Prove that for real numbers $x,y$ with $x< y$, there is a rational and an irrational between $x$ and $y$ in the following cases:

a) when $x< 0< y$;

b) when $x< y \le 0$.

For a) this is my proof: Let $r$ and $q$ be rational and irrational numbers, respectively. First I show $r$ and $q$ are in $(0,y)$ and then I show that $r$, $q$ are in $(x,0)$.

To show that $r$ and $q$ are in $(0,y)$ I use Case 1 from my class:

$(a,b)$ = $\{x\in\mathbb{R}|a< x< b\}$. By the "squeezing in" theorem, there exists $n\in\mathbb{N}$ so that $0 < 1/n < y$. So take $r = 1/n$. Consider $(1/n)(1/2^{1/2})<1/n<y$. Since $2^{1/2}>1$, $1/2^{1/2} < 1$. By the previous result, $(1/n)(1/2^{1/2})$ is irrational. Hence, $r$ and $q$ in $(0,y)$.

Now I want to show that $r$ and $q$ are in $(x,0)$. Since $x<0\implies- x>0$. By Case 1, I can find a rational $r\in(0,-x)$ and an irrational $q\in(0,-x)$. But, we can use Case 1 again to find $r,q\in(0,y-x)$ as $0< -x< y - x$.

Since $r,q\in(0,y)$ and $r,q\in(0,y-x)$, there is a rational and irrational number in $x< 0<y$.

So what do you guys think about my proof?

  • $\begingroup$ I think you are being asked to prove the "density of rationals and irrationals theorem" for these special cases. $\endgroup$ – Adriano Sep 13 '13 at 17:22
  • $\begingroup$ HINT: Archimedan Property $\endgroup$ – Don Larynx Sep 13 '13 at 17:23

Let's look at the case $x<0<y$. I'll assume that you can find a rational number in the interval $(x,y)$ by yourself, so let's look at the irrational case. The Archimedean property of the real numbers implies that there exists an integer $n$ such that $\vert y \vert >1/n.$ In particular, since $\sqrt{2}>1$, we have

$$\vert y \vert > z:=\frac{1}{\sqrt{2}n}.$$

The real number $z$ is irrational, and lies in the interval $(0,y) \subset (x,y)$.

The second case is slightly more difficult. As a hint, first try to find a rational number in your interval (for example, find an $n$ such that $\vert x -y \vert >1/n$, and consider the rational numbers with denominator $n$), and then try adding a sufficiently small irrational (scaling by a sufficiently small irrational would work as well).

  • $\begingroup$ For a) I am having difficulty proving that there is a rational in x< 0< y. $\endgroup$ – user87274 Sep 13 '13 at 20:34
  • 3
    $\begingroup$ But you've just written one! $\endgroup$ – awwalker Sep 13 '13 at 21:41

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