0

Why is the successor of n, n++, unequal to n?

I've been reading about Peano's axioms in Analysis 1 by Terence Tao recently (which I am thoroughly enjoying :)).

In one of the exercises I found myself with what seemed a self-evident contradiction: n++ = n. Though, I can't seem to find any mention in the text of the succesor of n being distinct from n itself.

Perhaps this is more of a philosophical, or semantic question; Peano's second axiom is phrased as follows in Analysis 1:

If n is a natural number, then n++ is also a natural number.

, which allows one to argue that n and n++ don't necessarily have to be distinct objects.

It would seem that this intends to mean:

If n is an element of the set of natural numbers, then there exists another (i.e. distinct) element n++ that is also an element of the set of natural numbers.

Is it fair to put the axiom this way? I'm guessing that Tao hasn't phrased it this way because sets are only defined formally in the next chapter.

Any ideas are appreciated!

Sayan Dutta
  • 8,831
Teun van Wezel
  • 11
  • The answer to this related question hints at an answer: https://math.stackexchange.com/questions/1077710/a-question-on-terence-taos-representation-of-peano-axioms – Ethan Bolker Apr 29 '22 at 21:25
  • @CarefreeXplorer yes that will do it! Thank you! – Teun van Wezel Apr 29 '22 at 21:36
  • This follows from the axioms that $n++ \neq 0$, that $m++=n++ \to m= n$, and the inductive axiom. Define $P(n)$ as $n++ \neq n$. By the first axiom you have $0++ \neq 0$, and so you have $P(0)$. With the inductive hypothesis that $P(k)$ (i.e. $k++ \neq k$, you now also have that $P(k++)$, since if you had $(k++)++ = k++$, then by the second axiom you would have $k++ = k$, contradicting the inductive hypothesis. So you now also have $P(k) \to P(k++)$. So with the inductive axiom, you have $\forall n \ P(n)$. – Bram28 May 03 '22 at 19:19

0 Answers0