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I came across the following problem:

Show that if $x$ and $y$ are real numbers with $x <y$, then there exists an irrational number $t$ such that $x < t < y$.

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 $\displaystyle \frac{m}{n} \leq x \leq \frac{m+1}{n} < y$.

This is essentially the proof for the denseness of the rationals. Instead of $\large \frac{m+1}{n}$ I need something of the form $\large\frac{\text{irrational}}{n}$. How would I get the numerator?

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+1 for showing some thought on the problem. – Ross Millikan Jun 22 '11 at 4:27
up vote 13 down vote accepted

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 $y$?

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@Arturo: We know that $x <M\left( \frac{\sqrt{2}}{n} \right) < M \left(\frac{y-x}{2} \right) < M(y-x)$. – Damien Jun 22 '11 at 4:00
@Damien: You don't know that $M=1$ works. Read what I wrote carefully: what property did I specify $M$ should have in order to pick it and not something else? – Arturo Magidin Jun 22 '11 at 4:05
How do you know my favourite positive irrational is $\sqrt{2}$? :-) – joriki Jun 22 '11 at 4:07
@joriki: I'm paraphrasing Hendrik Lenstra (-; In class, whenever he worked out an example, you had "You pick your favorite prime number, which is 37", or "you pick your favorite real number greater than 0, which is 1", etc. – Arturo Magidin Jun 22 '11 at 4:09
@Damien: For $r$ an integer smaller than $M$ (not just any $r$). So that means, for instance, that this is true for $r=M-1$; that is, $(M-1)\frac{\sqrt{2}}{n}\leq x$. So... – Arturo Magidin Jun 22 '11 at 4:11

Suggestion: I expect that you can use the fact that $\sqrt{2}$ is irrational.

From the denseness of the rationals, you know that there is a non-zero rational $r$ such that $$\frac{x}{\sqrt{2}} <r <\frac{y}{\sqrt{2}}.$$

Now it's essentially over. (I almost forgot to insist that $r$ be non-zero!)

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Bravo, this is very nice! I have always thought of it using the method I showed below. – ncmathsadist Jun 22 '11 at 14:38
@ncmathsadist: I prefer your solution, it gets closer to the heart of the matter, since one can then replace co-countable by any "almost all" notion, such as measure $1$. But I am told that recycling is a virtuous thing to do, so decided to recycle the density of the rationals. – André Nicolas Jun 22 '11 at 15:17
but how does it follow from here on that t is between our x and y? – smihael Jan 26 '13 at 18:36
Multiply by $\sqrt{2}$: we get $x\lt r\sqrt{2}\lt y$. since $r$ is a no-zero rational, $t=r\sqrt{2}$ is irrational. – André Nicolas Jan 26 '13 at 19:04

Choose any real numbers $a$ and $b$ with $a<b$. The interval $(a,b)$ is not denumerable. However, the rationals inside of it are so $(a,b) - \Bbb Q$ is nonvoid; it has an element. Hence every open interval contains an irrational. It follows immediately the irrationals are dense in the line.

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One way to show this would be to use the fact that the rationals are countable, whereas the interval $(x,y)$ is uncountable (these facts must be proven, though), and therefore $(x,y)$ must contain some irrational number $t$, which will satisfy $x<t<y$.

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If the rationals are dense then there are at least two distinct rationals $a$ and $b$ between $x$ and $y$.

Then $t=a \dfrac{\sqrt{2}}{2} + b \dfrac{2-\sqrt{2}}{2}$ is irrational and also between $x$ and $y$.

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