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I'm trying to prove that given $n\in \mathbb N$, there isn't a natural number such that $n\lt x\lt n+1$, using the axioms of the natural numbers and the definition of $\lt$ ($m\lt n$ iff $n=m+p$, $p\in \mathbb N$). I've already proved associativity, commutativity and cancellation law of the natural numbers. So we can use this to prove this question, I need help here.

By the way, anyone knows where can I find more exercises like this one in order to train the axioms and first definitions of numbers?

Thanks a lot

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Do you have additive inverses? – Clayton Jan 16 '13 at 4:04
@Clayton That's cancellation. Inquest, That's not what the OP is asking, though I admit I thought it was at first as well. – Gyu Eun Lee Jan 16 '13 at 4:07
@proximal: Thanks; I wasn't sure whether to consider multiplication or not. – Clayton Jan 16 '13 at 4:07
@Clayton that's my problem, I don't have additive inverses, if I had, it would be easy. – user42912 Jan 16 '13 at 4:17
up vote 1 down vote accepted

Edited to reflect new information.

Since $x > n$, if we assume $x \in \mathbb{N}$ then we can write $x = n + a$ for some $a \in \mathbb{N}$. Then $n < n + a < n + 1$.

Proceed by cases on $a$ to reach a contradiction for any $a \in \mathbb{N}$.

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What I did so far: since $n\lt x\lt n+1$, then there are natural numbers $a, b$ such that $n+1=x+a$ and $x=n+b$, am I in the right way? – user42912 Jan 16 '13 at 4:33
thank you a lot, now I got the contradiction! :) – user42912 Jan 16 '13 at 5:32

The key is to prove that there are no natural numbers between $0$ and $1$. You will then also need the following lemmas:

Lemma 1: If $a < b$ and $c \leq a$, $a-c < b-c$.

Lemma 2: If $a < b$ then $a^2 < b^2$.

Lemma 3: If $a < b$ and $c>0$, then $ac < bc.$

Let $S= \{a\in \mathbb{N} \vert 0 < a < 1\}$ and assume, for contradiction, that $S$ is nonempty. Since $\mathbb{N}$ is well-ordered, $S$ has a least element $b$. By Lemma 2, $$0 < b^2 < 1,$$ so $b^2 \in S$. Moreover, by Lemma 3, $b^2 < b$, a contradiction, since $b$ is the least element of $S$. Therefore $S$ is empty and there are no natural numbers between $0$ and $1$.

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Hint: Try by contradiction and use cancellation.

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I'm trying this, but I don't know how to use the cancellation in this case, since $n\lt x\lt n+1$, then there are natural numbers $a, b$ such that $n+1=x+a$ and $x=n+b$, am I in the right way? – user42912 Jan 16 '13 at 4:21

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