Hyperoperators
#3
What about defining the inverses of operations?
So, for example, we could write a-b as A(a,-b,1) (or whatever other notation you wish to use), but if we wanted to evaluate A(a,-b,1) purely in terms of its recursive definitions, we'd never reach a value. Perhaps it could then be written so that a-b = A(a,b,k) for some k?

Don't know what that k would be though. An intuitive answer would be something like -1, but A(a,b,-1) is simply b+1, so that doesn't work.

The same problem may arise for a/b, log[a](b), etc. Especially log[a](b) - whereas the others can be expressed in terms of their inverse function, log[a](b) is independent of a^b.

Sorry if these are old questions - I'm new here, and these things have been troubling me for a while. :/

By the way, how does one write in equation mode?
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Messages In This Thread
Hyperoperators - by Mr. Pig - 06/10/2010, 04:37 AM
RE: Hyperoperators - by bo198214 - 06/10/2010, 06:42 AM
RE: Hyperoperators - by 73939 - 06/20/2010, 02:41 AM
RE: Hyperoperators - by bo198214 - 06/20/2010, 12:26 PM
RE: Hyperoperators - by andydude - 06/20/2010, 05:46 AM

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