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
:duck:
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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
:duck:
Reread your question, -1 not eqauls iota. It's [iota*iota = -1] or [iota = -11/2].Quote:
Originally Posted by monu2010
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
And you'd be wrong! (-1)3 = -1, not 1!Quote:
Originally Posted by bobabot1
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)
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
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.
I said not... I was trying to prove that what he was doing was incorrectQuote:
Originally Posted by bobabot1