Density of Rationals in R: u>0, x<y

[Punkyc7]

if u>0 is any real number and x<y show there exist a rational number r such that x<ru<y. Hence {ru: r$\in$Q is dense in R.

I am not sure how to show that there exist a rational number. I was thinking this has something to do with the archimedian property. This is what I have triedx<$\frac{(x+y)}{2}$<y

Define r= $\frac{(x+y)}{2}$(1/n) when u>=1 and
r=$\frac{(x+y)}{2}$(n) when u<1

where n is an element of the naturals

but I don't think that works? Any suggestions?

 

[lippyka]

Yes, the idea is correct. Here is a more detailed proof:Let u>0 be any real number and x<y. We will show there exists a rational number r such that x<ru<y. Let r = (x+y)/2n where n is a natural number. Then ru = (x+y)/2, so x<ru<y. Since r is a rational number and ru is between x and y, this shows that there exist a rational number r such that x<ru<y. Hence {ru : r ∈ Q} is dense in R. To prove this, let a and b be two real numbers such that a < b. Then we can find an integer n such that u/n < b-a. Let r = (a+b)/2n. Then ru = (a+b)/2, which is between a and b. This shows that for any real numbers a and b, there exists a rational number r such that a<ru<b. Thus {ru : r ∈ Q} is dense in R.