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I am trying to prove that if $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $x<y$,then there exists a rational $p$ such that $x<r<y$.

Here is what I tried to do:

I am trying to show that given $x$,$y$,there exist $m$ and $n$ such that $x<r=\frac{m}{n}<y$ where $m$ and $n$ are naturals or $$nx<m<ny$$

By the Archimedean property, as $x<y$,

$y-x>0$ so $n(y-x)>1$ for some $n\in \mathbb{N}$.So, $ny>1+nx$.

There must also be an integer $m$ between $nx$ and $1+nx$ so that $nx<m\leq 1+nx$

Which gives $nx<m<ny$, which completes the proof.

Is this proof correct?

I just read Arkeet's comment and I think the statement "There must also be an integer $m$ between $nx$ and $1+nx$" needs a formal proof.I don't think I have one at hand.

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Looks correct. Are you able to justify why an integer $m$ with $nx < m \le 1+nx$ exists? –  arkeet Dec 10 '12 at 15:30
    
I think that is where I have some difficulty. –  Richard Nash Dec 10 '12 at 15:33
    
Let $m$ be the smallest natural number, which satisfies $nx < m$ (exists because of well ordering). Then $m - xn <= 1$ because otherwise is $xn < m-1$. –  manado Dec 10 '12 at 15:36
    
OK, the set of integers larger than $nx$ has a smallest element. I just noticed, however, that unless $x$ is nonnegative, $m$ may have to be negative (so not a natural number), so you might have to do a bit more work as the set of integers larger than $nx$ may not be a subset of $\mathbb{N}$. –  arkeet Dec 10 '12 at 15:38
    
For negatives you should say something. If $x$ and $y$ are negative, multiply by $-n$, etc. Otherwise $y$ is positive and we need to find $n$ and $m$, such that $0 < m \leq ny$. Should I remove my "answer"? –  manado Dec 10 '12 at 15:41
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2 Answers

up vote 1 down vote accepted

The part of the proof that you are missing (mentioned in Arkeet's comment) is essentially the existence of the integer part, that is if $x\in \mathbb{R}$ then $\exists k\in \mathbb{N}:k<x< k+1$. (apply this to $nx$ and you are done).

Proof of the existence of integer part:

Let $x\in \mathbb{R}$ and $S=\left\{ k\in \mathbb{Z} :k>x \right\}$. Then $S \subseteq \mathbb{Z}$ is bounded below by $x$ and so $S$ has a least element, $k_0\in \mathbb{Z}$. Then $k_0-1\notin S\Rightarrow k_0-1\le x$ while $k_0\in S\Rightarrow k_0>x$. Therefore by letting $k=k_0-1$ we have that $k<x<k+1$. That $k$ is called the integer part of $x$

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This seems valid for me as long as you rely on knowledge about the embedding of natural numbers into the real numbers and $x$ and $y$ are positive. You still need to say, what happens if $x < y \leq 0$ or $x < 0 < y$.

As an alternative to this approach, you could have used that $\mathbb Q$ is dense in $\mathbb R$ or that real numbers are suprema of subsets in $\mathbb Q$. It depends a little how you defined the real numbers or what you already assume to know.

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