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Fractions

Fractions. Section 1.3 (20). Objectives (20). Learn multiplication symbols and recognize factors Simplify fractions Multiply fractions Divide fractions Add and subtract fractions Change mixed numbers to improper fractions and vice versa. 1.3.1 Multiplication Symbols and Factors (20).

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Fractions

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  1. Fractions Section 1.3 (20)

  2. Objectives(20) Learn multiplication symbols and recognize factors Simplify fractions Multiply fractions Divide fractions Add and subtract fractions Change mixed numbers to improper fractions and vice versa

  3. 1.3.1 Multiplication Symbols and Factors(20) A parenthesis and no +/- sign normally means the value is multiplied (or raised to a power) The symbols: x , (more properly) × , and • also denote multiplication. Since in algebra uses a single letter (not like most programming languages) denote a symbol representing a value, sometimes mathematicians are lazy and just place the values side by side. Examples of multiplying x times y are: x y ; x • y ; x ( y ) ; ( x ) y ; ( x ) ( y ) ; x × y

  4. Factors(20) We normally limit ourselves to factors that are integers. Agreed 6 = ½ • 12, but we will say only that 6 is equal to 2 • 3 or 3 • 2 . There are integers that are called prime numbers. These are values that can only be factored into itself times 1. Examples of primes are 3, 7, 11, and 17. Examples of non-primes are 77 ( 7 • 11 ), 15 ( 3 • 5 ), and 25 ( 5 • 5 ) .

  5. Useful Hints about Factors Know the initial primes {2, 3, 5, 7, 11, 13, 17, 19, 23} Learn the products of prime numbers (at least through 13 • 13 ( 169 ). Note the only even prime is 2 . Therefore; any even number has a factor of 2 . Strange, but if your add the digits of any number (e.g. 456 {4 + 5 + 6 = 15 which is divisible by 3} and the sum is divisible by 3, the number has a factor of 3. Any number ending in 0 or 5 is divisible by 5

  6. Prime Factors Every integer can be factored in a unique (except for order) product of prime numbers. We can see easily that 24 = 6 • 4 . But writing it as a product of prime factors we get 24 = 2 • 2 • 2 • 3 ( better 23 • 3 ). Sometimes we will see when we are reducing fractions we can get away with 24 = 6 • 4, but it doesn’t hurt to break into prime factors. Example:

  7. Prime Factorization Start by taking 2’s out one at a time, then 3’s, then 5’s, etc. until there is only a 1 left. Example: 120 2) 120 2) 60 2) 30 3) 15 5) 5 1 120 = 2•2•2•3•5 or 23•3•5

  8. Algebra or Arithmetic(21) The book differentiates arithmetic and algebra. Essentially algebra has values that are unknown and have to be found. But that is nit-picking. If I had 2 quarters, 3 dimes, and 7 pennies, how much do I have? Couldn’t I just say x = 2 • .25 + 3 • .10 + 9 • .01

  9. 1.3.2 Simplify Fractions (21) Simple fractions are written as a numerator over a denominator. The denominator ≠ 0 . Examples: The first two fractions equal the same value. They are called equivalent fractions. A reduced fraction has no common factors in the numerator and denominator.

  10. Simplify Fractions (formal)(21) The book formalizes it with these steps: • Find the largest integer that is a factor into both the numerator and denominator. This is called the greatest common factor (GCF). • Divide the numerator and denominator by the GCF.

  11. Greatest Common Factor There is an easy way of finding the GCF if you have factored the numerator and denominator into prime factors. The GCF is the product of the min of each prime. Example: 120 / 912 23 • 3 • 5 / 24 • 3 • 19 GCF = 23 • 3 = 24 Therefore 120 / 912 reduces to 5 / 38

  12. Example Simplify: • • • •

  13. 1.3.3 Multiply Fractions(22) To multiply fractions, you multiply the numerators and the denominators. Note: Neither denominator can be zero. Then you have to reduce the products. But if you pulled out common factors from either numerator and either denominator, it will ease your simplification. Example:

  14. Examples

  15. Example An engine requires a mixture of 7/16 quart of oil for each quart of gasoline. You want to use 8 quarts of gasoline for the engine. How much oil is required? Given: 7/16 qt oil per qt gas 8 qt of gas Find: How much oil do you need How: number of qt of gas times oil/gas Solve: 7/16 • 8 = 7/2 • 1 = 7/2 or 3 1/2 Solution: You need 3 1/2 qt of oil • • • • •

  16. 1.3.4 Division of Fractions(23) Division is just flip-flop the divisor (second term) and multiply. a / b ÷ c / d = a / b • d / c Note that in this case, not only can neither denominator be zero, but the second numerator can not be zero either. Also, you have to invert before you simplify. 2 / 3 ÷ 1 / 6 = 2 / 3 • 6 / 1 = 4 / 1 = 4 • • • • •

  17. Examples 4 / 7 ÷ 7 / 8 4 / 7 • 8 / 7 can’t simplify 32 / 49 3 / 2 ÷ 6 / 7 3 / 2 • 7 / 6 1 / 2 • 7 / 3 7 / 6 or 1 1/6 • • • • •

  18. 1.3.5 Add or Subtract Fractions(23) When you multiply fractions, you can reduce the fractions, but as long as the denominators are not zero, you don’t care if they are same or different. But when you add or subtract fractions, you must have the same denominator. If they are different, you find the Least Common Multiple (LCM) { aka in this case, the Least Common Denominator (LCD) } You write each term as an equivalent fraction with the LCD, then add/subtract the numerators.

  19. Example You say amen if the denominators are already the same. Just add/sub tract the numerators. Don’t forget to simplify the result. 5 / 12 + 3 / 12 = (5 + 4) / 12 = 9 / 12 = 3 / 4 7 / 16 - 3 / 16 = (7 – 3) / 16 = 4 / 16 = 1 / 4 5 / 11 - 7 / 11 = (5 – 7) / 11 = -2 / 11 • • • • •

  20. Reality Most of the time, we will have different denominators. We have to find equivalent fractions for each term with the same denominator. To compute the LCD, we write each denominator as a product of primes, and multiply the highest power of each term. Example: If the denominators are 18 and 30 18 = 2 • 32 ; 30 = 2 • 3 • 5 The LCD = 2 • 32 • 5 = 90 • • • • •

  21. Example 7 / 8 - 1 / 6 8 = 23; 6 = 2•3; LCD = 23•3 = 24 7 / 8 • 3 / 3 - 1 /6 • 4 / 4 21 / 24 - 4 / 24 (21 – 4) / 24 17 / 24 7 / 18 + 23 / 30 18 = 2•32; 30 = 2•3 •5; LCD = 2•32•5 = 90 7 / 18 • 5 / 5 + 23 / 30 • 3 / 3 35 / 90 + 69 / 90 (35 + 69) / 90 104 / 90 52 / 45 or 1 7/45 • • • • •

  22. 1.3.6 Change Mixed Numbers to Improper Fractions and vice-versa(24) We’ve been changing improper fractions to mixed numbers in several problems earlier. Note that a true improper fraction is not incorrect. However you can’t have a mixture of a mixed number and improper fraction. An example would be 4 7/3 is not legit. You would have to write it either as 6 1/3 or 19/3.

  23. Mixed Numbers and Improper Fractions A mixed number is a combination of a whole number and a proper fraction. Examples: 4 3/7 -1 1/6 5 4/6, but it should be simplified to 5 2/3 An improper fraction has the numerator greater than or equal to the denominator. Examples: 17/6 -2/2, but it should be written as -1 -5/3 • • • • •

  24. Mixed Number to Improper Fraction To change a mixed number to an improper fraction, you multiply the whole portion times the denominator and add it to the numerator. Then write the fraction with this value over the denominator. Example: 4 2/7 [(4 • 7) + 2] / 7 (28 + 2) / 7 30 / 7 • • • • •

  25. Examples Change 6 5/7 to an improper fraction [( 6 • 7) + 5] / 7 (42 + 5) / 7 47 / 7 Change -4 2/3 to an improper fraction -[(4 • 3) + 2] / 3 -(12 + 2) / 3 -14 / 3 • • • • •

  26. Improper Fraction to a Mixed Number To convert an improper fraction to a mixed number, divide the denominator into the numerator. Write the value as the quotient. Example: 62 / 8 7 R 6 8 ) 62 56 6 7 6/8, simplify to 7 3/4

  27. Example You have a sidewalk that is 53 4/5 feet long. You add a 12 3/10 feet long. How long is the sidewalk now. Given: 53 4/5 feet originally 12 3/10 extension Find: How long is it now How: Add the original to the extension Solve: 53 4/5 = 53 8/10 10 3/10 = 10 3/10 63 11/10 = 64 1/10 Solution: The sidewalk is now 64 1/10 feet long • • • • •

  28. Example When you went to the doctor, you weighed 240 3/5 lbs. He told you to lose weight. Over the past 2 weeks, you lost 7 5/6 lbs. What is your current weight. Given: Original weight 240 3/5 lbs Lost 7 5/6 lbs Find: Current weight How: Subtract loss from original Solve: 240 3/5 = 240 18/30 - 7 5/6 = - 7 25/30 232 23/30 lbs Solution: Your current weight is 232 23/30 lbs • • • • •

  29. Objectives(20) Learn multiplication symbols and recognize factors Simplify fractions Multiply fractions Divide fractions Add and subtract fractions Change mixed numbers to improper fractions and vice versa

  30. Fractions Section 1.3 (20)

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