# 1=0.

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• 08-05-2008
The_Hermit
1=0.
assuming that x and y represent the same nonzero number:

x=y
x^2=xy
x^2 - y^2=xy - y^2
(x+y)(x-y)=y(x-y)
x+y=y
2y=y
2=1
1=0

what just happened here?
I honestly don't know...
• 08-05-2008
Thantos
if x=y then x-y = 0
so in the third step you are dividing by 0
• 08-05-2008
The_Hermit
oh, wow. good catch.
Thanks, man.
• 08-06-2008
indigo0086
My friend showed me this in high school, he was good at math so he thought he could trick us and I found it out fast. I do remember his being different, though. It still had the divide by zero thing.
• 08-06-2008
Thantos
Yeah I've seen this one enough to immediately know it is dividing by 0.
• 08-06-2008
QuantumPete
Quote:

Originally Posted by Thantos
Yeah I've seen this one enough to immediately know it is dividing by 0.

I think that's pretty much the only way to get 1=0. Undefined behaviour ;)

QuantumPete
• 08-06-2008
abachler
Well, you also violated a lot of conventions and rules about the seperation of equality and equivelancy.

x=y

this is a statement of equality which makes x and y dependant variables.

x^2=xy

this is a conjecture of equivelancy which only holds true for the case of x == y

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

this further assumes the case that x == y. If x and y are independant this statement is false.

(x+y)(x-y)=y(x-y)

You are mixing general solutions with special case solutions and presenting the special case solution as a general solution which is erroneous.
y(x-y) = xy - y^2
(x+y)(x-y) = x^2 - xy + xy -y^2 = x^2 - y^2 not xy - y^2

x+y=y
this holds true only for the case of x = 0

2y=y
this is true only for the case of y = 0

the rest is nonsense as a result of faulty logic. makign a statement which holds true only for a single case and then applying it as if it held for all cases (false generalization,).
2=1
1=0

for 0<x<9 all odd numbers are prime. TRUE

since odd numbers are prime and 9 is an odd number 9 is prime. FALSE.

that is what you are doing, assuming that because a statement (xy = y^2) holds true under restricted conditons (x = y), that it holds true without those restrictions.
• 08-07-2008
Sang-drax
Quote:

Originally Posted by QuantumPete
I think that's pretty much the only way to get 1=0. Undefined behaviour ;)

You could get it without dividing by zero this way:
1 = 1
sqrt(1)= 1
sqrt((-1)*(-1))= 1
sqrt(-1)*sqrt(-1)= 1
i * i = 1
-1 = 1
-1/2 = 1/2
0 = 1
• 08-07-2008
Sang-drax
Quote:

Originally Posted by abachler
Well, you also violated a lot of conventions and rules about the seperation of equality and equivelancy.

No, every step was correct except the step where he was dividing by zero.
• 08-07-2008
indigo0086
Quote:

Originally Posted by Sang-drax
You could get it without dividing by zero this way:
1 = 1
sqrt(1)= 1
sqrt((-1)*(-1))= 1
sqrt(-1)*sqrt(-1)= 1
i * i = 1
-1 = 1
-1/2 = 1/2
0 = 1

still, it's clear to see that you're mixing equality with equivalence. Also that last step seems to just be the magic trick in the sea of misdirection.
• 08-07-2008
QuantumPete
Quote:

Originally Posted by Sang-drax
sqrt(1)= 1

That's still cheating. :P But yes, you can do it without a divide by 0.

QuantumPete
• 08-09-2008
Stonehambey
Quote:

Originally Posted by abachler
Well, you also violated a lot of conventions and rules about the seperation of equality and equivelancy....

Well of course, why do you think that most mathematical theorems and proofs start with "Given that...", "assume that..." or "Let us define...". In a mathematical proof, it is ok to define a set of initial conditions and then proceed logically assuming these are true, so long as you say so. So the fallacy in this classic 1=2 proof, is not where you suggest.

For example, Pythagoras' theorem says something along the lines of "Given a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the two remaining sides." What you appear to be arguing is that this would not be true if it was not a right angled triangle, well of course! That's why we define it to be one in the first place.

Similarly, it's perfectly ok to say that x^2 - y^2 = xy - y^2 in this case, since the first thing we did in this proof was say that x=y.
• 08-09-2008
vart
When you get to this step:
Quote:

sqrt(-1)*
you get to the complex numbers - talking about complex numbers - you have 2 possibilities for the expression instead of one
you choose the wrong one to preserve the equility
• 08-13-2008
abachler
Quote:

Originally Posted by Sang-drax
You could get it without dividing by zero this way:
1 = 1
sqrt(1)= 1
sqrt((-1)*(-1))= 1 wrong
sqrt(-1)*sqrt(-1)= 1 wrong
i * i = 1 wrong
-1 = 1
-1/2 = 1/2
0 = 1

sqrt(-1) = i
i * i = -1
• 08-13-2008
cpjust
Quote:

Originally Posted by abachler
Quote:

Originally Posted by Sang-drax
You could get it without dividing by zero this way:
1 = 1
sqrt(1)= 1
sqrt((-1)*(-1))= 1 wrong
sqrt(-1)*sqrt(-1)= 1 wrong
i * i = 1 wrong
-1 = 1
-1/2 = 1/2
0 = 1
sqrt(-1) = i
i * i = -1

Actually, the first one is right (look at the parenthesis):
sqrt( (-1) * (-1) ) = sqrt( 1 ) = 1

The second one is where reality flies out the window since it uses the Product Rule for Radicals, but the product rule for radicals says:
Quote:

The nth root of a product is equal to the product of the nth roots. provided that all of the expressions represent real numbers.
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