2 = 1 ???? check the process .?

Question

consider x^2 - x^2 (x^2 means x squared)
x^2 - x^2 = x(x-x)
and, x^2 - x^2 = (x + x) (x - x)
so,x(x-x) = (x+x) (x-x).
(x-x)cancels.
so, x = x+x.
x = 2x.
1 = 2
but how is this possible???
if iam wrong, where and how??

Answer 1

x^2-x^2=0
x(x-x)=0
so x=0 or x-x=0
the sum terminates in this step and you cannot proceed further when the answer becomes zero
check out how this is possible?
10*0=0 and 5*0=0
equating 10*0=5*0
cancelling 0 on both sides
10=5?????

Answer 2

If we forget the fact that any number - that number = 0 (so is x^2-x^2)...

one of the rules in solving equations, is to NEVER cancel the unknowns, be it x, y, etc., because (1) unknowns represent solutions to be discovered, plus (2) you get what you've got (1 = 2). You can transfer, add, subtract, multiply, etc., but never cancel.

It's obvious that x-x = 0, so you could stop right here. But to continue your quest, let's solve x = 2x. It's easy:

transfer all knowns on one side, and the unknowns on the other:
0 = 2x-x
0 = x
x = 0

Check:
x = 2x
0 = 2*0
0 = 0

so the system works for x = 0 (referring to x = 2x), as well as for any x (referring to x-x), which again complies with the initial premise.

The alternative work:
You could solve this "philosophically", ie. to stop and think is there any number in integer system (or in any other), that would satisfy the relation x=2x. And the answer would again be 0, because only 0 gives 0 in a product, regardless of other factors and multipliers in it.

As for the relation x-x, this equals 0, or x = x, and the logic says that every number in every system (N, R, ...) equals to itself.

Answer 3

Cos x-x in line 2 means multiplying by 0. Theres a similar proof here.

let a = b

a² = ab Multiply both sides by a

a² + a² - 2ab = ab + a² - 2ab Add (a² - 2ab) to both sides

2(a² - ab) = a² - ab Factor the left, and collect like terms on the right

2 = 1 Divide both sides by (a² - ab)

Answer 4

Here is the flaw:

When you do x(x-x) = (x+x)(x-x)
and you attempt to cancel out x-x, it would SEEM correct and valid, but it's not

x - x is obviously 0

So you're trying to divide 0 from both sides of the equation, and dividing by 0 is not a valid calculation. Therefore, everything from that point on is false, making this "proof" invalid.

Answer 5

Gamefreak is correct. (x-x) is not a valid step since you were dividing both sides by 0 (x-x) = 0

All bets are off at that point

Answer 6

x-x=0
substitute x-x =0
x(0)=x(0)
0=0
true

Answer 7

x(x-x) does not equal (x + x) (x - x)

x(x-x) equals (x*x) - (x*x)

(by the way, I edited this after I caught a mistake)