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

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    $\begingroup$ 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 $w$? $\endgroup$ – André Nicolas Nov 2 '15 at 3:42
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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 $3$.

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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.

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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.

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