ags@pucc-i (Seaman) (02/23/84)
Now that we know how to prove that all natural numbers can be represented
with finitely many digits, would anyone care to prove that every natural
number is less than 2**n for some n?
I can think of an easy proof that begins by assuming that every number
has finitely many digits...
Seriously, neither of the above properties of natural numbers comes from
the definition. If you want to avoid circularity, you had better start
with the Peano Axioms. That is the only way you can know you are talking
about the natural numbers, and not about something else.
Here is a restatement of the Fifth Peano Axiom which I looked up last
night. This version does not mention sets:
(5) Let P be a property which is true of zero, and which is true of the
successor of any number which has the property. Then P is true of
all natural numbers.
--
Dave Seaman
..!pur-ee!pucc-i:ags
"Against people who give vent to their loquacity
by extraneous bombastic circumlocution."robertm@dartvax.UUCP (Robert P. Munafo) (02/25/84)
-
I don't understand what all this fuss is about. Every natural number
is finite, and so can be represented (in ordinary notation) in a finite
number of digits. This is true, N'est-ce pas? What am I missing?
Robert P. Munafo ...!{decvax,cornell,linus}!dartvax!robertmgrunwald@uiuccsb.UUCP (02/26/84)
#R:pucc-i:-21800:uiuccsb:9700024:000:260 uiuccsb!grunwald Feb 25 17:44:00 1984 Isn't a proof by induction rigorous enough? I'm not try to be difficult, but it seem that if one assumes the peano axioms for Naturals, and then applies the inductive proof, that would be sufficient. Clearly, one should reduce the problem back to the axioms.