Proof technique for between any two real numbers is an irrational number I'm proving that between any two real numbers is an irrational, and my first thought is to prove that between any two real numbers is a rational number, and then show that between any two rational numbers is an irrational. However, I don't think this would complete the proof because I will not have shown that between any two real numbers there are two rational numbers. Would the first part of my proof imply that there are an infinite number of rationals between any two reals, or do I need to show that separately? 

Also, I'm really trying to figure out this problem on my own, so I ask not for the answer, but a pointer in the right direction.

Thanks!
Here's my proposed solution:

1) First you show that between two rationals there is always an irrational number. Suppose that $q_1&lt;q_2$, then take for ex. $r=\frac{q_1}{\sqrt{2}}+q_2\left(1-\frac{1}{\sqrt{2}}\right)$, which obeys $q_1&lt;r&lt;q_2$. The number $r$ can be written as $a+b\sqrt{2}$, where $a, b\in\mathbb{Q}$, which implies that $r\in\mathbb{R}-\mathbb{Q}$.

2) Given any two real numbers $r_1&lt;r_2$, then $s_1=3r_1/4+r_2/4$ and $s_2=r_1/4+3r_2/4$ are in the order $r_1&lt;s_1&lt;s_2&lt;r_2$. If $s_1$ or $s_2$ is irrational then you're done. Otherwise, go to step 1).

For the infinite part, you could use the mapping $r(\lambda)=\lambda q_1+(1-\lambda)q_2$, where $\lambda\in(0, 1)$. Again $q_1&lt;r(\lambda)&lt;q_2$.
Suppose $x &lt; y$.

Define $f(t) = (1-t)x + t y = x + t(y-x)$.

Let $Q$ be any rational number in the interval $(0,1)$.

Let $Z$ be any irrational number in the interval $(0,1)$.

Then $ x &lt; f(Q), f(Z) &lt; y$ for all $Q$ and $Z$ as defined.

If $x$ and $y$ are both rational, then f(Z) is irrational.

If $x$ and $y$ are both irrational, then $f(t)$ is irrational where

$$t = 
\begin{cases}
   Z &amp; \text{If $y-x$ is rational.} \\
   Q &amp; \text{If $y-x$ is irrational.}
\end{cases}
$$

If one of $x,y$ is rational and the other is irrational, then $f(Q)$ is irrational.