Prove Dyadic Rationals are Dense in R

[Newtime]

Homework Statement

Prove that for any a,b \in R a,b < 0, there is a rational q of the form k/(2ⁿ) for n,k\in Z such that a < q < b.

Homework Equations

None.

The Attempt at a Solution

The thing is...I have a solution. Basically, you take the proof for the rationals being dense in R and literally substitude in 2ⁿ wherever there is an n. My question is...at the first step for proving the rationals are dense in R, we choose, by the Archimedean Property, a natural number n so that n > max(1/a, 1/(b-a)). So for this proof, we recognize that 2^n > n for all integers n and thus for all natural numbers n. However, in Analysis, "Recognizing" something is true is not enough...and I've got to prove it. And I can prove it..but by induction, which is in the section after this, which leads me to believe that I went about this proof the wrong way. Or do you think the author messed up and included this probem a bit too early in the book? Or is there simply another way to show that 2^n > n for all integers n?

If it's necessary to obtain help, I can type up my proof in its full form, but since it literally is (except for the step above that's giving me trouble) the proof for the density of rationals with one variable replaced throughout, I wanted to save myself the trouble.

Thanks in advance, this one has been nagging at me for a while.

 

[lurflurf]

0)Break the problem into steps.
1)Prove that for any a,b R a,b < 0, there is an integer n such that (2^n)(b-a)>1.
2)Prove that for any a,b R a,b < 0, there is an integer k such that (2^n)a<k<(2^n)b.
3)combine 1) and 2)

 

[Newtime]

0)Break the problem into steps.
1)Prove that for any a,b R a,b < 0, there is an integer n such that (2^n)(b-a)>1.
2)Prove that for any a,b R a,b < 0, there is an integer k such that (2^n)a<k<(2^n)b.
3)combine 1) and 2)

That seems like a much more efficient approach, thanks. The only problem is that for your first step, how would I go about proveing there is an integer n so that (2^n)(b-a) > 1 without induction? Using the Archimedean Principle, it would be straightforward to simply find an n so that n(b-a) > 1 but keep in mind this is the section in my book directly after the construction of the reals, we have not yet learned induction (at least, it hasn't come in he book yet) so this is where I'm a little lost... Also did you mean a,b > 0 for your steps 1 and 2? If not, I don't understand how negative reals aids the proof.

 

[Newtime]

bump.

any other ideas?

my professor has office hours tomorrow afternoon but i would just to get this figured out before then. I've been thinking more and so far I'm stuck in the same rut...trying to show 2^n > n for all natrual numbers without using induction...

 

[Billy Bob]

Technically, how would you even define 2^n without induction?

You could try (essentially) reproving the Archimedean Property. If x is real, prove there exists natural n s.t. x < 2^n. Suppose not. Then x is a u.b. for numbers of the form 2^m. Get l.u.b. s. Consider s-1. Etc. Eventually when you get s<2^j+1, use 1<2^j and 2^j+2^j=2^(j+1). This relies on believing 1<2^j I guess.

 

[mingk]

I'm trying to figure out this exact same proof. Did you happen to find out how to get this done? If so could you please help me out :S