Math 25: Advanced Calculus
UC Davis, Spring 2011
Math 25 — Solutions to Homework Assignment #4
1. Using the Archimedean Theorem, prove each of the three statements that follow the proof of that theorem
in section 1.7 of the textbook.
(a) No matter how large a real number
x
is given, there is always a natural number
n
larger.
Proof.
Suppose that there is some
x
such that no natural number is larger than
x
. Then
x
is an upper
bound for the set of natural numbers, which contradicts the Archimedean Property.
(b) Given any positive number
y
, no matter how large, and any positive number
x
, no matter how small,
there is some natural number
n
such that
nx > y
.
Proof.
Assume there is no such
n
. That is
nx
≤
y
for all
n
. Note that this is equivalent to
n
≤
yx

1
for all
n
, which contradicts item (a).
(c) Given any positive number
x
, no matter how small, once can always find a fraction
1
n
such that
1
n
< x
.
Proof.
Assume there is no such
n
. Then
1
n
≥
x
for all
n
. Note that this is equivalent to
n
≤
x

1
for all
n
, which contradicts item (a).
2. The completeness axiom can be used to construct numbers whose existence we only suspected before. As an
illustration of this idea, prove that the number
α
= sup
{
x
∈
R
:
x
2
<
2
}
exists in
R
and satisfies
α
2
= 2. (Hint: show that either of the assumptions
α
2
<
2 and
α
2
>
2 lead to a
contradiction.)
Proof.
We will use the following auxiliary result: if
x
2
< y
2
, then
x < y
where
x, y >
0. To see this, note
that
x
2
< y
2
is equivalent to
x
2

y
2
<
0 or (
x

y
)(
x
+
y
)
<
0. Since
x, y >
0, we have that
x
+
y
6
= 0, so
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 Winter '11
 Wiley
 Differential Equations, Calculus, Equations

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