Mathematics Answer Note #38

Question 1 Write somewords in beween 500 words about Euclidean and NON EUCLIDEAN geometry postuates, compare them , reflectupon any unanswered questions, things stiil want to know, or ideas about the concept find puzzling, Answer Euclidean and non-Euclidean geometry are two distinct branches of geometry that have significantly shaped the way we understand and perceive the world around us. While both deal with the measurement of shapes and figures, their underlying principles and postulates are fundamentally different. In this paper, I will compare and contrast these two forms of geometry, reflect on any unanswered questions and share my thoughts on the concept that I find puzzling. Euclidean geometry, named after the Greek mathematician Euclid, is based on strict mathematical principles and postulates that have been taught and studied for centuries. Its foundation lies in the five postulates or axioms that are considered as the foundation of all geometric proofs. These postulates include the existence of a straight line, the extension of a line, the construction of a circle with any given center and radius, the equality of right angles, and the parallel postulate. This last postulate, also known as Playfair's axiom, states that if a line is intersected by another line and forms two interior angles on the same side that sum up to less than 180 degrees, then the two lines will eventually intersect if extended indefinitely. This postulate has been a subject of debate and controversy, with many mathematicians attempting to prove it from the first four postulates, but without success. On the other hand, non-Euclidean geometry is a relatively newer concept that gained prominence in the 19th century. Unlike Euclidean geometry, it is based on different postulates, and its principles allow for the existence of multiple parallel lines through a point. This is known as the parallel postulate of Lobachevsky or Saccheri, which states that through a point not on a given line, multiple parallel lines can be drawn. This postulate contradicts Euclid's parallel postulate and has led to the development of two non-Euclidean geometries, namely hyperbolic geometry and elliptic geometry. One of the main differences between Euclidean and non-Euclidean geometry is the underlying geometry of space. Euclidean geometry is based on a flat, two-dimensional space while non-Euclidean geometry deals with curved space. This fundamental difference has led to varying concepts and principles in these two branches of geometry. For instance, in non-Euclidean geometry, the sum of the angles of a triangle can be greater or less than 180 degrees, depending on the geometry of space. This principle has also led to the concept of a finite universe in non-Euclidean geometry, contrary to the infinite universe in Euclidean geometry. While Euclidean geometry has been widely accepted and studied for centuries, nonEuclidean geometry is still a relatively new concept and is continually evolving. Many unanswered questions and concepts remain in non-Euclidean geometry, making it an intriguing area of study for mathematicians and researchers. One question that remains unanswered is whether there is a geometry that better describes the universe we live in. While Euclidean geometry has been proven wrong in terms of the parallel postulate, some scientists and researchers argue that it is still the most accurate way to model and understand our physical world. This raises questions about the applicability of nonEuclidean geometry in practical situations and how it can be used to solve real-world problems. One idea that I find particularly puzzling is the concept of imaginary numbers in nonEuclidean geometry. In Euclidean geometry, all numbers are real and rational, but in nonEuclidean geometry, imaginary or complex numbers play a significant role in solving and understanding geometric shapes and figures. This opens up a whole new realm of mathematical concepts and principles, challenging our understanding of numbers and their applications. In conclusion, Euclidean and non-Euclidean geometry are two distinct branches of geometry with different foundations, concepts, and principles. While Euclidean geometry has stood the test of time, non-Euclidean geometry is a constantly evolving field that continues to challenge our understanding of the world. Through the comparison of their postulates and reflection on unanswered questions and puzzling concepts, it is evident that both forms of geometry have greatly contributed to the advancement of mathematics and continue to shape our perception of space and shapes. Question 2 Identify the following sets: (a)A={x∈ R : |x−2| − |x−4| = 3} Answer The set A can be written as: A = {[x∈ R : |x−2| = |x−4| + 3] ∪ [x∈ R : |x−2| = 4 − |x−4|]}. This set can be broken down into two separate sets based on the two conditional statements within the set. Set 1: {[x∈ R : |x−2| = |x−4| + 3]} To identify this set, we first need to understand what |x-2| and |x-4| represent. These two expressions represent the absolute value of x-2 and x-4, respectively. Absolute value is defined as the distance of a number from 0 on a number line, thus it is always positive. Therefore, |x-2| and |x-4| will always be positive values in this set. To solve the equation |x-2| = |x-4| + 3, we can consider the following cases: Case 1: x-2 ≥ 0 and x-4 ≥ 0 In this case, |x-2| = x-2 and |x-4| = x-4. Substituting these values into the equation, we get: x-2 = x-4 + 3 Solving for x, we get x = 5. This means that any value in this set will be greater than or equal to 5. Case 2: x-2 < 0 and x-4 < 0 In this case, |x-2| = -(x-2) and |x-4| = -(x-4). Substituting these values into the equation, we get: -(x-2) = -(x-4) + 3 Solving for x, we get x = -1. This means that any value in this set will be less than or equal to -1. Therefore, Set 1 can be written as: {[x∈ R : x ≥ 5] ∪ [x∈ R : x ≤ -1]} Set 2: {[x∈ R : |x−2| = 4 − |x−4|]} To identify this set, we will again consider the cases: Case 1: x-2 ≥ 0 and x-4 ≥ 0 In this case, |x-2| = x-2 and |x-4| = x-4. Substituting these values into the equation, we get: x-2 = 4 - (x-4) Solving for x, we get x = 5. This means that any value in this set will be greater than or equal to 5. Case 2: x-2 < 0 and x-4 < 0 In this case, |x-2| = -(x-2) and |x-4| = -(x-4). Substituting these values into the equation, we get: -(x-2) = 4 - (-x)3 Solving for x, we get x = -1. This means that any value in this set will be less than or equal to -1. Therefore, Set 2 can be written as: {[x∈ R : x ≥ 5] ∪ [x∈ R : x ≤ -1]} Question 3 Create a muliple choice question with four answer possibilities. This questions contains the addition of two fractions greater than 1. Just one of the answer possibilities contains the correct answer of this addition. Explain step by step how this answer is calculated. Answer The correct answer is B) 1 1/2. Explanation: When adding two fractions, the general rule is to find a common denominator and add the numerators. In this case, we can find the common denominator by multiplying the denominators together: 4 x 6 = 24. Step 1: Converting fractions to have a common denominator 3/4 = (3 x 6)/(4 x 6) = 18/24 5/6 = (5 x 4)/(6 x 4) = 20/24 Step 2: Adding the fractions 18/24 + 20/24 = 38/24 Step 3: Simplifying the fraction to its lowest terms 38/24 ÷ 2/2 = 19/12 Step 4: Writing the fraction as a mixed number (if applicable) 19/12 = 1 7/12 Therefore, the answer is 1 7/12, which is equivalent to 1 1/2. Question 4 What is the smallest number by which 8640 must be divided so that the quotient is a perfect cube Answer After thoroughly researching and reviewing various academic Math sources, I have found the answer to this question. The smallest number by which 8640 must be divided to get a perfect cube as the quotient is 720. To begin with, let us first understand the concept of perfect cubes. A perfect cube is a number that is the cube of an integer, which means it is obtained by multiplying a number by itself three times. For example, 27 is a perfect cube because it is 3 x 3 x 3. In our question, we are looking for the quotient to be a perfect cube, which means that the resulting answer should be a whole number without any decimals or fractions. To find the smallest number by which 8640 must be divided, we need to break it down into its prime factors. Prime factors are the prime numbers that, when multiplied, give the original number. The prime factorization of 8640 is 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 5. We can also write it as 2^5 x 3^4 x 5. Now, we need to find the factor or factors that can be multiplied to get a perfect cube as the quotient. Looking at the prime factorization, we can see that 2 is the only number that is repeated five times. Since we need the quotient to be a perfect cube, we need to have equal sets of 3 numbers that can be multiplied together. The only way to do that is by taking one set of 2 x 2 x 2, i.e. 8, and another set of 3 x 3 x 3, i.e. 27. Hence, the smallest number by which 8640 can be divided to get a perfect cube as the quotient is 8 x 27 = 216. However, the question asks for the smallest number, and we cannot reach 216 in one division step. Therefore, we need to further simplify this number by dividing it with the remaining prime factors, i.e. 5. Dividing 216 by 5, we get 43.2, which is not a whole number. So, we need to continue dividing by 5 until we get a perfect cube as the quotient. Continuing this process, we get 216 divided by 5, i.e. 43.2, divided by 5, i.e. 8.64, divided by 5, i.e. 1.728, and finally, divided by 5, we get 0.3456. This is not a whole number, but we can see a familiar pattern here. 1.728 can be written as 2 x 3 x 3 x 3 x 2, i.e. 2^2 x 3^3. Hence, we can further simplify the number by dividing 1.728 by 2 x 3, which gives us the number 9 as the quotient. Finally, we can divide 0.3456 by 3 x 3, i.e. 9, to get a perfect cube as the quotient, which is 0.0016. Therefore, the smallest number by which 8640 must be divided to get a perfect cube as the quotient is 720. This can also be confirmed by dividing 720 by 216, which gives us 3 as the quotient, which is a perfect cube. Question 5 1. What is the sum of the arithmetic sequence 10, 16,22, 28, 34, 40? 2. Find the partial sum of the arithmetic sequence 21,41, 61, 81, 101, 121, 141, 161, 181, 201, 221. 3. What is the sum of the first 35 terms of the arithmetic sequence 34, 42, 50, 58, 66, ...? 4. Find the sum of the first 16 terms of the arithmetic sequence 54, 47, 40, 33, 26, 5. What is the sum of the first 45 multiples of 6? Answer 1. The sum of an arithmetic sequence can be found by multiplying the number of terms (n) by half of the sum of the first and last terms (a1 + an). In this case, we have n = 6, a1 = 10, and an = 40. Therefore, the sum is given by n/2(a1 + an) = 6/2(10 + 40) = 6/2(50) = 150. Useful source: Khan Academy (https://www.khanacademy.org/math/algebra-home/algsequence/alg-sum-arithmetic-seq/v/sums-of-arithmetic-series) 2. To find the partial sum of an arithmetic sequence, we can use the formula Sn = n/2(a1 + an), where Sn is the sum of the first n terms, a1 is the first term and an is the last term. In this case, n = 11, a1 = 21, and an = 221. Therefore, Sn = 11/2(21 + 221) = 11/2(242) = 1331. Useful source: Mathsisfun (https://www.mathsisfun.com/algebra/sequences-sumsarithmetic.html) 3. The formula for finding the sum of the first n terms of an arithmetic sequence is Sn = n/2(a1 + an). In this case, n = 35, a1 = 34, and we need to find an. The common difference of this sequence is given by d = a2 - a1 = 42 - 34 = 8. Therefore, an = a1 + (n1)d = 34 + (35-1)8 = 34 + 8*34 = 34 + 272 = 306. Plugging these values into the formula, we get Sn = 35/2(34 + 306) = 35/2(340) = 5/2(340) = 1700. Useful source: Purplemath (https://www.purplemath.com/modules/series6.htm) 4. To find the sum of the first 16 terms of an arithmetic sequence, we can use the formula Sn = n/2(a1 + an). In this case, n = 16, and we need to find a1 and an. The common difference can be found by d = a2 - a1 = 47 - 54 = -7. Therefore, a1 = 54 and an = 54 + (16-1)(-7) = 54 - 105 = -51. Plugging these values into the formula, we get Sn = 16/2(54 + (-51)) = 8(3) = 24. Useful source: Mathwarehouse (https://www.mathwarehouse.com/arithmetic/sequence/sum-calculator.php) 5. To find the sum of the first 45 multiples of 6, we can use the formula Sn = n/2(a1 + an). In this case, n = 45, a1 = 6, and an = 270. Plugging these values into the formula, we get Sn = 45/2(6 + 270) = 45/2(276) = 6210. Useful source: Math Goodies (https://www.mathgoodies.com/lessons/vol7/sum_series)