Prove that for every $xSo I came across this problem and here is my proof, is it valid? I really liked it so I wanted to share it.
We will only use two properties, $\mathbb{N}$ is not bounded from above and $\mathbb{N}$ is well ordered.
Just like many proofs of this problem, we begin with a positive interval.
Proposition:
If $x,y\in \mathbb{R}^+$ such that $x<y$, then there is a rational number in the interval $(x,y)$.
Proof:
Suppose $x,y\in\mathbb{R}^+$ such that $x<y$, since $\mathbb{N}$ is not bounded from above there is an $a\in\mathbb{N}$ such that $\frac{1}{n}<x$. Now notice that $0<y-x$, then we have such an $m\in\mathbb{N}$ that $\frac{\frac{1}{n}}{m}<y-x$ which is $\frac{1}{mn}<y-x$. Here notice that since $m\in\mathbb{N}$, $\frac{1}{mn}\le\frac{1}{n}<x$ implying $\frac{1}{mn}<x$ and the inequality $ \frac{1}{mn}<y-x$ can be read as:
$$x+\frac{1}{mn}<y$$
Now notice that there is a $c\in\mathbb{N}$ such that $\frac{1+c}{mn}>x$, since $\mathbb{N}$ is well ordered we can pick the smallest of such $c$'s, when we pick it that way we must have $\frac{1+(c-1)}{mn}\le x$, summarizing in:
$$\frac{c}{mn}\le x<\frac{c+1}{mn}$$
Now notice that if $\frac{c}{mn}\le x$ then $\frac{c}{mn}+\frac{1}{mn}\le x+\frac{1}{mn}$ and since $x+\frac{1}{mn}<y$ we have $\frac{c}{mn}+\frac{1}{mn}<y$ which is essentialy:
$$\frac{c+1}{mn}<y$$
Here we have found our rational number because we have picked our $c\in \mathbb{N}$ such that $\frac{1+c}{mn}>x$ implying:
$$x<\frac{c+1}{mn}<y$$
Since $c+1\in\mathbb{N}$ and $mn\in\mathbb{N}$, $\frac{c+1}{mn}\in\mathbb{Q}$ and $\frac{c+1}{mn}\in(x,y)$.
Now we can easily show that there is a rational number in any interval $(x,y)$
Theorem:
If $x,y\in \mathbb{R}$ such that $x<y$, then there is a rational number in the interval $(x,y)$.
Proof:
Suppose $x,y\in \mathbb{R}$ such that $x<y$, then since $\mathbb{N}$ is not bounded from above we have such an $a\in \mathbb{N} $ that $x+a>0$ and since $y>x$, $y+a>x+a>0$ also. Then by the proposition we have a rational number $q$ in the interval $(x+a,y+a)$. Here since $a\in\mathbb{N}\subset\mathbb{Q}$ and $q\in\mathbb{Q}$, $q-a\in\mathbb{Q}$ and $q-a\in(x,y)$ as desired.
Thank you for reading!
 A: Theorem :
If $x,y\in \mathbb{R}$ such that $x<y$, then there is a rational number in the interval $(x,y)$.
Proof :
Let $\;n=\left\lfloor\dfrac1{y-x}\right\rfloor+1\in\mathbb N\;\;,\;\;m=\lfloor nx\rfloor+1\in\mathbb Z\;.$
It results that $\;n>\dfrac1{y-x}\;,\;m>nx\;,\;$ moreover ,
$x\!=\!\dfrac{nx}n\!<\!\dfrac mn\!=\!\dfrac{\lfloor nx\rfloor\!+\!1}n\!\leqslant\!\dfrac{nx\!+\!1}n\!=\!x\!+\!\dfrac1n\!<\!x\!+\!(y\!-\!x)\!=\!y\;.$
Hence , there exists $\;\dfrac mn\in\mathbb Q\;$ such that $\;x<\dfrac mn<y\;.$
A: Since $y-x>0,\ $ by Archimedes $\ \exists\ n\in\mathbb{N}$ such that $n > \frac{1}{y-x}.\ $ Let $n$ be one such positive integer. Note also that $\ ny > nx+1.\ $
Let $k_1$ be the greatest integer $\leq nx.\ $ If $\ k_1+1 \leq nx,\ $ then this contradicts the definition of $k_1.$ Therefore $k_1 + 1> nx.$ Therefore we have:
$\ k_1\leq nx < nx + 1 < ny,\ \implies nx<k_1+1 \leq nx + 1 < ny,\implies x\overset{(\times\frac{1}{n})}{<} \frac{k_1+1}{n}<y,\ $ i.e. $\ \frac{k_1+1}{n} \in (x,y).$
