Set 4 -> algebra -online-help/radical-equations/rational-expressions-solver.html">Rational Numbers . Prove the fol lowing statements (a, b ∈Q).
Note that | -a| = |a|. Then |a - b| = |a + (-b)| ≤
|a| + | - b| = |a| + |b|.
It follows from a) that |a| = |b - (b - a)| ≤ |b| + |b
- a|, so |a| - |b| ≤ |a - b|.
Fact: u ≤ v and -u ≤ v implies
|u| ≤ v. From b), we know that |a|-|b| ≤
|a-b|. If we
inter change a and b , we have |b|-|a| ≤ |a-b|. Using Fact,
this implies that 
Set 5 -> Cauchy Sequences.
(14) a) Let {xk} and {yk} be two equivalent Cauchy sequences. Show that
equivalence of Cauchy sequences respects the properties of
Reflexivity: Show that 
We want to show that beyond a certain m, we will have
and for
k ≥ m). But that will always be true since 
for any k and 
Symmetry: Show that 
Same as above except that instead of |xk - xk|
= 0 we use the fact that |xk - yk| =
|yk - xk|.
Transitivity: Show that 
See Lemma 2.1.1 in Book.
(10) b) What kinds of real numbers are re presentable by
Cauchy sequences
of integers?
At some point in a Cauchy sequence (i.e. beyond a term x m,
for some m), all terms
are going to be very close to each other (i.e.
and where 1/n can certainly be smaller than 1, which is the distance between any
two
integers). Since the terms in the sequence are integers, the only way the
sequence can be
Cauchy is if, beyond the mth term, all terms are equal. For example, a Cauchy
sequence of
integers could be (1, 3, 4,-50, 14, 13,-4, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,
5, 5, 5, 5, ...), which means
that only integers can be represented by Cauchy sequences of integers.
(+10) *c) Suppose 
are
two sequences
of rational numbers. Define the shuffled sequence to be
Prove that the shuffled sequence is Cauchy if and only if {xk} and {yk}
are
equivalent Cauchy sequences.
(->) To show {zk} is Cauchy implies {xk}
and {yk} are equivalent, we will show that
when {xk} and {yk} are not equivalent then {zk}
is not Cauchy.
If {xk} and {yk} are not equivalent,
then for any fixed m, we can find a k s.t. |xk-yk| >
1/n for some n. Knowing that zk = xk and zk+1 =
yk, this means that |zk - zk+1| is also
> 1/n, which means that {zk} is not Cauchy.
(<-) Now, let’s as sume that {xk} and {yk}
are equivalent Cauchy sequences. This means
that 
, 
Since both sequences are Cauchy, we can
say that |xj - xk|
≤ 1/2n and |yj -
yk| ≤ 1/2n for j, k≥m.
Case 1: If zk = xk and zj
= xj , then |zk
- zj | = |xk
- xj | ≤ 1/2n <
1/n.
Case 2: If zk = yk and zj
= yj , then |zk
- zj | = |yk - yj
| ≤ 1/2n < 1/n.
Case 3: If zk = xk and zj
= yj , then
Case 4: If zk = yk and zj
= xj , then
(10) d) Can a Cauchy sequence of positive rational
numbers be equivalent to
a Cauchy sequence of negative rational numbers?
Yes. Consider the sequence of positive rational numbers
{1/k}. It is Cauchy since 
we can find an m s.t. 
We just need to take m
≥ 2n. We then
have 
We can similarly show
that the sequence of negative rational numbers {-1/k} is also Cauchy. (Both
sequences
represent the number 0). To show that the sequences are equivalent, we observe
that for
any k, the distance between the terms of both sequences, |1/k-(-1/k)| = 2/k,
will become
very small as 
. Formally,
, we can find m ≥
2n s.t. |1/k - (-1/k)| = 2/k and
since k ≥ m and therefore, k ≥ 2n, we have 
(15) e) Show that the rationals are dense in R by
showing that given any real
number x and error 1/n, there exists a rational number y such that
See Theorem 2.2.5 in Book.
(+10) *f) Show that there are an infinite number of
rational numbers in
between any two distinct real numbers.
First let’s show that between any two real numbers, there
is a rational number (which
was In-Class Example #1, Set 3, b), i.e., show that
Case 1: Suppose b - a > 1. Since the distance between a
and b is greater than 1, we
can actually find an integer between a and b: if a ∈Z, we then know that b > a +
1, so
take q = a + 1. If 
then take q = [a] + 1,
where the 
means “integer part” (e.g.
[3.14159] = 3).
Case 2: Suppose 
Since
b - a > 0 we know that (Axiom of Archimedes)
s.t. b - a > 1/k, which means that bk - ak >
1. So we can apply case 1 to ak and
bk instead of a and b and again, find an integer, say m s.t. ak < m < bk, which
can be
rewritten a < m/k < b. Since m/k is rational, we then take q = m/k.
Having shown that there is a rational number between any
two distinct real numbers,
we can certainly repeat the above procedure on a and q or on q and b. In fact,
nothing
restricts us from repeating this infinitely.
Set 6 -> Limits and Completeness.
(15) a) Let 
be a
sequence of real numbers such that 
and set 
Show that the sequence
converges.
To show that the sequence converges, we can show that it
is Cauchy. Fix 1/n. We need
to find a positive integer m s.t. 
. We will
use the fact that
(it’s a convergent series, which in fact is
= 1). Without any loss of generality,
assume that j ≤ k.
And for any fixed 1/n, we can find a big enough m s.t.
for any j, k≥m, which
shows that the sequence {yk} is Cauchy and thus converges. This
argument is sufficient to
deserve full marks. However, if we want to be formal, we need to show what value
of m
must be chosen. To find it, we can work backwards and find that
(with a little
algebra).
(+10) *b) Prove that if 0 < a < 2, then
Prove that the sequence
converges. What is the limit? (Help: if
then
If 0 < a < 2, then 
This shows that
so the sequence converges by Theorem 3.1.2 (Book). Denote
the sequence by 
Then
the sequence 
is the same as
. So the hint shows that
