Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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. –  kigen 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

3 Answers 3

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 $a \in \mathbb{N}$.

share|improve this answer
    
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 $1$.

share|improve this answer

Hint: Try by contradiction and use cancellation.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.