. For a more detailed
treatment of both the ﬂoor and ceiling see the book
Concrete Mathemat
ics
[5]. According to the definition of
x
we have
x
= max
{
n
∈
Z

n
≤
x
}
(4.1)
13
14
CHAPTER 4.
THE FLOOR AND CEILING OF A REAL NUMBER
Note also that if
n
is an integer we have:
n
=
x
⇐⇒
n
≤
x < n
+ 1
.
(4.2)
From this it is clear that
x
≤
x
holds for all
x,
and
x
=
x
⇐⇒
x
∈
Z
.
We need the following lemma to prove our next theorem.
Lemma 4.1.
For all
x
∈
R
x
−
1
<
x
≤
x.
Proof.
Let
n
=
x
.
Then by (4.2) we have
n
≤
x < n
+ 1.
This gives
immediately that
x
≤
x
, as already noted above. It also gives
x < n
+ 1
which implies that
x
−
1
< n
, that is,
x
−
1
<
x
.
Exercise 4.1.
Sketch the graph of the function
f
(
x
) =
x
for
−
3
≤
x
≤
3.
Exercise 4.2.
Find
π
,
π
,
√
2 ,
√
2 ,
−
π
,
−
π
,
−
√
2 , and
−
√
2 .
Definition 4.2.
Recall that the
decimal representation
of a positive in
teger
a
is given by
a
=
a
n
−
1
a
n
−
2
· · ·
a
1
a
0
where
a
=
a
n
−
1
10
n
−
1
+
a
n
−
2
10
n
−
2
+
· · ·
+
a
1
10 +
a
0
(4.3)
and the
digits
a
n
−
1
, a
n
−
2
, . . . , a
1
, a
0
are in the set
{
0
,
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
}
with
a
n
−
1
= 0. In this case we say that
the integer
a
is an
n
digit number
or
that
a
is
n
digits long.
Exercise 4.3.
Prove that
a
∈
N
is an
n
digit number where
n
=
log(
a
) +1.
Here log means logarithm to base 10.
Hint: Show that if (
4.3
) holds with
a
n
−
1
= 0
then
10
n
−
1
≤
a <
10
n
.
Then apply the
log
to all terms of this
inequality.
Exercise 4.4.
Use the previous exercise to determine the number of digits
in the decimal representation of the number 2
3321928
.
Recall that
log(
x
y
) =
y
log(
x
)
when
x
and
y
are positive.
Chapter 5
The Division Algorithm
The goal of this section is to prove the following important result.
Theorem 5.1 (The Division Algorithm).
If
a
and
b
are integers and
b >
0
then there exist unique integers
q
and
r
satisfying the two conditions:
a
=
bq
+
r
and
0
≤
r < b.
(5.1)
In this situation
q
is called the
quotient
and
r
is called the
remainder
when
a
is divided by
b
.
Note that there are two parts to this result.
One
part is the EXISTENCE of integers
q
and
r
satisfying (5.1) and the second
part is the UNIQUENESS of the integers
q
and
r
satisfying (5.1).
Proof.
Given
b >
0 and any
a
define
q
=
a
b
r
=
a
−
bq
Cleary we have
a
=
bq
+
r
.
But we need to prove that 0
≤
r < b
.
By
Lemma 4.1 we have
a
b
−
1
<
a
b
≤
a
b
.
Now multiply all terms of this inequality by
−
b
. Since
b
is positive,
−
b
is
negative so the direction of the inequality is reversed, giving us:
b
−
a >
−
b
a
b
≥ −
a.
15
16
CHAPTER 5.
THE DIVISION ALGORITHM
If we add
a
to all sides of the inequality and replace
a/b
by
q
we obtain
b > a
−
bq
≥
0
.
Since
r
=
a
−
bq
this gives us the desired result 0
≤
r < b
.
We still have to prove that
q
and
r
are uniquely determined. To do this
we assume that
a
=
bq
1
+
r
1
and
0
≤
r
1
< b,
and
a
=
bq
2
+
r
2
and
0
≤
r
2
< b.
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 Fall '09
 Logic, Number Theory, Prime number, gcd, lemmas