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1=2 ?
Topic Started: Dec 2 2008, 02:58 PM (1,390 Views)
Deleted User
Deleted User

I know this proof is false, but I'm daring you guys to find out why is it false, don't try to ask someone else as the answer is REALLY easy, you just need to think :) , you all get only one shot so think carefully

1)We have to numbers, X and Y, X=Y

2)Multiply both sides by X

X^2 = XY

3)subtract y^2 from each side

x^2 - y^2 = xy - y^2

4)(x + y)(x - y) = y(x - y)
Factorization :D

5)Divide by (X-Y)
x + y = y

6)Since Y=X, so

2Y = Y (Or 2X=X)

7) 2=1

I'm the pope :D , till proven otherwise :D
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Fission
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Uguu
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5.) Divide by (x - y)

You can not divide by 0!
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Deleted User
Deleted User

Damn it is it THAT obvious ?
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Fission
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Uguu
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To be honest, yes. At first I was going over your factoring, but then I went to the next line and was like "duh."
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Christian
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Always a Step Ahead
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Quote:
 
x + y = y

Since Y=X


There's no way Y can equal X when you have x+y=y. X would have be 0 and y would be any number including 0. Replacing X with 1 and Y with 2 would show 1+2=2 which is incorrect and vice versa is incorrect: 2+1=1.
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Fission
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Uguu
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I already found it out, and it's an earlier problem: dividing by x - y (0).
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Deleted User
Deleted User

Christian, that's the point showing that 1 is equal to two, saying that it's no way does NOT mean you have PROOF for it, Fission got it
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ElementalAlchemist
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And even before the division by 0...

xy-y^2 != x-y

You have to solve it as a quadratic...
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FinalKiller0
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OVER 1,000!!
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ElementalAlchemist
Dec 2 2008, 11:24 PM
And even before the division by 0...

xy-y^2 != x-y

You have to solve it as a quadratic...
I am confused by the above post.
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ElementalAlchemist
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You can't subtract that, you have to solve it another way.
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Fission
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Uguu
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(x + y) * (x - y) == x ^ 2 - y ^ 2
VVV
x * x - y * y + xy - xy = x * x - y * y = x ^ 2 - y ^ 2

y(x - y) = xy - y ^ 2
VVV
y * x - y * y = xy - y ^ 2

They both work out.
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Deleted User
Deleted User

Think again Alchemist, only wrong part is step #5
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ElementalAlchemist
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Oh, never mind. I brain-farted there and remembered step 4 differently, as if you subtracted or something.

Meh.
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IceMetalPunk
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We are all IMPerfect. Be proud!
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Mind if I post my own mathematical fallacy challenge?

For this one, you must know about modulo math. Basically, (x) mod y is the remainder when x is divided by y. Knowing that:

Code:
 
Proof that 5=1

1) (3) mod 4=3 : Fact
2) (2) mod 4=2 : Fact
3) (5) mod 4=1 : Fact
4) (3+2) mod 4 = (5) mod 4 : 3+2=5, so (3+2) mod 4 = (5) mod 4
5) [(3) mod 4] + [(2) mod 4] = (5) mod 4 : Distribute the "mod" function over (3+2)
6) 3 + 2 = 1 : Substitute using the first 3 facts.
7) 5=1 : Add

Obviously, 5=1.


How can 5=1? Or is there a problem in my proof? Can you find one? It's actually very simple, if you know your math :P .

-IMP ;) :)
Edited by IceMetalPunk, Dec 3 2008, 09:33 PM.
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Deleted User
Deleted User

IMP all I can read in your post is "This is some serious math #$!@# that you can't understand in a hundred years"

Nice thinking there O___O
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