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I know about the proof found here: Proof there is a rational between any two reals.

I wanted to know if this similar proof is also correct?

Assume $x > 0$. Since $y > x$, it follows $y-x>0$. There exists some $n\in \mathbb{Z}^+$ such that $\frac{1}{n}<\min(y-x,x)$. Define $B=\{\frac{k}{n}\mid nx\ge k\in\mathbb{Z}^+\}$ which is nonempty by construction and bounded above by $x$.

Thus, $\sup(B)=\beta$ exists and $x-\beta<\frac{1}{n}$ by construction. Furthermore, $\beta\in B$ by the well ordering principle (edit: this is wrong). Hence, there exists some $k\in\mathbb{Z}^+$ so $\frac{k}{n}=\beta$ where $\frac{k+1}{n}>x$. Observe $x<\frac{k}{n}+\frac{1}{n}<x+(y-x)=y$.

If $y<0$, use $x=-y$ and $y=-x$ in the above proof to get $\frac{-(k+1)}{n}$ as the rational between x and y. If $y>0$ and $x<0$, use $0$ as the rational.

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    $\begingroup$ A few notes on formatting: in math mode, spaces don't matter (and are ignored). The only spaces that are necessary are those after a \command and the next letter (also unnecessary if the next character isn't a letter, so "\beta>1" works as-is. Common math operations like sin, max, etc, can be made commands \sin, \max to look better. The "such that" bar is \mid, which gives you the correct spacing. To get an actual space in math mode, you need the "punctuation" commands \, \: \; or \! (negative space), or \quad (massive space), or even \qquad. Nice job on the formatting by the way. $\endgroup$
    – obscurans
    Commented May 23, 2020 at 2:18
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    $\begingroup$ \mathbb{single letter} gets you the fancy font you usually use for $\mathbb{R},\mathbb{N},\mathbb{Z},\mathbb{C}$, etc. $\endgroup$
    – obscurans
    Commented May 23, 2020 at 2:22
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    $\begingroup$ given two real number $a<b$, the distance between them may be small, but you can multiply them by some large enough integer $m$ so that there is an integer $n$ between them $ma<n<mb$, then from here $a<\frac{n}{m}<b$ $\endgroup$
    – user614287
    Commented May 23, 2020 at 2:23

1 Answer 1

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Proof looks fine although it's a bit circuitous to get the correct bounding number. You also want to start with "assume $0<x<y$" outright.

Note that $B$ is simply a finite bounded set so it easily contains its supremum. Also probably easier to write "$\exists k\in\mathbb{Z}^+$ such that $$\frac{k}{n}\leq x<\frac{k+1}{n}\text{".}$$

As mentioned in the comment, it's easier to look at $y-x$ from the beginning, and just scale it up large enough: consider $$m=\left\lceil\frac{1}{y-x}\right\rceil+1\text{,}$$ and notice that some fraction $\frac{n}{m}$ must lie strictly between $x$ and $y$, since $x+\frac{1}{m}<y$, therefore $n=\left\lceil mx\right\rceil+1$ works.

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