KNOWLEDGE OF NUMBER SENSE, CONCEPTS, AND OPERATIONS

DOMAIN I. COMPETENCY 1.0 MATHEMATICS KNOWLEDGE OF NUMBER SENSE, CONCEPTS, AND OPERATIONS Skill 1.1 Compare the relative value of real numbers (e.g., integers, fractions, decimals, percents, irrational numbers, and numbers expressed in exponential or scientific notation). We can express rational numbers as the ratio of two integers a b example, 2, - 4 5, 5 = 5 1., where b 0. For The rational numbers include integers, fractions, mixed numbers, and terminating and repeating decimals. We can express every rational number as a repeating or terminating decimal and represent it on a number line. Integers are positive and negative whole numbers and zero....-6, -5, -4, -, -2, -1, 0, 1, 2,, 4, 5, 6,... Whole numbers are natural numbers and zero. 0, 1, 2,,,4,5,6... Natural numbers are the counting numbers. 1, 2,, 4, 5, 6,... Irrational numbers are real numbers that we cannot be write as the ratio of two integers. These are infinite non-repeating decimals. Examples: 5 = 2.260.., pi = =.1415927... A fraction is an expression of numbers in the form of x y, where x is the numerator and y is the denominator, which cannot be zero. 7 is the numerator, 7 is the denominator If the fraction has common factors for the numerator and denominator, divide both by the common factor to reduce the fraction to its lowest form. 1 1 1 1 = = 9 1 Divide by the common factor 1

A mixed number has an integer part and a fractional part. 1 1 2, 5, 7 4 6 1 Percent = per 100 (written with the symbol %). Thus 10% 10 1 = =. 100 10 Decimals = deci = part of ten. To find the decimal equivalent of a fraction, use the denominator to divide the numerator as shown in the following example. Find the decimal equivalent of 7 10. Since 10 cannot divide into 7 evenly 7 0.7 10 = The exponent form is a shortcut method to write repeated multiplication. Basic form: n b, where b is the base and n is the exponent. b and n are both real numbers. n b indicates that we multiply the base, b, by itself n times. Examples: 4 = = 81 2 = 2 2 2= 8 4 ( 2) = ( 2) ( 2) ( 2) ( 2) = 16 Key exponent rules: 4 2 = (2 2 2 2) = 16 For a (nonzero), and m and n (real numbers): m n ( m n) 1) a a = a + Product rule m a ( mn) 2) = a Quotient rule n a ) a a m n = a a n m When we raise 10 to any power, the exponent tells us the number of zeroes in the product.

10 7 = 10,000,000 Caution: Unless the negative sign is inside the parentheses and the exponent is outside the parentheses, the exponent does not affect the sign. 4 ( 2) 4 2 implies that we multiply -2 by itself 4 times. implies that we multiply 2 by itself 4 times, then negate the answer. Scientific notation is a more convenient method for writing very large and very small numbers. It employs two factors. The first factor is a number between 1 and 10. The second factor is a power of 10. This notation is a shorthand for expressing large numbers (like the weight of 100 elephants) or small numbers (like the weight of an atom in pounds). Recall that: 10 n n = (10) Ten multiplied by itself n times. 0 10 = 1 Any nonzero number raised to power of zero is 1. 1 10 = 10 2 10 = 10 10 = 100 10 = 10 10 10 = 1000 (kilo) 1 10 = 1 10 (deci) 2 10 = 1 100 (centi) 10 = 1 1000 (milli) 6 10 = 1 1,000,000 (micro) Write 46,68,000 in scientific notation. 1) Introduce a decimal point and decimal places. 46,68,000 = 46,68,000.0000 2) Make a mark between the two digits that give a number between -9.9 and 9.9. 4 6,68,000.0000 ) Count the number of digit places between the decimal point and the mark. This number is the n -the power of ten. 7 So, 46,68,000 = 4.668 10

Write 0.0097 in scientific notation. 1) Decimal place is already in place. 2) Make a mark between and 9 to form a number between -9.9 and 9.9. ) Move decimal place to the mark ( hops). 0. 00 97 n Motion is to the right, so n of 10 is negative. Therefore, 0.0097.97 10 =. Skill 1.2 Solve real-world problems involving addition, subtraction, multiplication, and division of rational numbers (e.g., whole numbers, integers, decimals, percents, and fractions including mixed numbers). Properties are rules that apply for addition, subtraction, multiplication, or division of real numbers. These properties are: Commutative: You can change the order of the terms or factors as follows. For addition: For multiplication: a + b = b + a ab = ba Associative: You can regroup the terms as you like. For addition: For multiplication: a + (b + c) = (a + b) + c a(bc) = (ab)c This rule does not apply for division and subtraction. ( - 2 + 7) + 5 = - 2 + (7 + 5) 5 + 5 = - 2 + 12 = 10 ( - 7) 5 = ( - 7 5) - 21 5 = - 5 = - 105

Identity: A number that when added to a term results in that same number (additive identity); a number that when multiplied by a term results in that same number (multiplicative identity). For addition: a + 0 = a (zero is additive identity) For multiplication: a x 1 = a (one is multiplicative) 17 + 0 = 17-4 1 = - 4 The product of any number and one is that number. Distributive: This technique allows us to operate on terms within a parentheses without first performing operations within the parentheses. This is especially helpful when we cannot combine terms within the parentheses. a (b + c) = ab + ac 6 ( - 4 + 9) = (6-4) + (6 9) 6 5 = - 24 + 54 = 0 To multiply a sum by a number, multiply each addend by the number, then add the products.

Addition of whole numbers At the end of a day of shopping, a shopper had $24 remaining in his wallet. He spent $45 on various goods. How much money did the shopper have at the beginning of the day? The total amount of money the shopper started with is the sum of the amount spent and the amount remaining at the end of the day. 24 + 45 69 The original total was $69. The winner of a race took 1 hr. 58 min. 12 sec. on the first half of the race and 2 hr. 9 min. 57 sec. on the second half of the race. What was the winner s total time? Subtraction of Whole Numbers 1 hr. 58 min. 12 sec. + 2 hr. 9 min. 57 sec. Add these numbers hr. 67 min. 69 sec. + 1 min - 60 sec. Change 60 seconds to 1min. hr. 68 min. 9 sec. + 1 hr.-60 min.. Change 60 minutes to 1 hr. 4 hr. 8 min. 9 sec. Final answer At the end of his shift, a cashier has $96 in the cash register. At the beginning of his shift, he had $15. How much money did the cashier collect during his shift? The total collected is the difference of the ending amount and the starting amount. 96-15 81 The total collected was $81.

Multiplication of whole numbers Multiplication is one of the four basic number operations. In simple terms, multiplication is the addition of a number to itself a certain number of times. For example, 4 multiplied by is the equal to 4 + 4 + 4 or + + +. Another way of conceptualizing multiplication is to think in terms of groups. For example, if we have 4 groups of students, the total number of students is 4 multiplied by. We call the solution to a multiplication problem the product. The basic algorithm for whole number multiplication begins with aligning the numbers by place value with the number containing more places on top. 172 x 4 Note that we placed 122 on top because it has more places than 4 has. Next, we multiply the ones place of the second number by each place value of the top number sequentially. (2) 172 { x 2 = 6, x 7 = 21, x 1 = } x 4 Note that we had to carry a 2 to the hundreds column 516 because x 7 = 21. Note also that we add, not multiply, carried numbers to the product. Next, we multiply the number in the tens place of the second number by each place value of the top number sequentially. Because we are multiplying by a number in the tens place, we place a zero at the end of this product. (2) 172 x 4 {4 x 2 = 8, 4 x 7 = 28, 4 x 1 = 4} 516 6880 Finally, to determine the final product we add the two partial products. 172 x 4 516 + 6880 796 The product of 172 and 4 is 796.

A student buys 4 boxes of crayons. Each box contains 16 crayons. How many total crayons does the student have? The total number of crayons is 16 x 4. 16 x 4 64 Total number of crayons equals 64. Division of whole numbers Division, the inverse of multiplication, is another of the four basic number operations. When we divide one number by another, we determine how many times we can multiply the divisor (number divided by) before we exceed the number we are dividing (dividend). For example, 8 divided by 2 equals 4 because we can multiply 2 four times to reach 8 (2 x 4 = 8 or 2 + 2 + 2 + 2 = 8). Using the grouping conceptualization we used with multiplication, we can divide 8 into 4 groups of 2 or 2 groups of 4. We call the answer to a division problem the quotient. If the divisor does not divide evenly into the dividend, we express the leftover amount either as a remainder or as a fraction with the divisor as the denominator. For example, 9 divided by 2 equals 4 with a remainder of 1 or 4 ½. The basic algorithm for division is long division. We start by representing the quotient as follows. 14 29 14 is the divisor and 29 is the dividend. This represents 29 14. Next, we divide the divisor into the dividend starting from the left. 2 14 29 14 divides into 29 two times with a remainder. Next, we multiply the partial quotient by the divisor, subtract this value from the first digits of the dividend, and bring down the remaining dividend digits to complete the number. 2 14 29 2 x 14 = 28, 29 28 = 1, and bringing down the yields 1. - 28 1 Finally, we divide again (the divisor into the remaining value) and repeat the preceding process. The number left after the subtraction represents the remainder. 20

14 29-28 1-0 1 The final quotient is 20 with a remainder of 1. We can also represent this quotient as 20 1/14. Each box of apples contains 24 apples. How many boxes must a grocer purchase to supply a group of 252 people with one apple each? The grocer needs 252 apples. Because he must buy apples in groups of 24, we divide 252 by 24 to determine how many boxes he needs to buy. 10 24 252-24 12 The quotient is 10 with a remainder of 12. - 0 12 Thus, the grocer needs 10 boxes plus 12 more apples. Therefore, the minimum number of boxes the grocer can purchase is 11. At his job, John gets paid $20 for every hour he works. If John made $940 in a week, how many hours did he work? This is a division problem. To determine the number of hours John worked, we divide the total amount made ($940) by the hourly rate of pay ($20). Thus, the number of hours worked equals 940 divided by 20. 47 20 940-80 140-140 0 20 divides into 940, 47 times with no remainder. John worked 47 hours.

Addition and Subtraction of Decimals When adding and subtracting decimals, we align the numbers by place value as we do with whole numbers. After adding or subtracting each column, we bring the decimal down, placing it in the same location as in the numbers added or subtracted. Find the sum of 152. and 6.42. 152.00 + 6.42 188.642 Note that we placed two zeroes after the final place value in 152. to clarify the column addition. Find the difference of 152. and 6.42. Multiplication of Decimals 2 9 10 (4)11(12) 152.00 152.00-6.42-6.42 58 115.958 Note how we borrowed to subtract from the zeroes in the hundredths and thousandths place of 152.00. When multiplying decimal numbers, we multiply exactly as with whole numbers and place the decimal moving in from the left the total number of decimal places contained in the two numbers multiplied. For example, when multiplying 1.5 and 2.5, we place the decimal in the product places in from the left (.525). Find the product of.52 and 4.1..52 Note that there are total decimal places x 4.1 in the two numbers. 52 + 14080 1442 We place the decimal places in from the left. Thus, the final product is 14.42.

A shopper has 5 one-dollar bills, 6 quarters, nickels, and 4 pennies in his pocket. How much money does he have? Division of Decimals 5 x $1.00 = $5.00 $0.25 $0.05 $0.01 x 6 x x 4 $1.50 $0.15 $0.04 Note the placement of the decimals in the multiplication products. Thus, the total amount of money in the shopper s pocket is: $5.00 1.50 0.15 + 0.04 $6.69 When dividing decimal numbers, we first remove the decimal in the divisor by moving the decimal in the dividend the same number of spaces to the right. For example, when dividing 1.45 into 5. we convert the numbers to 145 and 50 and perform normal whole number division. Find the quotient of 5. divided by 1.45. Convert to 145 and 50. Divide..65 145 50 145 50.00 Note that we insert - 45-45 the decimal to 95 950 continue division. - 870 800 Because one of the numbers divided contained one decimal place, we round the quotient to one decimal place. Thus, the final quotient is.7.

Operating with Percents 5 is what percent of 20? This is the same as converting 5 20 to % form. 5 100 5 5 = = 25% 20 1 1 1 There are 64 dogs in the kennel. 48 are collies. What percent are collies? Restate the problem. 48 is what percent of 64? Write an equation. 48 = n 64 Solve. 48 64 = n n = 4 = 75% 75% of the dogs are collies. The auditorium was filled to 90% capacity. There were 558 seats occupied. What is the capacity of the auditorium? Restate the problem. 90% of what number is 558? Write an equation. 0.9n = 558 Solve. n = 558.9 n = 620 The capacity of the auditorium is 620 people. A pair of shoes costs $42.00. Sales tax is 6%. What is the total cost of the shoes? Restate the problem. What is 6% of 42? Write an equation. n = 0.06 42 Solve. n = 2.52 Add the sales tax to the cost. $42.00 + $2.52 = $44.52 The total cost of the shoes, including sales tax, is $44.52.

Addition and subtraction of fractions Key Points 1. You need a common denominator in order to add and subtract reduced and improper fractions. 1 7 1+ 7 8 2 + = = = 2 4 6 4+ 6 7 + = = 12 12 12 12 12 2. Adding an integer and a fraction of the same sign results directly in a mixed fraction. 2 2 2+ = 2 2 = 2 4 4. Adding an integer and a fraction with different signs involves the following steps. - find a common denominator - add or subtract as needed - change to a mixed fraction if possible 1 2 1 61 5 2 2 = = = = 1 Add 2 7 + 5 8 7 Add the whole numbers; add the fractions and combine the two results: 2 2 7 + 5 = (7+ 5) + ( + ) 8 7 8 7 (7 ) + (8 2) = 12 + (LCM of 8 and 7) 56 21+ 16 7 7 = 12 + = 12 + = 12 56 56 56

Perform the operation. 2 5 6 We first find the LCM of and 6 which is 6. 2 2 5 45 1 = 2 6 6 6 1 7 7 + 2 4 8 1 7 1 7 7 + 2 = ( 7+ 2) + ( + ) 4 8 4 8 ( 2+ 7) = ( 5) + 8 5 = ( 5) + ( ) 8 5 5 8 5 40+ 5 = ( 5) + = + = 8 1 8 8 8 5 = = 4 8 8 Divide 5 by 8 to get 4, remainder. Caution: A common error is... 1 7 2 7 9 7 + 2 = 7 + 2 = 5 4 8 8 8 8 Wrong. It is correct to add -7 and 2 to get -5, but adding 2 + 7 = 9 8 8 8 2 7 5 is wrong. It should have been + =. Then, 8 8 8 5 5+ = 4 as before. 8 8

Multiplication of fractions Using the following example: 1 5 4 6 1. Convert each number to an improper fraction. 1 (12+ 1) 1 5 = = 4 4 4 6 is already in reduced form. 2. Reduce (cancel) common factors of the numerator and denominator if they exist. 1 5 No common factors exist. 4 6. Multiply the numerators by each other and the denominators by each other. 1 5 65 = 4 6 24 4. If possible, reduce the fraction back to its lowest term. 65 24 Cannot be reduced further. 5. Convert the improper fraction back to a mixed fraction by using long division. 2 65 = 24 65 = 2 17 24 24 48 17 Summary of sign changes for multiplication: a. ( + ) ( + ) = ( + ) b. ( + ) ( ) = ( ) c. ( + ) ( ) = ( ) d. ( ) ( ) = ( + )

1 5 22 5 7 = Reduce like terms (22 and 11) 11 11 = 2 5 = 10 = 1 1 1 5 25 5 6 = 4 9 4 9 125 17 = = 6 6 1 Negative times a negative equals a positive. 4 7 Division of fractions = 1 = 4 7 28 1. Change mixed fractions to improper fractions. 2. Change the division problem to a multiplication problem by using the reciprocal of the number after the division sign.. Find the sign of the final product. 4. Cancel common factors if they exist between the numerator and the denominator. 5. Multiply the numerators together and the denominators together. 6. Change the improper fraction to a mixed number. 1 1 16 2 = 9 5 4 5 4 = 16 4 Reciprocal of 9 is 4 5 9 4 9. 64 19 = = 1 45 45

5 1 9 7 11 = 4 8 4 8 = 1 8 Reduce like terms. 4 9 = 1 2 = 2 1 1 1 5 25 2 4 = 2 6 2 6 = 5 6 Reduce like terms. 2 25 = 1 = 1 5 5 7 4 7 5 8 = 16 8 1 6 = 4 16 Reduce like terms. 8 7 4 2 = Negative times a negative equals a positive. 1 7 86 2 = = 12 7 7

Skill 1. Apply basic number theory concepts including the use of primes, composites, factors, and multiples in solving problems. GCF is the abbreviation for the greatest common factor. The GCF is the largest number that is a factor of all the numbers given in a problem. The GCF can be no larger than the smallest number given in the problem. If no other number is a common factor, then the GCF will be the number 1. To find the GCF, list all possible factors of the smallest number given (include the number itself). Starting with the largest factor (which is the number itself), determine if it is also a factor of all the other given numbers. If so, that is the GCF. If that factor doesn't work, try the same method on the next smaller factor. Continue until you find a common factor. That is the GCF. Note: There can be other common factors besides the GCF. Find the GCF of 12, 20, and 6. The smallest number in the problem is 12. The factors of 12 are 1,2,,4,6, and 12. 12 is the largest factor, but it does not divide evenly into 20. Neither does 6, but 4 will divide into both 20 and 6 evenly. Therefore, 4 is the GCF. Find the GCF of 14 and 15. Factors of 14 are 1,2,7 and 14. 14 is the largest factor, but it does not divide evenly into 15. Neither does 7 or 2. Therefore, the only factor common to both 14 and 15 is the number 1, the GCF. LCM is the abbreviation for least common multiple. The least common multiple of a group of numbers is the smallest number that all of the given numbers will divide into. The least common multiple will always be the largest of the given numbers or a multiple of the largest number. Find the LCM of 20, 0 and 40. The largest number given is 40, but 0 will not divide evenly into 40. The next multiple of 40 is 80 (2 x 40), but 0 will not divide evenly into 80 either. The next multiple of 40 is 120. 120 is divisible by both 20 and 0, so 120 is the LCM (least common multiple). Find the LCM of 96, 16, and 24. The largest number is 96. 96 is divisible by both 16 and 24, so 96 is the LCM.

Elly Mae can feed the animals in 15 minutes. Jethro can feed them in 10 minutes. How long will it take them if they work together? If Elly Mae can feed the animals in 15 minutes, then she could feed 115 of them in 1 minute, 215 of them in 2 minutes, x 15 of them in x minutes. In the same fashion Jethro could feed x 10 of them in x minutes. Together they complete 1 job. The equation is: x x + =1 15 10 Multiply each term by the LCD of 0: 2x + x = 0 x = 6 minutes Composite numbers are whole numbers that have more than 2 different factors. For example 9 is composite because besides factors of 1 and 9, is also a factor. 70 is also composite because besides the factors of 1 and 70, the numbers 2,5,7,10,14, and 5 are also factors. Prime numbers are whole numbers greater than 1 that have only 2 factors, 1 and the number itself. Examples of prime numbers are 2,,5,7,11,1,17, or 19. Note that 2 is the only even prime number. When factoring into prime factors, all the factors must be numbers that cannot be factored again (without using 1). Initially we can factor numbers into any 2 factors. Check each resulting factor to see if we can factor it again. Continue factoring until all remaining factors are prime. This is the list of prime factors. Regardless of what way the original number was factored, the final list of prime factors will always be the same. Remember that the number 1 is neither prime nor composite. Factor 0 into prime factors. Factor 0 into any 2 factors. 5 6 Now factor the 6. 5 2 These are all prime factors. Factor 0 into any 2 factors. 10 Now factor the 10. 2 5 These are the same prime factors even though the original factors were different.

Factor 240 into prime factors. Factor 240 into any 2 factors. 24 10 Now factor both 24 and 10. 4 6 2 5 Now factor both 4 and 6. 2 2 2 2 5 These are prime factors. We can also write this as 5. 2 4 Skill 1.4 Apply the order of operations with or without grouping symbols. We must always follow the Order of Operations when evaluating algebraic expressions. Follow these steps in order: 1. Simplify inside grouping characters such as parentheses, brackets, square roots, fraction bars, etc. 2. Multiply out expressions with exponents.. Do multiplication or division, from left to right. 4. Do addition or subtraction, from left to right. 5( b + 2) = 5b 10 = 27 5b 10 = 17 5b 2 4 2 2(4 2 ) = 2 4 2 2(4 6) = 2 4 2 2( 2) = 2 4 2 + 4= 2 4 8+ 4 = 2 2+ 4= 6 2= 26