I have read Edward Nelson's Warning signs of a possible collapse of contemporary mathematics a couple of times, it is a very interesting read, but I do not understand the conclusory paragraph. In particular I am interested in the final 'warning sign', beginning from the fable on page 8 and continuing through to the end (all relevant information for my question is copied below).

The belief that exponentiation, superexponentiation, and so forth, applied to numerals yield numerals is just that—a belief.

Note: we are working in the standard 7 axioms of PA for this question.

Numerals are defined as follows. $0$ is a numeral; if $x$ is a numeral, so is $\text{S}x$. In a standard way, we introduce addition: $$\begin{align} x+0 &=x,\\ x+\text{S}y &=\text{S}(x+y), \end{align}$$ multiplication: $$\begin{align} x\cdot 0 &=0,\\ x\cdot\text{S}y &=x+(x\cdot y), \end{align}$$ and exponentiation: $$\begin{align} x\uparrow 0 &=\text{S}0,\\ x\uparrow\text{S}y &=x\cdot(x\uparrow y). \end{align}$$ Then we define counting numbers, introducing two new axioms to our system:

  1. $0$ is a counting number.
  2. If $x$ is a counting number, then $\text Sx$ is a counting number.

additionable numbers:

$x$ is an additionable number in case for all counting numbers $y$, the sum $y + x$ is a counting number.

and finally, multiplicable numbers:

$x$ is a multiplicable number in case for all additionable numbers $y$, the product $y \cdot x$ is an additionable number.

We prove that additionable and multiplicable numbers are closed—i.e. if $x_1$ and $x_2$ are additionable (respectively, multiplicable) numbers, then $x_1+x_2$ is additionable (respectively, $x_1\cdot x_2$ is multiplicable).

But now we come to a halt. If we attempt to define 'exponentiable number' in the same spirit, we are unable to prove that if $x_1$ and $x_2$ are exponentiable numbers then so is $x_1 \uparrow x_2$. [...] The belief that exponentiation, superexponentiation, and so forth, applied to numerals yield numerals is just that—a belief.

It is at this point that I lose the author. If I understand correctly, he is saying that given two numerals $x_1$ and $x_2$ (i.e. $\text S\text S\text S\ldots 0=x_1$ for some number of $\text S$s, and similarly for $x_2$), then $$x_1\uparrow x_2=\underbrace{x_1\cdots x_1}_{x_2\text{-times}}$$ is not a numeral (i.e. we cannot definitely say that there is some number of $\text S$s for which $x_1\uparrow x_2=\text S\text S\text S\ldots 0$). If this is not what was intended, what is the author's point at the end here?

However, if I understand right, then I am not convinced; below I demonstrate my attempt to show this given some $x_1\uparrow x_2$. Can we not 'break this down' as follows? (From here I use the notation $\text S^k 0$ for the numeral $k$, that is the numeral $\underbrace{\text{SS}\ldots\text S}_{k\text{-times}}0$.)

$$\begin{align} x_1\uparrow x_2&=\underbrace{x_1\cdots x_1}_{x_2\text{-times}}\\ &=\underbrace{(\text S^{x_1} 0)\cdots (\text S^{x_1} 0)}_{(\text S^{x_2} 0)\text{-times}}\\ &=(\text S^{x_1} 0\cdot \text S^{x_1} 0)\cdot\underbrace{(\text S^{x_1} 0)\cdots (\text S^{x_1} 0)}_{(\text S^{x_2}0-2 )\text{-times}}, \end{align}$$ now apply the definition of multiplication to $\text S^{x_1} 0\cdot \text S^{x_1} 0$ to find $$\begin{align} \text S^{x_1} 0\cdot \text S^{x_1} 0&=\text S^{x_1} 0+(\text S^{x_1} 0\cdot \text S^{x_1-1} 0)\\ &=\text S^{x_1} 0+\text S^{x_1} 0+(\text S^{x_1} 0\cdot \text S^{x_1-2} 0)\\ &=\cdots=\underbrace{\text S^{x_1} 0+\text S^{x_1} 0+\cdots+\text S^{x_1} 0}_{x_1\text{-times}}+(\text S^{x_1} 0\cdot 0)\\ &=\underbrace{\text S^{x_1} 0+\text S^{x_1} 0+\cdots+\text S^{x_1} 0}_{x_1\text{-times}}. \end{align}$$ Do this a finite number of times to put $x_1\uparrow x_2$ in the form of an addition; at which point applying the definition of addition a finite number of times should yield $$x_1\uparrow x_2=\underbrace{\text{SS}\ldots\text S}_{x_1^{x_2}\text{-times}}0.$$ Now $x_1\uparrow x_2=\underbrace{\text{SS}\ldots\text S}_{x_1^{x_2}\text{-times}}0$ is a numeral by induction.

The claim 'that exponentiation, superexponentiation, and so forth, applied to numerals yield numerals is just that—a belief' is a strong one, what does Nelson mean and how does he conclude this?

Thank you!

  • 1
    $\begingroup$ It's just Nelson expressing his ultrafinitist position. Because there is no generally well-regarded grounding for ultrafinitism - unlike ordinary finitism, which has no problems with exponentiation, or constructivism, which also accepts exponentiation - there is not much to say other than: the argument comes from Nelson attempting to explain his personal viewpoint on the matter. $\endgroup$ – Carl Mummert Oct 24 '16 at 14:23
  • $\begingroup$ I kinda have to agree with @CarlMummert here. I see no reason that if he accepts the addition and multiplication schemes that he would find the exponentiation scheme unacceptable. It's literally is to multiplication exactly what multiplication is to addition. I suppose the only negative thing about is that, unlike addition and multiplication, exponentiation is not a commutative operator on 'counting numbers' (e.g. $a^b\neq b^a$ in general). $\endgroup$ – Justin Benfield Oct 24 '16 at 14:28
  • $\begingroup$ I think I see his reasoning at least (I may take issue with what it's predicated on), Thm 14 uses associativity, and addition and multiplication have that property, exponentiation does not. Hence the proof technique used for thm 14 no longer works for exponentiation. $\endgroup$ – Justin Benfield Oct 24 '16 at 14:38
  • $\begingroup$ A key part of his technique is to work with a theory - which is supposed to be a theory of natural numbers - so that the theory includes a concept of "counting number" with an axiom that $0$ is a counting number, and an axiom that the counting numbers are closed under successor, but the induction axiom implying that every number is a "counting number" is intentionally not included. Then he defines "additionable", etc., in terms of "counting numbers", which makes it harder to prove things about these concepts by induction. Overall the theory he is working with is very idiosyncratic. $\endgroup$ – Carl Mummert Oct 24 '16 at 14:46
  • 1
    $\begingroup$ Nelson sounds like the kooks at sci.math. With the usual 5 Peano Axioms + set theory you can construct, i.e. prove the existence of functions that seem to perfectly model the usual addition, multiplication, and exponentiation functions on the natural numbers. With only the possible exception of $0^0$, $x^y$ can be determined for any natural numbers $x$ and $y$. $\endgroup$ – Dan Christensen Oct 25 '16 at 3:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.