2

In Tao's Real Analysis he defines addition:

Let $m$ be a natural number. To add zero to $m$, we define $0+m \equiv m$. Now suppose inductively that we have define how to add $n$ to $m$. Then we can add $n++$ to $m$ by defining $(n++) + m \equiv (n+m)++$.

Later he states:

Right now we only have two facts about addition: that $0+m=m$, and that $(n++)+m = (n+m)++$.

Do we not also get from this simple addition? $n + m = z$ for example.

Chris
  • 863
  • What do you mean by "getting simple addition"? Tao does not say "nothing else can be done", he only states what has been observed so far. Finally, this is not a real analysis question, even though the title of the book is such. –  Dec 27 '15 at 21:41
  • what is z here? – TomGrubb Dec 27 '15 at 21:43
  • I do not understand the question... are you saying $n+m$ could be any natural number? – Forever Mozart Dec 27 '15 at 21:44
  • Yes, he just defines how addition is done, regarding Peano axioms. and your question unclear. Where is your doubts? –  Dec 27 '15 at 21:45
  • Sheesh this response makes me not want to ask here anymore. Yes there is ambiguity. I'll delete this. – Chris Dec 27 '15 at 22:53
  • 1
    Example: Prove 2+3=5. Using the first rule, you have 0+3=3. Apply the 2nd rule to this statement to obtain, 1+3=4. Apply the 2nd rule again to this statement in turn to obtain 2+3=5. PS: There is nothing ambiguous about your question. Don't let them get to you. There is no need to delete your question. – Dan Christensen Dec 31 '15 at 03:11

1 Answers1

3

If I understand correctly, the answer is yes: These axioms allow you to calculate $m+n$ for every pair of natural numbers $m, n$ (viewing $m, n$ as expressions of the form "$(...(0++)++)++)...)$").

To see this, consider the following example: $$2+2=[1+2]++=([0+2]++)++=[2++]++=(((0++)++)++)++.$$ (Here I use e.g. "$2$" as shorthand for "$(0++)++$" in the obvious way.)

The strategy to compute $m+n$ is: apply the second axiom $m$-many times to get $(0+n)++...++$, and then apply the first axiom once. It's a good exercise to show by induction that this always works. (Note that this proof will take place in the metatheory, since it relies already on a notion of "natural numbers".)

Note that a more common notation for "$x++$" is "$S(x)$"; I've used the former (since that's what the OP uses), but I strongly prefer the latter since it allows us to write things like "$SSSSSSSS(0)+SSSS(0)$" unambiguously and (relatively) shortly.

Noah Schweber
  • 245,398