Question and Answers #27 Fractions

Questions and Answers Sheet 27 Fractions Question #1 How do you convert 3 8 to a decimal? Answer: Answer: 3 8 = 0.375. Explanation: You can find it by long division, but let's try another way... Since 8 = 2 × 2 × 2and 10 = 2 × 5 , we know that 8 is a divisor of 1000 = 10 × 10 × 10 = 2 × 2 × 2 × 5 × 5 × 5 = 8 × 125 3 3⋅125 8 8⋅125 Therefore, = = 375 1000 = 0.375 Question #2 What is the simplified value of 1 8 + 4 18 ? Answer: Step 1 1 8 + = 4 18 and as LCD is 72, converting denominator to 72, we get 1×9 4×4 + 8 × 9 18 × 4 Step 2 = 9 16 + 72 72 = 25 72 Question #3 How do you find the fraction between two fractions 1 4 and 3 5 ? Answer: Well we can probably better visualize this if we convert each fraction to a decimal. 1 → 0.25 4 3 → 0.6 5 So now we ask ourselves, what decimal(s) is in between 0.25 and 0.6? Well, in reality there are infinitely many but that's a topic for another day. All we need is one decimal: 0.5 is in between 0.25 and 0.6 Now we just convert 0.5 into a fraction and that's 1 2 ! Question #4 3 How do you simplify 4 × 5 12 ? Answer: Step 1 3 4 × 5 12 = 3 5 × 2×2 2×2×3 = 3 5  2  2 2  2  3 Step 2 = = 5 2×2×2×2 5 16 Question #5 How do you write 4 3 5 as an improper fraction? Answer: First, you multiply 4 and 5 to get 20 and add it to the 3. This is the numerator. 3 4⋅5+3 5 5 4 = = 23 5 Question #6 What is the reciprocal of 2? Answer: Reciprocal means to "flip the fraction". So 2 is basically So if you simply flip 2 1 , it is 1 2 2 1 (2 over 1) if you write it as a fraction. , which can also be 0.5 Question #7 What is the simplified value of 2 9 1 7 5 2 8 6 + − + ? Answer: To simplify 2 9 1 7 5 2 8 6 + − + , we should first make all denominators equal. For this, we ought to know Least Common Denominator (LCD) of {9, 2, 8, 6} , which is 72. 2 Hence, 9 1 7 5 2 8 6 + − + = 2 × 8 1 × 36 7 × 9 5 × 12 + − + 9 × 8 2 × 36 8 × 9 6 × 12 = 16 36 63 60 + − + 72 72 72 72 = 16 + 36 − 63 + 60 72 = 49 72 Question #8 How do I convert a mixed number into a decimal? Answer: 𝑏 𝑏 𝑐 𝑐 If you have a mixed number f type 𝑎 , just convert As 𝑏 𝑐 into decimal form. is a rational number, it would be either be a limiting decimal or non-limiting but repeating decimal. Round it to the nearest decimal place if desired. 𝑏 Further, as 𝑐 < 0 , it will have just 0 to the left of the decimal portion. Now add a towards the left of decimal and you have your result. 3 3 8 8 For example in 5 , we have Hence 3 8 = 3×125 8×125 = 375 1000 = 0.375 = 5.375 or 5.38, if you round up to two places of decimals. 5 ln 3 , 7 5 = 0.714285714285714285714285 … 7 5 Hence, 3 = 3.714285714285714285714285... . or 3.7143 rounded up to four places of decimal. 7 Question #9 What is 3 4 as a decimal? Answer: For converting a fraction into decimal, we should first reduce the fraction to its lowest terms. As 3 4 is already in lowest term, we can skip it here. Further as factors of denominator 4 are only 2s, we should have terminating decimals (for terminating decimals, we should have only 2 and 5 as ' factors of denominator). In fact there are two 2s, hence to convert it into decimal we can multiply numerator and denominator by two 5s i.e. 5 × 5 . 3 3×5×5 4 4×5×5 Now dividing 3 by 4, we get = = 75 100 = 0.75 Question #10 When is the lcm of a fraction sum the actual denominator. 𝑎 𝑐 𝑥 𝑏 𝑑 𝑦 Consider a sum + = where each fraction is reduced. Alternatively using the familiar process of lowest common 𝑑 𝑏 𝑎 𝑐 𝑎⋅(𝑏,𝑑)+𝑐⋅(𝑏,𝑑) 𝑏 𝑑 [𝑏,𝑑] denominators, we have + = where (b,d) denotes the gcd and [b,d] denotes the lcm. My question is, when is it true that 𝑦 = [𝑏, 𝑑] ? For example, this does not hold for 5 6 + 1 14 = 38 42 = 19 21 𝑤ℎ𝑒𝑟𝑒 21 ≠ [14,6] Are there any simple necessary and sufficient conditions for 𝑦 = [𝑏, 𝑑] ? Edit It has been suggested that 𝑦 = [𝑏, 𝑑] ⟺ (𝑏, 𝑑) = 1 is a necessary and sufficient condition. I'm interested in either a proof or a counterexample if possible. Edit 2 After a bit of searching, I've found 1 24 + 1 16 = 5 48 as a counterexample. I am still looking for nice conditions for this to be satisfied and I feel that I should give an explanation of exactly what type of condition I am seeking. Angela has provided a necessary and sufficient condition, but it does not seem to be "simpler" than simply adding the fraction and seeing if it reduces. This is perhaps ambitious, but I am looking for a condition which is simple enough to use by inspection for simple fractions. Answer: let 𝑥 = gcd(𝑏, 𝑑) 𝑏 = 𝑥𝑒 𝑑 = 𝑥𝑓 𝑎 𝑐 𝑎𝑓 + 𝑐𝑒 + = 𝑏 𝑑 𝑒𝑓𝑥 𝑒𝑓𝑥 = 𝑙𝑐𝑚(𝑏, 𝑑) e,f are relative 's The fraction simplifies when gcd( af + ce, x ) =  1 which is when gcd(ad + bc, bd ) = gcd(b, d ) Question #11 What is the formula of converting fraction into addition of fractions? 4−𝑠 1 1 2 1 5 1 = − + (𝑠 − 3)(𝑠 − 2)(𝑠 + 1) 4 𝑠 − 3 3 𝑠 − 2 12 𝑠 + 1 Answer: Partial Fraction Decomposition 4−𝑠 𝐴 𝐵 𝐶 = + + (𝑠 − 3)(𝑠 − 2)(𝑠 + 1) 𝑠 − 3 𝑠 − 2 𝑠 + 1 ∴ 4 − 𝑠 = 𝐴(𝑠 − 2)(𝑠 + 1) + 𝐵(𝑠 − 3)(𝑠 + 1) + 𝐶(𝑠 − 3)(𝑠 − 2) To find the values of A, B, C , simply let s equal certain values 0, 1, 2,.. . (choosing specific values can make it extremely easy) and solve the ensuing equations. Question #12 Are these fractions all equal? Are the following expressions all equal to one another? 2𝑎𝑏 + 𝑐 𝑦 2𝑎𝑏 𝑐 + 𝑦 𝑦 2𝑎 2𝑏 𝑐 ∗ + 𝑦 𝑦 𝑦 2𝑎𝑏 + 𝑐 2𝑎𝑏 𝑐 2𝑎 2𝑏 𝑐 = + = × + 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 Answer: By starting with the first, and applying the distributive law, we get the second: 2𝑎𝑏 + 𝑐 1 1 1 2𝑎𝑏 𝑐 = (2𝑎𝑏 + 𝑐) = 2𝑎𝑏 + 𝑐 = + 𝑦 𝑦 𝑦 𝑦 𝑦 𝑦 The last one is not equal to the other two, because equal to 2𝑎𝑏 𝑦 2𝑎 𝑐 𝑦 𝑎 𝑦 𝑏 ⋅ 2𝑏 𝑦 𝑐 + . Notice that we used the fact that ⋅ = 𝑑 So the first two are equal, but not the third. Question #13 Common denominator with radicals Can you make these 2 fractions into 1? 2√9 − 2𝑥 − 2𝑥 √9 − 2𝑥 I thought you could make them into −2𝑥+18 √9−2𝑥 Answer: You're almost correct. 2√9 − 2𝑥 − = = = = = 2𝑥 √9 − 2𝑥 2(√9 − 2𝑥) 2 √9 − 2𝑥 2(9 − 2𝑥) √9 − 2𝑥 − − = 2√9 − 2𝑥 × 2𝑥 √9 − 2𝑥 2𝑥 √9 − 2𝑥 2(9 − 2𝑥) − 2𝑥 √9 − 2𝑥 18 − 4𝑥 − 2𝑥 √9 − 2𝑥 18 − 6𝑥 √9 − 2𝑥 Question #14 Evaluate fraction of sum So i have to evaluate this sum: √9 − 2𝑥 √9 − 2𝑥 − 𝑐 2𝑎⋅2𝑏 𝑦 𝑎𝑐 𝑦⋅𝑦 + = 2𝑥 √9 − 2𝑥 𝑏𝑑 . 𝑐 4𝑎𝑏 𝑦 𝑦2 + = 𝑐 4𝑎𝑏 𝑦 𝑦2 + , and + 𝑐 𝑦 is (usually) not 1−2−2 +4−2 −5−2 +7−2 −8−2 +10−2 −11−2 +⋯ 1+2−2 −4 −2 −5−2 +7−2 +8−2 −10−2 −11−2 +⋯ ∑∞[(3𝑛+1)−2 −(3𝑛+2)−2 ] 0 it has the form: ∑∞(−1) 𝑛 [(3𝑛+1)−2 0 +(3𝑛+2)−2 ] My current attempt : Trying to convert this into power series 𝑎(𝑛) = (3𝑛 + 1)−2 − (3𝑛 + 2)−2 𝑏(𝑛) = (3𝑛 + 1)−2 + (3𝑛 + 2)−2 Can a(n) and b(n) be the definite integral of certain polynomial function f(x) ? Maybe there is a better direction. Can someone give me a hint ? Answer: Let 𝐿𝑖2 (𝑧) = ∑∞ 𝑛=1 𝑧𝑛 𝑛2 , 𝐶𝑙2 (𝜃) = ∑∞ 𝑛=1 sin 𝑛𝜃 𝑛2 = 𝐼[𝐿𝑖2 (𝑒 𝑖𝜃 )] be dilogarithm and Clausen function, respectively. Then it is easy to see that the expression in question reduces to 𝐶𝑙2 (2𝜋/3) 𝐶𝑙2 (𝜋/3) . 1 By comparing the power series of both sides, we obtain 𝐿𝑖2 (𝑧) + 𝐿𝑖2 (−𝑧) = 𝐿𝑖2 (𝑧 2 ). 2 Now plugging 𝑧 = 𝑒 𝜋𝑖/3 gives 𝐿𝑖2 (𝑒 2𝜋𝑖/3 ) = 2𝐿𝑖2 (𝑒 𝜋𝑖/3 ) + 2𝐿𝑖2 (−𝑒 𝜋𝑖/3 ). Now taking imaginary part, we obtain 𝐶𝑙2 (2𝜋/3) = 2𝐶𝑙2 (𝜋/3) − 2𝐶𝑙2 (2𝜋/3), since −𝑒 𝜋𝑖/3 = 𝑒 −2𝜋𝑖/3 .Therefore we have 𝐶𝑙2 (2𝜋/3) 𝐶𝑙2 (𝜋/3) 2 = . 3 Question #15 Evaluate the expression 12 12 −10+50 + 22 22 −20+50 +⋯+ 802 802 −80+50 What should I do after that ? ∑ 𝑛2 𝑛2 − 10𝑛 + 50 Answer: You can do this by approximating the sum by an integral: 𝑛 ∫ 0 𝑥2 𝑑𝑥 𝑥 2 − 10𝑥 + 50 Mathematica says the integral is 𝑛 − 5 ln 5 0 + 5 ln(50 − 10𝑛 + 𝑛2 ) For n=80, this evaluates to 80 + ln 1 8424351793 = 103.637 This is only an approximation that is useful for large n. For n=80 , the sum itself is 104.203 . The general formula you posted is 𝑛2 𝑛2 −10𝑛+50 When n=80 , substituting n for 80 gives 802 802 −80⋅10+80 = 802 802 −800+50 For n=8 , this does give -80 in the denominator, but then, you won't have an 802 in the numerator. Question #16 Formula help with this equation 5 15 6 18 If 𝑥 = + Answer: − 10 12 , then (𝑥 − 1)3 =? 5 15 6 18 The expression 𝑥 = + 𝑥= − 10 12 may be simplified considerably: 5 15 10 + − 6 18 12 = 5 30 30 +( − ) 6 36 36 = 5 6 1 1 6 2 Thus, 𝑥 − 1 = − and 3(𝑥 − 1) = − (𝑥 − 1)3 = − 13 1 =− . 63 216 Question #17 Fractions and roots √18 + √98 + √50 + 4 2√2 I'm able to get to this part by myself: 15√2 + 2 2√2 But that's when I get stuck. The book says that the next step is 15√2 2√2 + 2 2√2 But I don't understand why you can take the 2 out of the original fraction, make it the numerator of its own fraction and having root of 2 as the denominator of said fraction. Answer: √18 + √98 + √50 + 4 2√2 3√2 + 7√2 + 5√2 + 4 2√2 = = 15√2 + 4 2√2 Now the standard procedure is to remove the radical in the denominator: 15√2 + 4 2√2 = 15√2 + 4 √2 2√2 √2 = 15√2 ⋅ √2 + 4√2 2√2 ⋅ √2 = 30 + 4√2 15 + 2√2 = 4 2 One can do it differently: set 𝑎 = √2 , so you can write 15√2+4 2√2 = 15𝑎 + 𝑎4 = 𝑎3 15 + 𝑎3 = 𝑎2 15 + 2√2 2 The final result can also be written by using 𝑎+𝑏 𝑐 𝑎 𝑏 𝑐 𝑐 = + so 15+2√2 2 = 15 2 + √2 Whether you want to do this last transformation depends on what you have to do with this number. Question #18 Fractions with radicals in the denominator I'm working my way through the videos on the Khan Academy, and have a hit a road block. I can't understand why the following is true: 6 = √85 6√85 85 Answer: With respect to handling square roots in the denominator: Say, for example, you have 𝑎 √𝑏 If you want a "square-root free" denominator, remember that you can multiply any fraction by the number 1 (or its equivalent) without changing its value. In this case, we want to multiply the given fraction by √𝑏 √𝑏 = 1 . Can you see why we would want to choose this particular representation of 1? Now, √𝑏 √𝑏 ⋅ 𝑎 √𝑏 = 𝑎⋅√𝑏 2 (√𝑏) = 𝑎⋅√𝑏 𝑏 And if the greatest common divisor of a and b > 1, then we can reduce \frac ab by dividing each of a and b by their 𝑎 greatest common divisor (i.e., reducing the fraction 𝑏 Now, we take a look at a slightly more complicated situation, as in your problem, where the denominator is itself a fraction given by 6 6√85 85 We can first multiply numerator and denominator by 85, then proceed as we did in the case above: 6 85 6 ⋅ 85 85 85 √85 85 ⋅ √85 = = = ⋅ = = √85 85 85 6√85 6 ⋅ √85 √85 √85 √85 85 ⋅ More simply, once we have reduced the expression to 85 √85 2 , we notice that 85 = (√85) , and so, this, together with the fact that we can then cancel a common factor of both the numerator and the denominator, we have: 85 √85 2 = (√85) √85 = √85 Note: When you have enough experience simplifying fractions and fractions with fractions with roots, etc., many of the steps which I've displayed explicitly above will be "second nature" for you, and you'll be able to manipulate such expressions with far less in the way of computations than I've shown above. Question #19 How do I use the partial fractions technique in this case? (𝑥 2 (𝑥 − 1) − 𝑥 + 1)(𝑥 + 1) Answer: The partial fraction decomposition takes the form 𝐹(𝑥) = (𝑥 2 2 𝑥−1 −𝑥+1)(𝑥+1) 2 = 𝑎 𝑥+1 + 𝑏𝑥+𝑐 𝑥 2 −𝑥+1 and notice that 𝑎 = 𝐹(𝑥)(𝑥 + 1)|𝑥=−1 = − and 0 = lim 𝑥𝐹(𝑥) = 𝑎 + 𝑏 → 𝑏 = and finaly 𝐹(0) = −1 = 𝑎 + 𝑐 → 𝑐 = − 3 𝑥→∞ 3 1 3 Question #20 How to simplify 4+2√6 √5+2√6 If 𝑥 = 5 + 2√6 , evaluate 𝑥 − 1/√𝑥 The answer written is 2√2 , but I can't figure my way out through the manipulations. I just know that I have the following: 4 + 2√6/√5 + 2√6 Answer: Squaring it, we get Remark Generally 40+16√6 5+2√6 =8 𝑛−1+√𝑛2 −1 = √2(𝑛 − 1) with analogous proof. √𝑛+√𝑛2 −1 For another example note 3+√11 √10+√99 = √2, by 𝑛 = 10 above.