For the second step, we consider
x
≤
0. Choose an integer
k
such that
k >

x

. Then we
apply the same argument above to the positive number
x
+
k
and
y
+
k
to conclude that
x
+
k < q < y
+
k
for some rational number
q
. Then
r
:=
q

k
satisfies
x < r < y
.
Corollary 0.4
Let
x, y
∈
R
with
x < y
. There exists an irrational number
w
such that
x < w < y.
Proof:
By Theorem 0.3, there exists a rational number
r
such that
x
√
2
< r <
y
√
2
.
It follows that
x < r
√
2
< y
, and
w
:=
r
√
2 is irrational.
Now we can show
Theorem 0.5
Let
p
be a prime number. Then
√
p
∈
R
, i.e., there is a positive
x
∈
R
such
that
x
2
=
p
.
27
Proof:
Step 1. Define
x
Let
S
:=
{
r
∈
R

r >
0
, r
2
< p
}
. The set
S
is nonempty.
Also, since
r
2
< p < p
2
, it implies that
r < p
, i.e.,
S
is bounded above. By the completeness
axiom,
x
:=
sup
(
S
) exists in
R
. It remains to show that
x
2
=
p
. It suffices to show that
neither
x
2
< p
nor
x
2
> p
can hold.
Step 2. Show
x
2
< p
cannot hold
We want to find a large integer
n
such that
3
x
2
<
x
+
1
n
2
< p,
i.e., to find an integer
n
such that
x
2
+
2
x
n
+
1
n
2
< p,
i.e.,
x
2
+
1
n
2
x
+
1
n
< p,
i.e., to find an integer
n
such that
x
2
+
1
n
2
x
+ 1
< p,
i.e.,
1
n
<
p

x
2
2
x
+ 1
.
Here we use
p > x
2
to imply
p

x
2
2
x
+1
>
0. In other words,
n >
2
x
+ 1
p

x
2
.
We have found the integer
n
.
Step 3. Show
x
2
> p
cannot hold
We want to find a large integer
m
such that
p <
x

1
m
2
< x
2
,
i.e., to find an integer
m
such that
p < x
2

2
x
m
+
1
m
2
< p,
3
If this is true, then
sup
(
S
)
≥
x
+
1
n
, which is a contradiction with the fact =
sup
(
S
).
28
i.e.,
p < x
2

1
m
2
x

1
m
< p,
i.e., to find an integer
m
such that
p < x
2

1
m
2
x

1
< p,
i.e.,
1
m
<
x
2

p
2
x

1
,
i.e.,
m >
2
x

1
x
2

p
.
We have found the integer
m
.
Appendix. Problems did in the class
3.3.1. True or false:
(a) If a nonempty subset of
R
has an upper bound, then it has a least upper
bound.
(b) If a nonempty subset of
R
has an infimum, then it is bounded.
(c) Every nonempty bounded subset of
R
has a maximum and a minimum.
(d) If
m
is an upper bound of
S
and
m
0
< m
, then
m
0
is not an upper bound
of
S
.
(e) If
m
=
inf
(
S
)
and
m
0
< m
, then
m
0
is a lower bound of
S
.
(f) For each real number
x
and each
>
0
, there exists
n
∈
N
such that
n
> x
.
Solution:
(a) Yes. By the definition of least upper bound.
(b) False. For example,
S
=
N
.
(c) False. For example,
S
= (0
,
1).
(d) False. For example,
S
= [0
,
2],
m
= 4 and
m
0
= 3. Here
m
0
< m
but
m
0
still is an
upper bound.
(e) True. By the definition of
inf
(
S
).
(f) True. If
x
≤
0, we are done. If
x >
0, by Theorem 3.3.10(a), there is
n
∈
N
such that
n >
x
, i.e.,
n
> x
.
3.3.6b. Suppose that
m
and
n
are both maxima of a set
S
. Prove
m
=
n
.
Proof:
We have
m
≤
n
29
because
n
=
max
(
S
) and
m
∈
S
, and
n
≤
m
because
m
=
max
(
S
) and
n
∈
S
. Then
m
≤
n
and
n
≤
m
⇐⇒
m
=
n.
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 Fall '08
 Staff