Prove there is an irrational between any two rationals

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Homework Statement



Prove that there exists an irrational between any two rationals.



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The Attempt at a Solution



How would one do this? So far I've proven there is an irrational between any rational and irrational, any irrational and rational, that there's a rational between any two reals, or a irrational between any two reals in the attempt; I realize that it can be taken as a case of the last two; but I haven't been able to start with just p,q such that p>q, and p and q rational, and arrive at this result. It's pissing me off.

Thanks in advance.
 
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Let p ,q be rationals with p>q

Let e be a positive real. Then for e < |p-q|, q+e is in the open interval (p, q). Due to the density of the irrationals in R (as you pointed out, between any two reals is a rational), there is a positive irrational, call it a, that is less than e. Now take q+a. What can you say about that value?
 
Say the two rationals are p and q.
define
x=[(p+q)/2]+[(p-q)/2]r/n
where r is any irrational and n is any integer such that
|r|<n

show
(1/2)(p+q-|p-q|)<x<(1/2)(p+q+|p-q|)
or
|2x-p-q|<|p-q|
and
x is irrational
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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