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You are not logged in. #1 20061005 15:56:17
Interesting proofsPost any proof that you find interesting here. It can be from any subject. Below is one of my favorites: "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #2 20061014 08:56:24
Re: Interesting proofsI'm not saying that his proof was correct or incorrect, but.... You can shear a sheep many times but skin him only once. #3 20061017 21:56:40
Re: Interesting proofs
Yes, I realised that after I put that second post up. I was just having a bit of a nobrainer moment Where each a_{ij} is one of the integers {0,1,...,9} representing the j^{th} decimal place of the i^{th} number in the list. Now, this list is supposed to include all the real numbers between 0 and 1. But suppose we construct a number x with the following algorithm: 1) let the number be: 2) go down our previous list and look at the n^{th} digit of the n^{th} number. 3) if the n^{th} digit of the n^{th} number is 0, let the n^{th} digit of x be 1. 4) if the n^{th} digit of the n^{th} number is any number other than 0, let the n^{th} digit of x be 0. We then have our number x well defined. It is, however, not equal to the first number in our list, by the way we have constructed it (because at least one of the digits is different). It is also not equal to the second, or the third... or in fact any other number on our list. Even if we put x at the start of our list, we can still construct another number using the same algorithm that would not be on the list, so our list will never be exhaustive. Bad speling makes me [sic] #4 20061018 19:09:58
Re: Interesting proofsTheorem: Last edited by Zhylliolom (20061018 19:11:23) #5 20070126 06:03:11
Re: Interesting proofsThis proof is by Zhylliolom: or For all natural numbers m and n. Proof: It is obvious that for any given natural numbers m and n, either or is true, since √2 is irrational and therefore cannot be equal to a rational number, and thus the result follows from trichotomy. Now consider the inequality where the ? denotes the unknown direction of the inequality. Since m and n are natural numbers, their sum is nonzero and positive, and thus we may multiply each side my m + n without reversing the inequality: Once again, n is a natural number and thus is nonzero and positive, so we may divide each side by n and maintain the direction of the inequality: Now subtract m/n and √2 from each side to obtain which gives (after dividing through by 1  √2, flipping the inequality, rationalizing, then flipping again so that we now have the original direction (these steps are left to the reader as an exercise)) which gives the direction of the inequality as the opposite of the "initial condition" m/n ¿ √2, where ¿ denotes the original (and opposite) inequality direction, so that if m/n < √2 then (m + 2n)/(m + n) > √2 or if m/n > √2 then (m + 2n)/(m + n) < √2. Thus, either or is true for natural numbers m and n. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #6 20070126 06:13:06
Re: Interesting proofsProve that card(A) < card(P(A)), where card(A) is the cardinality of A, and P(A) is the power set of A. Note that it must be that: And so T is in the range of f. Thus, there exists some t such that f(t) = T. First, assume that t is in T. Then it must be that: And so t is not in f(t) = T, so t is not in T. Contradiciton. Now assume that t is not in T. Then it must be that: So it must be that t is in f(t) = T, and so t is in T. Contradiction. Therefore, there must not exist such a function, and so card(A) < card(P(A)). "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #7 20070204 20:48:33
Re: Interesting proofsI had read somewhere that 26 is the only number in the universe jammed between a square and a cube. Character is who you are when no one is looking. #8 20070204 21:10:34
Re: Interesting proofsI'm willing to bet that with the increase in squares and cubes as you move up the number line, there will be more numbers that share 26's properties, after all, we have the whole number line to explore #9 20070204 21:31:14
Re: Interesting proofsThat couldn't have been said without a proof! I am sure there exists one Character is who you are when no one is looking. #10 20070204 23:01:20
Re: Interesting proofs65536³  1 = 16777216² + 1 Why did the vector cross the road? It wanted to be normal. #11 20070205 04:23:35
Re: Interesting proofsmathsyperson, I'm not sure what you're trying to do. You want to find two numbers x and x+2 such that one is a square and one is a cube. And yes, 26 is the only such number. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #12 20070205 04:54:00
Re: Interesting proofsOh, sorry. I tried to bruteforce it in Excel and it gave me that 65536³ and 16777216² were two apart. Something must have gone wrong with rounding errors. Apologies. Why did the vector cross the road? It wanted to be normal. #13 20070801 05:10:43
Re: Interesting proofsThis is something like the Mihailescu's theorem: http://en.wikipedia.org/wiki/Mih%C4%83i … 7s_theorem IPBLE: Increasing Performance By Lowering Expectations. #14 20070802 05:50:46
Re: Interesting proofs
Yes, your proof is good, I think, but Cantor's own proof was a lot sweeter, if you'll forgive me for saying so. Plus it requires (almost) no knowledge of mathematics! #15 20070815 10:55:28
Re: Interesting proofson the subject of interesting proofs, i was wondering.. Code:int max = f(a); for (int i = a +1; a <= b; i++) { if (f(i) > max ) max = f(i); } print(max); if the program outputs M, could that be considered a valid proof that M is the max? Obviously we depend on the computer to make correct comparisons and to calculate f(n) exactly, but if we are willing to assume it does, could that be considered proof? Last edited by mikau (20070815 11:01:51) A logarithm is just a misspelled algorithm. #16 20070815 11:05:17
Re: Interesting proofsYes, any problem with a finite number of computational cases can be proved using a computer. For example, The Four Color Theorem. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #17 20070815 11:09:00
Re: Interesting proofsthat's COOL! still..they claim the proof is not satisfactory becuase it requires a computer. Hmm... just the oppinion of the writer, perhaps? A logarithm is just a misspelled algorithm. #18 20070815 11:50:51
Re: Interesting proofsThe proof is most likely entirely valid, but it relies upon something we don't want to check, as it would be way too tedious. That points to not something wrong with the proof, but with us instead (i.e. not being a computer, which would be way cool). But many times in math, there are better proofs. For example: "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #19 20071126 06:59:49
Re: Interesting proofsI'm really happy I managed to figure out this proof last night, turned out it was exactly what my book did and I thought it was kind of cool. suppose are linearly dependent. now suppose we one by one remove every vector in that can be expressed as a linear combination of the others. When we are done we will have a linearly independent subset of consisting of m elements, call it we have the set is not empty for the vectors are all non zero so the linearly independent subset contains at least one element. Now take an element not in the set we have that but also we have that can be expressed as a linear combination of so where are the scalers of the linear combination. But also by the distributive property of matrix multiplication: thus so since are linearly independent, we have that all scalers are zero. Recall now that we have all eigenvalues are unique, and is not an element of so thus thus, since they are linearly dependent, and by the zero factor theorem, all scalers must be zero. But this implies that and we were given all vectors were nonzero. Hence the contradiction. So it must be linearly independent. Last edited by mikau (20071126 12:33:12) A logarithm is just a misspelled algorithm. #20 20071126 07:12:00
Re: Interesting proofsto do with computer proof, people dont want to trust that the computer program was correct because it wasn't done by a human, but to be honest i would more likely trust a well written computer program than a team of humans running through thousands upon thousands of scenarios where they would likely become lazy and make mistakes The Beginning Of All Things To End. The End Of All Things To Come. #21 20071126 09:37:10
Re: Interesting proofsBut the question is, do you realize what it is you are trusting when you trust a computer? "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #22 20071126 09:56:05
Re: Interesting proofsThe human brain can be viewed as a very complex computer, and we have no idea how it works. Why did the vector cross the road? It wanted to be normal. #23 20071126 12:30:58
Re: Interesting proofsblah, just fixed the latex in my proof. Have a look! A logarithm is just a misspelled algorithm. #24 20071207 06:55:58
Re: Interesting proofsFirst off, hiya . #25 20110109 07:23:25
Re: Interesting proofsganesh, 