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Anna Medvedovsky
medved@brandeis.edu
Math 23b / Spring 2009
HW #7 solutions
1.
Let
y > x
.
(a)
Prove that there is a natural number
n
such that
n
(
y

x
)
>
1
.
The numbers
y

x
and 1 are both positive reals, so the archimedean property tells
us that there exists an
n
∈
N
satisfying
n
(
y

x
)
>
1.
(b)
Prove that there is an integer
m
such that
m > nx
≥
m

1
.
We’ll be using the wellordering property to ﬁnd
m
. But the classic version of WOP
(any nonempty subset of the positive integers has a least element) will be tricky to
use here because we don’t know that
nx
is positive. There are a few ways to deal
with this problem, but the one we’ll take is to prove a lemma.
Lemma: If
S
is a nonempty subset of
Z
and
S
is bounded below, then
S
has a
least element.
Proof:
Let
r
be a lower bound for
S
.
First we need to ﬁnd an
integral
lower bound
b
for
S
.
If
r
is nonnegative, set
b
= 0
∈
Z
. Then
b
≤
r
, so
b
is also a lower bound for
S
: for any
s
∈
S
, we have
s
≤
r
≤
b
.
And if
r
is negative, use the archimedean property to ﬁnd a natural number
z
such that
z
·
1
>

r
, and set
b
=

z
. In this case
b < r
, so that
b
is again
a lower bound for
S
. Note that
b
∈
Z
.
Now deﬁne a set
T
=
{
s

b
:
s
∈
S
}
by translating
S
by
b
. I claim that
T
is
a nonempty subset of the positive integers. First oﬀ, since
S
⊂
Z
and
b
∈
Z
we
know that every element of
T
is an integer. Moreover, since
b
is a lower bound
for
S
, we have
s

b >
0 for all
s
∈
S
, so that in fact every element of
T
is
positive. Finally, since
S
is nonempty,
T
is nonempty also.
Since
T
is a nonempty subset of the positive integers, we know by wellordering
that
T
has a least element; call it
t
. Then
t
+
b
∈
S
, and I claim that
t
+
b
is the
least element of
S
. Indeed, for every
s
∈
S
, we have
s

b
∈
T
, so that
s

b
≤
t
.
Therefore
s
≤
t
+
b
for all
s
∈
S
, and
t
+
b
is the least element of
S
, as claimed.
Back to the problem at hand: ﬁnding
m
such that
m > nx
≥
m

1.
Let
S
=
{
s
∈
Z
:
s > nx
}
. Then
S
is a subset of
Z
, and
S
is clearly bounded below
by
nx
. So all we need to do to WOP this
S
is to show that
S
is nonempty. Again
we take cases. If
nx
≤
0, then 1
> nx
, so 1
∈
S
. And if
nx
is positive, then using
the archimedean property we can ﬁnd a natural number
s
such that
s
·
1
> nx
. Then
s
∈
S
. In either case,
S
is nonempty, so it has a least element, by the lemma above.
Let
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 Spring '09
 BongLian
 Math

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