Guest Posted September 13, 2008 Report Share Posted September 13, 2008 Assume x = y which means 1 = 1 x = y then multply both sides by x so... x^{2} = xy then subtract y^{2} so... x^{2} - y^{2} = xy - y^{2} then expand these quandratric equations to... ( x + y ) ( x - y ) = x ( x - y ) which is just another representation of the previous line. Now, we currently have ( x - y ) on both sides, so lets divide both sides by ( x - y ) to get what is left... (x+y)(x-y) = x(x-y) . (x-y) . . . . (x-y) That leaves us with the following x + y = x if we go back to the beginning and say that x equals y so that x = 1 and y = 1 then 1 + 1 = 1 OR 2 = 1 Logical except for the fact that you can not divide by x-y because if x=y then x-y=0 and you can not divide a number by nothing - ie, nothing goes into one an infinite number of times plus one. Ohh..!! I saw this before! It was on my test (Geometry) as an extra credit (I didn't get it... ) I also saw a similar problem in my geometry book (another extra credit) which described how 2X2=5 (something similar to that...) All of these are just breaking only one rule, which is that you can't divide by zero. However...I'm kind of confused on what people are talking about when it comes to things like .9999... ...Oh well... Link to comment Share on other sites More sharing options...

Guest Posted November 16, 2008 Report Share Posted November 16, 2008 (edited) Actually, I've seen a similar mathematical fallacy stated as the following. I like it better because the error is hidden a little better and the proof looks a little cleaner. The proof is pretty much the same and is identical to the problem posted at the begging of the thread in that the algebra is correct but the proof violates a basic rule of mathematics. Proof that any number 'a' is equal to a smaller number 'b': a = b + c a(a - b) = (b + c)(a - b) a^2 - ab = ab - b^2 + ac - bc a^2 - ab - ac = ab - b^2 - bc a(a - b - c) = b(a - b - c) a = b Edited November 16, 2008 by 95yj Link to comment Share on other sites More sharing options...

Guest Posted December 10, 2008 Report Share Posted December 10, 2008 x(x-2)=x-2 x(2-2)=2-2 0=0 Link to comment Share on other sites More sharing options...

Guest Posted March 28, 2010 Report Share Posted March 28, 2010 Alright, try this one on for size: X=.9999999... 10x=9.99999... (10x)-x=9 (9x)/9=1 x=1 .999999...=1 X=.9999999... 10x=9.99999... (10x)-x=9 10x = 9+x 10x-9 = x 10-9 = x/x 1 = 1 so, it always results in 1=1 Link to comment Share on other sites More sharing options...

Guest Posted April 29, 2010 Report Share Posted April 29, 2010 Alright, try this one on for size: X=.9999999... 10x=9.99999... (10x)-x=9 (9x)/9=1 x=1 .999999...=1 There is no problem with this statement; it is true. Think about it this way: (1/3)=.333333..... (2/3)=.666666..... (1/3)+(2/3)=.999999.... or (1/3)+(2/3)=(3/3)=1 The original question is wrong because he divides by (x-1) which would be division by 0. Link to comment Share on other sites More sharing options...

Guest Posted October 16, 2010 Report Share Posted October 16, 2010 x(x-2) = x-2 because it had defined x=2 so （x-2） can't be reduced Link to comment Share on other sites More sharing options...

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