$a > b+1 \Rightarrow a>x>b$?

Question

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.

Answer 1

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 $b\lt m\lt a$.

Answer 2

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.

Answer 3

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 $$pr \gt q \gt ps \implies r \gt \tfrac{q}{p} \gt p$$