# Prove that between any two rational numbers, there are two irrational numbers whose difference is rational

I've gone through how to prove that there is an irrational between any two rational numbers but this is somehow stumping me

• Once you have shown there is an irrational $c$ between $a$ and $b$, all you need to do is to add to $c$ a rational $w\lt b-c$. Can you show there is such a $w$? – André Nicolas Nov 2 '15 at 3:42

If you have proven that there is an irrational number between two rational numbers, then the next step I would take is to prove that a rational number plus an irrational number is also irrational, see http://educ.jmu.edu/~taalmala/235_2000post/235contradiction.pdf

Let's say $m$ and $n$ are both rational, $m < n$, and $x$ is an irrational number between them, that is, $m < x < n$. Calculate $n - x$, which is irrational, and then choose $k$ a rational number less than that. Thus $k < (n - m)$, $x + k$ is irrational and $m < x < (x + k) < n$.

For example, let's say we have found $\sqrt{7}$ is an irrational number between $2$ and $3$. Then $3 - \sqrt{7} \approx 0.3542486889354$, which we approximate to $\frac{7}{20}$. Then $\sqrt{7} + \frac{7}{20}$ is irrational and less than $3$.

Say the rationals you have are $a$ and $b$ where $a<b$.

First we will find two rational numbers, this is easy, $c={2a+b\over3}$ and $d={a+2b\over3}$ where $a<c<d<b$ and $d-c$ is obviously rational.

Now there should be a irrational $x$ between $a$ and $c$ since $a<c$, then by $x<c$, we have $x+d-c<b$. Also by $d>c$ we have $a<x+d-c$.

Also since $x$ is irrational and $d-c$ is rational, $x+d-c$ is irrational. So $x$ and $x+d-c$ are the irrational numbers we want.

You know that between any two rationals there is an irrational.

Let $a < b$ a, b rational.

We can find $a < \frac{a+b}2 = a + \frac{b - a}2 = b - \frac{b-a}2 < b$; $\frac{a + b} 2$ rational.

We can find $a < x < \frac{a+b}2 < b$; x irrational.

So we can find $a < x < \frac{a+b}2< x + \frac{b-a}2 =y <b$

y is irrational. $y -x = \frac{b-a}2$ is rational.