1=2

where is the error ???

1^2 = 1
2^2=2+2
3^2 = 3+3+3

so
x^2 = x+x+x+x+x+x+x.........x times

Differentiate both dides w.r.t x

2x = 1+1+1+1+1......x times

so
2x = x ( 1+1+1...xtimes = x)

Hence 1=2 ??????

:rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes:

you didn't take into account the "x times" on the right side
x+x+x+x+x+x+x.........x times = x^2
Dx^2=2x

No thats not true

x^2 = x+x+x+x+x+x..... xtimes

Ok

now differentiate both sides with respect to x
so
Dx^2 = D(x+x+x+x+x+x..... xtimes)

Ok

Now

Dx^2 = Dx+Dx+Dx+Dx+Dx+Dx..... xtimes

OK


finally

2x = 1+1+1+1+1+......xtimes (as Dx = 1)

or

2x = x

=> 1=2

of course it is true, you have to derivate on all x, including the "x times" you just treated as any parameter.

thats none sense,
the error lies in the differentiation,
its not possible because the
number of terms in x+x+x+x...x times
depends on x, so its none sense to
to assume that its differentiation is
1+1+1+1..... x times

i just wanted to show why
its none sense: (where k is constant)
kx,
derive it to get:
k
so what if k=x:
x*x= x^2
derive it to get:
2x
but x=k so its
2k

so k=2k, so 1=2

thats none sense because k is constant no more
so here K is not constant,
and there, the number of terms is not constant
its x times, see ??!

:D

Everyone knew there had to be an error, this discussion of 1 being equal to 2 has been all over these maths forums enough times.

"Society will fall, my bank account will be overdrawn and in credit simultaneously" blah blah blah blah...

Firstly:
"(where k is constant)" ... "so what if k=x"
These do not go down well together, esp. when taking derivatives, as taking the derivative of 'x*x= x^2' gives 0=0 if x=k=CONSTANT. And as we all know 0=0 :D

Oh, and does that mean:
x = 1+1+1+......+1 (x times)
.....Take derivatives.....
1 = 0 + 0 + 0....(x times)
1 = 0! :eek:

No, it doesn't work like that, because x is NOT constant, you can't just take out 'x times' like that...

Quote:

---

Originally posted by sql_lall
Firstly:
"(where k is constant)" ... "so what if k=x"
These do not go down well together, esp. when taking derivatives, as taking the derivative of 'x*x= x^2' gives 0=0 if x=k=CONSTANT. And as we all know 0=0 :D

---

im not sure if you got me right,
but i just said that to show
its none sense !!!
i see that we both agree on that
its none sense ;)

r u joking?? u cant derive 2 sides of an equation and expect them to be equal, thats the error

forget about everything and try this:

Code:

---

x^2 = x + x + x, at this point x = 0 or 3; differentiate both sides and get
2x = 1 + 1 + 1 or
2x = 3, at this point x = 3/2

---

anyway when u differentiate both sides of an equation that is a no-no

Quote:

---

Originally posted by dis1411
r u joking?? u cant derive 2 sides of an equation and expect them to be equal, thats the error

forget about everything and try this:

Code:

---

x^2 = x + x + x, at this point x = 0 or 3; differentiate both sides and get
2x = 1 + 1 + 1 or
2x = 3, at this point x = 3/2

---

anyway when u differentiate both sides of an equation that is a no-no

---

well this equation appears to hold true for all x so why not? They're after all equivalent expressions

:confused:

till now we have not reached to a conclusion

or do you mean agreement?

the conclusion is that when solving an equation you can't differentiate both sides

Quote:

---

the conclusion is that when solving an equation you can't differentiate both sides

---

Thats totally wrong.... You can do diff both sides of an equation and there is nothing wrong with that........

I am sure abut it.

sw_is_great: no you can't, if you have two different expressions on each side of the equation.
dis1411: this wasn't about solving but proving.
If you start out with the same expression, and differentiate you should arrive with the same derivate, however sw_is_great didn't differentiate the right side expression.

lets make this a little more simple, so you can understand :D

x^2 = 3x at this point x = 0 or 3; differentiate both sides

2x = 3 at this point x = 3/2

you say there is nothing wrong with x changing values??