x2-x2=(x-x)*(x+x)
if we devide the left side by x we get
x(x-x)=(x-x)*(x+x)
now devide both sides by (x-x) and we get
x=x+x
so
x=2x
devide both sides by x and we get
1=2
what was done wrong?
2007-05-09 19:45:09 · 6 answers · asked by mazen 1 in Science & Mathematics ➔ Mathematics
Dividing by x-x means dividing both the sides by zero.
Never do that or be prepared to face more nonsensical results like this.
2007-05-09 19:50:21 · answer #1 · answered by alien 4 · 1⤊ 0⤋
You begin to assume x^2-x^2=(x-x)*(x+x)
but,
(x-x)(x+x)= x^2+x^2-x^2-x^2= 0
These two sides of the equation are only equal if x=0, in which case you divided by 0 at step 2.
2007-05-09 19:52:16 · answer #2 · answered by Anonymous · 0⤊ 0⤋
You can't arbitrarily divide one side by x and not the other side. What you do to one side you must do the other.
It like saying
8=8
-Divide the left side by 4.
2=8
It just doesn't work that way. What you do to one side, you do to both sides.
2007-05-09 20:09:04 · answer #3 · answered by jasbest34 2 · 0⤊ 0⤋
you cannot divide both the sides by 0 i.e dividing both sides by (x-x) cannot be done
2007-05-09 19:52:00 · answer #4 · answered by pravesh doshi 1 · 1⤊ 0⤋
x2-x2-0
(x-x)*(x+x)
x-x cancel out leaving 0*(x+x) there
the anwer to this equasion is 0=0
2007-05-09 19:52:14 · answer #5 · answered by Kristenite’s Back! 7 · 0⤊ 0⤋
You divided by x - x which is zero ----- no good
2007-05-09 19:54:25 · answer #6 · answered by Gene 7 · 0⤊ 0⤋