Jay-qu Posted March 18, 2005 Report Share Posted March 18, 2005 x = 2x(x-1) = 2(x-1)x2-x = 2x-2x2-2x = x-2 x(x-2) = x-2 x = 1 it looks right but there is a mistake in there -> can you see it? Quote Link to comment Share on other sites More sharing options...
Tormod Posted March 18, 2005 Report Share Posted March 18, 2005 Well, this is apparently not a progression but just a lot of equations. So there is nothing wrong. The last line is just as isolated as the rest. Edit: I misread the equations, sorry. Still, the final line does not make sense at all. I don't see how you derive x=1 from this. Quote Link to comment Share on other sites More sharing options...
Tormod Posted March 18, 2005 Report Share Posted March 18, 2005 x2-x = 2x-2x2-2x = x-2 This should bex2-x = 2x-2x2 = 3x-2x=(3x-2)/2x=(6-2)/2x=2 And the last line is simply wrong...nowhere does x equate 1. Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted March 18, 2005 Author Report Share Posted March 18, 2005 i see where your coming from but if you sub in x = 1 then the equations also work except the one you pointed out... Quote Link to comment Share on other sites More sharing options...
Tormod Posted March 18, 2005 Report Share Posted March 18, 2005 Maybe I'm not getting the point. I thought you somehow resolved x=2 to become x=1... ;) I am NOT a math wiz... Quote Link to comment Share on other sites More sharing options...
C1ay Posted March 18, 2005 Report Share Posted March 18, 2005 x = 2x(x-1) = 2(x-1)x2-x = 2x-2x2-2x = x-2 x(x-2) = x-2 x = 1 it looks right but there is a mistake in there -> can you see it? x(x-1) does not equal 2(x-1) or x2-xx(x-1) = x^2-x x2-2x does not equal x-2 either The whole thing is riddled with errors.... Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted March 18, 2005 Report Share Posted March 18, 2005 The real point is that: x = 2 and x(x-1) = 2(x-1) are only equivalent if x is not 1. This resolves the paradox. There's nothing wrong with the rest of the algebra. Quote Link to comment Share on other sites More sharing options...
C1ay Posted March 18, 2005 Report Share Posted March 18, 2005 The real point is that: x = 2 and x(x-1) = 2(x-1) are only equivalent if x is not 1. This resolves the paradox. There's nothing wrong with the rest of the algebra. Huh? Doesn't x(x-1) = 2(x-1) resolve to x^2 - x = 2x - 2? I don't see how these can be equivalent except for x = 1 and x = 2. Quote Link to comment Share on other sites More sharing options...
zadojla Posted March 18, 2005 Report Share Posted March 18, 2005 x = 2x(x-1) = 2(x-1)x2-x = 2x-2x2-2x = x-2 x(x-2) = x-2 x = (x-2)/(x-2) <===== implied stepx = 1 The implied step is the problem. Because x = 2, this is equivalent to dividing by 0. This is a common trick in this kind of algebraic "paradox". Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted March 18, 2005 Report Share Posted March 18, 2005 Doesn't x(x-1) = 2(x-1) resolve to x^2 - x = 2x - 2?Yes. The trouble is not in that step but in the previous one. Or the one Zadojla points out. It's a matter of which way you go, or IOW the equivalence fails because, if x = 1 then (x - 1) is 0, if x = 2 then (x - 2) is 0. Multiplying both sides by zero doesn't give an equivalent equation, dividing both sides by zero doesn't make sense. Quote Link to comment Share on other sites More sharing options...
C1ay Posted March 18, 2005 Report Share Posted March 18, 2005 Yes. The trouble is not in that step but in the previous one. Or the one Zadojla points out. It's a matter of which way you go, or IOW the equivalence fails because, if x = 1 then (x - 1) is 0, if x = 2 then (x - 2) is 0. Multiplying both sides by zero doesn't give an equivalent equation, dividing both sides by zero doesn't make sense. I realize why the fallacy in the math. I was questioning your statement, "...are only equivalent if x is not 1." I've seen similar 1=2 problems. They most always rely on division by 0 somewhere in the stack. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted March 18, 2005 Report Share Posted March 18, 2005 I was questioning your statement, "...are only equivalent if x is not 1."I don't get your point, all the equations obtained after multiplication by (x - 1) are 0 = 0 in the x = 1 case. Hence they aren't fully equivalent to the equation x = 2. What, then, was your reason for questioning the statement? Quote Link to comment Share on other sites More sharing options...
C1ay Posted March 18, 2005 Report Share Posted March 18, 2005 I don't get your point... Nothing, it was just a nitpick on wording. IMO the statements x=2 and x(x-1) = 2(x-1) are only equivalent for x = 2. It didn't make since to me to say, "they are equivalent only if x does not equal 1" because they are not equivalent for other values of x either. Quote Link to comment Share on other sites More sharing options...
freQ Posted March 19, 2005 Report Share Posted March 19, 2005 x-"my butt" = ("my butt"/ pi) x ;) >_> o_Õ LMAO Quote Link to comment Share on other sites More sharing options...
C1ay Posted March 19, 2005 Report Share Posted March 19, 2005 x-"my butt" = ("my butt"/ pi) x ;) >_> o_Õ LMAO While all of that makes for an amusing equation I fail to see how it is relevant to the discussion. BTW, Welcome to the forum. Have you read the FAQ yet? Quote Link to comment Share on other sites More sharing options...
Aki Posted March 19, 2005 Report Share Posted March 19, 2005 This is very interesting. At first when I read this, I was like ??????, but now it makes sense Quote Link to comment Share on other sites More sharing options...
Bo Posted March 19, 2005 Report Share Posted March 19, 2005 interesting this all; especially the amount of discussion it stirs up ;) It is indeed impid that x = (x-2)(x-2). This equation means in simple terms: x=1, unless it is 2. since we required x to be 2, this equation does not imply that x is 1. Bo Quote Link to comment Share on other sites More sharing options...
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