Question and Answers #23 Fractions

Questions and Answers Sheet 23 Fractions Question #1 How do you convert 0.4 (4 being repeated) to a fraction? Answer: We first let 0.4 (4 being repeated) be x Since x is recurring in 1 decimal places, we multiply it by 10. 10𝑥 = 4.4 Next, we subtract them. 10𝑥 − 𝑥 = 4.4 = 0.4 9𝑥 = 4 Lastly, we divide both sides by 9 to get x as a fraction. 𝑥= 4 9 Question #2 Express 0.85 as a fraction? Answer: The decimal 0.85 can be converted to the fraction 85 100 , by definition of a decimal. This fraction however, is not in simplest form. The fraction can be simplified to 17 20 85 100 , as both the numerator (top) and denominator (bottom) of the fraction are divisible by 5. Therefore, your answer is 17 20 Question #3 3 2 4 3 Which fraction is bigger or ? Answer: To check which of the two factors is bigger, we should make the denominator of two factors same by finding GCD and then making each of the denominator equal to their GCD. Here in 3 4 and 2 3 denominators are 4 and 3 and their GCD is 12. Now let us make each denominator equal to 12. 3 4 = 3×3 3×4 = 9 12 and 2 4×2 8 = = 3 4 × 3 12 , 9 Now as 12 3 Hence 4 > 8 12 , we have 3 4 > 2 3 is bigger. Question #4 What is 6 10 in decimals ? Answer: 6 = 6 ÷ 10 = 0.6 10 6 = 6 × 10−1 = 0.6 10 Question #5 Is 0.6 a rational number? Answer: A rational number is a number, which can be described as a ratio of two integers. As it is a ratio of integers, it can be positive as well as negative. Further, any decimal fraction, which limits itself beyond the decimal point (such as 5.7) which does not go beyond tenth place) or has continuously repeats numbers (till infinity) beyond a certain place of decimal (such as 4.33333 … or 0.23124312341325134 ….) can be easily written as ratio of integers, however large and hence are rational. Now as 0.6 = 6 10 , is a ratio of two integers, it is a rational number. Question #6 How to find 𝑓 ′ (0)? 𝑓(𝑥) = 𝑥2 + 3 2𝑥 − 1 Answer: Explanation: As 𝑓(𝑥) = 𝑥 2 +3 2𝑥−1 and using quotient rule, 𝑓 ′ (𝑥) = 2𝑥 × (2𝑥 − 1) − 2 × (𝑥 2 + 3) (2𝑥 − 1)2 = 4𝑥 2 − 2𝑥 − 2𝑥 2 − 6 (2𝑥 − 1)2 = 2𝑥 2 − 2𝑥 − 6 (2𝑥 − 1)2 and 𝑓 ′ (0) = 2(02 −0−3) (2×0−1)2 Question #7 6 = − = −6 1 What is the fraction 4 9 simplified? Answer: The fraction can't be simplified because there is no common factor, or GCF, between 4 and 9. Question #8 How do you write the fraction 3 6 in simplest form ? Answer: We have to find a common factor that will divide into 3 and 6 and reduce the fraction. This process is called cancelling The only factor that is common to 3 and 6 is 3 3 3÷3 6 6÷3 𝑟⇒ = = 1 2 in somplest form This is usually written as follows. 3 31 1 = = 6 62 2 A fraction is in simplest form when no other factor apart from 1 will divide into the numerator/denominator. Question #9 Convert 1.7 into a fraction, please. Answer: Proper fraction: 1,7 = 1 7 10 = 17 10 Question #10 Write 78% as a fraction Answer: Divide 78 by 100 to get a fraction 78 100 Now, we can simplify it 78 100 = 39 50 Question #11 3 What is as a fraction? 2 Answer: As 1 2 3 2 is an improper fraction, we can rearrange it as a sum of two fractions: 2 1 1 2 2 2 + , that can be simplified to 1 + = 1 Well, that's your answer Question #12 Simplify 4 if it isn't in lowest terms. 10 Answer: 4 10 isn't in lowest terms, so we can simplify it (4 ÷ 2) 4 2 = = 10 (10 ÷ 2) 5 Question #13 2 Simplify − + 𝑥 5 𝑦 3 𝑦 − 2 𝑥 Answer: 5 2 𝑦 2 − + − 𝑥 3 𝑥 𝑦 Multiply the numerator and denominator of the middle fraction by y 2 5 2 =− + − 𝑥 3 𝑥 Move the negative signs = −2 5 −2 + + 𝑥 3 𝑥 Add the numerators of a fraction with the same denominator = −2 − 2 5 + 𝑥 3 Simplify the numerator = −4 5 + 𝑥 3 4 5 =− + 𝑥 3 Question #14 3 How do you simplify 92 ? Answer: Explanation: 3 Breaking 92 down into its component parts 3 2 is the same as 3 × 3 1 2 which is the same as 1 so 92 is the same as 92×3 1 2 ×3 1 3 Write this as: (92 ) 1 Consider the part of 92 This is the same as √9 = 3 1 3 So (92 ) = (√9) 3 33 = 27 However, it true to say that: √9 = ±3 and (±3)3 = ±27 Question #15 What is the derivative of 10𝑥 ? Answer: There is a rule for differentiating these functions 𝑑 𝑑𝑥 [𝑎𝑢 ] = (ln 𝑎) ⋅ (𝑎𝑢 ) ⋅ 𝑑𝑢 𝑑𝑥 Notice that for our problem 𝑎 = 10 and 𝑢 = 𝑥 so let's plug in what we know. 𝑑 𝑑𝑢 [10𝑥 ] = (ln 1 0) ⋅ (10𝑥 ) ⋅ 𝑑𝑥 𝑑𝑥 if 𝑢 = 𝑥 then, 𝑑𝑢 =1 𝑑𝑥 because of the power rule: so, back to our problem, which simplifies to 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑 𝑑𝑥 [𝑥 𝑛 ] = 𝑛 ⋅ 𝑥 𝑛−1 [10𝑥 ] = (ln 1 0) ⋅ (10𝑥 ) ⋅ (1) [10𝑥 ] = (ln 1 0) ⋅ (10𝑥 ) This would work the same if u was something more complicated than x. A lot of calculus deals with the ability to relate the given problem to one of the rules of differentiation. Often we have to alter the way the problem looks before we can begin, however that was not the case with this problem. Question #16 Simplifying long fractions 3 2 (8 + ) + (3 ) 4 3 2 7 (4 + − (1 ) 5) 8 The correct answer is (4 + Answer: 278 303 ) Combine the numerator into an improper fraction: 3 2 3 2 8 +3 =8+3+ + 4 3 4 3 = 132 9 8 + + 12 12 12 = 149 12 Do the same for the denominator: 2 7 2 7 4+ −1 = 4−1+ − 5 8 5 8 = 120 16 35 + − 30 40 40 = 101 40 Then divide both: 149 12 101 40 = = 149 101 ⋅ 40 12 149 10 ⋅ 101 3 1490 303 =4+ 278 303 Question #17 If 𝑎 𝑏 = 𝑐 𝑑 Why does 𝑎+𝑐 𝑏+𝑑 = 𝑎 𝑏 𝑐 = ? 𝑑 Answer: Sketch: If you have 𝑝+𝜆𝑝 𝑞+𝜆𝑞 (1+𝜆)𝑝 = (1+𝜆)𝑞 = 𝑝 𝑞 and 𝜆𝑝 𝜆𝑞 , then 𝑝 𝑞 provided 1 + 𝜆 ≠ 0 Question #18 Proof of one inequality 𝑎+𝑏+𝑐 ≤ 𝑎3 𝑏 3 𝑐 3 + + 𝑏𝑐 𝑐𝑎 𝑎𝑏 Answer: Using Cauchy-Schwarz Inequality twice: 𝑎4 + 𝑏 4 + 𝑐 4 ≥ 𝑎2 𝑏 2 + 𝑏 2 𝑐 2 + 𝑐 2 𝑎2 ≥ 𝑎𝑏 2 𝑐 + 𝑏𝑎2 𝑐 + 𝑎𝑐 2 𝑏 = 𝑎𝑏𝑐(𝑎 + 𝑏 + 𝑐) Question #19 How do we write 4.5 as a fraction? Answer: Explanation: Every number in decimal form, in which either decimals terminate or certain numbers after decimal point continue to repeat endlessly, can be written as a fraction. As the given number 4.5 terminates after first deciimal it can be written as fraction. Here we have 4.5 = 45 10 = 9 2 fractions Question #20 Bad Fraction Reduction That Actually Works 16 1 = 64 4 This is certainly not a correct technique for reducing fractions to lowest terms, but it happens to work in this case, and I believe there are other such examples. Is there a systematic way to generate examples of this kind of bad fraction reduction? Answer: It's easy to find them all. Suppose 10𝑎+𝑛 10𝑛+𝑏 = 𝑎 𝑏 . Thus (10𝑎 − 𝑏)𝑛 = 9𝑎𝑏 . Case 1: (9, n) = 1: 9 |10 a − b  9 | a − b  a = b  9 a n = 9 a 2  n = a = b (trivial) Case 2: (9, n) = 9:10 a − b = a b  a | b, 10 = (b / a ) (a + 1) soa, b = 1,5 or 4,8 which yields the solutions: Case 3: (9, 𝑛) = 3 : yields 19 95 16 64 = 1 5 and 49 98 1 26 4 65 = , and 1 = . Similar analysis of the remaining 2 2 = . 5