Download Math 142A Homework Assignment 3 Selected Solutions 2.2.4 Show

Math 142A Homework Assignment 3
Selected Solutions
2.2.4 Show that the set of irrational numbers fails to be closed.
Solution Consider the sequence { πn }. Observe that nπ is an irrational number for
every index n; for, if r = πn were rational, then n r would be a product of a
natural number and a rational number equal to the irrational number π, which
is pure balderdash. Next, observe that nπ → 0; for, given ǫ > 0, the archimedean
property of the real numbers assures us there is a natural number N such that
1
< πǫ so that πn < ǫ for every index n ≥ N. Since { nπ } is a sequence of irrational
N
numbers converging to the rational number 0, the set of irrational numbers fails
to be closed.
2.3.11 For a pair of positive numbers a and b, define sequences {an } and {bn } recursively
as follows: define a1 = a and b1 = b. If n is an index for which {an } and {bn } have
been defined, define
an+1 =
an + bn
2
and bn+1 =
p
an bn .
a. Use Exercise 10 to prove that for every index n ≥ 2,
an ≥ an+1 ≥ bn+1 ≥ bn .
b. Show that the sequences {an } and {bn } converge. Then show that {an } and {bn }
have the same limit.
Solution
√
n
≥ an bn = bn+1 for every index n. Thus
a. • By Exercise 10, an+1 = an +b
2
an ≥ bn for every index n ≥ 2.
n
n
= an −b
≥ 0 for every index n ≥ 2, by the above.
• an − an+1 = an − an +b
2
2
Thus, an ≥ an+1 for every index n ≥ 2.
√
p
an bn
• bn+1
=
= abnn ≥ 1 for every index n ≥ 2 since an ≥ bn for every
bn
bn
index n ≥ 2. Thus, bn+1 ≥ bn for every index n ≥ 2.
b. By part (a), {an } is monotonically decreasing and bounded below by b. Thus,
there is a number α for which an → α, by the monotone convergence theorem.
Similarly, {bn } is monotonically increasing and bounded above by a. Thus,
there is a number β for which bn → β, by the monotone convergence theorem.
n
n
→ α+β
and an+1 = an +b
for every index n ≥ 2, it follows that
Since an +b
2
2
2
α+β
α = 2 whence α = β.