again can 1=(-1) ??? with cube

[monu2010]

ok lets check this
1=1
1=cuberoot(1)
1=cuberoot(-1*-1*1)
1=cuberoot(-1)*cuberoot(-1)*cuberoot(1)
1=cuberoot(iota)*cuberoot(iota)*cuberoot(1)

cubing both sides

1=iota*iota*1
1=-1*1
1=-1

[Harsh Gupta]

1=cuberoot(-1)*cuberoot(-1)*cuberoot(1)
1=cuberoot(iota)*cuberoot(iota)*cuberoot(1)

Reread your question, -1 not eqauls iota. It's [iota*iota = -1] or [iota = -1^1/2].

[bobabot1]

Math is too strong to be proven wrong by something this simple.
You could say:
(-1)^3=1^3
but it does not mean
-1=1

[Dross]

You could say:
(-1)^3=1^3

And you'd be wrong! (-1)³ = -1, not 1!

[Mattywoo2]

Yes, the problem is between your 4th and 5th lines. You seem to suggest

cuberoot(-1) == cuberoot(iota)

which is clearly not true.

cuberoot(-1) = iota*(-1)^(-1/6)

or

cuberoot(-1) = (iota)^(2/3)

[bobabot1]

Dross: You're right. I meant to say that if you did the same thing to both sides (in this case cube them) that they would not be the same, thus proving 1 != -1

[zaza]

For goodness sakes. Can we not go down this road? Next up will be (-1)^4 = (1)^4 etc.




You cannot use the equality (-1)^2 = (1)^2 in any way, shape or form to argue that -1 = 1.

[bobabot1]

Dross: You're right. I meant to say that if you did the same thing to both sides (in this case cube them) that they would not be the same, thus proving 1 != -1

I said not... I was trying to prove that what he was doing was incorrect