# Proof that a rational exists between two real numbers

Without loss of generality let us assume $$x > 0$$.Hence we have $$(y-x)>0$$.

Now using the Archimedean principle we must have $$(y-x) > \frac{1}{n}$$

Hence $$(ny - nx)> 1$$.

$$ny> nx+1$$.

Now let us say $$m-1<nx<m$$.

Now we write $$m \le nx+1<ny$$.

Hence we have $$y > \frac{m}{n}$$ and $$x < \frac{m}{n}$$.So there exists a rational number between x and y.

3 years, 10 months ago

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Wait, you didn't define $$y$$ though! :P

- 3 years, 10 months ago

I thought the same thing. I'm sure he meant $$y \in \mathbb{Z}$$ such that $$y >x$$

- 3 years, 10 months ago

That's what it seems like. :P

- 3 years, 9 months ago