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Answered) Solutions to Principles of Mathematical Analysis (Walter Rudin) Jason Rosendale [email protected]/* */ December 19, 2011 This work was done as an...


Hello Expert13, can you help me with my homework? They seem to be getting harder and harder. #5, 6, and 7 I have solutions for that I can post to make it a bit easier. The questions are in 'rudin ch 7' and a more thorough answer has been given by another undergraduate student in that solutions manual. Both can be found online for free. The deadline is in about a week so you have plenty of time, I know some of the questions can be lengthy so let me know if you need more time or I should increase the price. Thank you and please let me know if you have any questions. :-)

Solutions to Principles of Mathematical Analysis (Walter Rudin)
Jason Rosendale
[email protected]
December 19, 2011
This work was done as an undergraduate student: if you really don’t understand something in one of these
proofs, it is very possible that it doesn’t make sense because it’s wrong. Any questions or corrections can be
directed to [email protected] Exercise 1.1a
Let r be a nonzero rational number. We’re asked to show that x 6∈ Q implies that (r + x) 6∈ Q. Proof of the
contrapositive:
7→ r + x is rational assumed → (∃p ∈ Q)(r + x = p) definition of rational → (∃p, q ∈ Q)(q + x = p) we’re told that r is rational → (∃p, q ∈ Q)(x = −q + p) existence of additive inverses in Q Because p and q are members of the closed additive group of Q, we know that their sum is a
member of Q.
→ (∃u ∈ Q)(x = u)
→ x is rational definition of rational By assuming that r +x is rational, we prove that x must be rational. By contrapositive, then, if x is irrational
then r + x is irrational, which is what we were asked to prove. Exercise 1.1b
Let r be a nonzero rational number. We’re asked to show that x 6∈ Q implies that rx 6∈ Q. Proof of the
contrapositive:
7→ rx is rational assumed → (∃p ∈ Q)(rx = p)
→ (∃p ∈ Q)(x = r −1 definition of rational
p) existence of multiplicative inverses in Q (Note that we can assume that r−1 exists only because we are told that r is nonzero.) Because r−1
and p are members of the closed multiplicative group of Q, we know that their product is also a
member of Q.
→ (∃u ∈ Q)(x = u)
→ x is rational definition of rational By assuming that rx is rational, we prove that x must be rational. By contrapositive, then, if x is irrational
then rx is irrational, which is what we were asked to prove. 1 Exercise 1.2
√
Proof by contradiction. If 12 were rational, then we could write it as a reduced-form fraction in the form of
p/q where p and q are nonzero integers with no common divisors.
7→ p
q = √ 12 assumed 2 → ( pq2 = 12)
→ (p2 = 12q 2 )
It’s clear that 3|12q 2 , which means that 3|p2 . By some theorem I can’t remember (possibly the
definition of ‘prime’ itself), if a is a prime number and a|mn, then a|m ∨ a|n. Therefore, since 3|pp
and 3 is prime,
→ 3|p
→ 9|p2
→ (∃m ∈ N)(p2 = 9m)
2 → (∃m ∈ N)(12q = 9m)
2 → (∃m ∈ N)(4q = 3m)
2 → (3|4q ) definition of divisibility
substitution from p2 = 12q 2
divide both sides by 3
definition of divisibility From the same property of prime divisors that we used previously, we know that 3|4 ∨ 3|q 2 : it
clearly doesn’t divide 4, so it must be the case that 3|q 2 . But if 3|qq, then 3|q ∨ 3|q. Therefore:
→ (3|q)
And this establishes a contradiction. We began by assuming that p and q had no common divisors, but we
have shown
√ that 3|p and 3|q. So our assumption must be wrong: there is no reduced-form rational number such
that pq = 12. Exercise 1.3 a
If x 6= 0 and xy = xz, then
y = 1y = (x−1 x)y = x−1 (xy) = x−1 (xz) = (x−1 x)z = 1z = z Exercise 1.3 b
If x 6= 0 and xy = x, then
y = 1y = (x−1 x)y = x−1 (xy) = x−1 x = 1 Exercise 1.3 c
If x 6= 0 and xy = 1, then
y = 1y = (x−1 x)y = x−1 (xy) = x−1 1 = x−1 = 1/x Exercise 1.3 d
If x 6= 0, then the fact that x−1 x =1 means that x is the inverse of x−1 : that is, x = (x−1 )−1 = 1/(1/x). Exercise 1.4
We are told that E is nonempty, so there exists some e ∈ E. By the definition of lower bound, (∀x ∈ E)(α ≤ x):
so α ≤ e. By the definition of upper bound, (∀x ∈ E)(x ≤ β): so e ≤ β. Together, these two inequalities tell us
that α ≤ e ≤ β. We’re told that S is ordered, so by the transitivity of order relations this implies α ≤ β. 2 Exercise 1.5
We’re told that A is bounded below. The field of real numbers has the greatest lower bound property, so we’re
guaranteed to have a greatest lower bound for A. Let β be this greatest lower bound. To prove that −β is
the least upper bound of −A, we must first show that it’s an upper bound. Let −x be an arbitrary element in −A:
7→ −x ∈ −A assumed →x∈A definition of membership in −A →β≤x β = inf(A) → −β ≥ −x consequence of 1.18(a) We began with an arbitrary choice of −x, so this proves that (∀ − x ∈ −A)(−β ≥ −x), which by definition
tells us that −β is an upper bound for −A. To show that −β is the least such upper bound for −A, we choose
some arbitrary element less than −β:
7→ α < −β assumed → −α > β consequence of 1.18(a) Remember that β is the greatest lower bound of A. If −α is larger than inf(A), there must be some
element of A that is smaller than −α.
→ (∃x ∈ A)(x < −α) (see above) → (∃ − x ∈ −A)(−x > α)
→ !(∀ − x ∈ −A)(−x ≤ α) consequence of 1.18(a) → α is not an upper bound of −A definition of upper bound This proves that any element less than −β is not an upper bound of −A. Together with the earlier proof
that −β is an upper bound of −A, this proves that −β is the least upper bound of −A. Exercise 1.6a
The difficult part of this proof is deciding which, if any, of the familiar properties of exponents are considered
axioms and which properties we need to prove. It seems impossible to make any progress on this proof unless we
1
can assume that (bm )n = bmn . On the other hand, it seems clear that we can’t simply assume that (bm ) n = bm/n :
this would make the proof trivial (and is essentially assuming what we’re trying to prove).
As I understand this problem, we have defined xn in such a way that it is trivial to prove that (xa )b = xab
1
when a and b are integers. And we’ve declared in theorem 1.21 that, by definition, the symbol x n is the element
1
n n
such that (x ) = x. But we haven’t defined exactly what it might mean to combine an integer power like n
and some arbitrary inverse like 1/r. We are asked to prove that these two elements do, in fact, combine in the
way we would expect them to: (xn )1/r = xn/r .
Unless otherwise noted, every step of the following proof is justified by theorem 1.21. 3 7→ bm = bm assumed 1
n 1 → ((bm ) )n = bm definition of x n 1 → ((bm ) n )nq = bmq
We were told that m
n =
mq = np. Therefore: p
q which, by the definition of the equality of rational numbers, means that 1 → ((bm ) n )nq = bnp
1 → ((bm ) n )qn = bpn commutativity of multiplication From theorem 12.1, we can take the n root of each side to get:
1 → ((bm ) n )q = bp
From theorem 12.1, we can take the q root of each side to get:
1 1 → (bm ) n = (bp ) q Exercise 1.6b
As in the last proof, we assume that br+s = br bs when r and s are integers and try to prove that the operation
p
works in a similar way when r and s are rational numbers. Let r = m
n and let s = q where m, n, p, q ∈ Z and
n, q 6= 0.
p
m
7→ br+s = b n + q
→ br+s = b
r+s →b mq+pn
nq = (b definition of addition for rationals mq+pn ) → br+s = (bmq bpn )
r+s →b r+s →b = (b
= (b r+s mq
mq
nq ) 1
nq )(b m
n 1
nq from part a 1
nq pn (b )
pn
nq legal because mq and pn are integers
1
nq ) corollary of 1.21
from part a p
q →b
= (b )(b )
r+s
→b
= (br )(bs ) Exercise 1.6c
We’re given that b > 1. Let r be a rational number. Proof by contradiction that br is an upper bound of B(r):
7→ br is not an upper bound of B(r) hypothesis of contradiction → (∃x ∈ B(r))(x > br ) formalization of the hypothesis By the definition of membership in B(r), x = bt where t is rational and t ≤ r.
→ (∃t ∈ Q)(bt > br ∧ t ≤ r)
It can be shown that b−t > 0 (see theorem S1, below) so we can multiply this term against both
sides of the inequality.
→ (∃t ∈ Q)(bt b−t > br b−t ∧ t ≤ r)
t−t → (∃t ∈ Q)(b r−t >b ∧ t ≤ r) theorem S2
from part b → (∃t ∈ Q)(1 > br−t ∧ r − t ≥ 0)
→ (∃t ∈ Q)(1−(r−t) > b ∧ r − t ≥ 0)
→1>b
And this establishes our contradiction, since we were given that b > 1. Our initial assumption must have
been incorrect: br is, in fact, an upper bound of B(r). We must still prove that it is the least upper bound of
B(r), though. To do so, let α represent an arbitrary rational number such that bα < br . From this, we need to
4 prove that α < r.
7→ bα < br
α −r →b b α−r α−r →b
→b hypothesis of contradiction
r −r <b b theorem S2 r−r <b from part b <1 from part b Exercise 1.7 a
Proof by induction. Let S = {n : bn − 1 ≥ n(b − 1)}. We can easily verify that 1 ∈ S. Now, assume that k ∈ S:
7→ k ∈ S hypothesis of induction → bk − 1 ≥ k(b − 1) definition of membership in S → bbk − 1 ≥ k(b − 1) we’re told that b > 1. k+1 − b ≥ k(b − 1) k+1 ≥ k(b − 1) + b →b
→b → bk+1 − 1 ≥ k(b − 1) + b − 1
→ bk+1 − 1 ≥ (k + 1)(b − 1)
→ k+1∈S definition of membership in S By induction, this proves that (∀n ∈ N)(bn − 1 ≥ n(b − 1)).
Alternatively, we could prove this using the same identity that Rudin used in the proof of 1.21. From the
distributive property we can verify that bn − an = (b − a)(bn−1 a0 + bn−2 a1 + . . . + b0 an−1 ). So when a = 1, this
becomes bn − 1 = (b − 1)(bn−1 + bn−2 + . . . + b0 ). And since b > 1, each term in the bn−k series is greater than
1, so bn − 1 ≥ (b − 1)(1n−1 + 1n−2 + . . . + 10 ) = (b − 1)n. Exercise 1.7 b
1 1 7→ n(b n − 1) = n(b n − 1)
1 1 → n(b n − 1) = (1 + 1 + . . . + 1)(b n − 1)
|
{z
}
1
n → n(b − 1) = (1 n times
n−1
n−2 +1 1 + . . . + 10 )(b n − 1)
1 It can be shown that bk > 1 when b > 1, k > 0 (see theorem S4). Replacing 1 with b n gives us the
inequality:
1 1 1 1 1 → n(b n − 1) ≤ ((b n )n−1 + (b n )n−2 + . . . + (b n )0 )(b n − 1)
Now we can use the algebraic identity bn − an = (bn−1 a0 + bn−2 a1 + . . . + b0 an−1 )(b − a):
1 1 → n(b n − 1) ≤ ((b n )n − 1)
1 → n(b n − 1) ≤ (b − 1) Exercise 1.7 c
7→ n > (b − 1)/(t − 1) assumed → n(t − 1) > (b − 1) this holds because n, t, and b are greater than 1
1
n → n(t − 1) > (b − 1) ≥ n(b − 1)
1
n → n(t − 1) > n(b − 1)
1
n → (t − 1) > (b − 1)
→t>b from part b
transitivity of order relations
n > 0 → n−1 > 0 would be a trivial proof 1
n 5 Exercise 1.7 d
We’re told that bw < y, which means that 1 < yb−w . Using the substitution yb−w = t with part (c), we’re lead
1
1
directly to the conclusion that we can select n such that yb−w > b n . From this we get y > bw+ n , which is what
1
1
w+
w
we were asked to prove. As a corollary, the fact that b n > 1 means that b n > b . Exercise 1.7 e
We’re told that bw > y, which means that bw y −1 > 1. Using the substitution bw y −1 = t with part (c), we’re
1
lead directly to the conclusion that we can select n such that bw y −1 > b n . Multiplying both sides by y gives us
−1
1
1
w
w−
b > b n y. Multiplying this by b n gives us b n > y, which is what we were asked to prove. As a corollary,
−1
1
1
the fact that b n > 1 > 0 means that, upon taking the reciprocals, we have b n < 1 and therefore bw− n < bw . Exercise 1.7 f
We’ll prove that bx = y by showing that the assumptions bx > y and bx < y lead to contradictions.
1
If bx > y, then from part (e) we can choose n such that bx > bx− n > y. From this we see that x − n1 is an
upper bound of A that is smaller than x. This is a contradiction, since we’ve assumed that x =sup(A).
1
If bx < y, then from part (d) we can choose n such that y > bx+ n > bx . From this we see that x is not an
upper bound of A. This is a contradiction, since we’ve assumed that x is the least upper bound of A.
Having ruled out these two possibilities, the trichotomy property of ordered fields forces us to conclude that
bx = y. Exercise 1.7 g
Assume that there are two elements such that bw = y and bx = y. Then by the transitivity of equality relations,
bw = by , although this seems suspiciously simple. Exercise 1.8
In any ordered set, all elements of the set must be comparable (the trichotomy rule, definition 1.5). We will
show by contradiction that (0, 1) is not comparable to (0, 0) in any potential ordered field containing C. First,
we assume that (0, 1) > (0, 0) :
7→ (0, 1) > (0, 0) hypothesis of contradiction → (0, 1)(0, 1) > (0, 0) definition 1.17(ii) of ordered fields We assumed here that (0, 0) can take the role of 0 in definition 1.17 of an ordered field. This is
a safe assumption because the uniqueness property of the additive identity shows us immediately
that (0, 0) + (a, b) = (a, b) → (0, 0) = 0.
→ (−1, 0) > (0, 0) definition of complex multiplication → (−1, 0)(0, 1) > (0, 0) definition 1.17(ii) of ordered fields, since we initially assumed (0, 1) > 0 → (0, −1) > (0, 0) definition of complex multiplication It might seem that we have established our contradiction as soon as we concluded that (−1, 0) > 0
or (0, −1) > 0. However, we’re trying to show that the complex field cannot be an ordered field
under any ordered relation, even a bizarre one in which −1 > −i > 0. However, we’ve shown that
(0, 1) and (0, −1) are both greater than zero. Therefore:
→ (0, −1) + (0, 1) > (0, 0) + (0, 1) definition 1.17(i) of ordered fields → (0, 0) > (0, 1) definition of complex multiplication This conclusion is in contradiction of trichotomy, since we initially assumed that (0, 0) < (0, 1). Next, we
assume that (0, 1) < (0, 0): 6 7→ (0, 0) > (0, 1) hypothesis of contradiction → (0, 0) + (0, −1) > (0, 1) + (0, −1) definition 1.17(i) of ordered fields → (0, −1) > (0, 0) definition of complex addition → (0, −1)(0, −1) > (0, 0) definition 1.17(ii) of ordered fields → (−1, 0) > (0, 0)
→ (−1, 0)(0, −1) > (0, 0) definition of complex multiplication
definitino 1.17(ii) of ordered fields, since we’ve established (0, −1) >
(0, 0) → (0, 1) > (0, 0) definition of complex multiplication Once again trichotomy has been violated.
Proof by contradiction that (0, 1) 6= (0, 0): if we assume that (0, 1) = (0, 0) we’re led to the conclusion that
(a, b) = (0, 0) for every complex number, since (a, b) = a(0, 1)4 + b(0, 1) = a(0, 0) + b(0, 0) = (0, 0). By the
transitivity of equivalence relations, this would mean that every element is equal to every other. And this is
in contradiction of definition 1.12 of a field, where we’re told that there are at least two distinct elements: the
additive identity (’0’) and the multiplicative identity (’1’). Exercise 1.9a
To prove that this relation turns C into an ordered set, we need to show that it satisfies the two requirements
in definition 1.5. Proof of transitivity:
7→ (a, b) < (c, d) ∧ (c, d) < (e, f ) assumption → [a < c ∨ (a = c ∧ b < d)] ∧ [c < e ∨ (c = e ∧ d < f )] definition of this order relation → (a < c ∧ c < e) ∨ (a < c ∧ c = e ∧ d < f )
∨(a = c ∧ b < d ∧ c < e) ∨ (a = c ∧ b < d ∧ c = e ∧ d < f ) distributivity of logical operators → (a < e) ∨ (a < e ∧ d < f ) ∨ (a < e ∧ b < d) ∨ (a = e ∧ b < f ) transitivity of order relation on R Although we’re falling back on the the transitivity of an order relation, we are not assuming what
we’re trying to prove. We’re trying to prove the transitivity of the dictionary order relation on C,
and this relation is defined in terms of the standard order relation on R. This last step is using the
transitivity of this standard order relation on R and is not assuming that transitivity holds for the
dictionary order relation.
→ (a < e) ∨ (a < e) ∨ (a < e) ∨ (a = e ∧ b < f ) p∧q →p → a < e ∨ (a = e ∧ b < f ) p∨p→p → (a, b) < (e, f ) definition of this order relation To prove that the trichotomy property holds for the dictionary relation on Q, we rely on the trichotomy
property of the underlying standard order relation on R. Let (a, b) and (c, d) be two elements in C. From the
standard order relation, we know that
7→ (a, b) ∈ C ∧ (c, d) ∈ C assumed → a, b, c, d ∈ R definition of a complex number → (a < c) ∨ (a > c) ∨ (a = c) trichotomy of the order relation on R → (a < c) ∨ (a > c) ∨ (a = c ∧ (b < d ∨ b > d ∨ b = d)) trichotomy of the order relation on R → (a < c) ∨ (a > c) ∨ (a = c ∧ b < d) ∨ (a = c ∧ b > d)
∨(a = c ∧ b = d) distributivity of the logical operators → (a, b) < (c, d) ∨ (a, b) > (c, d) ∨ (a, b) < (c, d) ∨ (a, b) > (c, d)
∨(a, b) = (c, d) definition of the dictionary order relation → (a, b) < (c, d) ∨ (a, b) > (c, d) ∨ (a, b) = (c, d)
And this is the definition of the trichotomy law, so we have proven that the dictionary order turns the 7 complex numbers into an ordered set. Exercise 1.9b
C does not have the least upper bound property under the dictionary order. Let E = {(0, a) : a ∈ R}. This subset
is just the imaginary axis in the complex plane. This subset clearly has an upper bound, since (x, 0) > (0, a) for
any x > 0. But it does not have a least upper bound: for any proposed upper bound (x, y) with x > 0, we see
that
x
(x, y) < ( , y) < (0, a)
2
So that ( x2 , y) is an upper bound less than our proposed least upper bound, which is a contradiction. Exercise 1.10
This is just straightforward algebra, and is too tedious to write out. Exercise 1.11
If we choose w = z
|z| z
and choose r = |z|, then we can easily verify that |w| = 1 and that rw = |z| |z|
= z. Exercise 1.12
√
√
Set ai zi and bi = z¯i and use the Cauchy-Schwarz inequality (theorem 1.35). This gives us 2 X n
n
X n √ p √ 2X p 2 zj z¯j ≤
| zj |
| z¯j | j=1 j=1
j=1
which is equivalent to
2 |z1 + z2 + . . . + zn |2 ≤ (|z1 | + |z2 | + . . . + |zn |)
Taking the square root of each side shows that
|z1 + z2 + . . . + zn | ≤ |z1 | + |z2 | + . . . + |zn |
which is what we were asked to prove. Exercise 1.13
|x − y|2
= (x − y)(x − y)
= (x − y)(¯
x − y¯) Theorem 1.31(a) = x¯
x − x¯
y − y¯
x + y y¯
= x¯
x − (x¯
y + y¯
x) + y y¯
= |x|2 − 2Re(x¯
y ) + |y|2 Theorem 1.31(c), definition 1.32 ≥ |x|2 − 2|Re(x¯
y )| + |y|2 x ≤ |x|, so −|x| ≥ |x|. ≥ |x|2 − 2|x¯
y | + |y|2 Theorem 1.33(d) 2 2 Theorem 1.33(c) = |x|2 − 2|x||y| + |y|2 Theorem 1.33(b) = |x| − 2|x||¯
y | + |y| = (|x| − |y|)(|x| − |y|)
= (|x| − |y|)(|¯
x| − |¯
y |) Theorem 1.33(b) = (|x| − |y|)(|x| − |y|) Theorem 1.31(a) 2 = ||x| − |y|| This chain of inequalities shows us that ||x| − |y||2 ≤ |x − y|2 . Taking the square root of both sides results
in the claim we wanted to prove.
8 Exercise 1.14
|1 + z|2 + |1 − z|2
= (1 + z)(1 + z) + (1 − z)(1 − z)
= (1 + z)(¯
1 + z¯) + (1 − z)(¯
1 − z¯) Theorem 1.31(a) = (1 + z)(1 + z¯) + (1 − z)(1 − z¯) The conjugate of 1 = 1 + 0i is just 1 − 0i = 1. = (1 + z¯ + z + z z¯) + (1 − z¯ − z + z z¯)
= (2 + 2z z¯)
= (2 + 2) We are told that z z¯ = 1 =4 Exercise 1.15
Using the logic and the notation from Rudin’s proof of theorem 1.35, we see that equality holds in the Schwarz
inequality when AB = |C|2 . This
have B(AB − |C|2 ) = 0, and from the given chain of equalities
P occurs when
2
we see that this occurs when
|Baj − Cbj | = 0. For this to occur we must have Baj = Cbj for all j, which
occurs only when B = 0 or when
C
aj = bj for all j
B
That is, each aj must be a constant multiple of bj . Exercise 1.16
We know that |z − x|2 = |z − y|2 = r2 . Expanding these terms out, we have
|z − x|2 = (z − x) · (z − x) = |z|2 − 2z · x + |x|2
|z − y|2 = (z − y) · (z − y) = |z|2 − 2z · y + |y|2
For these to be equal, we must have
−2z · x + |x|2 = −2z · y + |y|2
which happens when
1
[|x|2 − |y|2 ]
(1)
2
We also want |z − x| = r, which occurs when z = x + rˆ
u where |ˆ
u| = 1. Using this representation of z, the
requirement that r2 = |z − y|2 becomes
z · (x − y) = r2 = |z − y|2 = |x + rˆ
u − y|2 = |(x − y) + rˆ
u2 | = |x − y|2 + 2rˆ
u · (x − y) + |rˆ
u|2 = d2 + 2rˆ
u · (x − y) + r2
Rearranging some terms, this becomes
u
ˆ · (y − x) = d2
d
=
|y − x|
2r
2r (2) A quick, convincing, and informal proof would be to appeal to the relationship a · b = |a||b| cos(θ) where θ is the
angle between the two vectors; the previous equation then becomes
|ˆ
u||y − x| cos(θ) = d
|y − x|
2r Dividing by |y − x| and remembering that u
ˆ = 1, this becomes
cos(θ) = d
2r where θ is the angle between the fixed vector (y − x) and the variable vector u
ˆ. It’s easy to see that this equation
will hold for exactly one u
ˆ when d = 2r; it will hold for no u
ˆ when d > 2r; it will hold for two values of u
ˆ when
d < 2r and n = 2; and it will hold for infinitely many values of u
ˆ when d < 2r and n > 2. Each value of u
ˆ
corresponds with a unique value of z. More formal proofs follow.
9 part (a)
When d < 2r, equation (2) is satisfied for all u
ˆ for which
u
ˆ · (y − x) = d
|y − x| < |y − x|
2r By the definition of the dot product, this is equivalent to
u1 (y1 + x1 ) + u2 (y2 + x2 ) + . . . + uk (yk + xk ) = d
|y − x|
2r The only other requirement for the values of ui is that
q
u21 + u22 + . . . + u2k = 1 (3) (4) This gives us a system of two equations with k variables. As long as k ≥ 3 we have more variables than equations
and therefore the system will have infinitely many solutions.
part (b)
Evaluating d2 , we have:
d2 = |x − y|2
= |(x − z) + (z − y)|2
= [(x − z) + (z − y)] · [(x − z) + (z − y)] definition of inner product · = (x − z) · (x − z) + 2(x − z) · (z − y) + (z − y) · (z − y) inner products are distributive 2 2 = |x − z| + 2(x − z) · (z − y) + |z − y| definition of inner product · Evaluating (2r)2 , we have:
(2r)2 = (r + r)2 = (|z − x| + |z − y|)2
= |z − x|2 + 2|z − x||z − y| + |z − y|2 commutativity of multiplication If 2r = d then d2 = (2r)2 and therefore, by the above evaluations, we have
|x − z|2 + 2(x − z) · (z − y) + |z − y|2 = |z − x|2 + 2|z − x||z − y| + |z − y|2
which occurs if and only if
2(x − z) · (z − y) = 2|x − z||z − y|
From exercise 14 we saw that this equality held only if (x − z) = c(z − y) for some constant c; we know that
|x − z| = |z − y| so c = ±1; we know x 6= y so c = 1. Therefore we have x − z = z − y, from which we have
z= x+y
2 and there is clearly only one such z that satisfies this equation.
part (c)
If 2r < d then we have
|x − y| > |x − z| + |z − y|
which is equivalent to
|(x − z) + (z − y)| > |x − z| + |z − y|
which violates the triangle inequality (1.37e) and is therefore false for all z. 10 Exercise 1.17
First, we need to prove that a · (b + c) = a · b + a · c and that (a + b) · c = a · c + b · c: that is, we need to prove
that the distributive property holds between the inner product operation and addition.
P
a · (b + c) = ai (bi + ci )
P
= (ai bi + ai ci )
P
P
=
ai bi + ai ci
=a·b+a·c
P
(a + b) · c = (ai + bi )ci
P
= (ai ci + bi ci )
P
P
=
ai ci + bi ci
=a·c+b·c definition 1.36 of inner product
distributive property of R
associative property of R
definition 1.36 of inner product
definition 1.36 of inner product
distributive property of R
associative property of R
definition 1.36 of inner product The rest of the proof follows directly:
|x + y|2 + |x − y|2 = (x + y) · (x + y) + (x − y) · (x − y)
= (x + y) · x + (x + y) · y + (x − y) · x − (x − y) · y
=x·x+y·x+x·y+y·y+x·x−y·x−x·y+y·y
=x·x+y·y+x·x+y·y
= 2|x|2 + 2|y|2 Exercise 1.18
If x = 0, then x · y = 0 for any y ∈ Rk . If x 6= 0, then at least one of the elements x1 , . . . , xk must be nonzero:
let this element be represented by xa . Let xb represent any other element of x. Choose y such that: x
b i=a xa
yi =
−1 i = b 0
otherwise
We can now see that x·y = xa xxab +xb (−1) = xb −xb = 0. We began with an arbitrary vector x and demonstrated
a method of construction for y such that x · y = 0: therefore, we can always find a nonzero y such that x · y = 0.
This is not true in R1 because the nonzero elements of R are closed with respect to multiplication. Exercise 1.19
We need to determine the circumstances under which |x a| = 2|x − b| and |x − c| = r. To do this, we need to
manipulate these equalities until they have a common term that we can use to compare them.
|x − a| = 2|x − b|
|x − a|2 = 4|x − b|2
|x|2 − 2x · a + |a|2 = 4|x|2 − 8x · b + 4|b|2
3|x|2 = |a|2 − 2x · a + 8x · b − 4|b|2
|x − c| = r
|x − c|2 = r2
|x|2 − 2x · c + |c|2 = r2
|x|2 = r2 + 2x · c − |c|2
3|x|2 = 3r2 + 6x · c − 3|c|2
Combining these last two equalities together, we have 11 7→ |a|2 − 2x · a + 8x · b − 4|b|2 = 3r2 + 6x · c − 3|c|2
→ |a|2 − 2x · a + 8x · b − 4|b|2 − 3r2 − 6x · c + 3|c|2 = 0
→ |a|2 − 4|b|2 + 3|c|2 − 2x · (a − 4b − 3c) − 3r2 = 0
We can zero out the dot product in this equation by letting 3c = 4b − a. Of course, this also
determines a specific value of c. This substitution gives us the new equality: 4b a 2
2
2 → |a| − 4|b| + 3 − − 3r2 = 0
3
3
3
·
16
3
3·8
→ |a|2 − 4|b|2 +
|b|2 −
a · b + |a|2 − 3r2 = 0
9
9
9
→ 3|a|2 − 12|b|2 + 16|b|2 − 8a · b + |a|2 − 9r2 = 0
→ 4|a|2 4|b|2 − 8a · b − 9r2 = 0
→ 4|a − b|2 = 9r2
→ 2|a − b| = 3r
By choosing 3c = 4b − a and 3r = 2|a − b|, we guarantee that |x − a| = 2|x − b| iff |x − c| = r. Exercise 1.20
We’re trying to show that R has the least upper bound property. The elements of R are certain subsets of Q,
and < is defined to be “is a proper subset of”. To say that an element α ∈ R has a least upper bound, then,
is to say that the subset α has some “smallest” superset β such that α ⊂ β. We’re asked to omit property III,
which told us that each α ∈ R has no largest element.
√ With this restriction lifted, we have a new definition of
“cut” that includes cuts such as (−∞, 1] and (−∞, 2].
To prove that each subset of R with an upper bound must have a least upper bound, we will follow Step 3
in the book almost exactly....

 


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