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.


  • 2
    $\begingroup$ A hint for one possible approach: there exists at least one irrational number $\omega$. Any nonzero rational multiple of $\omega$ is still irrational. $\endgroup$
    – D_S
    Sep 18, 2017 at 22:33
  • $\begingroup$ Okay, so how would I use that in my proof? Sorry, I'm just not seeing the connection. $\endgroup$
    – nicostr
    Sep 18, 2017 at 22:37
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    $\begingroup$ Sure, your approach would work. If you know that between any two real numbers $a<b$ there is another real $c$ (rational or not) so that $a<c<b$. If you prove there is a rational between any two reals then you can find rationals $r_1,r_2$ satisfying $a<r_1<c<r_2<b$. Finally if you can find an irrational between any two rationals, then there is an irrational $q$ that satisfies $a<r_1<q<r_2<b$. That's a perfectly valid approach. $\endgroup$
    – adfriedman
    Sep 18, 2017 at 22:58

3 Answers 3


Here's my proposed solution:

1) First you show that between two rationals there is always an irrational number. Suppose that $q_1<q_2$, then take for ex. $r=\frac{q_1}{\sqrt{2}}+q_2\left(1-\frac{1}{\sqrt{2}}\right)$, which obeys $q_1<r<q_2$. 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 $r_1<r_2$, then $s_1=3r_1/4+r_2/4$ and $s_2=r_1/4+3r_2/4$ are in the order $r_1<s_1<s_2<r_2$. 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 $q_1<r(\lambda)<q_2$.


Try this one:

First we show that the sum of a rational number $r$ and an irrational number $i$ is irrational. This can be proved by the fact that $\mathbb{Q}$ is closed under addition. (If $i+r$ was rational, then $(i+r) + (-r) = i$ would become rational).

So our task becomes proving the existence of an arbitrarily small irrational number $s$. If $0<s<y-x$, $x+s$ then becomes an irrtional number between $x$ and y.

We require the existence of any positive irrational number. We may use $\sqrt{2}$. Then consider the sequence $\frac{\sqrt{2}}{n}$, which gets arbitrarily small as $n \to \infty$.

The only thing left now is to prove that $\frac{\sqrt{2}}{n}$ is irrational. The proof can be done by using a process similar to how we proved the sum of a ration number and an irrational number is irrational ($\mathbb{Q}$ is also closed under multiplication).


Suppose $x < y$.

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 < f(Q), f(Z) < 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

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

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


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