Height of Zero Tetration Problem
#2
(06/27/2010, 09:50 PM)moejoe Wrote: BUT, taking my 2nd example for why a^0=1 in exponentiation - this doesn't work at all with tetration because if you do the following:
"a^^k = y" and "a = y^^(1/k)"
...
So to me this all seems like some kind of paradox is happening but I hope that someone will be able to put this in a clear light.

I think your paradox emerges from the wrong assumption in your equations.
You dont get the inverse of f(x)=x^^k (i.e. the tetration root) by taking x^^(1/k).
This is the case for powers but not for tetration.

It works for x^k because we have \( x^{m\cdot n} = (x^m)^n \) for integer numbers.
If we extend this law to rational numbers we can set \( x=x^{\frac{1}{n} n} = (x^{\frac{1}{n}})^n \) which means that \( x^{\frac{1}{n} \) is the inverse of \( x^n \), i.e. \( y=x^{\frac{1}{n}} \) is the number such that \( y^n = x \).

All this consideration fails for tetration because we dont have the rule x^^(m*n)=(x^^m)^^n.


Messages In This Thread
Height of Zero Tetration Problem - by moejoe - 06/27/2010, 09:50 PM
Tetration root vs x^^(1/k) - by bo198214 - 06/28/2010, 03:26 AM
RE: Tetration root vs x^^(1/k) - by bo198214 - 06/28/2010, 06:24 AM
RE: Height of Zero Tetration Problem - by moejoe - 06/28/2010, 11:34 PM

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