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the logical hierarchy
#1
here i present what is according to me the " logical hierarchy "

i found it important to say , because it appears often on math forums and is usually stated or ordered in a way i disagree with.

no tetration or non-commutativity.

no ' popularized ' ackermann or buck.

but plain good old logic in my humble opinion.

most imporant is the existance of a single neutral element

f_n ( a , neutral ) = f_n ( neutral , a ) = a for all a !



1) a + b

2) a * b

3) a ^ log(b)

to see how i arrived at 3 :

a ^ log(b) = b ^ log(a) = exp( log(a) * log(b) )

4) exp ( log(a) ^ log(log(b)) )

to see how i arrived at 4 :

note that 3) is used upon log(a) and log(b).

etc etc


note that the neutral elements are

1) addition -> 0

2) multiplication -> 1

3) a ^ log(b) -> e

4) -> e^e

5) -> e^e^e

6) -> e^e^e^e

etc
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#2
Ya this hierarchy was already considered.
The main observation is that the real numbers with operations and are isomorphic to the positive real numbers with and (The isomorphism is ). I.e. we dont add really something new. Each two consecutive operations are isomorphic (i.e. behave completely the same as) to + and *.
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#3
bo198214 Wrote:Ya this hierarchy was already considered.
The main observation is that the real numbers with operations and are isomorphic to the positive real numbers with and (The isomorphism is ). I.e. we dont add really something new. Each two consecutive operations are isomorphic (i.e. behave completely the same as) to + and *.

right.

but some people always insist on

a + b

a * b

a ^ b

which is wrong.


glad we agree.

thread closed ?

regards

tommy1729
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#4
tommy1729 Wrote:but some people always insist on

a + b

a * b

a ^ b

which is wrong.

Its not wrong its a different hierarchy.

It follows the pattern: a[n+1](b+1)=a[n](a[n+1]b)
where [n] is the the nth operation.
For example:
a[2](b+1)=a[1](a[2]b) which corresponds to a*(b+1)=a+a*b
a[3](b+1)=a[2](a[3]b) which corresponds to a^(b+1)=a*(a^b)
and so tetration [4] satisfies:
a[4](b+1)=a[3](a[4]b)
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