Prove dyadic rationals are dense on [0,1]

[aortizmena]

Homework Statement

Prove that D={$$\frac{m}{2^{n}}$$ : n$$\in$$ N , m=0,1,2,...,2$$^{n}$$} (dyatic rationals set) is dense on [0,1] , i.e. if (a,b) $$\subset$$ [0,1] then (a,b) $$\bigcap$$ D $$\neq$$ emptyset

Homework Equations

The Attempt at a Solution

Is it wrong if I just state that because a,b$$\in$$$$\Re$$ we know that there exists m,n$$\in$$N, l=$$\frac{m}{2^{n+1}}$$, u=$$\frac{m}{2^{n-1}}$$ and r=$$\frac{m}{2^{n}}$$ such that l $$\leq$$ a $$\leq$$ r $$\leq$$ b $$\leq$$ u?

 

[Gib Z]

Why do you need the extremal inequalities? You should only need $a \leq r \leq b$. Also it seems like you are trying to say "Because a and b are real, its is obvious there are naturals m,n that allow us to conclude the result", which really is not substantial enough.

Instead, I would use the Archimedian property to argue where there exists $n \in \mathbb{N}$ such that $$b-a < \frac{1}{2^n}$$, then you a Pigenhole Principle type argument to see why D should intersect with (a,b) .