Introduction to Math Thinking: Assignment Insights

Introduction to Mathematical Thinking. Tutorial for Assignment One of the important aspects of mathematical thinking is that you first ask yourself, "What does this problem ask me?" You then think about what you know and what you need to do in order to reach your target. At first, do not apply a previous solution to your current problem because doing so may lead you in the wrong direction. For example, words are used to communicate ideas; however, this technique is not particularly useful when solving advanced mathematics problems about perfect squares or integers. 𝑃𝑟𝑜𝑣𝑒 𝑜𝑟 𝑑𝑖𝑠𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑒 𝑐𝑙𝑎𝑖𝑚 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑚, 𝑛 2 2 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑚 + 𝑚𝑛 + 𝑛 𝑖𝑠 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑎𝑟𝑒 When we consider integers, we can set m equal to n equal to zero. 𝑚= 𝑛 =0 Then m squared plus m n, plus m squared equals zero, which is zero squared. 2 2 2 𝑚 + 𝑚𝑛 + 𝑛 = 0 = 0 How can we have mn+1 equal to p-squared for sum p. 2 𝑚𝑛 + 1 = 𝑝 ? If mn = p2, that means mn = p2–1. And one of the rules drilled into us was that p2–1 is P–1, P+1. 2 𝑚𝑛 = 𝑝 − 1 = (𝑝 − 1)(𝑝 + 1) If A minus P minus 1P plus 1, then we could set M equal to P minus 1 and N equal to P plus 1. 𝑚 = 𝑝 − 1, 𝑛 = 𝑝 + 1 P would be m plus 1. 𝑝=𝑚 +1 To find the number n, first set p equal to m + 1. Then n is equal to p + 1. This gives us our n value—the number m + 2. 𝑝=𝑚 +1 𝑛 =𝑚 +2 P is m+1 and n is p+1. So n is m p+1. So given the m, we've already found the n. 𝑛=𝑚 +2 Then mn plus 1 Equals m, m plus 2 plus 1 which is m squared plus 2 m plus 1 which is m plus 1 squared. 𝑚𝑛 + 1 = 𝑚(𝑚 + 2) + 1 2 = 𝑚 + 2𝑚 + 1 = (𝑚 + 1) 2 We have explained why, given an m and n such that m is even and n is odd, we took m + 2. 2 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓(𝑛) = 𝑛 + 𝑏𝑛 + 𝑐, 𝑤𝑖𝑡ℎ 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑏, 𝑐, 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑓(𝑛)𝑖𝑠 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑛 𝑜𝑟 𝑒𝑙𝑠𝑒 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑠𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑓𝑎𝑙𝑠𝑒. As you might first assume, it is difficult to find a quadratic equation whose roots are all composite numbers. However, by considering that if a number is composite, then it must be the product of two other whole numbers, we can simply choose those two numbers as coefficients in our equation. 𝐹(𝑛) = (𝑛 + 1)(𝑛 + 2) And lo and behold that is indeed of the form n squared plus bn plus c, where b and c are positive. 2 𝐹(𝑛) = (𝑛 + 1)(𝑛 + 2) = 𝑛 + 3𝑛 + 2 We proved that the number n squared plus 3n + 2 is always composite, or not prime. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑖𝑓 𝑒𝑣𝑒𝑟𝑦 𝑒𝑣𝑒𝑛 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 2 𝑖𝑠 𝑎 𝑠𝑢𝑚 𝑜𝑓 𝑡𝑤𝑜 𝑝𝑟𝑖𝑚𝑒𝑠 (𝑡ℎ𝑒 𝐺𝑜𝑙𝑑𝑏𝑎𝑐ℎ 𝐶𝑜𝑛𝑗𝑒𝑐𝑡𝑢𝑟𝑒), 𝑡ℎ𝑒𝑛 𝑒𝑣𝑒𝑟𝑦 𝑜𝑑𝑑 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 5 𝑖𝑠 𝑎 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒 𝑝𝑟𝑖𝑚𝑒𝑠. Our assignment asked us to solve a conjecture about Goldbach's conjecture, a problem that has been around for hundreds of years and has not yet been solved. How can we build on this conjecture? The first observation is that if n is an odd number bigger than 5, then n is of the form 2k + 3 where k is bigger than 1. 𝑛 = 2𝑘 + 3 𝑘 >1 Although we typically think of numbers as being of the form 2k+1, rewrite this one as 2k+3 so that we can find prime numbers. If n is greater than 5 and odd, then any number greater than 5 (which is odd) is of the form 2k+3 where k is greater than 1. Since 2k is even—because k is greater than 1—then 2k will equal p+q where p and q are primes. 2𝑘 > 2 𝑏𝑦 𝐺𝐶 2𝑘 = 𝑝 + 𝑞 𝑝, 𝑞 𝑎𝑟𝑒 𝑝𝑟𝑖𝑚𝑒𝑠 In which case, n, which is 2k plus 3, is p plus q plus 3. 𝑛=𝑝 +𝑞 + 3 The sum of three primes is equal to an even number for every even number greater than 5. We will prove the conjecture by assuming Goldbach's Conjecture, which states that every even number greater than 5 is the sum of three primes. We will demonstrate how any number bigger than 5 can be broken down into a sum of at least two primes, one of which is 3. The fifth and sixth questions both ask you to prove a statement. The difference is that the fifth asks you to prove it using mathematical induction, and the sixth asks you to prove it using a direct proof. 𝑈𝑠𝑒 𝑡ℎ𝑒 𝑚𝑒𝑡ℎ𝑜𝑑 𝑜𝑓 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑡𝑜 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑛 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑛 2 We'll begin by expressing the sum of the first n odd numbers as an equation with a formula. Otherwise, we will have to deal with this expression: the sum of the first n odd numbers. We therefore need to write down a formula for this sum. The problem asks us to show that 1+3+5+···+n=n2 . 2 1 + 3 + 5 + … + (2𝑛 − 1) = 𝑛 The hardest part of this proof is finding the last term. To prove this by induction, we must first show that the formula is true for n=1. We can do so by showing that 1 squared equals 1, which is true. Next, we assume that our formula holds for all positive integers less than or equal to n. Then we need to prove that it also holds for n+1. To do this, we'll add 2n+1 to both sides of our formula and get 1+3+5+ ...+(2n -1)+2n+1+2n+1. This is the first n+1 term from our formula. 1 + 3 + 5 + … + (2𝑛 − 1) + (2𝑛 + 1) The first odd number is n plus 1. The second case in the induction is when n plus 1 is added to the right-hand side. And now, when I add 2n plus 1 to the right-hand side it becomes that. 2 1 + 3 + 5 + … + (2𝑛 − 1) + (2𝑛 + 1) = 𝑛 + 2𝑛 + 1 The formula used to solve this problem is known as a recursive equation, which works in the same way as formulas typically taught in high school. The only difference is that when solving for n + 1, one must take into account there is one more thing: the extra one.