Why is 1 = 2 ??????

[paadness]

X2 - X2 = X2 - X2 " 2 is square"

X(X-X) = (X+X)(X-X)

X = X+X

X=2X

1=2 wtf???

 

[deveraux]

Division by zero error. (X-X) = 0 hence you can't divide both sides of the equation by it.

 

[paadness]

No, i am not doing X-X to obtain Zero.

Ok beat this.

let,

(X-X) = A

now eq becomes

X(A)= (X+X)(A)

 

[Matthias99]

Originally posted by: paadness
No, i am not doing X-X to obtain Zero.

Ok beat this.

let,

(X-X) = A

now eq becomes

X(A)= (X+X)(A)

If (X-X) = A, then A must be zero, no matter what tricks you play with it, and dividing by A then implies division by zero and is undefined. At least if you're working with integers or real numbers. 'X(A) = (X+X)(A)' becomes '0*x = 0*(2*x)', otherwise known as '0 = 0'.

Thanks for playing.

 

[paadness]

but as far as i know, 0/0 can have 3 values.

Basically it involves limit and 0/0 is a dangerious thing. Le-Hopital rule has a solution to it if u have a denomitor.

In different situations we can treat this danger according to needs.

Still 1 = 2 trick is good enough to fool high school students

 

[deveraux]

De-Hopital's rule applies to a limit. There is no limit involved here. A does not approach 0, A is 0. Of course I could be wrong, but from what I understand from De-Hopital's rule, it doesn't apply here.

 

[krcat1]

X(X-X) does not equal X2-X2 if you set (X-X) = A

X(A) != (X+X)(A) unless A=0

 

[Smilin]

Originally posted by: paadness
but as far as i know, 0/0 can have 3 values.

Basically it involves limit and 0/0 is a dangerious thing. Le-Hopital rule has a solution to it if u have a denomitor.

In different situations we can treat this danger according to needs.

Still 1 = 2 trick is good enough to fool high school students

limit approaching zero isn't undefined. divide by zero is.

0/2 = 0, 0/1 = 0, 0/0 = undefined. No cheating

 

[deveraux]

Originally posted by: krcat1
X(X-X) does not equal X2-X2.

err ..yes it does

 

[jagec]

This is as bad as saying 5*infinity=infinity, therefore 1=5.

 

[blahblah99]

Does 1 = 0.99999999...?

 

[Fullmetal Chocobo]

Originally posted by: blahblah99
Does 1 = 0.99999999...?

It can be mathematically proven, yes. http://forums.anandtech.com/messageview...atid=50&threadid=1600758&enterthread=y But it can also be statistically proven that the sky is green...
Tas.

 

[DrPizza]

L'Hopital's rule applies to LIMITS

0/0 = undefined. ALWAYS. 0/0 is undefined.

However, if evaluating something like
lim (as x -> 0) sin(x) / x
then, since the value is approaching 0 in the numerator and 0 in the denominator, which is an indeterminate form of a limit, THEN you can apply L'Hopital's rule. (in which case, it becomes lim (as x -> 0) cos(x) / 1 which would evaluate to 1/1 or 1.

Now, you said something about 3 values?
lim (as x -> 0) sinx / x = 1
lim (as x -> 0) sin2x / x = 2
lim (as x -> 0) sin3x / x = 3
lim (as x -> 0) sin4x / x = 4

obviously, we can come up with a limit problem, the answer to which can be any real number we want.
There are other limits of the form 0/0 which evaluate to:
There is no limit, it increases without bound. (usually just abbreviated as +infinity; the math people know what you mean)
There is no limit, it decreases without bound. (usually just abbreviated as -infinity; again, the math people know what you mean)
and
There is no limit because it approaches different values depending on the direction the limiting value is approached from (DNE)

 

[paadness]

Originally posted by: DrPizza
L'Hopital's rule applies to LIMITS

0/0 = undefined. ALWAYS. 0/0 is undefined.

However, if evaluating something like
lim (as x -> 0) sin(x) / x
then, since the value is approaching 0 in the numerator and 0 in the denominator, which is an indeterminate form of a limit, THEN you can apply L'Hopital's rule. (in which case, it becomes lim (as x -> 0) cos(x) / 1 which would evaluate to 1/1 or 1.

Now, you said something about 3 values?
lim (as x -> 0) sinx / x = 1
lim (as x -> 0) sin2x / x = 2
lim (as x -> 0) sin3x / x = 3
lim (as x -> 0) sin4x / x = 4

obviously, we can come up with a limit problem, the answer to which can be any real number we want.
There are other limits of the form 0/0 which evaluate to:
There is no limit, it increases without bound. (usually just abbreviated as +infinity; the math people know what you mean)
There is no limit, it decreases without bound. (usually just abbreviated as -infinity; again, the math people know what you mean)
and
There is no limit because it approaches different values depending on the direction the limiting value is approached from (DNE)

:thumbsup:

 

[jman19]

Originally posted by: krcat1
X(X-X) does not equal X2-X2 if you set (X-X) = A

X(A) = (X+X)(A) unless A=0

X(A) = (X+X)(A) even if A=0...

 

[krcat1]

Originally posted by: jman19

Originally posted by: krcat1
X(X-X) does not equal X2-X2 if you set (X-X) = A

X(A) = (X+X)(A) unless A=0

X(A) = (X+X)(A) even if A=0...

X(A) = (X+X)(A) only if A=0

 

[jman19]

Originally posted by: krcat1

Originally posted by: jman19

Originally posted by: krcat1
X(X-X) does not equal X2-X2 if you set (X-X) = A

X(A) = (X+X)(A) unless A=0

X(A) = (X+X)(A) even if A=0...

X(A) = (X+X)(A) only if A=0

For some reason when I decided to reply I completely ignored the fact that A is set to (X-X). Anyway, "only if" doesn't ever pertain, as that IS the value of A.