# Understanding density of irrational numbers and Archemedian property

From

I know this much of the proof of the density of irrational numbers

"We know that $y-x>0$.
By the Archimedean property, there exists a positive integer $n$ such that $n(y-x)>1$ or $1/n < y-x$. There exists an integer $m$ such that $m \leq nx < m+1$ or $\displaystyle \frac{m}{n} \leq x \leq \frac{m+1}{n} < y$. "

Why does the following flow from the Archimedean property? As I know it the property states only what's above in the proof.

"Pick your favorite positive irrational, which is $\sqrt{2}$. By the Archimedean property, there exists $n$ such that $\frac{\sqrt{2}}{n}\lt \frac{y-x}{2}$. Again by the Archimedean property, we know there exists an integer $m$ such that $m\left(\frac{\sqrt{2}}{n}\right)\gt x$. Pick $M$ to be the least such $m$. Can you show that $M\left(\frac{\sqrt{2}}{n}\right)$ is strictly between $x$ and $y$? (Above quote By Arturo Magidin)

I've seen the Archimedean property used similarly in this post
Proof that the set of irrational numbers is dense in reals

"By the density of rational numbers, there exists a rational number $r \in (x, y)$.

Since $\frac{y - r}{2} > 0$, by the Archimedian Property there exists $n \in \mathbb{N}$ such that $\frac{y - r}{2} > \frac{1}{n}$. Then we have $x < r + \frac{\sqrt{2}}{n} < r + \frac{\sqrt{4}}{n} < y$. Now check that $s = r + \frac{\sqrt{2}}{n}$ is an irrational number sitting in $(x, y)$."(above proof written by Akech)

Thanks for explaining

The Archimedean property in $\mathbb{R}$say that , for all $x\in \mathbb{R}$ there exists an integer $n$ such that $n>x$.

As an immediate consequence we have that , given two real numbers $a, b$ with $a>0$, there exists an integer $n$ such that $na>b$ (use the property with $x=b/a)$.

In your case you have : $b=1$ and $a=y-x$, so: $n(y-x)>1$.

• Yes, I get that much, but how does it lead to "Pick your favorite positive irrational, which is $\sqrt{2}$. By the Archimedean property, there exists $n$ such that $\frac{\sqrt{2}}{n}\lt \frac{y-x}{2}$. Again by the Archimedean property, we know there exists an integer $m$ such that $m\left(\frac{\sqrt{2}}{n}\right)\gt x$."? – Forever_A_Student May 19 '15 at 19:49
• I've seen another poster also use $\frac{y-x}{2}$ as well, but. I haven't seen it in this form yet – Forever_A_Student May 19 '15 at 19:51
• The key fact is $na>b$ if $a\ne 0$. Chose $b=\sqrt{2}$ and $a=\frac{y-x}{2}$ ( with $y-x >0$) and you have the result. – Emilio Novati May 19 '15 at 19:58

If $$M$$ is the least $$m$$ such that $$m\left(\frac{\sqrt{2}}{n}\right)\gt x$$, then $$(M-1)\left(\frac{\sqrt{2}}{n}\right) \le x \lt M\left(\frac{\sqrt{2}}{n}\right)$$.

Since $$\frac{\sqrt{2}}{n}\lt \frac{y-x}{2}$$,

$$\begin{array}\\ y &\gt 2\frac{\sqrt{2}}{n}+x\\ &\ge 2\frac{\sqrt{2}}{n}+(M-1)\left(\frac{\sqrt{2}}{n}\right)\\ &= 2\frac{\sqrt{2}}{n}+M\left(\frac{\sqrt{2}}{n}\right)-\left(\frac{\sqrt{2}}{n}\right)\\ &= \frac{\sqrt{2}}{n}+M\left(\frac{\sqrt{2}}{n}\right)\\ &\gt M\left(\frac{\sqrt{2}}{n}\right)\\ \end{array}$$

Actually, it looks like all that is needed is $$\frac{\sqrt{2}}{n}\lt y-x$$.