Solution of Term Exam 1
- Prove directly from the postulates for the real numbers and from the
relevant definitions that if and , then . If
you plan to use a formula such as
you don't need to
prove it, but of course you have to be very clear about how it is used.
- Use induction to prove
that any integer can be written in exactly one of the following
two forms: or , where is also an integer.
- Prove that there is no rational number such that .
- If then means that and therefore .
Along with , and P11, it follows that . If
then again along with and P11 we get that , so in either
case is assured. Now
is by definition the same as , and therefore
Had been negative, then would have been positive and
by and P12 we'd have that
Equation (1). Hence is positive, and this by
definition means that .
- First we show using induction that every natural number (positive
integer) can be written in at least one of the forms or
for an integer . Indeed, for we write
as required. Now if is of the form for an integer , then
is of the second allowed form, and if is of the form
for some integer then
is of the first
allowed form, for is also an integer. Therefore if can be
written in either of the required forms then so is , and the
inductive proof is completed.
Let's deal with 0 and with the negative integers now. First,
so 0 is of the form . Next, if is an integer, then is a
positive integer and therefore or . In the former
case, and we are done. In the latter case,
and again we are done.
Finally, an integer cannot be of both forms at the same time, for if
we could write and with integer and , then
we'd have that
, which is
. But is an
integer and it is easy to show that is not twice an integer.
Having said all that, we can call the integers of the form ``even''
and the integers of the form ``odd'', and then every integer is
either even or odd but never both.
and hence if
is odd then so is . Therefore if is even then so is . Now
assume by contradiction that there is some rational number with
and write where and are integers and is the least
positive integer for which it is possible to present in this form. Now
hence is even hence is even hence we can
find an integer with . But then becomes
so is even and hence so is , so for
some positive (which of course is smaller than itself). But now
contradicting the minimality of . Therefore there is
no rational number with .
- Suppose . Are there any functions such that
- Suppose that is a constant function. For which functions does
- Suppose that
for all functions . Show
that is the identity function .
- Yes. For example, the identity function has this property.
- Suppose for all . Then
if and only if
iff . So satisfies
for all functions then in particular
for all constant functions . But then by the
previous part for all constant we have and this precisely
means that is the identity function .
Problem 3. Sketch, to the best of your understanding, the
graph of the function
(What happens for near 0? For large ? Where does the graph lie
relative to the graph of the function ?)
Solution. The first graph below shows (above the
axis) and (below the axis. The second shows the sum of the
two, the desired function :
For near 0 our function goes
to , for large it goes to . It is always below
but for large it is very near .
Problem 4. Write the definition of
and give examples to show that the
following definitions of
agree with the standard one:
- For all there is an
such that if
- For all
there is a such that if
Solution. The definition is: For every
there is a so that whenever
we have that
. The required examples:
- This is satisfied whenever is bounded and regardless of its
limit. Indeed, choose bigger than where is a bound
on , and
is always true.
- According to this definition, for example,
is false, and hence it cannot be equivalent to the standard
definition. Indeed, in this case
This imposes no condition on , so need not be smaller than
Problem 5. Suppose that is continuous at 0 and
for all . Show that is
continuous at 0.
Solution. At the inequality
, and hence . We claim
and hence that is continuous at
0. Indeed let
be given. Then
and by the
continuity of at 0 we can find a so that whenever
we have that
. But then if
and the definition of
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