User blog comment:Vel!/Yudkowsky on googology/@comment-24.103.234.74-20140328005126

is this number valid using caret tops E100{#&#&#&#&#&#&#&#&#&#& ... &#&#&#&#&#&#&#&#&#&#}100>100 w/100 #s

@65.26.80.144

No, not normally. Caret tops are only defined as part of my delimiter notation. However, given what I said earlier about applying to any binary operation ... yes technically that makes sense, although it was only meant to apply to infix operators (ExE in general is not infix because expressions like ' 100#100 ' are meaningless without an ' E ' appended at the beginning). Let a%b mean Ea{#&#&....&#&#}b w/100 #s. Then a%b>c means...

( ... (((a%b)%b)%b) ... )%b w/c b's

The result of this however is just to iterate into the base argument. This will not be as significant however as iterating in the 2nd argument as in...

a%(a%(a%( ... (a%b) ... )))

which is what ExE accomplishes just with the proto-hyperion. So in short a%b>c would be less powerful than a%b#c. To give you a concrete example consider...

E100#100, E100#100#2 , E100#100#3, etc.

vs.

E100#100, E(E100#100)#100, E(E(E100#100)#100)#100

It may seem like the second one is really powerful ... but it turns out that E(Ea#b)#c = Ea#(b+c).

Thus the above sequence reduces to ...

E100#100, E100#200 , E100#300 , E100#400 , etc.

But compare this to...

E100#100, E100#100#2 = E100#(E100#100)

The 2nd member of the sequence has E100#100 as the second argument! The first sequence will have to reach the (grangol/100)th member just to reach this!!!

Hence the caret top doesn't lend very much power as an operator notation in and of itself. It's power comes from applying it to ordinal notation. Remember that...

(3^3)^3 is less than 3^(3^3)

The caret top expands to a left-to-right associative chain, where as a proto-hyperion expands to a right-to-left associative chain. In general this is vastly more powerful.

(Sbiis.ExE)