I have the following statement

"If $x$ and $y$ are real numbers with $x < y$, then there exists an irrational number $z$ such that $x < z < y$.

So, I have begun by applying the density theorem that there exists a rational number $r$ such that $x < r < y$. The theorem also states that $r$ cannot be equal to $0$ but I am not sure why. Does it have something to do with the Archimedean Property?

Any help would be appreciated.


If $\frac{x+y}2$ is an irrational number then we can set $z=\frac{x+y}2$. If not, then let $n$ be a natural number that is large enough so that $(y-x)n>1$ (by the Archimedean property). Then set $z=\frac{x+y}2+\frac{\sqrt{2}}{100}\frac{1}n$. This is an irrational number because we assumed that $\frac{x+y}2$ is rational and $\sqrt{2}$ is irrational. I leave the verification that $x<z<y$ up to you.

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  • $\begingroup$ but I still don't get why this means that r cannot be equal to 0. $\endgroup$ – Michelle Drolet Oct 16 '17 at 15:52
  • $\begingroup$ @Michelle: What they may mean is that one can choose $r$ not to be $0$. Of course if $x$ is negative and $y$ positive, then $r=0$ will work fine, but maybe you want to have an $r\not=0$ and the theorem gives you that. $\endgroup$ – J.R. Oct 16 '17 at 15:54
  • $\begingroup$ Okay. Do you mind explaining why we can choose r ≠ 0. Maybe I'm really just not getting it $\endgroup$ – Michelle Drolet Oct 16 '17 at 15:56
  • $\begingroup$ You can use literally the same argument as in my answer.. if $\frac{x+y}2$ is not zero, then set $r=\frac{x+y}2$. If $\frac{x+y}2=0$, then .. $\endgroup$ – J.R. Oct 16 '17 at 16:00
  • $\begingroup$ then I'm assuming you set r not equal to 0 $\endgroup$ – Michelle Drolet Oct 16 '17 at 16:26

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