Analysis: No strictly increasing fn such that f(Q)=R

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Homework Statement


Prove that there is no strictly increasing function f: Q->R such that f(Q)=R. (Do not use a simple cardinality argument)


Homework Equations


The section involves montone functions, continuity and inverses. I believe the theorem to be used is that a monotone function on an interval has a continuous inverse, and the intermediate value theorem.


The Attempt at a Solution


In class, our professor said that f has a continuous inverse, but I'm not sure why exactly. From there, I realize you can use the intermediate value theorem to contradict the fact the inverse is continuous, by letting c belong to the irrationals and showing it is not in the image.


EDIT: I now have everything down to proving that a strictly increasing, onto function f:Q->R is continuous
 
Last edited:
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Meant to say I have it down to proving that a strictly increasing, onto function: f: R->Q is continuous
 
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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