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

Here is what I tried to do:

I am trying to show that given $x$,$y$,there exist $m$ and $n$ such that $x where $m$ and $n$ are naturals or $nx

By the Archimedean property, as $x,

$y-x>0$ 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

Which gives $nx, 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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    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"?2012-12-10

2 Answers 2

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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. (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. 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.