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Chapter 1 Section 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

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Presentation on theme: "Chapter 1 Section 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley."— Presentation transcript:

1 Chapter 1 Section 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2 Multiplying and Dividing Real Numbers Find the product of a positive number and a negative number. Find the product of two negative numbers. Identify factors of integers. Use the reciprocal of a number to apply the definition of division. Use the rules for order of operations when multiplying and dividing signed numbers. Evaluate expressions involving variables. Interpret words and phrases involving multiplication and division. Translate simple sentences into equations. 1 1 4 4 3 3 2 2 6 6 5 5 1.61.6 8 8 7 7

3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiplying and Dividing Real Numbers Slide 1.6- 3 The result of multiplication is called the product. We already know how to multiply positive numbers, and we know that the product of two positive numbers is positive. We also know that the product of 0 and any positive number is 0, so we extend that property to all real numbers. Multiplication by Zero says, for any real number x,.

4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1 Objective 1 Find the product of a positive and negative number. Slide 1.6- 4

5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the product of a positive number and a negative number. The product of a 3(−1) represents the sum. Also, and. These results maintain the pattern, which suggests the rule for Multiplying Numbers with Different Signs; For any positive real numbers x and y, and. That is, the product of two numbers with opposite signs is negative. Slide 1.6- 5

6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Solution: Multiplying a Positive and a Negative Number Find the product. Slide 1.6- 6

7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2 Objective 2 Find the product of two negative numbers. Slide 1.6- 7

8 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the product of two negative numbers. The rule for Multiplying Two Negative Numbers states that: Slide 1.6- 8 For any positive real numbers x and y, That is, the product of two negative numbers is positive. Example:

9 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Solution: Multiplying Two Negative Numbers Find the product. Slide 1.6- 9

10 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3 Objective 3 Identify factors of integers. Slide 1.6- 10

11 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Identify factors of integers. In Section 1.1, the definition of a factor was given for whole numbers. The definition can be extended to integers. If the product of two integers is a third integer, then each of the two integers is a factor of the third. The table below show several examples of integers and factors of those integers. Slide 1.6- 11

12 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4 Objective 4 Use the reciprocal of a number to apply the definition of division. Slide 1.6- 12

13 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use the reciprocal of a number to apply the definition of division. The quotient of two numbers is found by multiplying by the reciprocal, or multiplicative inverse. By definition, since and, the reciprocal or multiplicative inverse of 8 is and of is. Pairs of numbers whose product is 1 are called reciprocals, or multiplicative inverses, of each other. Suppose that k is to be the multiplicative inverse of 0. Then k · 0 should equal 1. But, k · 0 = 0 for any real number. Since there is no value of k that is a solution of the equation k · 0 = 1, the following statement can be made: 0 has no multiplicative inverse Slide 1.6- 13

14 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use the reciprocal of a number to apply the definition of division. (cont’d) The Definition of Division says that, for any real numbers x and y, with y ≠ 0, That is, to divide two numbers, multiply the first by the reciprocal, or multiplicative inverse, of the second. If a division problem involves division by 0, write “undefined.” In the expression, x cannot have the value of 2 because then the denominator would equal 0 and the fraction would be undefined. Slide 1.6- 14

15 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Solution: Using the Definition of Division Find each quotient, using the definition of division. Slide 1.6- 15

16 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Dividing Signed Numbers When dividing fractions, multiplying by the reciprocal works well. However, using the definition of division directly with integers is awkward. It is easier to divide in the usual way and then determine the sign of the answer. The quotient of two numbers having the same sign is positive. The quotient of two numbers having different signs is negative. Examples:,, and Slide 1.6- 16

17 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Find each quotient. Solution: Dividing Signed Numbers Slide 1.6- 17

18 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5 Objective 5 Use the rules for order of operations when multiplying and dividing signed numbers. Slide 1.6- 18

19 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Solution: Using the Rules for Order of Operations Perform each indicated operation. Slide 1.6- 19

20 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 6 Objective 6 Evaluate expressions involving variables. Slide 1.6- 20

21 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Evaluating Expressions for Numerical Values Solution: Evaluate if and. Slide 1.6- 21

22 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 7 Objective 7 Interpret words and phrases involving multiplication and division. Slide 1.6- 22

23 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Interpret words and phrases involving multiplication. The word product refers to multiplication. The table gives other key words and phrases that indicate multiplication in problem solving. Slide 1.6- 23

24 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Write a numerical expression for the phrase and simplify the expression. Three times the difference between 4 and −11. Three-fifths of the sum of 2 and −7. EXAMPLE 7 Solution: Interpreting Words and Phrases Involving Multiplication Slide 1.6- 24

25 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Interpret words and phrases involving division. The word quotient refers to division. In algebra, quotients are usually represented with a fraction bar; the symbol ÷ is seldom used. The table gives some key phrases associated with division. Slide 1.6- 25

26 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 8 Write a numerical expression for the phrase and simplify the expression. The product of −9 and 2, divided by the difference between 5 and −1. Solution: Interpreting Words and Phrases Involving Division Slide 1.6- 26

27 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 8 Objective 8 Translate simple sentences into equations. Slide 1.6- 27

28 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 9 Write the sentence in symbols, using x as the variable. Then find the solution from the list of integers between −12 and 12, inclusive. The quotient of a number and −2 is 6. Translating Sentences into Equations Solution: Here, x must be a negative number since the denominator is negative and the quotient is positive. Since, the solution is −12. Slide 1.6- 28


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