The following theorem is an extension of the Well-Ordering Axiom. It will be used to justify the definition of the greatest integer function
Theorem.
(a) Suppose S is a nonempty set of integers which is bounded below: There is an integer M such that
for all
. Then S has a smallest element.
(b) Suppose S is a nonempty set of integers which is bounded above: There is an integer M such that
for all
. Then S has a largest element.
Proof. (a) Suppose S is a nonempty set of
integers, and for all
. I'll consider two cases.
First, if , then
for all
. This shows that S is a subset of the
positive integers, so it has a smallest element by the Well-Ordering
Axiom.
Next, suppose . In this case, the idea is to
"translate" S to the right to get a subset of the positive
integers to which Well-Ordering can be applied.
Consider the set
If , then
This shows that the elements of are positive integers. By Well-Ordering,
has a smallest element y. Thus,
, and
for all
.
Since , I can write
for some
.
Now if , then
Thus, z is an element of S which is at least as small as any other element of S --- that is, z is the smallest element of S.
(b) Suppose S is a nonempty set of integers and for all
. Then
for all
, so the followng set is bounded below
By part (a), has a smallest element. Suppose
that y is the smallest element of
. Thus,
, and
for all
.
Since , I can write
, where
.
Now if ,
Thus, z is an element of S which is at least as large as any other
element of S --- that is, z is the largest element of S.
Definition. If x is a real number, then denotes the greatest integer
function of x. It is the largest integer less than or equal to
x.
It's probably obvious to you based on your experience with the real
numbers that there is such an integer . You might find justifying this a bit of a
challenge. Here's the idea.
For every real number x, the Archimedean Axiom
for the real numbers says that there is an integer n such that .
Lemma. For every , there is an integer m such that
.
Proof. Apply the Archimedean Axiom to to get an integer n such that
. Negating the inequality, I get
. Then
is an integer less than x.
Now go back to the greatest integer function. Let . Why is there a largest integer less than
or equal to x?
First, the lemma shows that there is an integer less than x, so the set S of integers less than or equal to x is nonempty.
Second, S is bounded above by x. By the Archimedean Axiom, there is
an integer n such that . Then n is also an
upper bound for S.
By part (b) of the theorem, S has a largest element. That element is
.
That was a lot of work to show that the function is actually defined, particularly since this fact was
probably obvious to you form the start!
When you learn about "ordinary" math like calculus, or number theory, you usually "start in the middle": A lot of things (that are hopefully plausible) are taken for granted. We don't normally go back to the very basic axioms, and perhaps this discussion helps you understand why we don't: We'd have to go through a lot of technicalities, and we wouldn't have the time to get to the ideas of calculus or number theory.
The following lemmas and examples should give you some ideas about how to work with the greatest integer function.
Example. Compute ,
, and
(Notice that is not
equal to -1.)
Example. Sketch a graph of .
Lemma. If x is a real number, then
Proof. By definition, . To show that
, I'll give a proof by
contradiction.
Suppose on the contrary that . Then
is an integer less than or equal to x, but
--- which contradicts the fact
that
is the largest integer less than
or equal to x. This contradiction implies that
.
Lemma. If and
, then
.
Proof. Suppose . I want to show that
.
Assume on the contrary that . Since
is the {\it greatest} integer which is less than or
equal to x, and since
is an integer which is
greater than
, it follows that
can't be less than or equal to x. Thus,
. But
, so
, which is a contradiction.
Therefore, .
Example. Let x be a real number and let n be
an integer. Prove that .
First, , so
. Now
is an integer less than or equal to
, so it must be less than or equal to the
greatest integer less than or equal to
--- which is
:
Next, , so
.
is an integer less than or equal to x.
Therefore, it must be less than or equal to the greatest integer less
than or equal to x --- which is
:
Adding n to both sides gives
Since and
, it follows that
.
Example. Consider the sequence
Show that it consists of two 1's, four 2's, six 3's, and so on:
Note that for , the expressions
and
are not
integers. In addition,
Thus, the interval
contains
integers.
Now
Hence, .
That is, if n is one of the integers between
and
, the
value of
is k. This shows that the given sequence consists of
two 1's, four 2's, six 3's, and so on.
Copyright 2019 by Bruce Ikenaga