Chapter 2 Exercises
 Let
be a function where S is either the set of
rational numbers or the set of real numbers. The range of f is the
set of all elements t in T such that t = f(s) for some s in S. Find the
range of the following functions:
 The identity function
given by i(x) = x for all x in S.
 The square function
given by
for all x
in S.
 The cube function
given by
for all x in S.
 The function defined by
.
 The function defined by
.
 A function
is said to be onetoone
(or injective) if f(x) = f(y) only happens when x = y. For each of
the functions of the last exercise, either show that the function is
onetoone, or give an example showing that it is not.
 Given a function
, and a subset U of S, there is
a function
defined by g(x) = f(x) for all x in U. We
say that the function g is the function induced on U by f. For each
of the functions of the last exercise which were not onetoone, find
subsets U of S such that the function induces onetoone functions on the
set U.
 A function
is said to be onto or surjective
if the range of f is the whole range space T. For each of the functions
in the first problem, determine whether or not the function is onto.
 A function
which is both onetoone and onto
defines a function
by g(y) = x if and only if f(x) = y.
So, if the graph of the function contains (x, y), the graph of the function
g contains (y, x).
 For each of the functions in the first problem which are onetoone and
onto, determine the inverse function.
 Each of the induced onetoone functions defined in the third
exercise can be considered to be functions mapping onto their range. Find the
inverse functions.
 A function
where S and T are subsets of the set of
real numbers is said to be increasing if f(x) < f(y) for all x and y
in S with x < y. It is said to be decreasing if f(x) > f(y) for
all x and y in S with x < y.
 For each of the functions in the first exercise, determine whether the
function is increasing, decreasing, or neither.
 Do the same for each of the induced functions in the third exercise.
 For each of the functions in the first exercise that are neither increasing
nor decreasing, determine subsets U on which the induced function is
increasing or decreasing.
 Give complete proofs of all the assertions of Proposition 3.
 Reexamine the rules in Chapter 1, section 1.1, 1.2, and 1.3. Identify
any results there which you cannot prove for any field F.
 Let n be a positive integer. We say that two integers a and b are
congruent modulo n if a  b is evenly divisible by n; this is written
.
 Show that congruence modulo n satisfies the common properties
of equality:

(reflexivity)
 If
, then
(symmetry)
 If
and
, then
(transitivity)
 Show that congruence modulo n is compatible with addition and multiplication, i.e.
 If
, then for every integer c, one has
.
 If
, then for every integer c, one has
.
 If
and
a + c equiv a + c text{ (mod n)}
a equiv b text{ (mod n)}
c equiv d text{ (mod n)} then for every integer c, one has
.
 Congruence modulo n partitions the set of integers into n subsets, such
that any two numbers in the same subset are congruent modulo. In particular,
the numbers 0, 1, ..., n  1 are all in different ones of these subsets.
The subsets are called congruence classes modulo n.
The above properties of compatibility of with addition and multiplication
mean that we can define addition on the congruence classes by simply
doing the operation on any of the integers in the congruence classes; which
integer one chooses in the congruence class does not matter because the
result will always be in the same congruence class.
Verify that all the field axions except for the existence of inverses
holds for the the set of congruence classes modulo n.
 Show that if p is prime, then for any integer a, the product of
a with any two different nonzero congruence classes modulo p cannot give the same congruence
class. So, if we multiply a times all the the nonzero congruence classes, we
get all the nonzero congruence classes back. In particular, one of these
must be the congruence class containing 1. So, if p is a prime, then
the congruence classes modulo p is a field.
 Show that if p is a prime, then even though the congruence classes
modulo p form a field, there is no way that it can be made into an
ordered field.
 The factorial function
is the product of all positive integers from
1 to n, i.e.
. When n = 0, one defines 0! = 1. Give
an inductive definition of
assuming that you have not yet defined the
product of n numbers.
 For every pair of nonnegative integers n and k with
, the
binomial coefficient
is defined to
be
. Use induction to
prove the identity:
This identity is the basis of
Pascal
's Triangle.
 The binomial theorem gives a formula for expanding any positive integer
power of a binomial. Prove this theorem:
Binomial Theorem: Let n be a nonnegative integer and a and b be
real numbers. Then
 Prove the following formulas by induction






 Although it is easy enough to verify by induction that each of the
previous formulas hold, it is not so obvious how one came up with the formula
to verify in the first place. This exercise will give you an idea of how to
do it.
 If f(x) is a nonzero polynomial of degree d, show that
is a
polynomial of degree d  1. (For the purposes of this result, it is convenient
to define the degree of the zero polynomial to be 1.)
 If f(x) = g(x) + h(x), show that
.
 Use the binomial theorem to calculate
.
 Suppose that for all
, one knows a polynomial
such that
. The last problem gave us
for i = 0, 1, 2, 3, and
4. Combine your work on this problem in order to show you how to
calculate
.
 Test out your method by using your result to find a formula
for
.
 Determine the flaw in the arguments:
 Theorem: Let n be a natural number. Any set of at most n elements
numbers contains at most one element.
Corollary: There are no finite sets with more than one element.
Proof of Theorem: Suppose the theorem is false. Let n be
the smallest natural number for which it is false. Clearly
since the theorem is true for n = 1. Let S be a set with no more than
n elements for which the assertion is false. Since n is the smallest natural
number for which the assertion is false, it is true for numbers less than n.
So the set S must have exactly n elements. Let the elements of S be
. Let
be the set containing every element of
S except for
; so its elements are
Since each
has less elements than S, the assertion must be true of
. But then we have
But since this is true for all choices of i, we have with a different
choice, say j
Since each of these leaves out only one of the
, we see that the two
sets of equalities together show that all the
are equal. So, all
the
are all equal. But then the assertion is true of S, contradicting
the choice of S.
 Define for numbers a and b:
Theorem: If a and b are two natural numbers with max(a, b) = n,
then a = b.
Proof: Prove the assertion by induction on n. The assertion for
n = 1 is clearly true. Assume that the assertion is true for n and try to
prove the assertion for n + 1. Let a and b be any two natural numbers such
that max(a, b) = n + 1. Then c = a  1 and d = b  1 satisfy
max(c, d) = n. But the assertion for n then implies that c = d. But
then a = c + 1 = d + 1 = b. So, the assertion is proved for n + 1. By
induction, the result is true for all natural numbers n.
Corollary: Any two natural numbers are equal.
 Let a, b, and c be natural numbers. Prove that if a and b are
relatively prime and a divides bc, then a divides c.
 Prove that the least common multiple of two natural numbers a and b is
.
 Prove that, if a natural number n is not divisible by any prime number
less than or equal to
, then n is a prime.
 Here is another proof of Euclid's Theorem that there are infinitely
many primes. The numbers
for
are called
Fermat numbers. He stated that they were all prime, but that has
since been proved false 
Euler was the first to factor
. The Fermat numbers are important
because
Gauss proved that a regular ngon can be constructed by straightedge and compass only
if and only if n is of the form
where p is either 4 or a prime of
the form
.
The known Fermat primes are 3, 5, 17, 257, and 65537. In 1732, Euler found the
factorization
. In the intervening
centuries,
for k = 6,7,8,9,10,11,12,13,14,15,16,18,19,23,36,38,39,55,63,
and 73 have been shown to be composite; no Fermat numbers beyond those originally
known to be prime have been shown to be prime. In case you are interested,
 if you don't think this is a large number,
calculate the number of truckloads of paper that would be required to write it out as a base
10 numeral.
 Show by induction that
 Use the last factorization to prove that
evenly divides
for all k > 0.
 Prove that
and
are relatively prime for all k > 0.
 Explain why this implies that Euclid's Theorem is true.
 Pythagorean Triples: From the most ancient times, it has been
known that a triangle with sides of length 3, 4, and 5 is a right triangle.
Almost equally ancient is the knowledge that a triangle with sides
of length 5, 12, and 13 is a right triangle. Of course, both of these
are consequences of the converse of the Pythagorean Theorem and the
computations
It is natural to ask if there are other right triangles whose sides
have natural number lengths. This means, we want to solve
where a, b, and c are required to be natural numbers. Given a solution,
(a, b, c) and any natural number r, it is trivial to verify that (ar, br, cr)
is also a solution; also, if a and b have a common factor f, then
(a/f, b/f, c/f) is also a solution. This makes it reasonable to define
a solution (a, b, c) to be primitive if a and b are relatively prime.
Then our problem is to find all primitive solutions.
 Letting x = a/c and y = b/c, we see that (x, y) is a point on the
unit circle
with rational number coordinates. In the section
on solving systems of simultaneous equations, one obtained a parametrization
of all solutions of this equation. Verify that if one allows the parameter t
to take on only rational solutions, then one obtains a parameterization of
all the solutions which are rational numbers.
 Determine conditions on t so that one has a parametrization of all
positive rational solutions.
 Let t = u/v where u and v are relatively prime natural numbers and
v > u. Then substitute these into the parametrization to obtain
Now, it is tempting, but not valid to conclude that the numerators and denominators
on each side are equal. However, one can conclude that there is a positive
rational number r such that
Explain why this is the case.
 Since r is a positive rational number, it can be written as r = s/t
where s and t are relatively prime natural numbers. Assume now that
(a, b, c) is a primitive solution and a is odd.
 Show that t has no odd prime factor.
 Show that v and u cannot both be odd.
 Show that t is not even. So, t = 1 and r = s is a natural number.
 Show that s = 1.
 One has shown that every primitive solution (a, b, c) with a odd
can be expressed in the form
where u and v are relatively prime natural numbers with v > u and exactly
one of u and v is even. Now verify that every (a, b, c) of this form
is a primitive solution with a odd.
 Having completely solved this equation, it would be natural
to ask for nonzero integer solutions of the equation
where n is a positive integer larger than 2. The assertion that
this equation has no solutions is called
Fermat's Last Theorem. Although Fermat provided no proof of the assertion,
one finds the following marginal note in his copy of
Diophantus's Arithmetica:
I have discovered a truly remarkable proof which this margin is too small to contain.
This challenge has generated some of the most beautiful mathematics of the
last three centuries, whole branches of mathematics have been created by some
of the world's best mathematicians to address the question of whether or not
Fermat's Last Theorem is true or false. This body of mathematics is vastly
more important than the actual resolution of the original assertion. However,
the theorem was finally proved in 1994 by
Wiles.
 Recall the equation
which came up in the problem of
bisecting the area of a trapezoid. Using the last problem as a model,
determine all the natural number solutions of this equation.
 The proofs of Corollaries 3 and 5 were based on the Euclidean Algorithm.
Let's see if we can't derive similar results without the Euclidean Algorithm.
 Let a and b be relatively prime integers. Show that the numbers
all lie in distinct congruence classes modulo b. Explain why this means
that ax + by = 1 has integer solutions.
 Let a and b be arbitrary nonzero integers. Show that the equation
ax + by = gcd(a, b) has integer solutions.
 Show that if
and
are solutions of ax + by = gcd(a, b),
then
and
for some integer
r. Conversely, if
is a solution and r is an integer, then
as defined here are also solutions.
 Let m and n be relatively prime natural numbers.
 Show that every sufficiently large integer q is of the form q = mx + ny
where x and y are nonnegative intgers.
 Find the greatest integer which cannot be so represented.
 Let p be a prime and a be an integer not a multiple of p.
. Show that
now two of these are in the same congruence class modulo p.
Explain why this means that ax + py = k has integer solutions for each
.
 Prove that
mod p.
 Fermat's Theorem: Show that for any prime p and any integer a which is not a multiple of
p, one has
mod p.
 Euler's Theorem: Show that for any nonzero integer n and and any
a relatively prime to n, one has
mod n
where
is the number of integers
which are
relatively prime to n.
 Show that in an Archimedean ordered field, if an infinite decimal has a limit,
then the limit is unique.
 Show that if F is an Archimedean ordered field in which every infinite
decimal in which every digit is even has a limit in F, then F is a field of
real numbers.
 Let F be the an ordered field, for example, you could take F to be
the field of rational numbers or the field of real numbers. Recall that
F(x) is the field of rational functions with coefficients in F. Its
elements are elements of the form
where p and q are
polynomials and q is nonzero. The leading coefficient of the
rational function
is the quotient of the leading
coefficient of p and the leading coefficient of q (or zero if p is zero).
One can define an order on F(x) by
if the leading coefficient of the first rational function is less than
the leading coefficient of the second rational function.
 Show that this definition makes F(x) into an ordered field.
 Show that F(x) is NOT Archimedean.
 Do infinite decimals in F(x) have unique limits?
 For every real number a and positive integer n, determine the number
of real
roots of a.
 Let a be a positive real number. Make a reasonable definition of
where p and q are integers. Does this quantity always exist?
If the function f defined by
an increasing
function of rational numbers?
 Show that the infinite sum
converges.
 The harmonic series is the
. The
harmonic series does not converge. The idea of
the proof is that the first term is clearly at least 1/2 as is the sum
of the next 2 terms, the sum of the succeeding 4 terms, and the sum of the
next 8 terms. Verify these assertions and see if you can show that the sum
is eventually larger than any number.
 Let F be an Archimedean ordered field in which every increasing bounded sequence of
numbers converges to an element of F. Show that F is a field of real numbers.
 Given any natural number b > 1, one can define a notion of infinite
decimals to the base b. Basically, one restricts the digits to lie between
0 and b  1 and interprets them using powers of b rather than powers of 10.
Show that if infinite decimals to the base b converge in an ordered field F,
then so do infinite decimals to the base 10 and conversely. In particular,
we could have defined real numbers using any other base than 10 and the
theory would be equivalent.
 Recall that if S is a subset of a ordered field F and b is in F, then
b is an upper bound of S if
for all s in S. b is said to
be a least upper bound of S if
for every upper bound c of S.
An ordered field is said to satisfy the least upper bound axiom if every
subset S of F with at least one upper bound in F has a least upper bound in F.
 Show that if F is an Archimedean ordered field which satisfies the least uppper bound
axiom, then F is a field of real numbers.
 Show that if F is a field of real numbers, then it satisfies the
least upper bound axiom.
 Make up a reasonable definitions for lower bound, greatest
lower bound, and the greatest lower bound axiom.
Prove results analogous to the last two parts of this exercise for these
notions.
 Let S be a subset of an ordered field F and b be an element of F. We
say that b is a limit point of S if for any
there is
an s in S such that
. A set S in F is said to be
bounded if there is a natural number M > 0 such that s < M for all s in S.
 Show that if F is an Archimedean ordered field in which every infinite bounded subset of F
has a limit point in F, then F is a field of real numbers.
 Show that if F is a field of real numbers, then every infinite bounded
subset S of F has a limit point in F.
 Show that the complex numbers form a field.
 Prove Proposition 15.
 Most of the arguments in this section work fine for other fields. Does
the argument for Proposition 14 work when the field of real numbers is
replaced by the field of rational numbers? What about Proposition 15?
 Does the proof of Proposition 14 work when the field of real numbers
is replaced by the field of complex numbers? If not, where does it break down?
 Show for every positive integer n that
The same sum with the sine function replaced with the cosine function is equal
to zero.
 The method used to calculate
in Chapter 1 can also be used to
calculate the area of a circle. Do this.
 Use the same method to calculate the lateral surface area of a
right circular cone. The formula is
where r is the radius
of the base and s is the distance from the vertex to any point on the edge of
the base. The quantity s is called the slant height.
 Use the result of the last exercise to show that the area of the
frustrum of a right circular cone is
where r is the average
of the radius of the top and bottom of the frustrum and s is the slant height.
If the frustrum is obtained by cutting the top off a cone, the slant height is
the distance along a line from the vertex to the base, from the point of
intersection with the top of the frustrum to the point of intersection with
the base.
 To calculate the surface area of a sphere, approximate a circle by
taking a regular inscribed 2ngon and spinning it about an axis. The figure
consists of several frustra of cones stacked on each other. You
calculate the total lateral area of these figures and then take the limit
to find the area of the sphere. The result is Archimedes famous result
that
where r is the radius of the sphere.
 Let
. So,
,
, and 1
are the cube roots of unity. Let a, b, and c be complex numbers, which
we also think of as being points in the complex plane. As such, the three
numbers determine a triangle.
 Suppose the three points a, b, and c are the vertices of an
equilateral triangle traversed in counterclockwise order. Show that
.
 Conversely, if
, then the three points
a, b, and c are the vertices of an equilateral triangle traversed in
counterclockwise order.
 How does the result change if you traverse the points in clockwise order?
 In the last section, for every positive real number a, one defined
a function with domain the rational numbers
.
Can you extend this in a natural way to function defined for all real
numbers?
 Already in Chapter 1, one noted that all first degree polynomials
define lines. This exercise will classify all curves defined by second degree
polynomials.
 The equation
for fixed values of s and p
has either 0, 1, or 2 solutions. If
and
are solutions, then
and
, i.e. the problem is equivalent to
finding numbers given their sum and product. To get a good graphical
view of how the number of solutions varies with s and p, we will fix
only one of the two parameters and graph the other as a function of x.
 Show that if s is fixed, then writing p as a function of x, gives
the graph of a parabola which opens downward. In particular, for a given
sum, one cannot find a pair of with a given product p unless p is sufficiently
small. Find the coordinates of the vertex of the parabola and interpret it
algebraically.
 Fix p and express s as a function of x. Graph the function and
interpet the graph in terms similar to that of the last part. Find a
rotation which shows that the graph is actually the graph of a conic
section. What kind of conic section is it?
 Show that by rotating through an appropriate angle, the general
second degree polynomial equation:
can be converted into a second degree polynomial in which the coefficient
of xy is zero.
 Show that by an appropriate translation, any second degree polynomial
in which the coefficient of xy is zero can be converted into one of the
form
 Show that after an appropriate rotation and translation, any second
degree polynomial equation can be converted into one of the following:
 An ellipse, hyperbola, or parabola
 Either one or two lines
 The empty set
Revised: July 5, 2001
All contents © copyright 2001 K. K. Kubota. All rights reserved
