MAT 137Y 20082009 Winter Session, Solutions to Problem Set Supplement #1
2
(a) Suppose
x
and
y
are numbers such that
y

x
>
1. Consider
b
x
c
. Then
b
x
c ≤
x
<
y
<
y
+
1. So
x
<
b
x
c
+
1
<
y
, so
k
=
b
x
c
+
1 is an integer such that
x
<
k
<
y
.
(b) Since
x
<
y
, consider the number 1
/
(
y

x
)
. Pick any positive integer
n
such that
n
>
1
/
(
y

x
)
.
Then
y

x
>
1
n
.
(c) Since there exists
n
∈
N
such that
y

x
>
1
n
, then
ny

nx
>
1. By part (a), there exists
k
∈
Z
such that
nx
<
k
<
ny
, or
x
<
k
n
<
y
. But
k
and
n
are integers, and
n
6
=
0, so
k
n
is rational. So there
exists a rational number
r
=
k
n
such that
x
<
r
<
y
, thereby showing that the rationals are dense.
(d) Now suppose we are given two rational numbers
r
and
s
such that
r
<
s
. Since
√
2
/
2 is irrational,
then
0
<
√
2
2
<
1
=
⇒
0
<
√
2
2
(
s

r
)
<
(
s

r
) =
⇒
r
<
r
+
√
2
2
(
s

r
)
<
s
.
Then
r
+
√
2
2
(
s

r
)
is irrational (otherwise
√
2
2
(
s

r
)
would be rational, which would imply that
√
2
2
is irrational since
s

r
is rational, a contradiction). Therefore, between any two rational
numbers there exists an irrational number.
(e) If
x
<
y
, then by part (b) there exists
r
0
∈
Q
such that
x
<
r
0
<
y
. But
r
0
∈
R
, so by part (b) there
exists
r
1
∈
Q
such that
r
0
<
r
1
<
y
. Thus,
x
<
r
0
<
r
1
<
y
. By part (c), there exists
t
6∈
Q
such
that
r
0
<
t
<
r
1
. Therefore
x
<
r
0
<
t
<
r
1
<
y
, or
x
<
t
<
y
, so there is an irrational number
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 Spring '08
 UPPAL
 2m, 2k, 2 j, 2 min, max S

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