0.1 Rationals are Dense in Reals
The rational numbers are dense in the real numbers, that is between any two real numbers we can always find a rational numbers (this is related to the property of dense order).
If \(x,y \in \mathbb{R}\) with \(x<y\) then there exists \(z \in \mathbb{R}\) such that \(x<p<y\).
\begin{proof} Since \(x<y\) we have \(y-x>0\) and from the [[archimedean-property|archimedean property]] of the reals we have a positive integer \(n\) such that \(n(y-x)>1\). Applying the same property again we obtain positive integers \(m_{1}\) and \(m_{2}\) such that \(m_{1}>nx\), \(m_{2}>-nx\). Then \(-m_{2}<nx<m_{1}\). Hence there is an integer \(m\) (with \(-m_{2}\leq m\leq m_{1}\)) such that \(m-1\leq nx < m\). If we combine these inequalities, we obtain \[
nx<m\leq 1+nx<ny.
\] Since \(n>0\) it follows that \[
x< \frac{m}{n}<y.
\] This proves the result taking \(p=\frac{m}{n}\).\end{proof}