Can anyone help me solve this question? I've been working on it for two days already.
Prove that for real numbers $x$ and $y$ with $x < y$, there is a rational and an
irrational between $x$ and $y$ in the following case when $x < y \le 0$.
Since $y \le 0$ is logically equivalent to $-y\ge 0$, we can use the squeezing in theorem to show that a rational $r$, and an irrational $q$ exist in the interval $(0,-y)$ which is the same as saying that there is a rational and an irrational in the interval $(y,0)$.
For the second part, I want to show that there exists a rational and irrational in the interval $(x,y)$. Since $x < y$ is logically equivalent to $-y<-x$, if we show that there is a rational and an irrational between $(-y,-x)$ then we're done. To show this I'm trying to use Archimedean Principle that there exists a rational $mr> -y$ and I am stuck here.
Best Answer
Assume $y-x>0$. By the Archimedean property, there is $n$ such that $n(y-x)=ny-nx>1$. But if two reals have a gap $>1$; there must be an integer between them. So...?
For irrationals, pick your favourite one in $(0,1)$ and transport it between two rationals by an appropriate map. That is, if $\kappa$ is irrational, and $r<s$ rationals, can you find rationals $\alpha\neq 0 ,\beta$ such that $r<\alpha\kappa+\beta<s$? Then use the first part to conclude. Note $\alpha\kappa+\beta$ is still irrational! (Why?)