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.

#Calculus #CosinesGroup #Goldbach'sConjurersGroup #TorqueGroup #JustForFun

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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

Finn Hulse - 6 years, 9 months ago

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

A Former Brilliant Member - 6 years, 9 months ago

That's what it seems like. :P

Finn Hulse - 6 years, 9 months ago

y belongs to real numbers, not integers.

VIVEK SABARAD - 4 months, 2 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$