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If I have $a,b \in \mathbb R$ such that $$a > b+1 $$ It is assured that $\exists\space x \in \mathbb Z: a>x>b$

Does this property have some special name? How can this be proved?

This idea isn't intuitive to me. Infact, it feels untrue. If so, when is it true.

Please help.

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    $\begingroup$ I guess taking $\lfloor b+1\rfloor$ will work though not sure $\endgroup$
    – kingW3
    Commented Sep 26, 2014 at 19:18
  • $\begingroup$ @kingW3: So, by what you're saying: $x \equiv \lfloor b+1\rfloor \equiv \lceil a \rceil$ $\endgroup$
    – Nick
    Commented Sep 27, 2014 at 10:47

3 Answers 3

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Since $a-b-1\gt 0$, there exists a real number $\alpha\gt 0$ such that $a=b+1+\alpha$. Then, since there exists an integer $m$ such that $m-1\le b\lt m$, we have $$b\lt m\lt m+\alpha\le b+1+\alpha =a,$$ which implies that $m$ satisfies $b\lt m\lt a$.

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  • $\begingroup$ You are using the fact that every real number lies in between integers? Why did you choose $m-1\le b\lt m$ instead of $m-1\lt b\le m$ ? $\endgroup$
    – Nick
    Commented Sep 26, 2014 at 19:33
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    $\begingroup$ I use that fact that every real number is either an integer or lies in between integers. You can use $m-1\lt b\le m$ if you want. You can get the same result if you use it. $\endgroup$
    – mathlove
    Commented Sep 26, 2014 at 19:36
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Draw this property in the Real Line. By the hypotessis, you have that a-b > 1, so, you can find an integer that is between them, in fact, integer part of b as your x works for your problem, in any case.

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    $\begingroup$ Only if $b$ is not itself an integer! $\endgroup$
    – user164587
    Commented Sep 26, 2014 at 19:08
  • $\begingroup$ sorry, is the integer part of b, plus 1 $\endgroup$
    – Julio
    Commented Sep 26, 2014 at 19:32
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The referenced proof is correct, but it can be easily made more clear.

Suppose $r,s \in \mathbb R$ and $r \gt s$ then $r - s \gt 0$. Then $\frac1{r-s} \gt 0 \implies \frac2{r-s} \gt 0$. (That was the change.)

Now, there exists an integer $p$ such that $$p \gt \frac2{r-s} \implies p(r-s) \gt 2 \implies pr \gt ps +2$$

Then there exists another integer $q$ such that $$pr \gt q \gt ps \implies r \gt \tfrac{q}{p} \gt p$$

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