$x, y\in R$, and $y>x$, prove: there exist $p\in Q$, such that $x<p<y$


Since $y>x$ is equvalent with $y-x>0$, by Archimede's property, there exists positive integer $n$, such that


why not $n (y - x) > 0$ since $y - x > 0$, how it becomes $1$ suddenly

and there exists positive integers $m_1,m_2$, such that $> n x,m_2>-n x$,i.e.

$$\begin{align*}-m_2<n x<m_1.\tag{2}\end{align*}$$

[How to think up this?, confusing at Minus $m_2>-n x$]

so, there exists positive integer $m\left(-m_2\leq m\leq m_1\right)$, such that

$$\begin{align*}m-1\leq n x<m.\tag{3}\end{align*}$$

following I understand,keypoint is how to understand and construct (3)

Hence $n x < m \leq 1 + n x < n y$.

Because n$>$0, we have $x<\frac{m}{n}<y$.

$p=\frac{m}{n}$ satisfy the condition.

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    $\begingroup$ Did you look up "Archimedean property?" $n(y-x)>0$ for all natural $n$. But there also must be some $n$ such that $n(y-x)>1$. That follows from the Archimedes property. $\endgroup$ – Thomas Andrews Jul 24 '13 at 6:38
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    $\begingroup$ Add any positive number to itself enough times and, no matter how small it originally was, eventually we will get to numbers greater than $1$. $\endgroup$ – anon Jul 24 '13 at 6:40
  • $\begingroup$ @ThomasAndrews ok, so change $1$ to any other positive numbers. That is also true? $\endgroup$ – User19912312 Jul 24 '13 at 6:42
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    $\begingroup$ For some $n$, it will be true. Not the same $n$ as for $1$ necessarily. $\endgroup$ – Thomas Andrews Jul 24 '13 at 6:44
  • $\begingroup$ @ThomasAndrews Hi, can you help me understand and construct (3). $\endgroup$ – User19912312 Jul 24 '13 at 7:08

For any positive real $w$, and $z$ another real there is $n\in\Bbb N$ such that $nw>z$. We're applying the claim with $w=y-x>0,z=1$.

Recall that given any real number $x$, we define $n=\lfloor x\rfloor$ to be the greatest integer such that $n\leq x <n+1$.

Now, I claim that if $b-a>1$, there exists an integer $m'$ with $a< m'<b$. Indeed, let $m=\lfloor a\rfloor$. Then since $-m \geq -a$ we must have $b-m\geq b-a>1\implies b> m+1$ so, since $a< m+1$, $$a< m+1< b$$

and the claim is proven with $m'=m+1$. Now, since $nx-ny>1$, there must exist an integer $m'$ such that $ny<m'<nx$, so...?

  • $\begingroup$ Hi, can you help me understand and construct (3). $\endgroup$ – User19912312 Jul 24 '13 at 7:07
  • $\begingroup$ @spuorg-imes I added something. $\endgroup$ – Pedro Tamaroff Jul 24 '13 at 7:29

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