Quote:
damn, my math skills is getting rusty....
so are your english skills...
well anyways your way off... it has nothing to do with ellipses.. nobody else seems to be trying it though.. oh wells
i guess i get to keep my bits thenComputer Science Canada another quick math question for 100 |
| Author: | SilverSprite [ Fri Jun 27, 2003 2:47 pm ] |
| Post subject: | another quick math question for 100 |
This one isnt quite as hard as darknesses question but since i wont put my bits to any good use.. here we go.. prove that 6x^2 + 2y^2 = z^2 has no natural solutions |
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| Author: | AsianSensation [ Fri Jun 27, 2003 3:39 pm ] |
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I don't know if Im right or not(probably not), but here is mine solution: 6x^2 + 2y^2 = z^2 assuming the statement is true, i'll try to prove it by using contradiction: first take out a common factor of 2 from the left side: 2(3x^2 + y^2) = z^2 if the above statement is true, with x,y,z belonging to Integers, then: 3x^2 + y^2 must be an odd power of 2. let n be a number belonging to the set of positive integers 3x^2 + y^2 = 2^(2n+1) therefore, 0 = 2^(2n+1) - 3x^2 - y^2 0 = 3x^2 + y^2 - 2^(2n+1) 0 = x^2 + y^2 + 2x^2 - 2^(2n+1) 0 = x^2 + y^2 + 2(x^2 - 2^(2n)) 0 = x^2 + y^2 + 2((x - 2^n)(x + 2^n)) now since x^2, y^2, and x + 2^n are all positive, therefore, for the equation to be true, x - 2^n must be negative. therefore, x - 2^n < 0 x < 2^ n sub that statement back into 3x^2 + y^2 = 2^(2n+1), and we get: 3(2^(2n)) + y^2 < 2^(2n+1) but this statement is obviously false, because 3(2^(2n)) > 2^(2n+1) therefore, contradiction arises, and there does not exist integer solutions to 6x^2 + 2y^2 = z^2 |
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| Author: | SilverSprite [ Fri Jun 27, 2003 4:02 pm ] | ||
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when you sub in x<2^n .. wouldnt you get 3(2^(2n)) + y^2 > 2^(2n+1) ? or am i wrong? that makes your proof false.. unless someone can tell me i'm wrong.. i hate working with these inequalities:S |
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| Author: | SilverSprite [ Fri Jun 27, 2003 4:03 pm ] |
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try again.. if you like.. anybody else? come on it isnt that hard.. |
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| Author: | AsianSensation [ Fri Jun 27, 2003 4:05 pm ] |
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am I even close, or am I like way off? (still working on it......) |
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| Author: | SilverSprite [ Fri Jun 27, 2003 4:08 pm ] |
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i actually don't remember the solution.. i just remember doing it for night math last year.. |
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| Author: | SilverSprite [ Fri Jun 27, 2003 4:11 pm ] |
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I have realised something else.. it is false for you to assume that 3x^2 + y^2 = 2^(2n+1) |
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| Author: | SilverSprite [ Fri Jun 27, 2003 4:17 pm ] |
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yeah i just got it.. you were on the right track at the second line.. after that everything went bad |
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| Author: | AsianSensation [ Fri Jun 27, 2003 4:19 pm ] |
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lol, i just realized that too, so instead of assuming 2^something, i'll assume it's a perfect square multiplied by 2 |
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| Author: | SilverSprite [ Fri Jun 27, 2003 4:24 pm ] |
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yes that would be fair.. |
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| Author: | SilverSprite [ Fri Jun 27, 2003 10:34 pm ] |
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nobody?? come on whoa fine the new prize is all my bits.. |
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| Author: | Amailer [ Sat Jun 28, 2003 12:35 am ] |
| Post subject: | IS THIS IT?: |
There is no solution because they are all different variables |
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| Author: | SilverSprite [ Sat Jun 28, 2003 7:19 am ] |
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haha..what? |
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| Author: | Amailer [ Sat Jun 28, 2003 7:36 am ] |
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Oh nothing....i just did some crap....duh im not in grd 12 but hehe i need bits! lol |
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| Author: | AsianSensation [ Sat Jun 28, 2003 10:39 am ] |
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Does this problem involves ellipse? x^2 / (2(n^2)/ 3) + y^2 / (2(n^2)) = 1 cause when you graphed it, you get a ellipse, but how do you prove there isn't integer solutions to a ellipse? Or maybe I am on a completely different approach then what is required? (probably the latter...) |
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| Author: | AsianSensation [ Sat Jun 28, 2003 11:14 am ] |
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that's it, I give up, I can't find a proof to this...... it seems like a simple problem, but I just can't seem to do it.....damn, my math skills is getting rusty.... |
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| Author: | SilverSprite [ Sat Jun 28, 2003 11:30 am ] |
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Quote: damn, my math skills is getting rusty....
so are your english skills... well anyways your way off... it has nothing to do with ellipses.. nobody else seems to be trying it though.. oh wells i guess i get to keep my bits then |
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| Author: | AsianSensation [ Sat Jun 28, 2003 11:31 am ] |
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post the solution, or rather, if you don't want to post it, PM me, I want to see what I did wrong, or what I didn;t do. |
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| Author: | Crono [ Sat Jun 28, 2003 7:35 pm ] |
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aights, since this is a comp sci forum, i guess u guyz could try n use turin 2 solve tis problem, it'll b real easy if u kno the approach, but i guess the math way is "cool" 2...haha |
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| Author: | bugzpodder [ Thu Jul 03, 2003 11:14 am ] |
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my turn: 6x^2 + 2y^2 = z^2 assuming (x,y,z) is a primitive solution (ie gcd(x,y,z)=1). otherwise you can take out a common factor gcd(x,y,z)^2 and so forth 6x^2=z^2-2y^2 in mod 6 0=z^2-2y^2 (mod 6) so z^2=2y^2 (mod 6) if y=0 (mod 6), z^2=0 (mod 6) y=1(mod 6), z^2=2 (mod 6) y=2 .., z^2=2 (mod 6) y=3..., z^2=0 (mod 6) y=4..., z^2=2 (mod 6) y=5..., z^2=2 (mod 6) you can verify that there is no solution to z^2=2 (mod 6) and of course, z^2=0 (mod 6) implies that z=6z' for some integer z' and y=3y' 6x^2=36(z')^2+18(y')^2 x^2=6(z')^2+3(y')^2 and it follows that x=3x' but gcd(x,y,z)=gcd(3x',3y',6z')>=3 ... we assumed it to be 1. contradiction |
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| Author: | Andy [ Thu Jul 03, 2003 5:12 pm ] |
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wow go bugz.. |
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| Author: | SilverSprite [ Thu Jul 03, 2003 5:25 pm ] |
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dodge_tomahawk wrote: wow go bugz..
haha that means you dont understand a word he said lol |
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| Author: | Andy [ Thu Jul 03, 2003 5:28 pm ] |
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shut up shut up... that means he's too good that i dun understand what he said |
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| Author: | SilverSprite [ Thu Jul 03, 2003 5:34 pm ] |
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dodge_tomahawk wrote: shut up shut up... that means he's too good that i dun understand what he said
nono it just means that you dont know what he said nothing more.. i'll be quoting bugz bugzpodder wrote: ..its common sense.. |
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| Author: | bugzpodder [ Thu Jul 03, 2003 7:43 pm ] |
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so do i get the 100 bits? ^.^' or is that just like 100 in jeopardy? |
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| Author: | eggplant_burger [ Thu Jul 03, 2003 9:06 pm ] |
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instead of using my math skillz (which i don't really have, my only 12 course is calculus, so far) to solve this rpoblem, I have decided to use my programming skillz. Here it is: var doneyet : boolean := false for x : 1 .. maxnat for y : 1 .. maxnat for z : 1 .. maxnat if 6 * (x ** 2) + 2 * (y ** 2) = z ** 2 then doneyet := true put "Eureka!!", skip, "I have found it", skip, skip, "when x=", x, " y=", y, " z=", z, " then it is true", skip, skip, "MUWAHAHAHAHAHAHA" end if end for end for end for if doneyet = false then put "Wow, there really ISN'T a natural solution... Trippy" end if That'll prove to you that there is no natural solution. And it was fun to write. |
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| Author: | bugzpodder [ Thu Jul 03, 2003 10:16 pm ] |
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the limitations of maxnat is only a computerized limitation. there is no limiations to natural numbers, and therefore, unfortunately, it is not a valid proof |
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| Author: | eggplant_burger [ Thu Jul 03, 2003 10:45 pm ] |
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that's a piddling argument. it can be proven that it is a valid proof. since there are 2 variables to a degree greater than 1 on the left side, and only 1 variable to a degree greater than 1 on the right side, the left will increase more quickly than the right. So if they are already not equal at a point where the values of the left and right side fall into a pattern, they will never be equal. I think that the maximum possible natural number in Turing (4294967294) is reasonably high enough to assume that a pattern will be established. It's like using limits... you don't quite show it exactly, but it's so close it doesn't make a difference. |
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| Author: | PaddyLong [ Fri Jul 04, 2003 12:11 am ] |
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no... it's not a propper proof... it's only showing it for n number of cases rather than for all possible |
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| Author: | SilverSprite [ Fri Jul 04, 2003 1:07 am ] |
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hmm i think he was just kidding? |
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| Author: | PaddyLong [ Fri Jul 04, 2003 10:53 am ] |
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SilverSprite wrote: hmm i think he was just kidding?
doesn't seem like it :/ |
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| Author: | bugzpodder [ Fri Jul 04, 2003 11:15 am ] |
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eggplant_burger wrote: that's a piddling argument. it can be proven that it is a valid proof.
since there are 2 variables to a degree greater than 1 on the left side, and only 1 variable to a degree greater than 1 on the right side, the left will increase more quickly than the right. So if they are already not equal at a point where the values of the left and right side fall into a pattern, they will never be equal. I think that the maximum possible natural number in Turing (4294967294) is reasonably high enough to assume that a pattern will be established. It's like using limits... you don't quite show it exactly, but it's so close it doesn't make a difference. you may pull it off if its only one variable (after you show exactly when RHS>LHS for all x>x0, and test all solution less than or equal to x0), but with these multiple variables, no. |
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| Author: | eggplant_burger [ Thu Jul 10, 2003 10:41 pm ] |
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Fine, then I'm wrong. Whatever. |
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| Author: | Corybu [ Fri Oct 03, 2003 1:12 am ] |
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Im no math genious, but is the answer 7? Lol, j/k guys. This is crazy, I dont know how any of you could rack your brains to try and figure this out... When will you ever use it? (outside of forum debates) My math skills are pretty basic... maybe slitly above average, but if any of you understand what is in this thread, you scare me, lol. |
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