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2 = 1?
Dec 24, 2006 21:13:11 GMT -8
Post by howardh on Dec 24, 2006 21:13:11 GMT -8
x^2 - x^2 = x^2 - x^2 (x-x)(x+x) = x(x-x) (x-x)(x+x)/(x-x) = x(x-x)/(x-x) x+x = x 2x = x 2x/x = x/x 2 = 1
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PhoenixFlare500
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2 = 1?
Dec 24, 2006 21:28:38 GMT -8
Post by PhoenixFlare500 on Dec 24, 2006 21:28:38 GMT -8
I saw this a long time ago. Don't remember the answer but I think it has something to do with one of the lines being wrong because of a mathematical rule that you overlook.
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2 = 1?
Dec 24, 2006 22:01:16 GMT -8
Post by ga on Dec 24, 2006 22:01:16 GMT -8
technically, that first line should result in 0=0
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2 = 1?
Dec 24, 2006 22:09:25 GMT -8
Post by dpgunit on Dec 24, 2006 22:09:25 GMT -8
this isn't really a riddle.
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2 = 1?
Dec 24, 2006 22:12:57 GMT -8
Post by howardh on Dec 24, 2006 22:12:57 GMT -8
its close enuf to a riddle
>>technically, that first line should result in 0=0 it is, you're supposed to find the mistake in it
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2 = 1?
Dec 24, 2006 22:17:24 GMT -8
Post by ga on Dec 24, 2006 22:17:24 GMT -8
the second line is asymmetrical. and it should be symmetrical every line that follows from that just correctly works on the mistake of the second line.
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dxlightning
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2 = 1?
Dec 24, 2006 22:37:30 GMT -8
Post by dxlightning on Dec 24, 2006 22:37:30 GMT -8
You can never divide both sides by X, in case X = 0. Therefore everything past (x-x)(x+x) = x(x-x) in line 3 is null and void. That's one of the first things we learned in Calculus.
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PhoenixFlare500
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I like chocolate[ss:LostPeon's Gray][ss:LostPeon's Gray]
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2 = 1?
Dec 25, 2006 10:36:06 GMT -8
Post by PhoenixFlare500 on Dec 25, 2006 10:36:06 GMT -8
That's the one, I just couldn't think of it.
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2 = 1?
Feb 19, 2007 1:02:03 GMT -8
Post by fatshinobi on Feb 19, 2007 1:02:03 GMT -8
I found another way where 2=1?
assume the formulas v=at and s=(at²)/2 are true. solving for the variable a in both formulas, we get 1)a=v/t and 2) a=(2s)/t² Because v=s/t and a=a, we can combine the equations to form v/t= 2v/t multiply both sides by the variable 't', then divide by 'v', and you get 1=2.
It does have a similar problem as the first thing, but I think that there was some sort of logic behind none of the variables being zero (or negative, because of measurements?)
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2 = 1?
Feb 26, 2007 0:10:06 GMT -8
Post by garious on Feb 26, 2007 0:10:06 GMT -8
the only possible way this problem could ever work is if x = 1... even then half the processes to solve the problem equate to 0. ... except the final resolutions... whereas 2x(1)*2*=x(1)*1* or 2=1... this is not a hard problem, just a misdirection of the applied steps of algebraic calculation
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