the logical hierarchy

[tommy1729]

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

[bo198214]

Ya this hierarchy was already considered.
The main observation is that the real numbers with operations \( a+b \) and \( a*b \) are isomorphic to the positive real numbers with \( a*b \) and \( a^{\log(b)} \) (The isomorphism is \( \exp \)). I.e. we dont add really something new. Each two consecutive operations are isomorphic (i.e. behave completely the same as) to + and *.

[tommy1729]

Ya this hierarchy was already considered.
The main observation is that the real numbers with operations \( a+b \) and \( a*b \) are isomorphic to the positive real numbers with \( a*b \) and \( a^{\log(b)} \) (The isomorphism is \( \exp \)). 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

[bo198214]

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)