Math 301/ENAS 513: Homework 2 Solution
Triet Le
September 21, 2007
1. Show that for any reals
x
and
y
with
x < y
, there are a rational
r
and an irrational
t
such that
x < r < y
and
x < t < y
.
Solution 1.
Suppose that
x >
0
. Let
z
=
y

x >
0
and let
r
0
>
0
be any rational
number. By
O5
, there exist a positive integer
n
such that
r
0
< nz
and
r
0
< nx
. This
implies
0
<
r
0
n
< z
and
r
0
n
< x
. Now let
m
be the smallest positive integer such that
mr
0
n
> x
. Thus
(
m

1)
r
0
n
≤
x
. We have,
x <
mr
0
n
=
(
m

1)
r
0
n
+
r
0
n
≤
x
+
r
0
n
< x
+
z
=
y.
Therefore,
r
=
mr
0
n
is a rational such that
x < r < y
.
Replacing
r
0
above by any positive irrational, say
√
2
, and follow the same construction
as above, we would obtain an irrational
t
such that
x < t < y
.
Now suppose that
x
≤
0
, we then replace
x
and
y
by
x

m
and
y

m
, where
m
is
any integer less than
x
. Now
0
< x

m < y

m
. Follow the same construction as
above, we obtain a rational
r
and an irrational
t
such that
x

m < r < y

m
and
x

m < t < y

m
. Thus
r
+
m
and
t
+
m
are the rational and irrational that solve
the original problem.
2. Show that the Archemedean axiom
O5
follows from the Least Upper Bound Property
O6
, together with the other axioms for the reals.
Solution 2.
Let
a, b
∈
R
be both positive. The the set
A
=
{
n
∈
N
:
n
·
a <
=
b
}
is either empty or bounded. If
A
is empty (when
a > b
), pick any
n
∈
N
, we have
b < n
·
a
. If
A
is not empty (when
a
≤
b
), then the least upper bound
N
= sup
{
A
}
exists and in
N
. Thus
(
N
+ 1)
·
a > b
.
3. True or False: For every
± >
0 there are positive integers
m
and
n
such that the
inequality

√
n

√
m

π

< ±
is true.
1
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View Full DocumentSolution 3.
Suppose
√
n

√
m
≈
π
. Let
k
=
n

m
. Then
k
=
n

m
= (
√
n

√
m
)(
√
n
+
√
m
)
≈
π
(
√
n
+
√
m
≈
π
(
π
+ 2
√
m
) =
π
2
+ 2
π
√
m.
Let
k
= [
π
2
+ 2
π
√
m
]
to be the integer part of
π
2
+ 2
π
√
m
. Then we have

√
n

√
m

π

=
±
±
±
±
q
m
+ [
π
2
+ 2
π
√
m
]

√
m

π
±
±
±
±
≤
±
±
±
±
q
(
√
m
+
π
)
2
+ 1

(
√
m
+
π
)
±
±
±
±
=
±
±
±
±
±
²
q
(
√
m
+
π
)
2
+ 1

(
√
m
+
π
)
³
p
(
√
m
+
π
)
2
+ 1 + (
√
m
+
π
)
p
(
√
m
+
π
)
2
+ 1 + (
√
m
+
π
)
±
±
±
±
±
=
1
p
(
√
m
+
π
)
2
+ 1 + (
√
m
+
π
)
Now for any
± >
0
, pick
m
large enough so that
1
√
(
√
m
+
π
)
2
+1+(
√
m
+
π
)
< ±
. This implies
that

√
n

√
m

π

< ±,
with
n
=
m
+ [
π
2
+ 2
π
√
m
]
.
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 Fall '07
 TrietLe
 Math, Rational number, binary expansion

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