Proving Density of Dyadic Rationals in Q | Rational Numbers

[cragar]

Homework Statement

Prove that the dyadic rationals are dense in Q.
That is the rationals of the form $\frac{m}{2^n}$
m is an integer and n is a natural

The Attempt at a Solution

Let's say we have two arbitrary rationals x and y. where x<y
Now I will pick a rational smaller than x such that it is of the form
$\frac{s}{2^k}$ and i will call this P ,
now I will pick a rational larger than y that is of the same form
and i will call it O .
Now I will add P and O together and then divide by 2, find the midpoint
Now this new rational has a denominator that is a power of 2 because
everything we did had a denominator of 2. Now I will keep doing this,
I will keep finding mid points between these sets of rationals
that I created and I might have to pick the left or right one and then
keep finding the midpoints. Eventually i will get in between x and y.
I realize this is informal but Is my general idea in the right direction.

 

[Dick]

Homework Statement

Prove that the dyadic rationals are dense in Q.
That is the rationals of the form $\frac{m}{2^n}$
m is an integer and n is a natural

The Attempt at a Solution

Let's say we have two arbitrary rationals x and y. where x<y
Now I will pick a rational smaller than x such that it is of the form
$\frac{s}{2^k}$ and i will call this P ,
now I will pick a rational larger than y that is of the same form
and i will call it O .
Now I will add P and O together and then divide by 2, find the midpoint
Now this new rational has a denominator that is a power of 2 because
everything we did had a denominator of 2. Now I will keep doing this,
I will keep finding mid points between these sets of rationals
that I created and I might have to pick the left or right one and then
keep finding the midpoints. Eventually i will get in between x and y.
I realize this is informal but Is my general idea in the right direction.

You can probably prove it in a less elaborate way. If x<y then y-x is positive and there must be an n such that 1/2^n is less than y-x, yes?

 

[cragar]

yes I could do it that way. Thats the cool thing about pure math is that it is very creative.