Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I: | (69) |
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**Problem 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 .

**Solution. **

- 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 , contradicting 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.

- First,
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
and hence
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 .

**Problem 2. **

- 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 .

**Solution. **

- Yes. For example, the identity function has this property.
- Suppose for all . Then
if and only if
iff
iff
iff . So satisfies
iff
.
- If
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

**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
do not
agree with the standard one:

- For all there is an such that if , then .
- For all there is a such that if , then .

**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 means . This imposes no condition on , so need not be smaller than .

**Problem 5. ** Suppose that is continuous at 0 and
and that
for all . Show that is
continuous at 0.

**Solution. ** At the inequality
reads
, and hence . We claim
that
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
then
and the definition of
is satisfied.

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Dror Bar-Natan 2002-10-23