I am trying to prove that if $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $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 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 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.
If $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $x
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0Looks correct. Are you able to justify why an integer $m$ with $nx < m \le 1+nx$ exists? – 2012-12-10
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0I think that is where I have some difficulty. – 2012-12-10
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0Let $m$ be the smallest natural number, which satisfies $nx < m$ (exists because of well ordering). Then $m - xn <= 1$ because otherwise is $xn < m-1$. – 2012-12-10
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0OK, 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 $\mathbb{N}$. – 2012-12-10
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0For 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
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 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
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.