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| It's Paradoxical! | |
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| Topic Started: Dec 3 2006, 05:00 AM (746 Views) | |
| Admin | Dec 3 2006, 05:00 AM Post #1 |
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What I will do for you now is find a real solution to the equation .Impossible, you might say. Well, watch. First, we square both sides: ![]() And then write it out: ![]() Cancel.... ![]() ![]() And we have actually solved for ! -------------- ------------- ------------ ----------- ---------- --------- -------- ------- ------ ----- ---- --- -- - Your job: find the one thing that prevents the answer I found from being a valid solution. (You'll sorta notice that -3/2 dosen't work if you put in in the original equation) |
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| Noah | Dec 3 2006, 05:34 PM Post #2 |
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Neophyte
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This is probably really obvious but i dont see how you ended up with +2X on one side and +4X on the other. |
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| appu_xavier | Dec 3 2006, 05:51 PM Post #3 |
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Poincaré Conjecture
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Andre, for some odd reason i don't think you're allowed to square both sides since it's an impossible equation. Unless, of course, you're using wrong (rong) numbers. |
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| Admin | Dec 3 2006, 07:06 PM Post #4 |
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You can still square the "equation" even if it's impossible. But your job is to tell me why -3/2 is not a valid answer. My steps all valid.....except one. What was it, and why did it remove any possibility that -3/2 was the solution? Also Noah, if you square (x + y), you get .
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| KRK | Dec 3 2006, 11:45 PM Post #5 |
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Poincaré Conjecture
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Implement -3/2 into the original problem, and you get -.5=.5 The original problem is impossible by glance so you can do anything to it you want. You are saying x is one more than x By the reflexive property, x=x, not x+1=x |
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| Admin | Dec 4 2006, 12:37 AM Post #6 |
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I know. The problem is "flawed" to begin with. Yet, by using completely valid steps, I can turn this wrong equation into a valid one. So, show me why I can turn this equation into a valid one simply by squaring it. |
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| KRK | Dec 4 2006, 01:13 AM Post #7 |
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Poincaré Conjecture
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Since, you are saying x+1=x+2, and you square that, you might as well multiply both by x+1-still the same as squaring, so you get xsquared+2x+1=xsquared+3x+2 That reduces to x=-1 Another "solution" When you find you can do that two different ways, the problem is automatically flawed |
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| Admin | Dec 4 2006, 02:06 AM Post #8 |
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Okay, I wasn't clear enough. You are supposed to show me WHY, when you square this flawed equation, that it becomes a correct equation. |
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| Edwardted | Dec 5 2006, 01:17 AM Post #9 |
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Twin Prime Conjecture
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you squared both sides. thats where the problem is. in the left half you multiply by x+1 but on the right you multiply by x+2... u have to multiply both sides by teh same amount |
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| Admin | Dec 5 2006, 01:27 AM Post #10 |
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Are you telling me it's illegal to square both sides of an eqaution? |
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| Edwardted | Dec 5 2006, 01:34 AM Post #11 |
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Twin Prime Conjecture
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yes unless its x=x, 2=2, etc etc etc etc etc etc etc |
| You cannot win without losing. | |
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| Edwardted | Dec 5 2006, 01:40 AM Post #12 |
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Twin Prime Conjecture
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Oh and also your squaring goes like this. (x+1)^2=(x+2)^2 which is this (x+1)(x+1)=(x+2)(x+2) which goes against some property |
| You cannot win without losing. | |
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| Admin | Dec 5 2006, 02:00 AM Post #13 |
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No, this dosen't go against any properties (Though the original equation does...). |
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| KRK | Dec 5 2006, 02:01 AM Post #14 |
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Poincaré Conjecture
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You thereby turn it INTO a legal one. |
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| Admin | Dec 5 2006, 02:10 AM Post #15 |
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Yes...you're getting close.... |
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5:18 PM Nov 27