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?
$\endgroup$ 05 Answers
$\begingroup$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!)
$\endgroup$ 4 $\begingroup$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 $y$?
$\endgroup$ 12 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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 $x<t<y$.
$\endgroup$ $\begingroup$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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