**Homework 6 Due ****Wednesday
Feb 8, 2006**** (revised)**

Read a_n as “a sub n” and a_n_j as “a sub n sub j”. The latter indicates j is a subscript of n, which is a subscript of a.

6.1

We create a list of **positive**
rational numbers as m/k where k, m are positive integers.

They do not have to reduced to lowest order.

There is one such number for which s = k+m = 2, namely 1/1.

There are two such numbers for which s = k+m = 3, namely 1/2 and 2/1

There are 3 for which k+m = 4: 1/3, 2/2, 3/1. Of course 2/2 is a duplicate of 1/1.

Make a list of all positive rational numbers listing all those with a given s before moving on to s+1,

and for a given s, list those in order of increasing m.

Let a_n be the n-th number in the list.

Let x be an arbitrary real number.

**6.1 a Show that if x ≥
0, then for each ε > 0, there is an n such that | x – a_n | < ε**

**6.1 b
We** now describe an approach to creating a subsequence
that may converge to x.

**Show this
construction below produces a sequence (b _{j}) and this converges to x if and only if x ≥
0.**

Create an increasing sequence of subscripts n_j and the corresponding subsequence (b_j =a_n_j) as follows:

n_1 = 1 and b_1 = a_1.

n_2 is the first integer > n_1 for which |x – a_n_2| ≤ |x – a_n_1|

and b_2 = a_n_2.

Proceed in that manner as follows:

n_j+1 is the first integer for which |x – a_n_j+1| ≤ |x – a_n_j|

Let (b_j) be the subsequence (a_n_j).

Hint: Have you made clear what happens if x is rational?

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Homework 7 **Due**** ****Friday
Feb 10, 2006**

p33 #2

p37 #1

p 42 #3

p46 #s 1& 2 (these has 4 parts each)