Analysis Questions URGENT!!

1. A dyadic rational is a number of the form k/(2^n) for some k, n belongs to Z (integers). Prove that if a and b are real numbers and a < b, then there exists a dyadic rational q such that a < q < b.

2. Prove each of the following statements:
a) 2n+1<2^n for n = 3,4,...
b) n<2^n for n = 1,2,...

3, Prove that (2^n)+3^n is a multiple of 5 for all odd n belongs to N (natural numbers)

4.Prove that f is 1 on 1 on E and find f(E). f(x) = 3x-7, E = R (real numbers)

5. Prove that set of rational lattice points in space - that is, the set Q^3 := {(x,y,z): x,y,z belongs to Q} - is countable.

Please split up the questions in the future!
------------------
1) Let's modify this a little bit. The main idea is to replace the denominator with 2^n everywhere in the classic proof about the density of Q in R.

We want to show that for any a, b in R with a < b,
we can produce integers k, n such that a < k/2^n < b.

Assume without loss of generality that a, b > 0.
[If a ≤ 0 and b > 0, consider the subinterval (b/2, b), and find a dyadic rational
there using the discussion below.
If a < 0 and b ≤ 0, then by below we can find a dyadic rational c in (-b, -a);
multiply that by -1 to get one in (a, b).]
---------------------
(i) By Archimedes, pick n large enough so that 1/2^n < b - a.

(ii) So, a < k/2^n < b <==> a * 2^n < k < b * 2^n.

With n now chosen, the idea is to choose m to be the smallest positive integer
greater than a * 2^n. i.e, pick k in N so that k - 1 ≤ a * 2^n < k. (**)

The right half of (**) implies that a < k/2^n, giving half the desired inequality.

Next, rewriting 1/2^n < b - a in the form a < b - 1/2^n, use the left half of (**)
k ≤ a * 2^n + 1 < (b - 1/2^n) * 2^n + 1 = b * 2^n.

Thus, k < b * 2^n ==> k/2^n < b.

(iii) Putting this altogether, we get a < k/2^n < b, as required.
------------------------------