Can You Prove There Are Infinite Rationals Between Two Real Numbers?

Shing
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Homework Statement


If x and y are arbitrary real numbers. x>y. prove that there exist at least one rational number r satisfying x<r<y, and hence infinitely.

The Attempt at a Solution


well, I have done my proof, but comparing to the solution offered by http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-014Calculus-with-Theory-IFall2002/1C8FA521-FDCE-491B-8689-955B04A4A4A2/0/pset2solutions.pdf" (*1), I have a bit doubt about whether my proof is precise enough or not.

anyway, here it is:

x,y belong to R, x<y
let|x-y|&gt;\varepsilon
let n belongs Z, n>1
obviously,\varepsilon satisfies x&lt;x+\frac{\varepsilon}{n}&lt;y
as there exist infinite numbers for n,
therefore, infinite r satisfy x<r<y

thanks for reading =)
 
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Shing said:

Homework Statement


If x and y are arbitrary real numbers. x>y. prove that there exist at least one rational number r satisfying x<r<y, and hence infinitely.


The Attempt at a Solution


well, I have done my proof, but comparing to the solution offered by http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-014Calculus-with-Theory-IFall2002/1C8FA521-FDCE-491B-8689-955B04A4A4A2/0/pset2solutions.pdf" (*1), I have a bit doubt about whether my proof is precise enough or not.

anyway, here it is:

x,y belong to R, x<y
let|x-y|&gt;\varepsilon
let n belongs Z, n>1
obviously,\varepsilon satisfies x&lt;x+\frac{\varepsilon}{n}&lt;y
as there exist infinite numbers for n,
therefore, infinite r satisfy x<r<y

thanks for reading =)

If \epsilon is not a rational number, then is \epsilon/n rational?

If |x-y|&gt;\epsilon then can you find a rational number such that \epsilon is larger than this rational number? The rest of your arguments can be used provided you find this rational number.
 
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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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