×

# 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, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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

- 3 years, 5 months ago

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

- 3 years, 5 months ago

That's what it seems like. :P

- 3 years, 4 months ago