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#1
 Dec 9 2006, 07:23 PM
In light of the topic, 2+2=5 methought I'd create my own puzzle: [hr=shade]Let x = y[/hr] Therefore x² = xy Therefore x² - y² = xy - y² If we factorise these equations we end up with (x + y)(x - y) = y(x - y) We now have a common term on both sides "(x - y)," which we will divide off to make our problem easier to solve x + y = y But x = y so y + y = y thus 2y = y Then, if we divide off the common value "y," 2 = 1 [hr=shade]Simple! Where's the flaw?[/hr] 
 
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#2
Dec 9 2006, 09:59 PM
I remember this from last year in my math class. It was in the step when you canceled (x-y) I believe, because (x-y) is 0, therefore you are multiplying by 0 which would be 0. You can't cancel (x-y). Good one!
This post has been edited by husker: Dec 9 2006, 11:28 PM
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#3
Dec 10 2006, 12:03 AM
Well like husker said (x-y) would equal zero and it's the fact that you can't divide by zero not multiply like husker said. So that would really mean that it ends up as 0 = 0 because if you plug in the 0 of (x - y) , that would multiply with the rest of the equation to make 0 because it is all multiplication, no addition or subtraction except between parentheses. There may be a way though so keep working at it.
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#4
Dec 10 2006, 05:37 AM
As I have said in similar topic 2+2=5, I would again say that this type of questions are raised by people with little knowledge of Mathematics. As explained by Husker and Plenoptic as x and y are equal therefore x-y is zero and whenever zero comes it cant be cancelled from both sides, if this is possible then all numbers will be equal to each other, like this. 0=0 or, 0*a=0*b i.e. a=b where a and b are any number. therefore zeros from both sides cannot be cancelled.
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#5
Dec 10 2006, 09:03 AM
Congrats! The answer was you can't divide by 0 or x - y. You can prove anything when you divide by 0.
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#6
Dec 20 2006, 04:36 PM
Here is another way of proving that 1 = 2 QUOTE Everyone knows this: -1/1= 1/-1 Now we will square root both sides: √-1/1 = √1/-1 Now we break up the roots: √-1 √1 --- = --- √1 √-1 The square root of a negative 1 is i and the square root of 1 is 1. In other words: i/1 = 1/i Now we multiply the entire thing by 1/2: i/2 = 1/2i Now let's add 3/(2i) to this to make the math easier. i/2 + 3/2i = 1/2i + 3/2i Now we can multiply the entire thing by i: i(i/2 + 3/2i) = i(1/2i + 3/2i) So now we expand this beast: 1^2/2 + 3i/2i = i/2i + 3i/2i We know that the square root of -1 is i, so i^2 must be -1.: -1/2 + 3i/2i = i/2i + 3i/2i Now we simplify the i's -1/2 + 3/2 = 1/2 + 3/2 Let's calculate this thing: 2/2 = 4/2 And so 1 = 2 That does work as far as I can see. And with this proof you can prove that dividing by zero is possible, because anygivennumber^0 = 1 Source I know the source isnt reliable, but the 1-=2 thing I posted does work as far as I can see. This post has been edited by Terciel Silvi: Dec 20 2006, 06:37 PM
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#7
Dec 20 2006, 08:09 PM
1 will never equal 2 by definition. So I'm going to bother with the second proof because there's definately a flaw somewhere. If it was correct then there would be a flaw in our addition, multiplication, etc. Remember you can't divide by variables unless you can prove they aren't 0 like y = x^2 + 1 where x is a real number (no i's  )
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