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I am reading Tao's book on Analysis in which the first two axioms apropos natural numbers are,

  1. 0 is a natural number.

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

As a motivation Prof. Tao states prior to these axioms,

On the other hand, incrementing seems to be a fundamental operation, not reducible to any simpler operation; indeed, it is the first operation one learns on numbers, even before learning to add. Thus, to define the natural numbers, we will use two fundamental concepts: the zero number 0, and the increment operation.

Question :

  1. Does incrementing here, mean incrementing by the 'quantity' 1

    or

  2. if incrementing implies just going to the next successor or 'next thing', whatever it may be?

I will be grateful if the answer is dumbed down as much as possible. Assume no other mathematical/logic knowledge besides the aforementioned text by Prof. Tao.

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  • Successor function: "the successor function or successor operation sends a natural number to the next one. " – Mauro ALLEGRANZA Sep 07 '23 at 07:40
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    It means going to the next one. It will be a theorem that this is equivalent to adding $1$, once addition has been defined. – Qiaochu Yuan Sep 07 '23 at 07:51
  • Bear in mind that this is a description of the motivation for the approach being taken - whether the motivation is captured by the description will emerge - or there may be extra rules necessary to capture motivation (eg no gaps). So the motivation is simply counting or ordering - being able to find the next object in the sequence. Motivation does not require precision. – Mark Bennet Sep 07 '23 at 09:09
  • This question may be useful. – Jam Sep 07 '23 at 12:32

1 Answers1

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With Axiom1 & Axiom2 , we have to take $n++$ to indicate some "Successor" Natural Number , not necessarily the "Next" Natural Number.

Here , $0++$ might be the Next Natural Number ( which we know as $1$ ) , it might be the Previous Natural Number ( which we know as $-1$ ) , it might jump over a Number ( eg $0.4$ & $0.5$ ) , it might even be $0$ itself !

Preventing going back to $0$ ( & going back to other Natural Numbers ) requires Axiom 3 & Axiom 4.

Preventing jumping over some Natural Numbers will require Axiom 5 to claim that $++$ gives all Natural Numbers & nothing is jumped over.

In other words , we can not have $n++ = 0$ & $n++ = n$ & $5++=9$ & $86++=42$ , which are allowed by Axiom 2.

When we Define Addition , we might claim $n++ \equiv n+1$.

"Does incrementing here, mean incrementing by the 'quantity' 1"
Incrementing means getting the "Successor" , what-ever that may be.
Eventually , it can be shown that it is Equivalent ( Isomorphic ) to Adding 1 : $n++=n+1$.

"if incrementing implies just going to the next successor or 'next thing', whatever it may be?"
Yes , it gives the "Successor". Eventually , it is shown that the "Successor" is actually the "Next" thing , though it requires Axiom 3 , Axiom 4 & Axiom 5 to achieve that.

Knowing the outcomes & the requirements "apriori" , the Descriptive Names "Successor" & "Increment" were chosen to indicate the $++$ , otherwise , hypothetically it could have been $X(n)$ or $P(n)$ or $n\uparrow$ or $n\#$ which are not very indicative of the Intent.

When we try plugging in various numbers like $(0.4)$ & $(0.5)$ & $(1/\sqrt{2})$ & $(e-2)$ & $(\pi-3)$ between $0$ & $1$ , then $0++ \equiv 0+1 = 1$ is jumping over these Numbers , but it is not jumping over Natural Numbers.

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