Homework 6 Due
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.
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 (bj) 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?
Homework 7 Due
p 42 #3
p46 #s 1& 2 (these has 4 parts each)