor, 23). 23× 3 = 69, which also leaves a remainder that is too big. Since 23×5 = 115, which is bigger than the dividend, 100, we know that the best estimate for the quotient is 4. Now 23 × 4 = 92. We subtract 92 from 100, leaving the remainder 8, which is less than the divisor, as it should be. To check our calculations, we verify that 23 × 4 + 8 = 100. (In this check, the multiplication is done before the addition – this is the standard Page 31 order of operations, which we will say more about later.) Let’s try another simple division problem. Example 25. Find the quotient and remainder of the division 162÷ 42. Solution. Putting the dividend and divisor into the long division form, we have 1 6 24 2 Let’s estimate the number of subtractions that we’ll need to perform. 40 goes into 160 four times, so, perhaps four is a good guess. But 4 × 42 = 168, which is too big (bigger than the dividend). So we lower our estimate by 1. We get 3 × 42 = 126, which is less than the dividend, so this must be the right choice. 1 6 24 2 3 −1 2 6 Subtracting, we obtain the remainder 1 6 24 2 3 −1 2 6 3 6 which is less than the divisor, as it should be. Thus, the quotient is 3 and the remainder is 36. As a check, we verify that 3× 42 + 36 = 162. (Multiplication before addition.) Page 32 306015 204 − 30 060 − 60 0 We have reached the end of our dividend – there are no more digits to bring down. Hence, our division stops. The quotient is 204, and the remainder is 0. We check by verifying that 15×204+0 = 3060. 1.5.3 Exercises Express each exact division as an equivalent multiplication. 1. 88÷ 22 = 4 2. 42÷ 6 = 7 3. 51÷ 17 = 3 4. 168÷ 14 = 12 5. 96÷ 12 = 8 Find the quotient and remainder of each division. 6. 37÷ 11 7. 712÷ 101 8. 3007 ÷ 110 9. 3333 ÷ 111 10. 3456 ÷ 241 11. 457÷ 41 12. 578÷ 19 13. 317÷ 21 14. 907÷ 201 15. 712÷ 21 Page 35 1.6 Division and 0 Can 0 be the divisor in a division problem? If you think so, then answer this: how many times does 0 go into 7? Thinking back to our definition of division as repeated subtraction, the question is “how many times can 0 be subtracted from 7 before we arrive at a negative number?” The answer is “as many times as you like!,” since subtracting 0 has no effect on 7, or any other number. For this reason, we say that division by 0 is undefined. On the other hand, it is easy to answer the question “how many times does 7 go into 0?” The answer is 0 times! You can’t take away any 7’s (or any other positive numbers) from 0 without obtaining a negative number. So 0÷ 7 = 0. 1.7 Order of operations We often do calculations that involve more than one operation. For example 1 + 2× 3 involves both addition and multiplication. Which do we do first? If we do the multiplication first, the result is 1 + 6 = 7, and if we do the addition first, the result is 3 × 3 = 9. Obviously, if we want the expression 1 + 2× 3 to have a definite and unambiguous meaning, we need a convention or agreement about the order of operations. It could have been otherwise, but the convention in this case is: multiplication before addition. With this convention, when I write 1 + 2 × 3, you know that I mean 7 (not 9). The precedence of multiplication can be made explicit using the grouping symbols () (parentheses): 1 + (2× 3) = 1 + 6 = 7. If one of us insists that the addition be done first, we can do that by re-setting the parentheses: (1 + 2)× 3 = 3× 3 = 9. Thus grouping symbols can be used to force any desired order of operations. Common grouping symbols, besides parentheses, are brackets, [ ], and braces, {}. The square root symbol √ is also a grouping symbol. For example √ 4 + 5 = √ 9 = 3. The √ symbol acts like a pair of parentheses, telling us to evaluate what is inside (in this case, the sum 4 + 5) first, before taking the square root. The order of operations is: 1. operations within grouping symbols first; 2. exponents and roots next; 3. multiplications and divisions (in order of appearance) next; Page 36 4. additions and subtractions (in order of appearance) last. “In order of appearance” means in order from left to right. Thus in the expression 2 + 5− 3, the addition comes first, so it is evaluated first, 2 + 5− 3 = 7− 3 = 4, while in the expression 8− 6 + 11, the subtraction is done first because it comes first, 8− 6 + 11 = 2 + 11 = 13. Example 27. Evaluate 6 · 5− 4÷ 2 + 2. Solution. All four operations appear here. There are no grouping symbols, exponents or roots. Following the order of operations, as well as the order of appearance, we do the computations in the following order: multiplication, division, subtraction, addition. The computations are as follows: 6 · 5− 4÷ 2 + 2 = 30− 4÷ 2 + 2 = 30− 2 + 2 = 28 + 2 = 30 In the next two examples, we use the same numbers and the same operations, but we insert grouping symbols to change the order of operations. As expected, this changes the final result. Example 28. Evaluate 6 · (5− 4)÷ 2 + 2. Solution. The grouping symbols force us to do the subtraction first. After that, the usual order of operations is followed. 6 · (5− 4)÷ 2 + 2 = 6 · 1÷ 2 + 2 = 6÷ 2 + 2 = 3 + 2 = 5 Example 29. Evaluate 6 · 5− 4÷ [2 + 2]. Page 37 1.8.1 Exercises Find the average of each of the following multi-sets of numbers. 1. {1, 2, 3, 4, 5, 6, 7} 2. {13, 13, 19, 15} 3. {206, 196, 204} 4. {85, 81, 92} 5. A baseball team had 7 games cancelled due to rain in the 2010 season. The number of cancelled games in the 2002-2009 seasons were 5, 6, 2, 10, 9, 4, 6, 5. What was the average number of cancelled games for the 2002-2010 seasons? 6. Suppose the average of the multi-set {20, 22,N, 28} is 25, where N stands for an unknown number. Find the value of N. 1.9 Perimeter, Area and the Pythagorean Theorem Squares and rectangles are examples of polygons – closed shapes that can be drawn on a flat surface, using segments of straight lines which do not cross each other. “Closed” means that the line segments form a boundary, with no gaps, which encloses a unique “inside” region, and separates it from the “outside” region. Example 33. Which of the following figures are polygons? (a) (b) (c) (d) (e) Solution. (b) is not a polygon since not all of its sides are straight lines. (c) is not a polygon because it does not have a unique ”inside” region. (d) is not a polygon because it is not closed. There are two useful numerical quantities associated with a polygon: the perimeter, which is the length of its boundary, and the area, which is (roughly speaking) the “amount of space” it encloses. Perimeters are measured using standard units of length such as feet (ft), inches (in) , meters (m), centimeters (cm). Areas are measured using square units, such as square feet (ft2), square inches (in2), square meters (m2), square centimeters (cm2). To find the perimeter of a polygon, we simply find the sum of the lengths of its sides. Example 34. Find the perimeter of each polygon. Assume the lengths are measured in feet. Page 40 3 4 1 5 5 7 5(a) 6 1 1 1 (b) Solution. Adding the lengths of the sides, we find that the perimeter of polygon (a) is 3 + 4 + 1 + 5 + 5 = 18 ft. Similarly, the perimeter of the polygon (b) is 7 + 5 + 6 + 1 + 1 + 1 = 21 ft. Finding the area of a polygon can be complicated, but it is quite simple if the polygon can be divided up into rectangles. A rectangle is a four-sided polygon with two pairs of opposite parallel sides, and “square” corners. The square corners – also known as right angles – imply that the paired opposite sides have the same length. The length of the shorter pair of sides is often called the width and denoted w , and the length of the longer pair is called the length and denoted l . length = l width = w l w The small squares in the corners are there to indicate that the polygon is a rectangle. Example 35. Estimate the area of the rectangle below, if the square in the upper left corner is 1 unit on a side. length = l width = w l w 1 1 Page 41 Solution. If the corner square is 1 unit on a side, the area of the rectangle is the number of squares of that size that fit inside the rectangle. As the next figure shows, 21 such square units fit inside, in 3 rows of 7. The width (w) is evidently 3 units, and the length (l) 7 units. The area is the product 3× 7 = 21 square units. length = 7 width = 3 7 3 The example illustrates a general fact: the area of a rectangle is the product of the length and the width. If A denotes the area, the formula is A = l · w . It is also clear that the perimeter of a rectangle is the sum of twice the length and twice the width, since to travel all the way around the boundary of the rectangle, both the length and the width must be traversed twice. If P denotes the perimeter, the formula is P = 2l + 2w . Example 36. In the figure below we have drawn an L-shaped polygon on a grid. Find the area and perimeter of the polygon. Take the unit of area to be one of the large squares of the grid, and assume these squares are 1 cm on a side. Page 42 6 5 3 2 1 5 Solution. There are some missing side lengths (for example, the vertical side on the upper right), but the area can be determined without them, using subtraction. We can visualize the polygon as a large rectangle of length 6 ft. and width 5 ft, out of which a rectangular “bite,” of length 3 ft and width 2 ft, has been taken. 6 5 3 “bite” 2 1 5 The area of the large rectangle (bite included) is 6 ·5 = 30 ft2, and the area of the bite alone is 3 ·2 = 6 ft2. Thus the area of the original polygon is 30 − 6 = 24 ft2. To find the perimeter, we need the missing side lengths. For the vertical side on the upper right, we reason as follows. The left edge is 6 ft, and so the total length of all the right vertical edges must also be 6 ft. The two known vertical edges on the right have lengths 1 and 3, and the unkown edge must Page 45 make up the difference. So its length must be 6− (1 + 3) = 2 ft. The other missing side (the short horizontal one) is 2 ft. Adding up all the side lengths, starting (arbitrarily) with the bottom edge and proceeding counter-clockwise, yields the perimeter: 5 + 1 + 2 + 3 + 2 + 2 + 5 + 6 = 26 ft. If we cut a rectangle in two by drawing a diagonal from one corner to the opposite corner, we get two right triangles, each having exactly the same size and shape, and therefore the same area. Right triangles are interesting in their own right, and we often consider them in isolation, without reference to the rectangle they came from. The side of a right triangle that is opposite the right angle (the longest side) is known as the hypotenuse. The two shorter sides are called the legs. In the figure below, the legs are labeled a and b, and the hypotenuse is labeled c . a b c A famous formula, called the Pythagorean theorem, states that, in any right triangle with legs of length a and b, and hypotenuse of length c , the following relation holds: a2 + b2 = c2. With this, if we know the lengths of any two of the three sides, we can find the length of the third side. For example, we can easily obtain a formula for the length of the hypotenuse in terms of the lengths of the legs. c = ! a2 + b2. Page 46 Since the area of a right triangle is exactly half the area of the rectangle it came from, it follows that the area of a right triangle is the product of the lengths of the legs, divided by 2. If A denotes the area, and the lengths of the legs are a and b as in the figure, the formula is A = (a · b)÷ 2. Example 38. Find the area and perimeter of a triangle whose legs have length 3 feet and 4 feet. Solution. Putting a = 4 and b = 3, we find the area using the formula A = (a · b)÷ 2 = (4 · 3)÷ 2 = 12÷ 2 = 6 ft2. To find the perimeter, we first find the length of the hypotenuse using the formula c = ! a2 + b2 = ! 42 + 32 = √ 16 + 9 = √ 25 = 5 ft. Thus the perimeter is a + b + c = 4 ft +3 ft +5 ft = 12 ft. Now we can find the area and perimeter of any polygon that can be divided up into rectangles and right triangles. Example 39. Find the area and perimeter of the polygon below. The large squares of the grid measure 1 centimeter (cm) on a side. 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 Solution. Drawing two vertical lines, we can divide up the polygon into two right triangles, T1 and T2, and a rectangle, R . 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 T1 T2R Page 47 | | | | | | | | | | | 0 1 √ 2 π- √ 2-π 2 3 4 5−1−2−3−4−5 2.1.1 Exercises Find the absolute value of the following numbers: 1. 12 2. −5 3. −π 4. √ 5 5. − √ 5 6. 0 7. −260 Find two numbers that have the given absolute value: 8. 99 9. 5 10. π 11. √ 7 2.2 Inequalities and signed numbers The inequality symbols < (less than) and > (greater than) are used to state size comparisons between numbers. It is intuitively clear that 2 < 5, and 15 > 10 because these are comparisons between nonnegative numbers. Can we make sense of the inequality symbols when negative numbers are involved? It seems reasonable to say that any negative number is less than 0, and therefore less than any positive number. How do we compare two negative numbers? For example, is −4 greater or less than −2? We laid out the number line so that numbers increase as we move to the right, and decrease as we move to the left. For consistency, we extend that rule to the negative side. The rule of thumb is “left is less” on the number line. Page 50 “Left is Less” on the number line: M < N if and only if M is left of N | | | | | | | | | | | 0 1 N M< <M N 2 3 4 5−1−2−3−4−5 With this rule, we see that −4 is less than −2 −4 < −2. Recall that √ 2 is less than √ 3, and that both are greater than 1 but less than 2. (If you’ve forgotten why, review the discussion in Section 1.4.3.) | | | | | 0 √ 2 √ 3 1 2−1−2 In symbols, 1 < √ 2 < √ 3 < 2. Do these relationships remain true if all the signs are changed? On the number line, 1 is left of 2, but, on the negative side, −2 is left of −1. Since “left is less,” it follows that −2 < −1. For the same reason, − √ 3 < − √ 2. | | | | | 0 - √ 2- √ 3 1 2−1−2 In symbols, −2 < − √ 3 < − √ 2 < −1. The mirror analogy helps here: when viewed in a mirror, your right hand ‘becomes’ a left hand – the thumb switches from one side to the other. Putting everything on one number line, | | | | | 0 √ 2 √ 3- √ 2- √ 3 1 2−1−2 we see a chain of inequalities, mirror-symmetric (0 acts as the mirror), except for the presence of negative signs on the left: −2 < − √ 3 < − √ 2 < −1 and 1 < √ 2 < √ 3 < 2. If you prefer a rule of thumb to a visual aid, just remember this: Page 51 The relations “less than” and “greater than” are reversed when signs are changed. Example 40. Is −π greater than or less than −3? Solution. On the ‘positive side,’ 3 < π. This relation is reversed when the signs are changed. −π < −3. Visually, −π lies (slightly) to the left of −3 on the number line. | | | | | | | | | | | 0 1 π-π 2 3 4 5−1−2−3−4−5 −π < −3 but 3 < π. 2.2.1 Exercises Insert the appropriate inequality symbol (< or >). 1. −7 0 2. 0 −7 3. −5 −7 4. 5 −7 5. √ 3 2 6. − √ 3 −2 7. √ 3 √ 2 8. − √ 3 − √ 2 Page 52 When two signed numbers are added • if the numbers have opposite signs, 1. the sign of the sum is the sign of the num- ber with the larger absolute value; 2. the absolute value of the sum is the differ- ence between the two individual absolute values (larger − smaller). • if the two numbers have the same sign, 1. the sign of the sum is the common sign of the summands; 2. the absolute value of the sum is the sum of the individual absolute values. Example 43. Add 15 + (−18). Solution. The numbers have opposite signs, so the sign of the sum is the same as the sign of the number with the larger absolute value (−18), i.e., −. The absolute value of the sum is the difference | −18 | − | 15 |= 18− 15 = 3. Thus the sum is −3. This reasoning is summarized as 15 + (−18) = −(18− 15) = −3. Example 44. Add −7 + (−9). Solution. The numbers have the same sign, −, so that is also the sign of the sum. The absolute value of the sum is the sum of the individual absolute values, | −7 | + | −9 |= 7 + 9 = 16. Thus the sum is −16. The reasoning is summarized as −7 + (−9) = −(7 + 9) = −16. Example 45. Add −36 + 49. Solution. The numbers have opposite signs, and the number with the larger absolute value (49) deter- mines the sign of the sum (+). The absolute value of the sum is the difference | 49 | − | −36 |= 49−36. Thus −36 + 49 = +(49− 36) = +13 = 13. Page 55 2.3.1 Exercises Add. 1. −6 + 19 2. 12 + (−4) 3. −34 + (−28) 4. 266 + (−265) 5. −133 + (−93) 6. −1001 + 909 Use an appropriate signed number addition for the following. 11. Find the temperature at noon in Anchorage if the temperature at dawn was −11◦ F and, by noon, the temperature had risen by 36◦ F. 12. Find the height (in feet above ground level) of an elevator which started 30 feet below ground level and subsequently rose 70 feet. 2.3.2 Opposites, Identity There are exactly two numbers at any given non-zero distance from 0, one negative and the other positive. Pairs of numbers such as {−5, 5}, {− √ 2, √ 2}, {−π,π}, which are unequal but equidistant from 0, and hence have the same absolute value, are called opposites. (0 is the only number which is its own opposite, having absolute value 0.) To find the opposite of a nonzero number, we simply change its sign. Example 46. (a) The opposite of 11 is −11. (b) The opposite of −17 is 17. Since there are only two possible signs, the opposite of the opposite of a number is the number we started with. The opposite of the opposite of N is N: −(−N) = N. Example 47. The opposite of the opposite of −5 is −(−(−5)) = −5. Suppose a nonzero number is added to its opposite, for example, 3+(−3). Our rule for adding signed numbers with opposite signs doesn’t quite work since it is not clear what the sign of the sum should be. But in fact it doesn’t matter: the absolute value of the sum is 0, which has no sign. Accordingly, −3 + 3 = 0. The tug-of-war picture helps here: if two people of exactly the same weight and strength pull in opposite directions, the rope doesn’t move at all! Page 56 The sum of a number and its opposite is 0: N + (−N) = 0. Example 48. (a) 25 + (−25) = 0. (b) −153 + 153 = 0. Recall that 0 is the additive identity for ordinary addition because, when 0 is added to a number, the result is the identical number, i.e, the number does not change. This remains true for signed numbers. Example 49. −4 + 0 = −4. 2.3.3 Exercises Find the following: 1. The opposite of 65 2. The opposite of −257 3. The sum of 99 and its opposite. 4. The sum of −π and its opposite. 5. The opposite of the opposite of −31. 6. The sum of −5 and the opposite of 5. 7. −258 + (−(−258)) 8. −91 + (−91) 9. −38 + 0 10. 0 + 55 11. 4 + 223 + (−223) 2.3.4 Associativity Another important property of addition, associativity, extends to signed number addition. Associativity of addition means that when three or more numbers are added, it doesn’t matter how you associate them into groups for addition: x + y + z = (x + y)+ z = x + (y + z). Recall that this property allowed us to add long columns of nonnegative numbers. We can make use of column addition with signed numbers, too, by associating and adding all the positive numbers, and, separately, associating and adding the absolute values of all the negative numbers. Then we add the two subtotals, treating the subtotal associated with the negative numbers as negative. This way, we apply the signed number rule just once, at the end. Example 50. Add: 43 + (−5) + (−135) + 69 + (−134) + 158 + (−162) Page 57 Formerly “impossible” subtractions, such as 7− 12 can now be easily performed. 7− 12 = 7 + (−12) = −(12− 7) = −5. Examples like this show that subtraction is not commutative: Changing the order of subtraction changes the result to its opposite. In symbols, for any two signed numbers A and B , B − A = −(A− B). Example 55. 22− 100 = −78 = −(100 − 22). We often need to find the difference of two unequal quantities. By convention, difference is given as a positive quantity. If the two quantities are A and B , their difference is either A − B or B − A (whichever is positive). Intuitively, difference = larger− smaller. Formally, the difference of A and B is defined to be the absolute value of A− B . Example 56. Find the difference in temperature between −2◦ F and 50◦F. Solution. Intuitively, higher temperature− lower temperature = 50− (−2) = 52, so the temperature difference is 52◦ F. Formally, | −2− 50 |=| −52 |= 52. Example 57. The summit of Mount Whitney in California is 14505 feet above sea level. Not far away, in Death Valley, the lowest point is 282 feet below sea level. What is the altitude difference between Mount Whitney’s summit and the lowest point in Death Valley? Solution. Assigning a negative altitude to a point below sea level, the difference is 14505 − (−282) = 14505 + 282 = 14787 feet. Page 60 2.4.1 Exercises Perform the subtractions. 1. Subtract 31 from 7 2. 98− 100 3. 65− (−640) 4. Subtract 55 from 24 5. Subtract −53 from 68.6 6. −888− (−111) 7. Subtract 22 from 5 8. −87− 23 9. −87− (−23) 10. A snowball at −5◦ C is heated until it melts and then boils. If water boils at 100◦C , by how much did the temperature of the snowball rise? 11. A sunken car is salvaged from the bottom of a lake. The elevation of the lake bottom is −66 feet. The car is lifted by a crane to a height 79 feet above lake level. Through what vertical distance was the car lifted? 2.5 Multiplying Signed Numbers Multiplication of positive numbers was defined in terms of repeated addition. For example, 3× 4 = 4 + 4 + 4 = 12. There is no problem extending this definition to the product of a negative number by a positive number. 3× (−4) = (−4) + (−4) + (−4) = −12. Products such as (−2)× 5 pose no problem if, for consistency, we define signed number multiplication to be commutative (we do): (−2)× 5 = 5× (−2) = (−2) + (−2) + (−2) + (−2) + (−2) = −10. The general rule is: The product of two numbers with opposite signs is the negative of the product of their absolute values. Example 58. Find the products (a) 7× (−11) and (b) (−12) × 5. Page 61 Solution. We take the negative of the product of the absolute values in each case. (a) 7 × (−11) = −(7× 11) = −77. (b) (−12) × 5 = −(12 × 5) = −60. When it comes to the product of two negative numbers, our intuition fails. It makes no sense to “repeatedly” add a number to itself when the number of repeats is negative! How should we define (−2)× (−3)? It is best to give up on intuition and let consistency rule. Look at the pattern below: 4× (−3) = −12 = the opposite of 4× 3 3× (−3) = −9 = the opposite of 3× 3 2× (−3) = −6 = the opposite of 2× 3 1× (−3) = −3 = the opposite of 1× 3 0× (−3) = 0 = the opposite of 0× 3 (−1)× (−3) = = ? (−2)× (−3) = = ?? It looks like the pattern “ought” to continue as follows: (−1)× (−3) = 3 the opposite of (−1)× 3 (−2)× (−3) = 6 the opposite of (−2)× 3 We make the following general definition: The product of two negative numbers is the (positive!) product of their absolute values. Example 59. Find the product (−8)× (−12). Solution. (−8)× (−12) = 8× 12 = 96. Note that our definition is consistent with what we already know about positive numbers: the product of two positive numbers is also the product of their absolute values. The definition now applies whenever numbers with the same sign are multiplied. [OPTIONAL: If you find it hard to accept that the product of two negative numbers is positive, the following discussion might help. An important axiom (fact) of arithmetic states that multiplication “distributes” over addition. That is, for any three numbers, A, B and C , A(B + C ) = AB + AC . If signed number arithmetic is to be consistent with the arithmetic of nonnegative numbers, this axiom must continue to hold. In particular, if A and B are positive, (−A)(B + (−B)) = (−A)(B) + (−A)(−B). Since the left hand side equals 0 (why?) so does the right hand side. It follows that (−A)(−B) must be the opposite of (−A)(B) = −(−AB) = AB . In other words, it must be true that (−A)(−B) = AB .] Here is a summary of the rules for multiplying signed numbers. Page 62 2.6 Dividing Signed Numbers The rules for dividing signed numbers are exactly analogous to the rules for multiplying them. When two signed numbers are divided (and the divisor is nonzero) • if the dividend and divisor have the same sign, the quotient is positive • if the dividend and the divisor have opposite signs, the quo- tient is negative. In both cases, the absolute value of the quotient is the quotient of the individual absolute values. Notice that we have made no mention of remainders here. To do so would require a new definition of the quotient; this is not worth the trouble. When a division of signed numbers is not exact (i.e, when the remainder is not zero) it is much better to treat the division as a fraction. We’ll do that in the next chapter. For now, we consider only exact divisions. In this case, as with positive numbers, we can restate division in terms of multiplication, and that explains the rules above. For example, because 6× 4 = 24, we say that 24÷ 6 = 4 (or 24÷ 4 = 6.) Similarly, we say that 24÷−6 = −4, because 24 = (−6)× (−4). Example 66. Express the statement −72 = −8 · 9 as an exact division of signed numbers. Solution. We can write −72÷ (−8) = 9 or we can write −72÷ 9 = −8. Example 67. The division 63÷ (−7) is exact. Find the quotient. Solution. Since −63 = 9(−7), 63÷ (−7) = −9, that is, the quotient is −9. We remind you that 0 cannot be the divisor in any division problem: repeated subtraction of 0 has no effect on any number (recall the discussion in Section 1.6). This remains true for signed numbers. Example 68. (−23)÷ 0 is undefined. Page 65 It remains true that 0 can be the dividend. Example 69. 0÷(−12) = 0. Note that this can be restated in terms of multiplication as 0×(−12) = 0. We summarize the properties of 0 with respect to division. For a nonzero signed number N • 0÷ N = 0 • N ÷ 0 is undefined. 0÷ 0 is also undefined. 2.6.1 Exercises Perform the divisions, or state that they are undefined. 1. (−24)÷ (−8) 2. (−24)÷ 8 3. 66÷ 0 4. 30÷ (−6) 5. (−30)÷ 6 6. 0÷ (−1000) 7. 50÷ (−4) 8. −95÷ 0 2.7 Powers of Signed Numbers Since exponents indicate repeated multiplication, there is no problem applying exponents to signed numbers. (−4)3 = (−4)(−4)(−4) = −64 and (−2)4 = (−2)(−2)(−2)(−2) = 16. Example 70. Evaluate (−5)3, and (−5)4. Solution. (−5)3 = (−5)(−5)(−5) = (25)(−5) = −125. (−5)4 = (−5)(−5)(−5)(−5) = (25)(25) = 625. Notice that every pair of negative factors has a positive product, by the rules for multiplying signed numbers with the same sign. It follows that Page 66 A power of a negative number is • negative if the exponent is odd, • positive if the exponent is even. When applying exponents to signed numbers, it is essential to put parentheses around the number, including its sign. For example, the square of −3 is written (−3)2 = (−3)(−3) = 9. Without parentheses, as in −32, the exponent applies only to 3, not to −3, and this is not a power of −3. Rather, −32 represents the opposite of 32, so −32 = −9. The behavior of 0 in an exponential expression, in particular, the interpretation of 0 as an exponent, extends unchanged to signed numbers. (You may wish to review Section 1.4 on powers of whole numbers in Chapter 1.) For a nonzero signed number N, • N0 = 1 • 0N = 0 if N is positive (00 is undefined.) Example 71. (−3)0 = 1. 04 = 0. 00 is undefined. 2.7.1 Exercises Find the value of each expression. 1. 82 2. (−8)2 3. −82 4. −63 5. (−6)3 6. 05 7. −(−33) Page 67 Page 70 Chapter 3 Fractions and Mixed Numbers Fractions are expressions such as 3 19 or more generally a b , where a and b are whole numbers, and b ̸= 0. For now, we assume that a and b are both positive (we will introduce negative fractions in due course.) The numbers a and b are called terms. The term on top is called the numerator, and the term on the bottom is called the denominator. The horizontal line in the middle is called the fraction bar. Sometimes to save space, we write fractions in one line, using a slash instead of the fraction bar, putting the numerator on the left and the denominator on the right: a b = a/b. When reading a fraction out loud, we attach the suffix th or ths to the denominator, saying “a bths” for a/b. Example 74. The fractions 1 8 and 6 13 are spoken “one eighth,” and “six thirteenths,” respectively. If the denominator is 2 or 3, we say “halves” or “thirds” (not twoths or threeths!), respectively. If the denominator is 4, we sometimes say “fourths,” and sometimes, ”quarters.” Example 75. 5/2 is spoken “five halves.” 4/3 is spoken “four thirds.” 3/4 is spoken “three fourths,” or “three quarters.” Sometimes, instead of using the th suffix, we just say “a over b.” Thus 4/9 can be spoken as “four ninths,” or just ”four over nine.” 3.1 What positive fractions mean Positive fractions represent parts of a whole in a precise way. In the fraction a/b, the denominator b represents the number of equal parts into which the whole has been divided, and the numerator represents how many of those parts are being taken into account. For example, if a rectangle is divided up into 3 equal parts, and 2 of those parts are shaded, then the shaded portion represents 2 3 (two-thirds) of the whole rectangle. 71 × × × × 2 3 We can divide any convenient figure into equal parts (not just a rectangle), to represent a fraction. For example, the circle below has been divided into 5 equal “wedges,” and three of them are shaded. So the picture represents the fraction 3 5 . Example 76. Use a square to represent the fraction 5/16. Solution. Since 16 is a perfect square, it’s easy to make a square 4 units on a side, and divide that into 16 small squares of equal size. Then we shade 5 of them (any 5 will do) to represent 5/16. 3.2 Proper and Improper Fractions A fraction is called proper if its numerator has smaller absolute value than its denominator. A positive proper fraction represents less than one whole. An improper fraction has a numerator that has Page 72 8. Write five fractions which are equal to 1. 9. Write five fractions which are equal to 0. 10. Write 15 as a fraction. 3.3 Mixed Numbers We have seen that a positive improper fraction represents either a whole number or a whole number plus a proper fraction. In the latter form, it is called a mixed number. Even though it is a sum, the + sign is omitted. Thus 3 1 2 represents 3 “wholes” plus an extra 1 2 . 3.3.1 Converting an improper fraction into a mixed or whole number Suppose we have an improper fraction numerator denominator . To convert this to a mixed number, we simply perform the division numerator÷ denominator, obtaining a quotient and a remainder. Then we use these to build the mixed number as follows: the quotient is the whole number part; the remainder over the divisor is the fractional part. Note that the fractional part obtained in this way is always proper, because the remainder is always smaller than the divisor. Summarizing: numerator denominator = quotient remainder divisor Example 78. Write the improper fraction 13/3 as a mixed number. Solution. Dividing 13 by 3 yields a quotient of 4 and a remainder of 1. The corresponding mixed number has whole number part 4 and fractional part 1/3. Thus 13 3 = 4 1 3 . If the remainder is 0, the mixed number is actually just a whole number. Example 79. Write the improper fraction 84/7 as a mixed number. Page 75 Solution. Dividing 84 by 7 yields a quotient of 12 and a remainder of 0. Thus 84 7 = 12 0 7 = 12, a whole number. Here are some everyday uses of mixed numbers. Example 80. Five hikers want to share seven chocolate bars fairly. How many bars does each hiker get? Solution. For fairness, each of the five hikers should get the same amount: exactly one fifth of the chocolate. There are seven bars, so the amount (in chocolate bars) that each hiker should get is 7 5 = 1 2 5 . Each hiker gets 12 5 chocolate bars. Example 81. Julissa waited 5 minutes for the train on Monday, 2 minutes on Tuesday, and then 4, 8, and 3 minutes on Wednesday, Thursday, and Friday, respectively. What was her average wait time for the week? Solution. Recall that the average of a set of numbers is their sum, divided by the number of numbers. In this case, we want the average of the five numbers {5, 2, 4, 8, 3}, which is 5 + 2 + 4 + 8 + 3 5 = 22 5 = 4 2 5 . Her average wait time was 42 5 minutes. (Extra credit: how long is two fifths of a minute, in seconds?) 3.3.2 Exercises Convert the following improper fractions into mixed numbers: 1. 19 3 2. 11 2 3. 135 5 4. 99 98 5. 77 5 Use mixed numbers to answer the following questions: 6. Three boys share four sandwiches fairly. How many sandwiches does each boy get? 7. What is the average of the set of numbers {11, 14, 9, 12}? 8. During the first seven days of January, Bill travels 31, 37, 46, 31, 77, 50 and 40 miles respectively. What is his average daily travel distance? Page 76 3.3.3 Converting a mixed or whole number to an improper fraction Sometimes we need to change a mixed number back into an improper fraction. The key fact, again, is that the fraction a a is always equal to 1, for any number a except 0. For example, 3 3 = 1. Example 82. Convert the mixed number 21 3 to an improper fraction, using the fact that 3 3 = 1. Solution. Think of the whole number 2 in the following way: 2 = 1 + 1 = 3 3 + 3 3 . It follows that the mixed number 21 3 can be written 3 3 + 3 3 + 1 3 . The figure below should make it clear that this sum is equal to the improper fraction 7 3 . 1 = 3 3 + 1 = 3 3 + 1 3 = 7 3 (The figure also demonstrates that fractions with the same denominator add up to a new fraction, with the same denominator, and a numerator which is the sum of all the old numerators. We’ll say more about adding fractions later. ) The procedure in the previous example is easily turned into a general formula. The mixed number N p q is equal to the improper fraction N · q + p q . Remember to follow the order of operations (multiplication before addition) when evaluating N · q+ p. Example 83. Convert the mixed number 82 3 into an improper fraction, using the boxed rule. Page 77 a shaded vertical rectangles is divided into d parts of which c are differently shaded. Hence there are a · c doubly shaded smaller rectangles, so that a · c is the new numerator. The rule is a b of c d = a · c b · d . We use this rule to define multiplication of fractions, replacing “of” by the multiplication symbol · (or ×): a b · c d = a · c b · d In words: The product of two fractions is the product of the numerators over the product of the denominators. Example 86. Find the product 3 8 · 5 7 . Solution. Using the boxed rule, 3 8 · 5 7 = 3 · 5 8 · 7 = 15 56 . It is easy to see that fraction multiplication is just an expanded version of whole number multipli- cation, since every whole number is a fraction with denominator 1. Thus the boxed rule applies when one (or both!) of the multiplicands is a whole number. Example 87. What is two thirds of five? That is, find the product 2 3 · 5. Solution. Writing 5 as the fraction 5/1 and using the fraction multiplication rule, we get 2 3 · 5 1 = 2 · 5 3 · 1 = 10 3 . Example 88. One third of a twelve-member jury are women. How many women are on the jury? Solution. One third of twelve equals 1 3 · 12 1 = 12 3 = 4. There are four women on the jury. You may have noticed that in our very first example with rectangles, 3 4 · 2 3 = 6 12 , exactly half of the rectangle ends up doubly-shaded (6 is half of 12). So it is correct (and simpler) to say that 3 4 · 2 3 = 1 2 . Does this contradict our rule? No, because 1/2 and 6/12 represent the same number. We’ll explain this, and examine some other technical details regarding the representation of fractions, in the next sections. Then we’ll continue with the arithmetic (division, addition, subtraction) of fractions. Page 80 3.4.1 Exercises Find the following products. (Just use the fraction product rule – no need to draw rectangles.) 1. 1 3 · 5 7 2. One half of one half 3. Two thirds of one third 4. 3 4 · 3 4 5. 3 · 5 8 6. " 2 3 #2 7. 1 2 · 7 8 · 3 3.5 Equivalent Fractions Fractions which look very different can represent the same number. For example, the fractions 2 4 , 5 10 , 6 12 , and 50 100 all represent the number 1 2 . What property do all these fractions share? Each has a denominator that is exactly twice its numerator; the simplest fraction with this property is 1 2 . Fractions which represent the same number are called equivalent, and we use the equal sign to indicate this. Thus, for example, 1 2 = 50 100 . There is an easy way to tell when two fractions are equivalent. We give it here because it is so simple and pleasing, but we postpone the explanation until we discuss proportions. a b = c d if (and only if) a · d = b · c. In words: two fractions are equivalent if (and only if) their cross-products are equal. A cross-product is the product of the numerator of one fraction and the denominator of the other. Starting with a given fraction, we can generate equivalent fractions easily, using the fact that 1 is the multiplicative identity, and that 1 = c c for any non-zero c . Then a b = a b · 1 = a b · c c , and so we have Page 81 The Fundamental Property of Fractions: If we multiply both the numerator and denominator of a fraction by any nonzero c, then the new fraction is equivalent to the original one, and represents the same number: a b = a · c b · c . Example 89. Write two fractions equivalent to 2 3 . Solution. We use the fact that 2 3 = 2 · c 3 · c for any nonzero c . Picking two values for c , say, 6 and 7, we get two fractions equivalent to 2/3: 2 3 = 2 · 6 3 · 6 = 12 18 and 2 3 = 2 · 7 3 · 7 = 14 21 . Of course, other choices of c would have produced other fractions equivalent to 2/3. 3.5.1 Cancellation and Lowest Terms The boxed rule above produces equivalent fractions with higher (larger) terms. It is sometimes possible to go the other way, producing lower (smaller) terms. If both numerator and denominator have a common factor – a whole number greater than 1 which divides them both with zero remainder – we can “cancel” it by division. The two quotients become the terms of an equivalent fraction with lower terms. For example, the numerator and denominator of 18 24 have the common factor 6. Thus 18 24 = 18÷ 6 24÷ 6 = 3 4 . In general, For any non-zero c, a b = a ÷ c b ÷ c . This method of obtaining lower terms is called cancellation or cancelling out. It is often indicated as follows: 18 24 = ✚✚❃ 3 18 ✚✚❃ 4 24 = 3 4 . This is useful short-hand, but it has one disadvantage: the common factor (6, in this case) is not made visible. To ensure accuracy, you can show the common factor explicitly, before cancelling it. Thus, explicitly, 18 24 = 3 · 6 4 · 6 = 3 · ✁✁✕6 4 · ✁✁✕6 = 3 4 . Page 82 are called composite. (Note that 1 is a special case according to these definitions: it is neither prime nor composite!) The first few composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, ... How do we know that these numbers are composite? Because each has at least one factor, other than 1 and itself. For example, 15 has factors 3 and 5, that is, 15 = 3 · 5. Every composite whole number has a unique prime factorization, which is its expression as the product of its prime factors, listed in order of increasing size. For example: 4 = 2 · 2 = 22 6 = 2 · 3 8 = 2 · 2 · 2 = 23 9 = 3 · 3 = 32 10 = 2 · 5 12 = 2 · 2 · 3 = 22 · 3 14 = 2 · 7 15 = 3 · 5 16 = 2 · 2 · 2 · 2 = 24 (3.1) To find prime factorizations, we use repeated division with prime divisors. Start by testing the number for divisibility by the smallest prime, 2 (a number is divisible by 2 if its final digit is even: 0, 2, 4, 6 or 8). If it is divisible by 2, we divide by 2 as many times as possible, until we arrive at a quotient which is no longer divisible by 2. We then repeat the procedure, starting with the last quotient obtained, and using the next larger prime, 3 (a number is divisible by 3 if the sum of its digits is divisible by 3 – did you know that?). Repeating again (if necessary) with the next larger prime, we eventually arrive at a quotient which is itself a prime number. At that point, we are almost finished. We collect all the primes that were used as divisors (if the same prime has been used more than once, it should be collected as many times), together with the final (prime) quotient. The product of all these numbers is the prime factorization. Example 94. Find the prime factorization of 300. Solution. 300, being even, is divisible by 2, so we start by dividing by 2. The steps are as follows: 300÷ 2 = 150; 150÷ 2 = 75; 75 is not divisible by 2; move on to 3 75÷ 3 = 25; 25 is not divisible by 3; move on to 5 25÷ 5 = 5; 5 is a prime number; stop. The primes that were used as divisors were 2, 2, 3, 5. Note that 2 was used twice, so it is listed twice. The final prime quotient is 5. The prime factorization is the product of all those prime divisors and the final prime quotient: 300 = 2 · 2 · 3 · 5 · 5 or 22 · 3 · 52. We can easily check our work by multiplying the prime factors and verifying that the product obtained is the original number. Page 85 3.6.1 Exercises Find the prime factorizations of the following numbers. Check the results by multiplication. 1. 60 2. 48 3. 81 4. 360 5. 85 6. 154 7. Which of the numbers above is divisible by 3? 3.6.2 Finding the GCF Using prime factorizations, it is easy to find the greatest common factor of a set of numbers. Example 95. Find the greatest common factor (GCF) of the two-number set {a, b}, where a and b have the following prime factorizations. a = 24 · 32 · 7 and b = 2 · 34 · 11. Solution. First, look for the common prime factors of a and b: they are 2 and 3. Note that 7 and 11 are not common, and therefore cannot be factors of the GCF. Now look at the powers (exponents) on 2 and 3. A small power of any prime is a factor of any larger power of the same prime. The smallest power of 2 that appears is 2 = 21 (in the factorization of b), and the smallest power of 3 that appears is 32 (in the factorization of a). It follows that the GCF is the product of the two smallest powers of 2 and 3. Thus GCF{a, b} = 21 · 32 = 18. Notice that the actual values of a and b (which we could have determined by multiplication) were not used – only their prime factorizations. Here is a summary of the procedure: To find the GCF of a set of numbers: 1. find the prime factorization of each number; 2. determine the prime factors common to all the numbers; 3. if there are no common prime factors, the GCF is 1; otherwise 4. for each common prime factor, find the smallest exponent that appears on it; 5. the GCF is the product of the common prime factors with the exponents found in step 4. Page 86 Example 96. (a) Find the GCF of the set {60, 135, 150}. (b) Find the GCF of the subset {60, 150}. Solution. (a) Following the boxed procedure: 1. the prime factorizations are 60 = 2 · 2 · 3 · 5 = 22 · 3 · 5 135 = 3 · 3 · 3 · 5 = 33 · 5 150 = 2 · 3 · 5 · 5 = 2 · 3 · 52 2. the common prime factors are 3 and 5; 3. (does not apply to this example); 4. the smallest exponent on 3 is 1 (in the factorizations of 60 and 150); the smallest exponent on 5 is also 1 (in the factorizations of 60 and 135); 5. the GCF is 31 · 51 = 15. (b) For the two-number subset {60, 150}, the common prime factors are 2, 3 and 5. The smallest exponent on all three factors is 1. So the GCF is 21 · 31 · 51 = 30. Can you explain why the GCF in part (b) is bigger than in part (a)? 3.6.3 Exercises Find the GCF of each of the following sets of numbers: 1. {72, 48} 2. {72, 48, 36} 3. {72, 36} 4. {48, 36} 5. {36, 15} 6. {36, 14} 7. {15, 14} 3.6.4 Cancelling the GCF for lowest terms Knowing that the GCF of {60, 150} = 30 allows us to reduce the fraction 60 150 to lowest terms in one step: we simply cancel it out. Thus, 60 150 = ✟✟✯ 2 60 ✟✟✯ 5 150 = 2 5 (cancelling the GCF, 30). Recall that this is short-hand for 60÷ 30 150÷ 30 = 2 5 . Page 87 Solution. The numerator of the first fraction (3) has a common factor with the denominator of the third fraction (9), so the product is equal to ✚❃ 1 3 4 · 8 5 · 10 ✚❃ 3 9 = 1 4 · 8 5 · 10 3 . Continuing on, the numerator of the second fraction (8) has a common factor with the denominator of the first fraction (4), so the product is equal to 1 ✚❃ 1 4 · ✚ ❃ 2 8 5 · 10 3 = 1 · 2 5 · 10 3 . Finally, the numerator of the third fraction (10) has a common factor with the denominator of the second fraction (5), so (omitting the factor 1) the product is equal to 2 ✚❃ 1 5 · ✚ ✚❃ 2 10 3 = 2 1 · 2 3 . No further cancellation is possible. The final answer is now a simple product 2 1 · 2 3 = 4 3 = 1 1 3 , which is already in lowest terms. These cancellations could have been done in a different order, or (carefully) all at once. Mixed numbers are multiplied by simply converting them into improper fractions. Example 99. Find the product 23 8 · 11 4 · 22 3 . Express the result as a mixed number. Solution. Rewriting each mixed number as an improper fraction, we have the product 19 8 · 5 4 · 8 3 . Cancelling 8’s, we have 19 · 5 4 · 3 = 85 12 = 7 1 12 . Example 100. A gas tank with a 131 2 -gallon capacity is only one third full. How much gas is in the tank? Solution. We need to find the product 1 3 · 131 2 . Converting 131 2 to the improper fraction 27 2 , we have 1 3 · 13 1 2 = 1 3 · 27 2 . Cancelling the common factor 3, 1 ✁✁✕ 1 3 · ✚✚❃ 9 27 2 = 9 2 . Converting 9 2 to a mixed number, we conclude that the tank contains 41 2 gallons of gas. Page 90 3.7.1 Exercises Find the products, using pre-cancellation where possible. Check that the answers are in lowest terms. Express any improper fractions as mixed numbers. 1. 4 5 · 7 12 2. 32 45 · 9 16 · 15 6 3. 12 · 5 8 · 2 3 4. 9 22 · 21 12 · 2 4 7 5. 2 3 of 24. 6. 3 4 of 2 3 of 50. 7. 2 2 3 · 1 3 4 . Use multiplication to answer the following questions. 8. What is the area of a rectangle with length 21 2 feet and width 2 3 10 feet? 9. A 121 2 -gallon fish tank is only three-fifths full. How many gallons of water must be added to fill it up? 10. 22 3 cups of beans are needed to make 4 bowls of chili. How many cups are needed to make 8 bowls? 1 bowl? 3 bowls? 3.8 Adding and Subtracting Fractions If I eat a third of a pizza for lunch, and another third for dinner, then I have eaten two thirds in total. That is, 1 3 + 1 3 = 2 3 . Similar logic applies whenever we add two (or more) fractions with the same denominator – we simply add the numerators, while keeping the denominator fixed: a c + b c = a + b c . A very similar rule holds for subtraction of fractions with the same denominator: Page 91 a c − b c = a − b c . Fractions with the same denominator are called like fractions. Example 101. Here are three examples involving addition or subtraction of like fractions: 1 5 + 2 5 = 1 + 2 5 = 3 5 13 7 − 2 7 = 13 − 2 7 = 11 7 = 1 4 7 7 8 − 5 8 = 7− 5 8 = 2 8 = 1 4 . Notice that we reduced to lowest terms where necessary, and changed improper fractions to mixed numbers. 3.8.1 Exercises Add or subtract the like fractions as indicated. Reduce the final answers to lowest terms if necessary. Change improper fractions to mixed numbers. 1. 1 5 + 3 5 = 2. 7 5 + 13 5 = 3. 11 15 + 13 15 + 8 15 = 4. 20 51 − 3 51 = 5. 5 13 − 4 13 = 6. 11 25 + 9 25 − 3 25 = 7. 109 7 − 11 7 = 8. 13 2 − 1 2 = 9. 10 7 + 6 7 − 11 7 = 10. 11 25 − 2 25 + 8 25 = Page 92 The LCM can also be found using prime factorizations. This is useful when the numbers are rather large. Example 107. Find the LCM of the two-number set {a, b}, where a and b have the following prime factorizations. a = 24 · 32 · 7 and b = 2 · 34 · 11. Solution. The LCM must be a multiple of both numbers, so that it must be divisible by the highest power of every prime factor that appears in any one of the factorizations. The prime factors of a and b are 2, 3, 7 and 11. The highest powers that appear are 24, 34, 71, 111. The LCM is the product of these powers: LCM{a, b} = 24 · 34 · 7 · 11. This is a rather large number, but it is the smallest which is a multiple of both a and b. Notice that the actual values of a and b, and their LCM (which we could calculate by multiplication) were not needed – only their prime factorizations. Here is a summary of the procedure: To find the LCM of a set of numbers: 1. find the prime factorization of each number; 2. for each prime factor, find the largest exponent that appears on it in any of the factorizations; 3. the LCM is the product of the prime factors with the exponents found in step 2. Compare this procedure with the procedure for finding the GCF of a set of numbers. There are similarities and significant differences. 3.8.4 Exercises Find the LCM of the following sets of numbers. Use the prime factorization method for larger numbers. 1. LCM{25, 10} 2. LCM{48, 60} 3. LCM{10, 15, 25} 4. LCM{8, 12} 5. LCM{9, 6, 12} 6. LCM{60, 168} Page 95 7. LCM{51, 34, 17} 8. LCM{15, 12} 9. LCM{18, 8} 10. LCM{3, 4, 5} 11. LCM{4, 14} 12. LCM{2, 5, 9} 13. LCM{32 · 52 · 11, 34 · 72} 3.8.5 The LCD To add unlike fractions so that the sum is in lowest terms, we use the LCM of their denominators. This is such a useful number that it has a special name: the LCD or Least Common Denominator. Example 108. Find the LCD of the fractions 1 8 , 3 10 , and 1 18 . Solution. The LCD of the fractions is the LCM of their denominators, LCM{8, 10, 18}. Looking at the prime factorizations 8 = 23, 10 = 2 · 5, 18 = 2 · 32, and taking the highest power of each prime that occurs, we see that the LCM is 23 · 32 · 5 = 360. This is LCD of the fractions. Example 109. Find the sum of the unlike fractions 1 8 + 3 10 + 1 18 . Solution. The LCD is the LCM from the previous example: 360. Now we observe that 360 = 8 · 45 = 10 · 36 = 18 · 20. Thus 1 8 = 1 · 45 8 · 45 , 3 10 = 3 · 36 10 · 36 and 1 18 = 1 · 20 18 · 20 . It follows that 1 8 + 3 10 + 1 18 = 1 · 45 8 · 45 + 3 · 36 10 · 36 + 1 · 20 18 · 20 (3.2) = 45 + 108 + 20 360 = 173 360 . Since the 173 is not divisible by 2, 3, or 5, the fraction is in lowest terms. Page 96 Practically the same method works for subtracting unlike fractions. Example 110. Find the difference 14 25 − 2 10 , using the LCD. Reduce to lowest terms, if necessary. Solution. The LCD is LCM{25, 10} = 50. Now 50 = 25 · 2 = 10 · 5. So 14 25 = 14 · 2 25 · 2 = 28 50 , and 2 10 = 2 · 5 10 · 5 = 10 50 . Thus, the difference of the two fractions is 28 50 − 10 50 = 28− 10 50 = 18 50 . The latter fraction is not in lowest terms, since the GCF of 18 and 50 is 2. Cancelling the GCF, we get 18 50 = ✚✚❃ 9 18 ✚✚❃ 25 50 = 9 25 . Example 111. Subtract 1 3 − 1 4 , and reduce to lowest terms if necessary. Solution. The LCD is 12. Changing both fractions to equivalent fractions with denominator 12, we get 1 3 = 1 · 4 3 · 4 = 4 12 and 1 4 = 1 · 3 4 · 3 = 3 12 . Thus, 1 3 − 1 4 = 4 12 − 3 12 = 4− 3 12 = 1 12 . The latter fraction is already in lowest terms, so we are done. Example 112. Find the sum 4 5 + 3 4 , reduce to lowest terms, and express the answer as a mixed number. Solution. The LCD{4, 5} = 20, so that 4 5 + 3 4 = 4 · 4 5 · 4 + 3 · 5 4 · 5 = 16 20 + 15 20 = 31 20 = 1 11 20 . 3.8.6 Exercises Add or subtract the following fractions as indicated, reduce to lowest terms if necessary, and change improper fractions to mixed numbers. 1. 1 5 + 3 6 2. 7 5 + 13 3 3. 15 1 + 2 3 Page 97