Solution of Term Exam 1
Problem 1. Find formulas for
,
and
in terms of
. (You may use any
formula proven in class; you need to quote such formulae, though you
don't need to reprove them).
Solution. (Graded by Shay Fuchs) Using the formulas
and
and taking
we get
Dividing the numerator and denominator by
this
becomes
Likewise using
we get
Finally, dividing these two formulas by each other we get
Problem 2.
 Let be a natural number. Prove that any
natural number can be written in a unique way in the form ,
where and are integers and .
 We say that a natural number is ``divisible by '' if is
again a natural number. Prove that is divisible by if and only if
is divisible by .
 We say that a natural number is ``divisible by '' if is
again a natural number. Is it true that is divisible by if and only
if
is divisible by ?
Solution. (Graded by Brian Pigott)
 We prove this assertion (without uniqueness) by induction. If
write (if ) or (if ). In either case
the assertion is proven for . Now assume can be written in the
form , where and are integers and . If
then and so
is a formula of the
desired form for . Otherwise and so
, and again that's a formula of the
desired form for . This concludes the proof that every natural
number can be written in the form , where and are
integers and . Now assume it can be done in two ways; i.e.,
assume
where , , and are
integers and
. But then
and so
and so
. But
is an integer and so
is an integer. From
it follows that
and so
and so the integer must be 0. Thus
and
so . But then the equality
implies
and so and we see that the pair is
unique.
 An integer is divisible by iff is an integer iff
with an integer . Now if is divisible by then
with an integer and then
. So is
also times an integer (the integer ), and so is also
divisible by . Assume now that is not divisible by . By the
previous part with integer and and with .
Had been 0 we'd have had that is divisible by
contrary to assumption. So or . In the former case
, but then by the uniqueness of writing
as it follows that , so cannot be written in
the form , so is not divisible by . In the latter
case
and for the same reason again we
find that is not divisible by . So if is divisible by
so is , and if is not divisible by so is .
 No it's not true. Example: is not divisible by but
is divisible by .
Problem 3. A function is defined for
and has the graph plotted above.
 What are , and ?
 Let be the function . What are , and
?
 Are there any values of for which ? How many
such 's are there? Roughly what are they?
 Plot the graph of the function . (The general shape of your plot
should be clear and correct, though numerical details need not be precise).
 (5 points bonus, will be given only to excellent solutions and
may raise your overall exam grade to 105!) Plot the graphs of the
functions and .
Solution. (Graded by Derek Krepski)
 By inspecting the graph, , and .

,
and
.
 means . Denoting we must have ,
and inspecting the graph we find that . Thus . Inspecting
the graph we find that there are two values of for which this happens
and they are approximately and .
 and 5.:
Problem 4.
 Define ``
'' and ``
''.
 Prove that if
and
then
.
 Prove that if
then
and
.
 Draw the graph of some function for which
and
.
Solution. (Graded by Shay Fuchs)
 ``
'' means that for every
there is a so that whenever
we have that
, while ``
'' means that for
every
there is a so that whenever
(i.e., whenever
) we have that
 Let
be given. Using
choose
so that whenever
we have that
. Using
choose
so that whenever
we have that
. Set
and assume
. If then
and by the
choice of it follows that
. If then
and by the choice of it follows
that
. So in any case,
as
required.
 Let
be given. Using
choose
so that whenever
we have that
.
But then if
then certainly
so by the choice
of we get
. Thus
. A
similar argument shows that also
.

Problem 5. 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. (Graded by Derek Krepski)
 This is satisfied whenever there exists a constant so that
for all and regardless of the limit of . Indeed,
choose bigger than where is a constant as in the
previous sentence (for example, if is , then can be
chosen to be ), and then
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
.
The results. 89 students took the exam; the average
grade was 59 and the standard deviation was about 18.5.
Dror BarNatan
20041020