Talk:-dex

I mentioned this before: neither (a^^b)^c nor (a^c)^^b is equivalent to a^c#b. One could define (n)-dex as E100#n, which would agree with most examples. However your definition is not equivalent.

You state:

grangoldex = g(2,100,g(3,grangol,10))

but this is actually (10^^grangol)^100 = 10^(100*10^^(grangol-1)) = 10^10^(2+10^^(grangol-2)) < 10^10^10^(1+10^^(grangol-3)) < ... < E(1+10^^0)#grangol = E(1+1)#grangol = E2#grangol < E100#grangol = grangoldex.

This value of g(2,100,g(3,grangol,10)) is actually smaller than E100#100#2, the true value of a grangoldex. Likewise, g(2,100,g(3,googol,10)) = (10^^googol)^100 is not a googoldex. (10^^googol)^100 is less than E2#googol < E100#googol = googoldex. Remember (a^^b)^c != a^a^a^...^a^c w/b a's. Instead (a^^b)^c = a^(c*a^^(b-1)), which is much smaller in most cases. Thus most expressions in E# can not be feasibly expressed in Up-arrow notation, and consequently in g-notation.

To provide you with a concrete example consider the following: Let's say we have (2^^4)^3. If your correct then this should evaluate to 2^2^2^2^3 or 2^2^2^8 or 2^2^256. This is a very large value. However the way we would actually evaluate (2^^4)^3 is by first working out the paratheses and then cubing the result. This gives us:

 (2^^4)^3 = (2^2^2^2)^3 = (2^2^4)^3 = (2^16)^3 = (65,536)^3. Now cube this result. You will get 281,474,976,710,656 (about 281 trillion). There is no way that this tiny number can be anywhere near 2^2^256. To prove this compute 2^48. You will find you get exactly the same decimal expansion of 281,474,976,710,656. This agrees with multiplying the exponents so that (2^16)^3 = 2^(16*3) = 2^48.

 Now consider the following...

 2^48 < 2^64 = 2^2^6 << 2^2^256

 There is no comparison. (2^^4)^3 does not equal 2^2^2^2^3, not even close! In the same way (10^^grangol)^100 is much smaller than 10^10^10^...^10^10^10^100 w/grangol 10s. Sbiis Saibian (talk) 00:23, July 10, 2014 (UTC)