Download The Real and Complex Number Systems

Chapter 1
The Real and Complex Number
Systems
The first chapter of Principles of Mathematical Analysis lays the foundations for the remainder of the book by motivating the need for the field of real numbers, and by showing
the proofs of some of the familiar operations in the real field (e.g., logarithms, exponentiation, and roots). Rudin begins the chapter stating that we shall assume knowledge of the
axioms which govern the arithmetic of the integers. Based on these axioms, Rudin shows
proofs for some of the minor rational field identities and uses them to motivate the proofs
for the existence and uniqueness of operations in the real field. The chapter ends with a
discussion of the complex field and various other Euclidean spaces.
I’ve divided the exercises at the end of the chapter into three sections. In the first section,
the problems deal with understanding the need for the field of real numbers, understanding
some of the properties that are a consequence of having ordered sets, and the existence
and uniqueness proofs for some of the familiar real-valued functions. The existence and
uniqueness of these real-valued functions are proved using supremums and infinums. The
second section contains problems relating to the complex field, whereas the the third section
contains problems relating to Euclidean spaces having three or more dimensions. In these
two sections, many of the proofs follow by means of algebraic manipulation. The final
exercise from Chapter 1 is based off of the construction of the real numbers given in the
Rudin’s Chapter 1 appendix. Because the proof for this question is very close to a subset
of the proof given in the Appendix, I only outline the rationale behind it.
1.1
Ordered sets, fields, and the real field
The exercises in this section cover problems 1 through 7 from Chapter 1. Let R, Q, and Z
represent the sets of all real numbers, rational numbers, and integers respectively.
Problem 1.1
If r is rational (r 6= 0) and x is irrational, prove that r + x and rx are irrational.
This problem is fairly straight forward. We need to prove that any choice of combinations
for a rational r and an irrational x always yield an irrational value of r + x and rx. Since
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Chapter 1 Solutions
since irrational numbers by definition are the opposite of rational numbers, this problem is
best proved by contradiction.
SOLUTION: Since r is rational, it can be represented by m/n, for some m, n ∈ Z
with n being non-zero. Suppose that the sum r + x was rational. If this were the case,
it could be represented by p/q, for some p, q ∈ Z with q being non-zero. Hence, we
would have that
m
p
np − mq
r+x=
+x= ⇒x=
(1.1)
n
q
np
which contradicts the fact that x is irrational (since the field of integers is closed under
multiplication and addition). Similarly, for rx, we would have that
rx =
m
p
np
x= ⇒x=
n
q
mq
(1.2)
which again contradicts the fact that x is irrational.
Problem 1.2
Prove that there is no rational number whose square is 12.
There is a similar example in the beginning of Chapter 1 for 2 instead of 12. The proof
is almost the same (i.e., by contradiction). If there was a rational number whose square
was 12, then we’d be able to show it as an irreducible fraction of integers. The prime
factorization of 12 is (3)(22 ), hence any multiple of 12 should be divisible by 3, 2, and 4.
We can use this to our advantage to end up with a contradiction similar to the example in
the beginning of the chapter.
SOLUTION: Let’s assume that there existed a rational number p ∈ Q whose square
was 12. Then this would mean that there would exist m, n ∈ Z, both non-zero, such
that p = m/n where p2 = 12 and the fraction m/n is irreducible. Consequently,
p2 =
m2
= 12 ⇒ m2 = 12n2 = (3)(22 )n2
n2
(1.3)
Equivalently, we have that m2 can be represented by the factorization (3)(22 )(n2 ).
Because of the equality, we see that m2 is divisible by 4, meaning that m is divisible
by 2. We can thus form the replacement m = 2m0 , such that m2 = 4m0 2 . This leaves
us with
m20 = 3n2
(1.4)
Hence the problem reduces to there existing a rational number whose square is 3.
Because of the equality, we must have that 3 is a prime factor of m20 , and as a result
3 must be a prime factor of m0 . Hence 9 is a factor of m20 , and by the equality again
we have that 9 must also be a factor of 3n2 . Thus we have that 3 is a prime factor
of n2 , which can only be true if 3 is a prime factor of n. In summary, we have that
3 is a prime factor of both n and m0 , thus it is also a factor of m. This contradicts
our initial assumption of irreducibility of the fraction m/n, hence there cannot exist a
rational number whose square is 12.
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Chapter 1 Solutions
Problem 1.3
Prove proposition 1.14. Specifically, for any field F with x, y, z ∈ F,
1. If x 6= 0 and xy = xz, then y = z.
2. If x 6= 0 and xy = x, then y = 1.
3. If x 6= 0 and xy = 1, then y = 1/x.
4. If x 6= 0, then 1/(1/x) = x.
This is straight forward and follows exactly in the same manner as Proposition 1.14. Once
Part 1 is proven (using properties of fields), Parts 2 and 3 follow as special cases. Furthermore, Part 4 follows as a special case of Part 3.
SOLUTION (Part 1): Since x 6= 0, there exists a multiplicative inverse 1/x ∈ F
such that x(1/x) = 1. Thus, we have that y = (x)(1/x)(y) = (1/x)(xy). If we assume
that xy = xz, thus y = (1/x)(xz) = (x)(1/x)(z) = z. Therefore, it must follow that
y = z.
SOLUTION (Part 2): Apply the results of Part 1 with z = 1.
SOLUTION (Part 3): Apply the results of Part 1 with z = 1/x.
SOLUTION (Part 4): Recall that by the existence of a multiplicative inverse for
x, we have that (1/x)(x) = 1. This form is exactly the same as the form of Part 3,
hence we have that x = 1/(1/x) (1/x playing the role of x and x playing the role of y).
Problem 1.4
Let E be a nonempty subset of an ordered set; suppose α is a lower bound of E and
β is an upper bound of E. Prove that α ≤ β.
Again, very straightforward. This falls right out of the definition of lower and upper bounds.
SOLUTION: Since α is a lower bound of E, we must have that α ≤ x for all x ∈ E.
Similarly, we must have that x ≤ β for all x ∈ E, since β is an upper bound of E.
Thus, for all x ∈ E, the inequality α ≤ x ≤ β must hold, implying that α ≤ β.
Problem 1.5
Let A be a nonempty set of real numbers which is bounded below. Let −A be the
set of all numbers −x, where x ∈ A. Prove that
inf A = − sup(−A)
Because of the relationship between A and −A, this question has the potential to become
confusing. One must take care in the proof to explicitly differentiate between the elements
of A and the elements of −A. As the question words it, we start with assuring the existence
of inf A. We then state the definition of inf A and rephrase the resulting inequalities in
terms of the elements of −A. This is done by negating all of the inequalities. We then end
up with the conditions for the supremum of −A.
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SOLUTION: Since A ⊂ R is nonempty and bounded from below, we have that
α = inf A exists. This yields, by definition, the following two statements.
1. For all x ∈ A, α ≤ x
2. For all x ∈ A, there does not exist an α0 ∈ R such that α < α0 ≤ x for all x ∈ A
Equivalently, these two properties can be rephrased as follows.
1. For all x ∈ A, −x ≤ −α
2. For all x ∈ A, there does not exist an −α0 = α00 ∈ R such that −x ≤ α00 < −α
In terms of the elements z ∈ −A these two properties can be rephrased as follows.
1. For all z ∈ −A, z ≤ −α
2. For all z ∈ −A, there does not exist an α00 ∈ R such that z ≤ α00 < −α
Hence, −α must be the supremum of −A, or equivalently that inf A = − sup(−A).
Problem 1.6
Fix b > 1.
1. If m, n, p, q are integers with n > 0 and q > 0, and r = m/n = p/q, prove that
(bm )1/n = (bp )1/q
Hence it makes sense to define br = (bm )1/n .
2. Prove that br+s = br bs if r, q ∈ Q.
3. If x is real, define B(x) to be the set of all numbers bt , where t ∈ Q and t ≤ x.
Prove that
br = sup B(r)
when r is rational. Hence it makes sense to define
bx = sup B(x)
for every real x.
4. Prove that bx+y = bx by for all real x and y.
The first part of this problem is solved by a few straight forward applications of Theorem
1.21. The difficulty in the proof, similar to the last problem, resides in the fact that the
notation can get out of hand.
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SOLUTION (Part 1): Consider the quantity y = (bm )1/n . By the uniqueness
assertion of Theorem 1.21, we know that y is the unique solution (i.e., root) of the
equation y n = bm , and similarlya , y nq = bmq . Given that r = m/n = p/q, we have
that mq = np. Hence y is also the unique solution to y nq = bnp , or rather, y q = bp .
Applying the uniqueness assertion one more time yields that y = (bp )1/q , and hence we
must have that (bm )1/n = (bp )1/q .
a n
y = ((bm )q )1/q is the unique solution to y nq = bmq .
The second part of the problem is also straight forward. Now that we can reduce rational
numbers in exponents, this becomes an exercise in algebra and an application of the corollary
of Theorem 1.21 (page 11).
SOLUTION (Part 2): Since r and s are rational, we can represent them both by
r = m1 /n1 and s = m2 /n2 for some integers m1 , m2 , n1 , n2 , wher n1 > 0, n2 > 0. Thus
the sum r + s is also rational, and can be represented
r+s=
m1 m2
m1 n2 + m2 n1
m
+
=
=
n1
n2
n1 n2
n
(1.5)
where m = m1 n2 + m2 n1 and n = n1 n2 are still integers. Because of what we proved
in (1), we can write bm/n = (bm )1/n , and as a result
br+s = bm/n = (bm )1/n
(1.6)
Since m is an integer, the product bm can be performed in any order (associative law
of multiplication). Hence, it makes sense to separate the operations into the first m1 n2
terms and the second m2 n1 terms. Specifically, we have that bm = (bm1 n2 )(bm2 n1 ). We
can now apply the corollary on page 11 of Rudin directly, yielding
(bm1 n2 bm2 n1 )1/n1 n2 = bm1 /n1 bm2 /n2 = br bs
(1.7)
Thus, we conclude that br+s = br bs .
The proof of part 3 is really a Lemma for the proof for part 4. Part 3 is a standard format
for a supremum proof. If you recall, Rudin has only stated that n1 ≤ n2 if and only if
bn1 ≤ bn2 for any n1 , n2 ∈ Z provided that b > 1. The extension to the fact that p ≤ q if
and only if bp ≤ bq for any p, q ∈ Q when b > 1 is fairly straightforward. For simplicity, we
will assume that this ordering holds for rational exponents.
0
SOLUTION (Part 3): Clearly, B(r) ⊂ R and bounded from above (e.g., br is an
upper bound for any r0 > r, as long as r is finite). Hence, sup B(r) exists for any finite
r ∈ Q. It is clear that br is an upper bound of B(r), since bt ≤ br for every rational
t ≤ r (an extension of Proposition 1.18(b), page 8). We will prove that br = sup B(r)
by contradiction. Suppose that br is not the least upper bound for B(r). This would
mean that there exists an α < br such that bt ≤ α for all rational t ≤ r. However,
this cannot be true for t = r since α < br . Hence α cannot be an upper bound of
B(r), which contradicts the supposition that br is not the least upper bound of B(r).
Therefore, we must have that br = sup B(r).
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Furthermore, we should also have that bx = sup B(x) since the rationals are dense in
the reals.
There are a few different ways that one can approach a proof for part 4. The goal is to take
B(x + y) and rewrite the elements using the identity from part 2. We then should be able
to take the supremum of that new representation, and find it equal to bx by . I’ve approached
this by representing B(x + y) in terms of a sum of two rational variables.
Again, we will assume that x ≤ y if and only if bx ≤ by for some b > 1. This can easily
be proven using the ordering of b with rational exponents, and using the sets B(x) from
part 3.
SOLUTION (Part 4): From part 3, we know that bx+y = sup B(x + y), where
B(x + y) = {bt : t ∈ Q, t ≤ x + y} for any x, y ∈ R. We can write B(x + y) in terms of
a sum of two rational variables, t = t1 + t2 , such that
B(x + y) = {bt1 +t2 : t1 , t2 ∈ Q, t1 + t2 ≤ x + y}
(1.8)
Notice that B(x + y) will contain the same elements, bt = bt1 +t2 , if t1 ≤ x and t2 ≤ y.
Combined with the identity from part 2, we have that
B(x + y) = {bt1 bt2 : t1 , t2 ∈ Q, t1 ≤ x, t2 ≤ y}
(1.9)
CLAIM: bx by = sup B(x + y)
This proof follows similar to part 3. We first prove that this is true for the
case when x and y are rational (i.e., x = rx and y = ry ). We know that since
brx = sup B(rx ), the inequality bt1 ≤ brx must hold for all rational t1 ≤ rx . We
must also have that bt1 bry ≤ brx bry . Since bry = sup B(ry ), the inequality bt2 ≤ bry
must hold for all rational t2 ≤ ry . Thus bt1 bt2 ≤ brx bry holds for all rational t1 ≤ rx
and t2 ≤ ry , and therefore brx bry is an upper bound of B(rx + ry ).
Now suppose that brx bry was not the supremum of B(rx + ry ). This would mean
that there exists an α ∈ R such that α < brx bry and bt1 bt2 ≤ α for all rational
t1 ≤ rx and t2 ≤ ry . However, this cannot be true for the case when t1 = rx
and t2 = ry . Hence α cannot be an upper bound of B(rx + ry ). Therefore,
brx bry = sup B(rx + ry ). Since this is true for any rational rx and ry , and since the
rationals are dense in the reals, we must also have that bx by = sup B(x + y).
Therefore, bx by = bx+y .
Problem 1.7
Fix b > 1 and y > 0. Prove that there is a unique real x such that bx = y by
completing the following outline (i.e., the logarithm of y to the base b).
1. For any positive integer n, bn − 1 ≥ n(b − 1).
2. Hence, b − 1 ≥ n(b1/n − 1).
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Chapter 1 Solutions
3. If t > 1 and n > (b − 1)/(t − 1), then b1/n < t.
4. If w is such that bw < y, then bw+1/n < y for sufficiently large n; to see this,
apply part (3) with t = yb−w .
5. If bw > y, then bw−1/n > y for sufficiently large n.
6. Let A be the set of all w such that bw < y. Show that x = sup A satisfies
bx = y.
7. Prove that this x is unique.
The approach I take to constructing the logarithm is a little different than the order presented above. Actually, the construction/proof follows similarly to the construction of the
n-th root of every real x > 0, as given in Theorem 1.21 on page 10. The heart of the proof
is in parts 4 through 6. The goal is to prove that bx < y and bx > y cannot be true if x
is the supremum of A as defined in part 6. Parts 1 through 3 form a Lemma that allows
us to do this. Finally, part 7 is straight forward and follows similarly to Theorem 1.21. As
in problem 1.6, we assume that for any rational r and s, br < bs if and only if r < s, and
similarly for the reals since the rationals are dense in the reals.
SOLUTION: Before we begin the proof, we provide the following Lemma.
Lemma 1: For any real t > 1 and b > 1, there exists an integer n > (b − 1)/(t − 1) such
that b1/n < t.
As in the proof for Theorem 1.21, consider the identitya bn − an = (b − a)(bn−1 +
bn−2 a + · · · + an−1 ) for the real numbers a and b, and for any positive integer
n > 0. Letting a = 1 yields bn − 1 = (b − 1)(bn−1 + bn−2 + · · · + 1). We know
that for any b > 1 and positive integer n > 0, bk ≥ 1 for all k ∈ {0, 1, . . . , n − 1}.
Hence, bn − 1 ≥ n(b − 1). Since b > 1, we have that b1/n > 1, and hence the
inequality can be realized as b − 1 ≥ n(b1/n − 1). Now, consider any real t > 1.
If we choose n such that n > (b − 1)/(t − 1), or rather n(t − 1) > b − 1, then
n(b1/n − 1) ≤ b − 1 < n(t − 1). Hence, b1/n < t.
Consider A = {w ∈ R : bw < y}. Clearly, A is bounded from above by any real w0
0
such that y < bw , hence x = sup A exists. Suppose that bx < y, rather 1 < yb−x .
Applying Lemma 1 with t = yb−x yields that b1/n < yb−x , or equivalently bx+1/n < y,
for sufficiently large n. However, (x + 1/n) ∈ A, which cannot be true since x is an
upper bound A. Now suppose that y < bx , rather 1 < y −1 bx . Applying Lemma 1 with
t = y −1 bx yields that b1/n < y −1 bx , or equivalently y < bx−1/n , for sufficiently large n.
Hence, (x − 1/n) is an upper bound of A that is less than x, which cannot be true since
x is the least upper bound of A. Therefore, we must have that bx = y.
Finally, it is clear that x must be unique since x1 < x2 implies that bx1 < bx2 for any
real x1 , x2 .
a
This is easily shown by long dividing bn − an by b − a.
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Chapter 1 Solutions
1.2
Complex Fields and Orders
The exercises in this section cover problems 8 through 15 from Chapter 1. Let i =
Problem 1.8
√
−1.
Prove that no order can be defined in the complex field that turns it into an ordered
field.
This is an easy proof by counterexample, and follows as an application of Proposition 1.18
(d) on page 8.
SOLUTION: Suppose that there exists an order on the complex field C that turns
it into an ordered field. By proposition 1.18 (d), this would mean that x2 > 0 for any
non-zero x ∈ C. However, choosing x = i yields x2 = i2 = −1 < 0 (Theorem 1.28,
page 14) which can never be true. Hence there cannot be an order on the complex field
which would turn it into an ordered field.
Problem 1.9
Suppose z = a + bi, w = c + di. Define z < w if a < c, and also if a = c but b < d.
Prove that this ordering ( i.e., Lexicographic ordering) turns the set of all complex
numbers into an ordered set. Does this ordered set have the least-upper-bound
property?
Although this solution is straightforward, the proof is tedious and repetitive. To show
that the lexicographic ordering turns the complex numbers into an ordered set, we need to
show that it satisfies the two properties given in Definition 1.5 (page 3). The lexicographic
ordering however, doesn’t produce an ordered set with the least-upper-bound property. To
show this, we need to construct a non-empty bounded set and show that by the lexicographic
ordering that we can never choose a least upper bound.
SOLUTION: To show that the lexicographic ordering turns the set of complex numbers C into an ordered set, we first prove the two properties of Definition 1.5 (page 3).
Denote the lexicographic ordering as ≺. For any z, w ∈ C, the lexicographic ordering
clearly defines z ≺ w and w ≺ z. The event that z = w occurs when a = c and b = d.
Clearly only one of z ≺ w, z = w, and w ≺ z can be true, hence (i) of Definition 1.5 is
satisfied. In addition to z, w ∈ C, consider v ∈ C such that v = e + f i. If z ≺ w, then
either a < c or a = c and b < d. If w ≺ v, then either c < e or c = e and d < f . The
following can be observed.
1. Suppose a < c, and c < e. Then a < e, and hence z ≺ v.
2. Suppose a = c and b < d, and c < e. Then a < e, and hence z ≺ v.
3. Suppose a < c, and c = e and d < f . Then a < e, and hence z ≺ v.
4. Suppose a = c and b < d, and c = e and d < f . Then a = e and b < f , and hence
z ≺ v.
Therefore, (ii) of Definition 1.5 must be true, thereby proving that ≺ turns C into an
ordered set.
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Chapter 1 Solutions
Now suppose C had the least-upper-bound property under ≺. Under Definition 1.10
(page 4), this would mean that any set E ⊂ C which is non-empty and bounded from
above ensures the existence of sup E ∈ C. Consider E = {(a + bi) ∈ C : a < 0, b ∈ R}.
Clearly, E is non-empty and bounded from above. In fact, the set E 0 = {(a0 + b0 i) ∈
C : a0 ≥ 0, b0 ∈ R} are all upper bounds of E according to ≺. However, E 0 has no
smallest element according to ≺. Hence sup E 6∈ C, and therefore C cannot have the
least-upper-bound property under ≺.
Problem 1.10 Suppose z = a + bi, w = u + vi, and
1/2
|w| + u
a=
,
2
b=
|w| − u
2
1/2
,
Prove that z 2 = w if v ≥ 0 and that (z)2 = w if v ≤ 0. Conclude that every
complex number (with one exception) has two complex square roots.
An exercise in complex algebra. Again, straight forward.
SOLUTION: Evaluating z 2 yields z 2 = (a2 − b2 ) + (2ab)i. We have that
(a2 − b2 ) = u, (2ab) = |w|2 − u2
1/2
(1.10)
Knowing that |w|2 = u2 + v 2 yields (2ab) = (v 2 )1/2 , which has a unique positive
solution of v. Hence, z 2 = u + vi = w, provided that v ≥ 0. Evaluating (z)2 yields
(z)2 = (a2 − b2 ) − (2ab)i, and hence (z)2 = u − vi = w, provided that v ≤ 0. Hence,
every complex number (excluding zero) has two complex square roots (that is, some
complex number and its conjugate).
Problem 1.11 If z is a complex number, prove that there exists an r ≥ 0 and a complex number
w with |w| = 1 such that z = rw. Are w and r always uniquely determined by z?
This is again just an exercise in complex algebra.
SOLUTION: Let z = a + bi. Given any non-zero z ∈ C, one can express it as
b
a
2
2 1/2
z = a + bi = (a + b )
+
i =r·w
(1.11)
(a2 + b2 )1/2 (a2 + b2 )1/2
Clearly, r ≥ 0 and |w| = 1. Notice that for all non-zero z, this representation is unique.
When z = 0 however, as long as r = 0 we can choose any w ∈ C such that |w| = 1.
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Chapter 1 Solutions
Problem 1.12 If z1 , . . . , zn are complex, prove that
|z1 + z2 + · · · + zn | ≤ |z1 | + |z2 | + · · · + |zn |
This is nothing more than Theorem 1.33 (e) applied n − 1 times.
SOLUTION: Let w = z2 + · · · + zn . Applying Theorem 1.33 (e) for the sum |z1 + w|
yields
|z1 + z2 + · · · + zn | = |z1 + w|
≤ |z1 | + |w|
= |z1 | + |z2 + · · · + zn |
(1.12)
If we redefine w = |z2 + · · · + zn | and apply Theorem 1.33 (e) to the result of (1) until
w = |zn |, the solution follows (i.e., a total of n − 1 applications of Theorem 1.33 (e)).
Problem 1.13 If x, y are complex, prove that
|x| − |y| ≤ |x − y|
There are many ways to arrive at this, but we want to try to stick with information we have
already proven either in the previous chapters, or in the previous problems. Rearranging
this yields us with two inequalities to prove: |x| ≤ |x − y| + |y| and |y| ≤ |x − y| + |x|. We
can arrive at both of these by using Theorem 1.33 (e).
SOLUTION: Let z1 = x − y and z2 = y. Applying Theorem 1.33 (e), we have that
|z1 + z2 | ≤ |z1 | + |z2 |
|x| ≤ |x − y| + |y|
|x| − |y| ≤ |x − y|
(1.13)
Similarly, let z3 = x and z4 = y − x. Applying Theorem 1.33 (e) again, we have that
|z3 + z4 | ≤ |z3 | + |z4 |
|y| ≤ |x| + |y − x|
|y| − |x| ≤ |y − x|
−(|x| − |y|) ≤ |x − y|
Therefore, |x| − |y| ≤ |x − y|.
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Chapter 1 Solutions
Problem 1.14 If z is a complex number such that |z| = 1, that is, such that zz = 1, compute
|1 + z|2 + |1 − z|2
This is straight forward, using the definition of the complex norm.
SOLUTION: Consider |1 + z|2 . This can be rewritten as
|1 + z|2 = (1 + z)(1 + z)
= (1 + z)(1 + z)
= 1 + z + z + zz
=2+z+z
(1.15)
Similarly, |1 − z|2 can be rewritten as
|1 − z|2 = (1 − z)(1 − z)
= (1 − z)(1 − z)
= 1 − z − z + zz
=2−z−z
(1.16)
Hence, |1 + z|2 + |1 − z|2 = 4.
Problem 1.15 Under what conditions does equality hold for the Schwarz inequality?
Now, I know a lot of you reading this question probably went and researched the Schwarz
inequality, found Cauchy-Schwarz, and saw that equality only holds when the two vectors
are linearly dependent. This is true here, since the Complex field is essentially a special
case of a 2D vector space. However, we haven’t gotten to vector spaces yet. Our goal is to
arrive at a conclusion very similar to this, but do it using only the terminology we know up
until this point. To do this, I’m going to attempt to expand both sides of the inequality,
see what is common, and see what we have to do to make what isn’t common equal.
SOLUTION: For reference, the Schwarz inequality is given for two sets of complex
numbers {a1 , . . . , an } and {b1 , . . . , bn } as
2
X
n
n
X
X
n
2
a
b
≤
|a
|
|bj |2
j j
j
j=1
j=1
j=1
(1.17)
For simplicity, let Nn = {1, 2, . . . , n}. Starting with the left side and expanding the
summations via Theorem 1.31 (a) and (b), we have
2 


X
X
X
aj bj = 
aj bj 
aj bj  
j∈Nn
j∈Nn
j∈Nn
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Chapter 1 Solutions

=

X
aj bj  
j∈Nn
=
X

X
aj bj 
j∈Nn
|aj |2 |bj |2 +
j∈Nn
X
(aj bj )(ak bk )
(1.18)
j,k∈Nn
k6=j
Similarly expanding the right side of the inequality yields
X
X
X X
|aj |2 |bk |2
|aj |2
|bj |2 =
j∈Nn
j∈Nn
j∈Nn k∈Nn
=
X
|aj |2 |bj |2 +
j∈Nn
X
|aj |2 |bk |2
(1.19)
j,k∈Nn
k6=j
If equality holds for the Schwarz inequality, then (2) and (3) must be equal. This
implies that
X
X
(aj bj )(ak bk ) =
|aj |2 |bk |2
(1.20)
j,k∈Nn
k6=j
j,k∈Nn
k6=j
which can only be true if b1 = αa1 , b2 = αa2 , . . . , bn = αan for some real valued α.
Hence equality will hold for the Schwarz inequality only when bj = αaj for all j ∈ Nn
and for any real α.
1.3
Euclidean Spaces
The exercises in this section cover problems 16 through 19 from Chapter 1. Let Rk be a
vector space of k dimensions over the real field. Any x ∈ Rk can be represented as a set of
coordinates x = (x1 , x2 , . . . , xk ).
Problem 1.16 Suppose k ≥ 3, x, y ∈ Rk , |x − y| = d > 0, and r > 0. Prove:
1. If 2r > d, there are infinitely many z ∈ Rk such that
|z − x| = |z − y| = r
2. If 2r = d, there is exactly one such z.
3. If 2r < d, there is no such z.
How must these statements be modified if k is 2 or 1?
There are many ways to approach this, with some of the easiest methods making use of
topology concepts. Unfortunately, we haven’t gotten topology yet in Rudin. So we’ll need
to rely on the information presented on Euclidean space thus far for this proof (which is
12
c 2014, Donald J. Bucci
Chapter 1 Solutions
not a lot). Since Rudin gives us a requirement on the dimension of the space, and since he
implies that things may change if the dimension is low enough, we are probably going to
be looking at a solution to a set of equations.
The two constraints on the construction of z ∈ Rk are that |z − x| and |z − y| are equal,
and furthermore that they are both equal to some r > 0. These can be represented by the
following system of equations.
|z − x| − |z − y| = 0
(1.21)
|z − x| + |z − y| = 2r
(1.22)
Notice that Equation (1.21) is equivalent to |z − x|2 − |z − y|2 = 0. Expanding this
yields
2
2
|z − x| − |z − y| =
k
X
(zj − xj )2 − (zj − yj )2
j=1
=
k
X
(zj xj − 1/2x2j ) − (zj yj − 1/2yj2 )
j=1
=
k
X
(xj − yj )(zj − 1/2(xj + yj )) = 0
(1.23)
j=1
Notice that Equation (1.22) is equivalent to (|z − x| + |z − y|)2 − |x − y|2 = 4r2 − d2
(since |x − y| = d). Expanding this yields
(|z − x| + |z − y|)2 − |x − y|2 = |z − x|2 + 2|z − x||z − y| + |z − y|2 − |x − y|2
= |z − x|2 + 2r2 + |z − y|2 − |x − y|2
= 2r2 +
k
X
(zj − xj )2 + (zj − yj )2 − (xj − yj )2
j=1
= 2r2 + 2
k
X
zj2 − zj (xj + yj ) + xj yj
j=1
= 2r2 + 2
k
X
zj2 − zj (xj + yj ) + xj yj + 1/4(xj − yj )2 − 1/4(xj − yj )2
j=1
= 2r2 + 2
k
X
zj2 − zj (xj + yj ) + 1/4(xj + yj )2 − 1/4(xj − yj )2
j=1
= 2r2 + 2
k
X
(zj − 1/2(xj + yj ))2 − 1/4(xj − yj )2
j=1
= 2r2 − 1/2d2 + 2
k
X
(zj − 1/2(xj + yj ))2 = 4r2 − d2
j=1
⇒
k
X
(zj − 1/2(xj + yj ))2 = r2 − 1/4d2
j=1
13
(1.24)
c 2014, Donald J. Bucci
Chapter 1 Solutions
Notice that Equation (1.23) and Equation (1.24) describe a system of two polynomial equations with k variables.If there existed a z ∈ Rk such that |z − x| = |z − y| = r, it must
satisfy Equation (1.23) and Equation (1.24). We will now use the system determined above
to prove the results of the problem.
SOLUTION (Part 1): Suppose that 2r > d. Consider the following three selections
of a strictly positive integer k.
• Suppose that k ≥ 3. In this case,Equation (1.23) and Equation (1.24) describe a
system of two equations with k ≥ 3 variables. Hence the system is underdetermined. This system must have at least one solution for some z ∈ Rk , otherwise
Theorem 1.37 (f) would not hold true for a general z ∈ Rk . Since the system
is underdetermined and has at least one solution, it must therefore have infinite
number of solutions.
• Suppose that k = 2. In this case,Equation (1.23) and Equation (1.24) describe
a system of two equations with k = 2 variables. This system again must have
at least one solution for z, otherwise Theorem 1.37 (f) would not hold true for a
general z ∈ Rk . For a given x, y, r, and d, we should be able to solve for one of
the two variables in Equation (1.23), substitute it into Equation (1.24), and end
up with a quadratic equation. This will give us exactly two solutions for z.
• Suppose that k = 1. In this case,Equation (1.23) and Equation (1.24) describe a
system of two equations with k = 1 variable. Hence the system is overdetermined.
Equation (1.23) reduces to z = 1/2(x + y), which implies that equation (2) is
equal to 0 and as a result that 2r = d. This contradicts the supposition of 2r > d,
and hence a solution cannot exist.
SOLUTION (Part 2): Suppose that 2r = d. Equation (1.24) is then equal to 0,
which can only be true of z = 1/2(x + y). It is trivial to see that this z is the only
solution of the system, regardless of k.
SOLUTION (Part 3): Suppose that 2r < d. Equation (1.24) is then equation to a
number less than zero. However, there are clearly no values of z ∈ Rk that satisfy this
requirement, regardless of k a .
a
This can also be proven by showing that 2r > d would violate the triangle inequality.
Problem 1.17 Prove that
|x + y|2 + |x − y|2 = 2|x|2 + 2|y|2
Interpret this geometrically, as a statement about parallelograms.
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c 2014, Donald J. Bucci
Chapter 1 Solutions
This is very straightforward, since we are able to commutatively distribute and factor inner
products like we would with multiplication1 .
SOLUTION: Expanding |x + y|2 + |x − y|2 yields
|x + y|2 + |x − y|2 = (x · x + 2x · y + y · y) + (x · x − 2x · y + y · y)
= 2|x|2 + 2|y|2
(1.25)
This result is the so called parallelogram law. For a parallelogram with sides defined
by the vectors x and y, the two diagonals are given by (x + y) and (x − y). This law
means that the sum of the norm squared of each diagonal of a parallelogram is equal
to the sum of two times the norm squared of each side.
Problem 1.18 If k ≥ 2 and x ∈ Rk , prove that there exists y ∈ Rk such that y 6= 0, but x · y = 0.
Is this also true for k = 1?
This is also straight forward. We just need to show that we can construct such a y to prove
its existence.
P
SOLUTION: Consider a y ∈ Rk defined such that y1 = −1/x1 ( nj=2 xj yj ) for any
y2 , . . . , yn ∈ R. Clearly, x · y = 0 and y 6= 0, thus proving existence of such a y for
k ≥ 2. For k = 1, it is trivial to see that such a y will exist if and only if x = 0.
Problem 1.19 Suppose a, b ∈ Rk . Find c ∈ Rk and r > 0 such that
|x − a| = 2|x − b|
if and only if |x − c| = r.
This problem looks convoluted at first glance. The if and only if condition means that both
criteria must be simultaneously satisfied for all x ∈ Rk . If we approach this in a manner
similar to Problem 1.16, then we’ll arrive at the correct result.
SOLUTION: We know that |x − a| = 2|x − b| if and only if 4|x − b|2 − |x − a|2 = 0.
Expanding this yields
4|x − b|2 − |x − a|2 = 4(x · x − 2x · b + b · b) − (x · x − 2xa + a · a)
= 3x · x − 2x · (4b − a) + (4b · b − a · a) = 0
(1.26)
Similarly, we know that |x − c| = r if and only if 3|x − c|2 − 3r2 = 0. Expanding this
yields
3|x − c|2 − 3r2 = 3(x · x − 2x · c + c · c) − 3r2
1
These properties are a direct consequence of how the inner product is defined. Although we haven’t
necessarily proved this, Rudin has used this fact in the proof of Theorem 1.37 (e).
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c 2014, Donald J. Bucci
Chapter 1 Solutions
= 3x · x − 2x · (3c) + (3c · c − 3r2 ) = 0
(1.27)
For Equation (1.26) and Equation (1.27) to always be true (i.e., regardless of any
x ∈ Rk ), we must have that
3c = 4b − a
2
3c · c − 3r = 4b · b − a · a
(1.28)
(1.29)
Substituting for c in Equation (1.29) using Equation (1.28) yields
1/3(4b − a) · (4b − a) − 3r2 = 4b · b − a · a ⇒
16b · b − 8a · b + a · a − 9r2 = 12b · b − 3a · a ⇒
4b · b − 8a · b + 4a · a = 9r2 ⇒
4|b − a|2 = (3r)2 ⇒
2|b − a| = 3r
(1.30)
Therefore, the values of c and r must be as defined in Equation (1.28) and Equation (1.30).
1.4
Modifying the Definition of Cuts in Constructing the
Real Field
Problem 1.20 With reference to the Appendix, suppose that property (III) were omitted from the
definition of a cut. Keep the same definitions of order and addition. Show that
the resulting ordered set has the least-upper-bound property, that addition satisfies
axioms (A1) to (A4) (with a slightly different zero-element!) but that (A5) fails.
I’m only going to discuss my rationale in answering this question. This is because the
majority of the steps that would be involved in the proof would be minimally different from
the original proof. The cuts that Rudin is talking about in constructing the real numbers
from the rationals are just Dedekind cuts. A Dedekind cut is a partitioning of the set
of rational numbers into two sets, call them A and B. These two sets are designed such
that A ∪ B = Q and such that A has no greatest element. If the smallest element of B
is rational, then the cut can be associated with that rational number. Otherwise, the cut
defines a unique irrational number that does not exist in either A or B.
By removing property (III) from the definition of a cut in Step 1 (Chapter 1 Appendix,
page 17), we introduce the possibility that a cut may have some maximum element. Property (III) is not used in Step 2, so it remains unchanged. Step 3 proves (III) in order to
prove that R as constructed from these cuts would have the least-upper-bound property.
Since (III) is removed, its role in Step 3 can just be omitted. The proofs for the addition
axioms (A1) through (A3) in Step 4 omit any reference to (III). To satisfy axiom (A4), the
zero element must change to the set of all negative rational numbers, including zero. The
part of axiom (A5) that proves 0∗ ⊂ α + β fails for ν = 0.
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