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What properties of the positive integers and $1$ guarantee that decreasing by $1$ eventually leads to $1$?

$f(x)=x-1$

Show that for any positive integer $m>1$, there exists $n$ such that $f^n(m)=1$

I have seen it simply stated that $f(x)<x\quad\forall x\implies\exists\quad n\mid f^n(x)=1$

But that's not formally sufficient, is it? We need a few other things. For example, this argument doesn't state its reliance upon the properties of $\mathbb{N}$ or $1$, but if we were in $\mathbb {Q}$ we might reduce the number $3$ repeatedly by $\frac{2}{3}$ and we would never arrive at $0$.

Firstly, we need to state that $1$ is in some sense the minimum resolution of the integers under addition, in order to show that no step of reducing by $1$ can leapfrog any integer.

Secondly (or perhaps this encapsulates the first) we need to know that the integers are ordered by the function $x\to x+1$ and that $1$ is the least of them by that ordering.

Thirdly we need to state that there are countably many whole numbers, to ensure that we will eventually arrive at $1$.

What is the conventional way of stating these properties, particularly the first?

Bill Dubuque
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it's a hire car baby
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  • Perhaps you can use $\mathbb Z= \langle 1\rangle$. – GuPe May 08 '17 at 14:59
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    Follows from the principle of induction, often stated as an axiom, but which in turn follows from the well-orderedness of the natural numbers. (If I understand your question properly.) – Brian Tung May 08 '17 at 15:00
  • @BrianTung it's particularly the property of $1$ that subtraction by $1$ cannot leapfrog any integer that I'm interested in. For example in $\mathbb{Q}$, starting with $10.5$, we can subtract $1$ repeatedly and never arrive at $1$. Or in $\mathbb{N}$ we can subtract $2$ repeatedly and never arrive at $1$. – it's a hire car baby May 08 '17 at 15:05
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    Do you mean "For all $m$, there exists $n$ ..."? If so, then you can explicitly take $n=m-1$ and you are done. If you mean "For some $m$ ..." then take $m=17$ and $n=16$. For the more general ideas in here, read up on well-orderings. – Ned May 08 '17 at 16:23
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    @RobertFrost: Note that $\mathbb{Q}$ is not well-ordered (with the usual topology), while $\mathbb{N}$ is. That is central to the distinction, I think. – Brian Tung May 08 '17 at 16:47
  • @BrianTung oh ok, thanks. – it's a hire car baby May 08 '17 at 16:49

1 Answers1

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Firstly, as already pointed out by Ned, you are asking the wrong question; I suggest you start on an introductory logic textbook and learn the meaning of quantifiers and note the difference between: $ \def\nn{\mathbb{N}} $

(A) $\exists m \in \nn_{>1}\ \exists n \in \nn\ ( f^n(m) = 1 )$.

(B) $\forall m \in \nn_{>1}\ \exists n \in \nn\ ( f^n(m) = 1 )$.

(A) is what you have in your third line, which is trivial because $2 \in \nn_{>1}$ and $0 \in \nn$ and $f^0(2) = 1$. (B) is what you are actually asking.

Secondly, it is a trivial application of induction, but as your first time encountering formal inductive proofs, it would be instructive for you to write out the argument in full. Here is the outline for you to fill in:

Define $P(m) \equiv ( m>0 \to \exists n \in \nn\ ( f^n(m) = 1 ) )$.

Then $P(0)$ because $\cdots$.

Given any $m \in \nn$ such that $P(m)$:

  $m = 0$ or $m > 0$.   [If you don't have this fact, prove it by induction!]

  If $m = 0$:

    $P(m+1)$ because $f^0(1) = \cdots$.

  If $m > 0$:

    $\exists n \in \nn\ ( f^n(m) = 1 )$.

    Let $c \in \nn$ such that $f^c(m) = 1$.

    Then $c+1 \in \nn$ and $f^{c+1}(m+1) = \cdots$.

    Thus $\exists n \in \nn\ ( f^n(m+1) = 1 )$.

  Therefore $P(m+1)$.

Therefore by induction $\forall m \in \nn\ ( P(m) )$.

And being countable is irrelevant, by the way. The concept generalizes to well-orderings, which are totally-ordered collections such that every element can be decreased only finitely many times (but in a general well-order there is no notion of subtraction). Depending on the specific foundational system, there may be uncountable well-orderings. For the general construction of arbitrarily long well-orderings in ZFC set theory, see this post.

user21820
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  • Thanks for the proof. I think it's a case of practice logic rather than learning logic. Am I right in thinking my $3$rd line would be fixed by $f(x)<x\quad\forall x\implies\forall x\exists\quad n\mid f^n(x)=1$ ? An oversight. – it's a hire car baby May 09 '17 at 11:29
  • @RobertFrost: Well I consider it learning until you reach the point of 100% precision. And no it won't be fixed by your suggested line. There is a specific syntax in first-order logic; you cannot anyhow put quantifiers anywhere you like. Also, you must use restricted quantifiers to specify their domains. Finally, use \to or \Rightarrow for implication in LaTeX, not \implies (it's the wrong symbol). – user21820 May 09 '17 at 11:51
  • @RobertFrost: By "third line" I meant "Show that for some positive integer $m>1$, there exists $n$ such that $f^n(m)=1$", which is what both Ned and Bram28 already pointed out but you didn't respond to them on that. – user21820 May 09 '17 at 11:53
  • I can't see Bram28 but Brian Tung mostly answered it with the well-ordered property of the integers. But that needs to be augmented with the fact that $1$ is the smallest increment. If you substituted $2$ in your proof it wouldn't be true. I'm enquiring what the name is for that property of $1$ over the integers, that it's the smallest step between $2$ numbers. – it's a hire car baby May 09 '17 at 12:33
  • @RobertFrost: Sorry it was Brian; I somehow thought it was Bram28 since he answers a lot of logic questions... I don't think there is more to say than that you can see precisely what properties of natural numbers are used where in the proof. Like if you tried with decreasing by $2$ it is clear it fails because induction requires you to prove $P(m) \to P(m+1)$ for each natural $m$. – user21820 May 09 '17 at 13:51
  • And well-ordering is a completely separate thing from subtraction. Well-ordered collections are simply well-ordered and there may be completely no notion of subtraction. It just so happens that naturals have both subtraction and well-ordering. – user21820 May 09 '17 at 13:52