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 The Mathematical Proof Thread Kernel Sohcahtoa The Outsider   Religious Views: non-religious and non-theist Posts: 1091 Threads: 25 Joined: September 5, 2016 Reputation: 39 RE: The Mathematical Proof Thread December 17, 2017 at 4:39 pm (This post was last modified: December 17, 2017 at 4:41 pm by Kernel Sohcahtoa.) Earlier this month, I did a proof that was pretty cool (IMO).  As a result, I thought I'd share it here.  As always, I'll put my post in hide tags.  In addition, I'd like to clarify  that my writing is not meant to be arrogant: my mathematical writing style is how I've learned to communicate mathematics (I definitely have lots of room for improvement). With that said, here is the following exercise from my mathematical proofs book by Gary Chartrand: Prove that if a,b, and c are positive real numbers, then (a+b+c)(1/a + 1/b + 1/c) ≥ 9 Hint/Analysis of Proof Let’s work backward from the conclusion and see where we end up: I will denote each backward statement with a B and a respective number.  First, let’s notice that the expression (1/a + 1/b +1/c) can be re-written as (bc+ac+ab)/abc.  Thus, the conclusion can be re-written as B1: (a+b+c)(bc+ac+ab)/abc ≥ 9.  Now, multiplying both sides of B1 by abc gives B2: (a+b+c)(bc+ac+ab) ≥ 9abc.  Expanding the left side of B2 via algebra gives B3: [abc + (a^2)c + (a^2)b] + [(b^2)c + abc + a(b^2)] +  [b(c^2) + a(c^2) + abc]= 3abc + (a^2)c + (a^2)b + (b^2)c + a(b^2) + b(c^2) + a(c^2) ≥ 9abc.  Subtracting both sides of B3 by 9abc and rearranging terms on the left side gives B4:[a(b^2) + a(c^2)] + [(a^2)b + b(c^2)] + [(a^2)c + (b^2)c] – 6abc ≥ 0.   Notice that –6abc= –2abc + (–2abc) + (–2abc), so we will put each –2abc term into a bracket in B4 to get B5: [a(b^2) – 2abc + a(c^2)] + [(a^2)b – 2abc + b(c^2)] + [(a^2)c – 2abc + (b^2)c] ≥ 0. Factoring out the common variable in each bracket in B5 gives B6: a[(b^2) – 2bc + (c^2)] + b[(a^2) – 2ac + (c^2)] + c[(a^2) – 2ab + (b^2)] ≥ 0.  Factoring the expressions in each bracket in B6 gives B7: a(b – c)^2 + b(a – c)^2 + c(a – b)^2 ≥ 0.  Now, since a,b, and c are positive real numbers (via the hypothesis), we know that the inequality in B7 is non-negative. Now, turning to the forward process, we can reach our last result in the backward process (B7) by making use of some properties of real numbers and the hypothesis that a,b, and c are positive real numbers (Note, I will label each statement in the forward process with an A and a respective number).  To that end, since a,b, and c are real numbers (positive real numbers in this case), then A1: b – c, a – c, and a – b are real numbers.  So, via the properties of real numbers, it follows that A2: (b – c)^2, (a – c)^2, and (a – b)^2 are non-negative real numbers. Since a,b, and c are positive real numbers, then A3: a(b – c)^2, b(a – c)^2, and c(a – b)^2 are non-negative real numbers  Thus, it follows that A4: a(b – c)^2 + b(a – c)^2 + c(a – b)^2 ≥ 0.  Hence, since we’ve reached our B7 statement, we have a proof, and writing the proof will consist of retracing our backward steps in a forward direction.  Please note that, since we’ve included a lot of the fine details in our analysis, we can write a more condensed version of the proof. Proof (condensed version) Suppose a,b, and c are positive real numbers.  Then b – c, a – c, and a – b are also real numbers, and by the properties of real numbers, (b – c)^2, (a – c)^2, and (a – b)^2 are non-negative real numbers.  Consequently, it follows that a(b – c)^2 + b(a – c)^2 + c(a – b)^2 ≥ 0 (1).  Now, expanding the left side of inequality (1) and using algebra gives [a(b^2) + a(c^2)] + [(a^2)b + b(c^2)] + [(a^2)c + (b^2)c] – 6abc ≥ 0 (2).  Adding 9abc to both sides of inequality (2) and rearranging terms on the left side of inequality (2) gives (a^2)c + (a^2)b + (b^2)c + a(b^2) + b(c^2) + a(c^2) + 3abc = [abc + (a^2)c + (a^2)b] + [(b^2)c + abc + a(b^2)] +  [b(c^2) + a(c^2) + abc] ≥ 9abc (3).  Factoring the left side of inequality (3) yields (a+b+c)(bc+ac+ab) ≥ 9abc (4).  Finally, dividing the right side of inequality (4) by abc and dividing the expression (bc+ac+ab) on the left side of inequality (4) by abc gives (a+b+c)(1/a + 1/b + 1/c) ≥ 9, which is the desired result.  Hence, the proof is complete.  P.S. I've included an analysis of proof in order to illustrate the thought processes that are involved in constructing a proof (please note that I had more scratch work for each of the statements listed in the analysis; thus, these steps were the result of me taking the time to work things out on paper).  For me at least, I gain an understanding of a proof via the analysis of proof and not by the condensed version.  Well, thanks for your time and attention,  and I hope that more people will post cool math stuff in this thread. Also, there was no solution for this exercise in the book, so the work that I have posted here is entirely my own. polymath257 Member   Religious Views: Atheist Posts: 1255 Threads: 4 Joined: January 3, 2018 Reputation: 19 RE: The Mathematical Proof Thread January 3, 2018 at 4:31 pm (This post was last modified: January 3, 2018 at 4:32 pm by polymath257. Edit Reason: Added comment ) (September 21, 2017 at 4:07 pm)LastPoet Wrote: When you analyse an algebraic division Ring, a set E with with 2 operations called '+' and '*' with the axioms of the algebraic goup (E.+) previously proven with commutivity a+b = b+a, the exitence of one element we called 0 where for all x in E: x+0 = x, with the set not being empty or singular; Given the properties of the operation '*': 1: there exists a member we call u where for all set members a, a*u = a 2: for all members of E, a*b = b*a 3: for all a,b,c in E, a*(b+c) = a*b+a*c With these in mind prove that a*0=0 and while you're at it prove that 0 is not equal this 'u' mentioned on the axioms. We are assuming that (E,+) is a group, so if x is in E, there is a y in E with x+y=0. Lemma 1: Suppose that x+x =x. Then x=0 Pf: Let y be such that x+y =0. Then 0=x+y =(x+x )+y =x+(x+y)=x+0 =x. Now, for any a, a*0=a*(0+0)=a*0 + a*0. Applying the lemma shows a*0=0. Next, if u=0, then for any a, a=a*u=a*0=0. Hence E is a singleton (only 0). Note: we do not need to know a*b=b*a for this result. polymath257 Member   Religious Views: Atheist Posts: 1255 Threads: 4 Joined: January 3, 2018 Reputation: 19 RE: The Mathematical Proof Thread January 9, 2018 at 7:14 pm (December 17, 2017 at 4:39 pm)Kernel Sohcahtoa Wrote: Earlier this month, I did a proof that was pretty cool (IMO).  As a result, I thought I'd share it here.  As always, I'll put my post in hide tags.  In addition, I'd like to clarify  that my writing is not meant to be arrogant: my mathematical writing style is how I've learned to communicate mathematics (I definitely have lots of room for improvement). With that said, here is the following exercise from my mathematical proofs book by Gary Chartrand: Prove that if a,b, and c are positive real numbers, then (a+b+c)(1/a + 1/b + 1/c) ≥ 9 Hint/Analysis of Proof Let’s work backward from the conclusion and see where we end up: I will denote each backward statement with a B and a respective number.  First, let’s notice that the expression (1/a + 1/b +1/c) can be re-written as (bc+ac+ab)/abc.  Thus, the conclusion can be re-written as B1: (a+b+c)(bc+ac+ab)/abc ≥ 9.  Now, multiplying both sides of B1 by abc gives B2: (a+b+c)(bc+ac+ab) ≥ 9abc.  Expanding the left side of B2 via algebra gives B3: [abc + (a^2)c + (a^2)b] + [(b^2)c + abc + a(b^2)] +  [b(c^2) + a(c^2) + abc]= 3abc + (a^2)c + (a^2)b + (b^2)c + a(b^2) + b(c^2) + a(c^2) ≥ 9abc.  Subtracting both sides of B3 by 9abc and rearranging terms on the left side gives B4:[a(b^2) + a(c^2)] + [(a^2)b + b(c^2)] + [(a^2)c + (b^2)c] – 6abc ≥ 0.   Notice that –6abc= –2abc + (–2abc) + (–2abc), so we will put each –2abc term into a bracket in B4 to get B5: [a(b^2) – 2abc + a(c^2)] + [(a^2)b – 2abc + b(c^2)] + [(a^2)c – 2abc + (b^2)c] ≥ 0. Factoring out the common variable in each bracket in B5 gives B6: a[(b^2) – 2bc + (c^2)] + b[(a^2) – 2ac + (c^2)] + c[(a^2) – 2ab + (b^2)] ≥ 0.  Factoring the expressions in each bracket in B6 gives B7: a(b – c)^2 + b(a – c)^2 + c(a – b)^2 ≥ 0.  Now, since a,b, and c are positive real numbers (via the hypothesis), we know that the inequality in B7 is non-negative. Now, turning to the forward process, we can reach our last result in the backward process (B7) by making use of some properties of real numbers and the hypothesis that a,b, and c are positive real numbers (Note, I will label each statement in the forward process with an A and a respective number).  To that end, since a,b, and c are real numbers (positive real numbers in this case), then A1: b – c, a – c, and a – b are real numbers.  So, via the properties of real numbers, it follows that A2: (b – c)^2, (a – c)^2, and (a – b)^2 are non-negative real numbers. Since a,b, and c are positive real numbers, then A3: a(b – c)^2, b(a – c)^2, and c(a – b)^2 are non-negative real numbers  Thus, it follows that A4: a(b – c)^2 + b(a – c)^2 + c(a – b)^2 ≥ 0.  Hence, since we’ve reached our B7 statement, we have a proof, and writing the proof will consist of retracing our backward steps in a forward direction.  Please note that, since we’ve included a lot of the fine details in our analysis, we can write a more condensed version of the proof. Proof (condensed version) Suppose a,b, and c are positive real numbers.  Then b – c, a – c, and a – b are also real numbers, and by the properties of real numbers, (b – c)^2, (a – c)^2, and (a – b)^2 are non-negative real numbers.  Consequently, it follows that a(b – c)^2 + b(a – c)^2 + c(a – b)^2 ≥ 0 (1).  Now, expanding the left side of inequality (1) and using algebra gives [a(b^2) + a(c^2)] + [(a^2)b + b(c^2)] + [(a^2)c + (b^2)c] – 6abc ≥ 0 (2).  Adding 9abc to both sides of inequality (2) and rearranging terms on the left side of inequality (2) gives (a^2)c + (a^2)b + (b^2)c + a(b^2) + b(c^2) + a(c^2) + 3abc = [abc + (a^2)c + (a^2)b] + [(b^2)c + abc + a(b^2)] +  [b(c^2) + a(c^2) + abc] ≥ 9abc (3).  Factoring the left side of inequality (3) yields (a+b+c)(bc+ac+ab) ≥ 9abc (4).  Finally, dividing the right side of inequality (4) by abc and dividing the expression (bc+ac+ab) on the left side of inequality (4) by abc gives (a+b+c)(1/a + 1/b + 1/c) ≥ 9, which is the desired result.  Hence, the proof is complete.  P.S. I've included an analysis of proof in order to illustrate the thought processes that are involved in constructing a proof (please note that I had more scratch work for each of the statements listed in the analysis; thus, these steps were the result of me taking the time to work things out on paper).  For me at least, I gain an understanding of a proof via the analysis of proof and not by the condensed version.  Well, thanks for your time and attention,  and I hope that more people will post cool math stuff in this thread. Also, there was no solution for this exercise in the book, so the work that I have posted here is entirely my own. OK, good. Now, suppose that x_1 ,...x_n >0. Show that (1/x_1 + 1/x_2 +...+1/x_n )(x_1 +x_2 +...x_n )>=n^2. (you did the n=3 case). Kernel Sohcahtoa The Outsider   Religious Views: non-religious and non-theist Posts: 1091 Threads: 25 Joined: September 5, 2016 Reputation: 39 RE: The Mathematical Proof Thread July 3, 2018 at 3:06 am Thanks for your last post, Polymath.  I must admit that I kept getting stuck during each of my attempts and decided to let it go for the time being.  However, this evening, I noticed that your post also appeared as an exercise in one of my math books, so I decided to give it another try.  That said, for anyone interested, I will post the result to be proved and then post the proof in hide tags.  Also, in my proof, I make use of another fact/result, so I will post that fact and my proof of it (in hide tags) above my main proof. That said, should anyone read the proofs below, please feel free to point out any errors/typos that I have made; I'm very appreciative of any feedback that you can offer.  Thanks Result: for every two positive real numbers a and b, it follows that a/b +b/a ≥ 2.  Proof Suppose a and b are positive real numbers.  Then (a–b)^2 ≥ 0 which is equivalent to a^2 –2ab +b^2 ≥ 0 which is equivalent to a^2 + b^2 ≥ 2ab.  Dividing both sides of the foregoing inequality by ab (recall that a and b are both positive reals, so ab is positive) gives (a^2 + b^2)/ab ≥ 2 which is equivalent to a^2/ab + b^2/ab ≥ 2 which is equivalent to a/b + b/a ≥ 2.  The proof is complete.    Main Result to Prove: for every n ≥ 1 positive real numbers a_1,a_2,…,a_n, it follows  that (a_1 +…+ a_n)*(1/a_1 +…+ 1/a_n) ≥ n^2 Proof The proof will be by induction.  For n=1, we have a_1*(1/a_1) = 1  ≥ 1^2=1, which is true.  Now, assume that (a_1 +…+ a_k)*(1/a_1 +…+ 1/a_k) ≥ k^2 where k ≥ 1.  We must establish that (a_1 +…+ a_k+1)*(1/a_1 +…+ 1/a_k+1) ≥ (k+1)^2.  To that end, observe that (a_1 +…+ a_k+1)*(1/a_1 +…+ 1/a_k+1) = [(a_1 +…+ a_k) + a_k+1]*[(1/a_1 +…+ 1/a_k) + 1/a_k+1] = (a_1 +…+ a_k)*(1/a_1 +…+ 1/a_k) + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + (a_k+1)*(1/a_k+1)= (a_1 +…+ a_k)*(1/a_1 +…+ 1/a_k) + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + 1 ≥ k^2 + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + 1 . Thus, so far, we have obtained the inequality (a_1 +…+ a_k+1)*(1/a_1 +…+ 1/a_k+1) ≥ k^2 + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + 1 (1) (note that the result in inequality (1) follows from the induction hypothesis). Now, let’s focus on the expressions [(a_1 +…+ a_k)*1/a_k+1] and [(1/a_1 +…+ 1/a_k)*a_k+1] in inequality (1).  We can rewrite each expression as follows: rewrite [(a_1 +…+ a_k)*1/a_k+1] as (a_1/a_k+1 +…+ a_k/a_k+1) (2)  and [(1/a_1 +…+ 1/a_k)*a_k+1 as (a_k+1/a_1 +…+ a_k+1/a_k) (3).  Now adding the first term in  expression (2) with the first term in expression (3) followed by adding the second term in expression (2) with the second term in expression (3) and continuing this addition up to the kth terms in expression (2) and expression (3) gives [(a_1/a_k+1) + (a_k+1/a_1)] + [(a_2/a_k+1) + (a_k+1/a_2)] + …+ [(a_k/a_k+1) + (a_k+1/a_k)] (4). Now, for every two positive real numbers a and b, it follows that a/b + b/a ≥ 2.  Now, using the first sum in expression (4) as an example, since a_1 and a_k+1 are two positive real numbers, then it follows that (a_1/a_k+1) + (a_k+1/a_1) ≥ 2; thus, for any two positive real numbers a_i and a_k+1, where  1≤ i  ≤ k, it follows that (a_i/a_k+1) + (a_k+1/a_i) ≥ 2.  Consequently, adding each (a_i/a_k+1) with each corresponding (a_k+1/a_i), which is how expression (4) is set up, yields k sums, and so , it follows that [(a_1/a_k+1) + (a_k+1/a_1)] +…+ [(a_k/a_k+1) + (a_k+1/a_k)] ≥ 2k. So, we have established that [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] = [(a_1/a_k+1) + (a_k+1/a_1)] +…+ [(a_k/a_k+1) + (a_k+1/a_k)] ≥ 2k (5). Now, via result (5), it follows from inequality (1) that (a_1 +…+ a_k+1)*(1/a_1 +…+ 1/a_k+1) ≥ k^2 + [(a_1 +…+ a_k)*1/a_k+1] + [(1/a_1 +…+ 1/a_k)*a_k+1] + 1 ≥ k^2 + 2k +1= (k+1)^2. Hence, by the principle of mathematical induction, for every n ≥ 1 positive real numbers a_1, a_2,…, a_n, it follows that (a_1 +…+ a_n)*(1/a_1 +…+ 1/a_n) ≥ n^2.  The proof is complete. Kernel Sohcahtoa The Outsider   Religious Views: non-religious and non-theist Posts: 1091 Threads: 25 Joined: September 5, 2016 Reputation: 39 RE: The Mathematical Proof Thread July 3, 2018 at 12:47 pm My apologies for this post.  I'm posting a slight clarification in hide tags. In the beginning of my proof, I make the following assumption: "now, assume that (a_1 +…+ a_k)*(1/a_1 +…+ 1/a_k) ≥ k^2 where k ≥ 1." It would have been better if I had also remarked that k is an integer, rather than leaving it to the reader to infer this (this actually bothered me when I posted this, but I decided to let it go). Kernel Sohcahtoa The Outsider   Religious Views: non-religious and non-theist Posts: 1091 Threads: 25 Joined: September 5, 2016 Reputation: 39 RE: The Mathematical Proof Thread July 6, 2018 at 6:40 pm Prove Bernoulli’s Identity: For every real number x > -1 and every positive integer n, (1 + x)^n ≥  1 + nx Proof Let y > -1 be an arbitrary real number.  We will demonstrate that for every positive integer n, (1 + y)^n ≥ 1 + ny.  We will establish this by induction.  To that end, let n=1.  Then (1 + y)^1 = 1 + y ≥ 1 + (1)y = 1 + y, which is true.  Assume that (1 + y)^k ≥ 1 +ky for some positive integer k.  We must show that (1 + y)^(k+1) ≥ 1 + (k +1)y.  Now, (1 + y)^(k+1) = [(1 + y)^k]*(1 + y) ≥ (1 + ky)*(1 + y) (1) = 1 + y +ky + ky^2 ≥ 1 + y + ky (2) = 1 + (k + 1)y (please note that we made use of the induction hypothesis to reach result (1), and we made use of the fact that k is a positive integer and y^2 is non-negative to obtain result (2)).  So, by the principle of mathematical induction, it follows that for every positive integer n, (1 + y)^n ≥ 1 + ny.  Hence, since y > -1 is an arbitrary real number, then it follows that for every real number x > -1 and every positive integer n, (1 + x)^n ≥ 1 + nx.  The proof is complete. Succubus Huge Member   Religious Views: Unsaved Posts: 2340 Threads: 21 Joined: May 5, 2017 Reputation: 26 RE: The Mathematical Proof Thread July 6, 2018 at 7:34 pm (July 6, 2018 at 6:40 pm)Kernel Sohcahtoa Wrote: Prove Bernoulli’s Identity: For every real number x > -1 and every positive integer n, (1 + x)^n ≥  1 + nx Proof Let y > -1 be an arbitrary real number.  We will demonstrate that for every positive integer n, (1 + y)^n ≥ 1 + ny.  We will establish this by induction.  To that end, let n=1.  Then (1 + y)^1 = 1 + y ≥ 1 + (1)y = 1 + y, which is true.  Assume that (1 + y)^k ≥ 1 +ky for some positive integer k.  We must show that (1 + y)^(k+1) ≥ 1 + (k +1)y.  Now, (1 + y)^(k+1) = [(1 + y)^k]*(1 + y) ≥ (1 + ky)*(1 + y) (1) = 1 + y +ky + ky^2 ≥ 1 + y + ky (2) = 1 + (k + 1)y (please note that we made use of the induction hypothesis to reach result (1), and we made use of the fact that k is a positive integer and y^2 is non-negative to obtain result (2)).  So, by the principle of mathematical induction, it follows that for every positive integer n, (1 + y)^n ≥ 1 + ny.  Hence, since y > -1 is an arbitrary real number, then it follows that for every real number x > -1 and every positive integer n, (1 + x)^n ≥ 1 + nx.  The proof is complete. Looks sound enough to me. It's amazing 'science' always seems to 'find' whatever it is funded for, and never the oppsite. Drich. Fireball Rapscallion    Religious Views: Atheist Posts: 5184 Threads: 16 Joined: September 8, 2015 Reputation: 76 RE: The Mathematical Proof Thread July 6, 2018 at 8:37 pm I was taking 2nd semester calculus (spring of '78) and the prof called on students to go to the board and solve a problem from the homework set. I had done all the homework, and had to do the proof that a particular function was continuously differentiable. Having just finished a refresher on college algebra the semester previous, I whipped that proof right out. I was pleased with myself, for sure. You should be, too, for your proof! Rewarding, isn't it? I never thought I'd live long enough to become a grumpy old bastard. Here I am, killing it! « Next Oldest | Next Newest »

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