5. Letn beanatural number. Prove that
13+23+-+rt3 =(1+2+-»+n)3.
6. Letnt and n be natural numbers.
a. Prove that the sum. m +n. also is a natural number. (Hint: Fix m and deﬁne S(n)
to be the statement that m + n is a natural number.)
b. Prove that the pr
MATH 142A
Homework 6 Solutions
Jonathan Conder
Section 3.4
5. The sequence cfw_f n +
1
n
f (n) does not converge to 0, because
f
n+
1
n
f (n) = 3n +
3
1
+
> 3n
n n3
for every index n N.
8. If f : I R is dened by f (x) :=
f
ba
xa ,
then
a+
ba
3n
f
for ev
Math 142A
Homework 7 Partial solutions
Problem 4.1.6
Solution f is differentiable at h(x0 ), so there is L such that for each > 0 there is > 0 such that
if 0 < |y h(x0 )| < , then
f (y) f (h(x0 )
L .
y h(x0 )
Of course, then L = f 0 (h(x0 ). Since h is
Math 142A
Homework 6 Partial solutions
Problem 3.6.8
There exists no surjection from a countable to an uncountable set, thus, no such f exists. But in
this class we did not discuss (un)countability, so we need a different solution.
Solution:
If such f exi
Math 142A
Homework 3 Partial solutions
Problem 2.1.17
Solution: Assume that limn an = + and let > 0. Then there exists N N such that
an > 1 for each n N , and so 0 < 1/an < for each n N . Thus, limn an = 0.
Conversely, assume that limn 1/an = 0 and let C
Math 142A
Homework 5 Partial solutions
Problem 3.3.11
Solution
a) Let x R. As I is unbounded below and above there exist y1 , y2 I such that y2 < x < y1 . As
I is convex [y2 , y1 ] I and so x I. Thus, R I and so I = R.
b) Suppose that I is bounded below b
Math 142A
Homework 1 Partial solutions
Problem 1.1.2
a) False. Consider the set S = (, 2). We see that sup S = 2 but 2
/ S and thus has no maximum.
b) True. It is clear that 0 is a lower bound for S and thus inf S 0.
c) Let M be an upper bound for S, and
Math 142A
Homework 2 Partial solutions
Problem 1.3.8
If n = 1, then (1 + x)1 1 + x.
Let n N and suppose that (1 + x)n 1 + nx. Then it follows that
(1 + x)n+1 = (1 + x)n (1 + x)
(1 + nx)(1 + x)
= 1 + nx + x + nx2
= 1 + (n + 1)x + nx2
1 + (n + 1)x,
where
1.1 #2, 11, 13, 14, 15. 1.2 #1, 2, 3, 4(a), 6. 1.3 #7, ll, 14, 17, 18.
2.1 #1, 2, 6, 8, 11.
1.1.2. (a) False. S = [0, 1).
(b) True. If :13 E S, then a: > 0, so that :1: Z O. (and S is nonempty)
Hence 0 is a lower bound for 5'. Hence 0 S inf S.
(c) True. s
Leudar et al.: Hostility towards refugees 187
ARTICLE
Hostility themes in media, community
and refugee narratives
IVAN LEUDAR AND JACQUELINE HAYES
U N IVERSITY OF MANC H ESTER, U K
J I R N E K V A P I L
CHARLES UNIVERSITY, CZECH REPUBLIC
JOHANNA TURNER BA
HOMEWORK #2 SOLUTIONS
Solution to Problem 1.2.6 Let b R be an upper bound for S. Suppose that b < a. Since Q is dense in R, then there
exist c (b, a) Q. Thus, c S, which implies that b is not an upper bound of S. This in a contradiction. So b a.
This impl
HOMEWORK #1 SOLUTIONS
Solution to Problem 1.1.2(a) False. The set (0, 1) is bounded above but does not have a largest number.
(b) True. Since 0 is a lower bound of S and inf S is the greatest lower bound, so 0 inf S.
(c) True. sup S is an upper bound of S
MIDTERM 1 SOLUTIONS
Problem 1. (10 points) Prove the following inequality (using induction)
n
X
k < (n + 1)2 , n 1.
k=1
Solution
Let S(n) be the statement
n
X
k < (n + 1)2 . We will show that S(n) is true
k=1
for all n by induction.
For n = 1, we have 1 <
Practice Midterm 1
Problem 1. Prove the following inequality (using induction)
n
X
k < (n + 1)2 , n 1.
k=1
Note: if you use any formula, you need to prove it!
Problem 2. a)
of a set S being dense in R.
Writedown the
definition
b) Prove that 5Q + 3 = cfw
Midterm 1
Math 142A, Lecture C
Fall 2016
Time allowed: 50mins
(40 points total)
Name:
PID:
1
1. (3 points) State the Completeness Axiom for R.
2. Carefully state the following denitions:
(a) (4 points) Dene what it means for a sequence cfw_an to converge
Math 142A (Fall 2016): Midterm Exam 2 Solutions
1. Circle the correct answer true (T) or false (F). No justifications are needed. (1 pt each)
Solution:
(1) If In is an interval and In+1 In for each n N, then there is at least one x R which
belongs to all
Math 142A (Fall 2016): Midterm Exam 1 Solutions
1. Circle the correct answer true (T) or false (F). No justifications are needed. (1 pt each)
Solution:
(1) A nonempty and bounded above A R has either a maximum or a minimum.
F
(2) If A B R and A, B are non
MATH 142A
Homework 4 Solutions
Jonathan Conder
Section 2.3
2. (a) Dene sn := n +
(1)n
n
sn+1 sn = (n + 1) +
for each index n. If n N and n 2 then
(1)n
(1)n+1
n
= 1 + (1)n+1
n+1
n
so sn < sn+1 . Since s1 = 0 <
(1)n
3n
1
(b) Dene sn := n2 +
cfw_sn is not m
MATH 142A
Homework 7 Solutions
Jonathan Conder
Section 4.1
1. (a) This is false: the absolute value function is continuous, but not dierentiable at 0.
(b) This is true, by Proposition 4.5.
(c) This is false: the absolute value function is not dierentiable
MATH 142A
Homework 3 Solutions
Jonathan Conder
Section 2.1
1. (a) This is false. For example, the sequence cfw_(1)n = cfw_1, 1, 1, 1, . . . does not converge, but cfw_(1)n )2
is constant (and converges to 1).
(b) This is false. For example the sequences
MATH 142A
Homework 5 Solutions
Jonathan Conder
Section 3.1
3. If x0 < 0 and cfw_xn is a sequence in R which converges to x0 , then there is an index N such that xn < 0 for all
indices n N. It follows that cfw_f (xN +n ) = cfw_x2 +n converges to x2 = f (
MATH 142A
Homework 2 Solutions
Jonathan Conder
Section 1.2
1. (a) This is false. For example (0, 1) contains no integers, by Proposition 1.6.
(b) This is false. For example (1, 0) contains no positive numbers.
(c) This is true; in fact Q \ Z is dense in R
MATH 142A
Chapter 3 Review
Jonathan Conder
Continuity
1. Which of the following functions are continuous?
(a) f : R R dened by f (x) := |x|
(b) f : R \ cfw_0 R dened by f (x) :=
(c) f : R \ cfw_1 R dened by f (x) :=
1
x
x2 1
x1
(d) f : R R dened by
(e) f
MATH 142A
Exam Solutions
Jonathan Conder
1. (a) True, by induction.
(b) True, because the cube root is continuous (Theorem 3.29).
(c) False. For example, every subsequence of cfw_n is unbounded.
(d) False. For example, if f (x) =
(e) True. For example, if
MATH 142A
Midterm 2 Solutions
Jonathan Conder
1. (a) This is true (in fact f is uniformly continuous): if > 0 and x, y Z satisfy |x y| < 1, then x = y and
hence |f (x) f (y)| = 0 < .
(b) This is true, by the extreme value theorem (take a maximum for f on
MATH 142A
Homework 1 Solutions
Jonathan Conder
Section 1.1
2. (a) This is false. For example (0, 1) is bounded above by 1, but given an element x (0, 1) there is a larger
element, such as x+1 .
2
(b) This is true, because 0 is a lower bound for S, and inf
MATH 142A
Midterm 1 Solutions
Jonathan Conder
1. (a) This is false, because 1 Q.
(b) This is false; for example N has no upper bound.
(c) This is true. Pick s S, and note that inf S s sup S.
(d) This is true (it was a homework problem).
(e) This is true,
MATH 142A
Homework 8 Solutions
Jonathan Conder
Section 4.2
1
1
4. If x, y I and x < y then 1 + x < 1 + y and hence f (x) = 1+x > 1+y = f (y). Therefore f is strictly decreasing.
1
1
1
By the quotient rule, f is dierentiable. Moreover 1 = 1+0 > f (x) > 1+1
4.
I! not (1), WE need to prove (2) must he the me:
Since 5 is not eornpact, so there E (1,.) c S with no convergent subsequenoe
or the subsequenoe converges to ii point outside 5, Considering (l)does riot hupe
pen, (2,.) is bounded Then by Theon 2.33, th
Math 142A
Homework 4 Partial solutions
Problem 2.4.12
Here issome motivation before giving the formal solution. Notice that if x2 x c = 0 then
x = 1 21+4c and as xn 0, we need only consider the root a = 1+ 21+4c . Thus, we want to show
that if xn > a the