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The problem is,

Prove that $n<m{++}\leq n{++}\iff m=n$.

Using only the followings,

  1. Peano Axioms (see the axioms here).

  2. Definition of Addition: Let $m$ be a natural number. We define, $0 + m = m$ and suppose we have inductively defined the addtion $n + m$ then we define, $(n{++})+m=(n+m){++}$. Where $n{++}$ is the successor of $n$.

  3. Commutativity, Associativity and Cancellation Laws of Addition.

  4. Definition of Positivity: A natural number $n$ is said to be positive if $n\neq 0$.

  5. Definition of $\ge$ and $>$: Let $m$ and $n$ be two natural numbers. We say $n\ge m$ or $m\le n$ if there exists some natural number $a$ such that $n=m+a$. We say $n>m$ or $m<n$ if $n\ge m$ (or $m\le n$) but $n\ne m$.

While trying to prove that for natural numbers $a$ and $b$ if $a<b$ then $a{++}\le b$ I found out that to prove that proposition I had to assume the fact that (speaking loosely) there exists no natural number between any natural number and its successor. I thought that this should be an axiom because even if it weren't true it wouldn't contradict the Peano Axioms. However, I am skeptic of my assertion and so I decided to post it as a problem. I will be glad if someone can show me a proof of this assertion using only the statements I have given.

Added:-

Note that, this question mainly focuses on the fact that whether the Peano Axioms are sufficient to let us conclude that, loosely speaking,

There exists no natural number in between a natural number and its successor.

And so we can claim that the natural number that the axioms are sufficient to construct our "wanted" natural numbers.

Please note that before adding any more answer to the question, please read the chain of comments below Sebastian G's answer.

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1 Answers1

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First prove by induction: $$ \begin{array}{ll}(1) && \forall n :n=0 \lor n=1 \lor n>1\\ (2) &&\forall n : n= 0 \lor \exists z : z++=n \end{array}$$

Let $P(n,m):= n<m++\leq n++ \Rightarrow m=n$

Now show the statement $\forall n\forall m P(n,m)$ by induction on $n$:

Base Case $(n=0)$: $\forall m P(0,m)$

Let $m$ be arbitrary and assume $0<m++\leq 0++=1$

By (1) $m++=1$ so $m=0$

Inductive step: $\forall m P(n,m) \Rightarrow \forall m P(n++,m)$

By inductive assumption we have $\forall m :n < m++ \leq n++ \Rightarrow m=n$.

We have to show $\forall m :n++ < m++ \leq (n++)++ \Rightarrow m=n++$

For that let m be arbitrary and assume $ n++ < m++ \leq (n++)++$

By commutativity of addition/cancellation laws: $ n < m \leq n++$

By (2) pick $z$ such that $z++=m$ (it is easily shown that $m \neq 0$)

But by inductive assumption $n < z++ \leq n++ $ implies $z = n$

Therefore $m=z++=n++$

abc
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  • I don't understand the sixth line. Instead of $m{++}$, I think it should be $(m{++}){++}$. –  Dec 28 '14 at 12:46
  • @user170039 No, Let $ P(n,m) = n < m++ \leq n++ \Rightarrow m=n$, you want to show $ \forall m \forall n P(n,m)$ I did this by showing $\forall m P(n,m)$ by induction over n. So the base case is $\forall m P(0,m)$ and the inductice step is $ \forall m P(n,m) \Rightarrow \forall m P(n++,m)$ – abc Dec 28 '14 at 14:55
  • If you apply induction on $n$ then $m$ must be fixed. Then the transition from $P(n,m)$ to $P(n{++},m)$ isn't very clear to me. It will be good if you can explain every step a bit elaborately. –  Dec 29 '14 at 04:25
  • @user170039 the transition goes from $\forall m P(n,m)$ to $\forall m P(n++,m)$ see edits – abc Dec 29 '14 at 05:00
  • Just one more question. How do you prove the base case of $(1)$? Let me elaborate what I have done. Basically $(1)$ equivalent to $\forall n: n=0\lor n\geq 0{++}$. "We apply induction on $n$. We let, $P(n):n=0\lor n\ge 0{++}$. Now note that $P(0)$ holds since in this case $n=0$. So now we inductively assume $P(n)$ to be true, i.e., $n=0\lor n\ge 0{++}$. To prove $P(n{++})$, i.e., $n{++}=0\lor n{++}\ge 0{++}$ we note that by Axiom 1.3 $a{++} \ne 0$." After that I can't proceed. Can you help? –  Dec 29 '14 at 05:47
  • And besides the assumption of $(1)$ to prove my problem seems circular to me. How do you know (speaking informally) that there doesn't exist any natural number in between $0$ and $1$? Note that it is also the base case for my problem as in $(1)$ of yours. –  Dec 29 '14 at 06:04
  • @user170039 It's not circular, you prove the special case (1) first and then having done this proceed to the more general case.

    This is for example how you prove commutativity, too. First prove inductively that m+0=0+m and use this as base case for inductively proving the general case m+n=n+m.

    The proof of (1) is very easy. the case n=0 is trivial. so assume (1) is true for n: 1.case n=0, then n++=1 so (1) is true for n++ 2.case n=1 then n++>1 so (1) is true for n++ 3.case n>1 then n++>n>1 so (1) is true for n++

     – abc Dec 29 '14 at 06:12
  • That doesn't address my question that how do we know that there doesn't exist any natural number in between $0$ and $1$? The inductive step that if $n\ne 0$ then $n\ge 1$ can only be written if we are sure that the "next" natural number after $0$ must be $1$. And besides in case of Commutativity we do prove the way you told but in this case $m$ is assumed to be some arbitrary but fixed natural number. It's more rigorous to apply the principle of Double Induction to it but since the processes are exactly similar, it is usually skipped. –  Dec 29 '14 at 06:15
  • Because we just proved it, i don't see the problem – abc Dec 29 '14 at 06:16
  • Thats the definition of "1", 1 is defined to be the successor of 0, i.e. 0++ – abc Dec 29 '14 at 06:19
  • In terms of Peano Axioms, $1$ is defined to be the number $0{++}$ (correct me if I am wrong) but how does this imply that there doesn't exist any natural number in between $0$ and $1$? –  Dec 29 '14 at 06:23
  • It doesnt, but I proved it above using this definition. You have written "The inductive step that if $n \neq 0$ then $n \geq 1$" this is wrong, thats not what the inductive step is. The inductive step is as follows: IF it is true for some number n that if$n \neq 0$ then $n \geq 1$ THEN this is also true for its successor n++ – abc Dec 29 '14 at 06:27
  • Yes. I have also noted that I have written that line wrong. But when I wanted to edit it, it was already 5 minutes. Anyway, in your statement "IF it is true for some number $n$ that if $n≠0$ then $n≥1$ THEN this is also true for its successor $n++$.", I object regarding the part that "...if $n≠0$ then $n≥1$..". –  Dec 29 '14 at 06:34
  • You have to note its a conditional statement! Its NOT claiming it is true for some n that " $n \neq 0 \Rightarrow n \geq 1$ " but it is claiming if it WERE true for some n THEN it must be true for n++ – abc Dec 29 '14 at 06:44
  • and since obviously $ 0 \neq 0 \Rightarrow 0 \geq 1 $ Axiom 2.5 gives you (1) – abc Dec 29 '14 at 06:51
  • Actually I am having trouble to prove that there doesn't exist any natural number $k$ such that $0<k<0{++}$. –  Dec 29 '14 at 06:53
  • That's exactly what (1) tells you. k has to be equal to 0, 1 or strictly above 1. – abc Dec 29 '14 at 07:00