Problem Set 2
Marcel Dupont
October 12, 2015
2-13
(a) Prove that 6 is irrational.
Proof. We first want to show that
(i) If n2 is an even number, then n is also even
We will prove by contradiction. Suppose that n is odd. Then n = 2k + 1 where k N . Now,
n2
Two problems on limits illustrated.
1. Let a R. To nd limxa x2 .
Note : The procedure here is mostly, but not fully adapted from the
class discussion of limx3 x2 . The inequality part is being handled with
the triangle inequality, which happens to be more
Some denitions for Chapter 3
Denition of injectivity : Let f be a function. Then f is said to be
injective, or equivalently one-one, if :
for all x, y R, if f (x) = f (y) then x = y.
If this is the rst time you are seeing this concept, please think throug
MA 161
Hour Test 1
Autumn 2011
Name:
Student ID #:
Instructor: Vadakkumkoor Sandeep Varma
Instructions
1. Write your name on each page. Well, at least on the rst page.
2. There are 6 problems on 8 pages.
3. Anything you write needs justication. For instan
Math 161, Section 43
Some partially done notes
We assume (no more than) that we are given :
a set R;
for each (a, b) R R (i.e., each ordered pair (a, b) of elements in R) a
unique element of R called a + b and a unique element of R called a b;
and
a se
Notes on Negation, on maximum of a subset of R, and open and
closed sets.
I. On negation
Let us illustrate negation with an example. The negation of the sentence
All humans are mortal na
vely (though still correctly) would be Not all
humans are mortal, bu
MA 161
Hour Test 1
Autumn 2010
Name:
Student ID #:
Instructor: Vadakkumkoor Sandeep Varma
Instructions
1. Write your name on each page. Well, at least on the rst page.
2. There are 6 problems on 8 pages.
3. Anything you write needs justication. For instan
Lemmas and Remarks from Lecture 2, Wednesday, September 28
Lemma 1.5 : For all a R, a 0 = 0.
Lemma 1.6 : For all a, b R, a (b c) = a b a c.
Remark 1.2 (a) : For all a, b R, a (b) = (a b).
Remark 1.2 (b) : For all a, b R, (a) b = (a b).
Remark 1.2 (c) : Fo
Lemmas discussed on Friday, September 30
Lemma 1.9 : For all a R, |a| = | a|.
Lemma 1.10 : For all a, b R, (a + b) = (a) + (b).
Lemma 1.11 (i) : For all a, b R, |a + b| |a| + |b|.
Lemma 1.11 (ii) : For all a, b R, |a| |b| |a b.
1
MA 161, Section 43
Final Examination
Autumn 2010
Name:
Student ID #:
Instructor: Vadakkumkoor Sandeep Varma
Instructions
1. There are 9 problems on 12 pages (including cover sheet and rough work sheet).
2. Anything you write needs justication. For instanc
Math 161, Section 21
Some problems to practise for Hour Test 1
Note : These problems are intended to help you test and strengthen your
conceptual grasp of Lessons 1-5, and there is no guarantee that the exam
problems will be similar to these.
1. Pre-absol
Math 161, Section 21
Some problems to practise nding supremum/inmum with
Find (and as always, justify!) sup A and inf A for the following sets A :
1. A = cfw_x2 | a < x < b, where a, b are real numbers with a < b.
2. A = cfw_x | x Q, x2 < 2.
Note : Here y
MA 161
Hour Test 2
Autumn 2010
Name:
Student ID #:
Instructor: Vadakkumkoor Sandeep Varma
Instructions
1. Write your name on each page, if possible.
2. There are 5 problems on 8 pages (including cover sheet and the last sheet).
3. Anything you write needs
Problem Set 8
Marcel Dupont
February 7, 2016
10-20
Suppose that f (n) (a) and g (n) (a) exists. Prove Leibnizs formula:
n
(k)
P
n
(f g)(n) (a) =
(a) g (nk) (a).
k f
k=0
Proof. We will proceed by induction on n. By the product rule,
(f g)0 (a) = f 0 (a) g
Problem Set 5
Marcel Dupont
January 7, 2016
7-12 b)
Prove the following more general fact: If g is continuous on [0, 1] and g(0) = 0,
g(1) = 1 or g(0) = 1, g(1) = 0, then f (x) = g(x) for some x.
Proof. Two cases.
For g(0) = 0 and g(1) = 1
If f (0) = 0 or
Problem Set 4
Marcel Dupont
October 26, 2015
6-1
(iv)
Proof. f (x) is not continuous because its domain is only x Q, so there is no continuous function
F with domain of rationals such that F (x) = f (x)
6-14
(a)
Proof. Since g and h are continuous at a, l
Problem Set 6
Marcel Dupont
February 7, 2016
8-14 (a)
Consider a sequence of closed intervals I1 = [a1 , b1 ], I2 = [a2 , b2 ], . Suppose that an an+1
and bn+1 bn for all n. Prove that there is a point x which is in every In .
Proof. We define A := cfw_an
Problem Set 9
Marcel Dupont
December 2, 2015
11-38
Prove that if
a0
1
+
a1
2
+ +
2
an
n+1
= 0, then a0 + a1 x + + an xn = 0 for some x in (0, 1).
n+1
x
, then f 0 (x) = a0 + a1 x + + an xn . f (0) = 0 and we
Proof. If f (x) = a0 x + a12x + + ann+1
a0
a1
a
Problem Set 3
Marcel Dupont
February 7, 2016
5-39
(v) find limx x2 + 2x x if it exists.
Proof. In problem 5-32, we showed that limx x1 = 0. We also know that a limit of a constant is
just the p
constant itself,
so limx 2 = 2 and limx 1 = 1. A variant of T
Problem Set 1
Marcel Dupont
October 9, 2015
1.
(i) If ax = a for some number a 6= 0, then x = 1
Proof. Assume a 6= 0. By (P7), there exists some number a such that a1 a = 1. Now
ax = a
1
a
ax = a1 a
(a1 a)x = (a1 a)
by (P5)
1x=1
by (P7)
x=1
by (P6)
(ii)
MA 161
Exam 2
Autumn 2011
Name:
Student ID #:
Instructor: Vadakkumkoor Sandeep Varma
Instructions
1. Write your name on each page, if possible.
2. There are 5 problems on 6 pages.
3. Anything you write needs justication. For instance, if a problem says Fi
Solution to Quiz 5
Question : Find the supremum of the set :
A=
1
1
|nN .
n
Solution : First you will have to guess the answer to be 1 - to gure this
out you can write the rst few or several terms in the rough space and notice
the trend.
Okay, so let us p
Math 161, Section 21
Some problems to practise for Hour Test 2
Note : These problems are intended to help you test and strengthen
your conceptual grasp of Lessons 6-8, and there is no guarantee that the
exam problems will be similar to these.
1. Continuit
Learn to state all denitions and theorems. You should write these in full
sentences. I was very lenient in grading these during the hour tests, for reasons of morale as you were only getting used to rigor, but this time I plan to
be more serious. Please w
Homework 26 (Recommended, but not mandatory)
Never ever due
First do : Suppose f (x) = n x (= x1/n ) for all x > 0. Prove using Theorem 12.5 that f is dierentiable on (0, ), with f (x) = (1/n)x(1/n)1 for all
x (0, ). If n is odd, and f (x) = n x for all x
Homework 25
Due Wednesday, November 30
1. Page 213, #54 (a)
Important Note : It is recommended that you also do 54 (b), (c),
(d) and 56. At any rate, be familiarze yourself with their statements,
as I may ask them for exam - as you have seen them from hig
Homework 13
Due Friday, October 28
Note on what you may use for this homework : For problems in
Chapter 7, you may assume that if f (x) is a polynomial or if f = sin, then
f is continuous at every point of R, i.e., f is continuous on R, and then a
problem
Homework 10
Due Wednesday, October 19
Note :
Yes, there are a lot of problems here, but these problems are
going to be extremely useful in later chapters. This is a long homework it is understandable if you dont nish everything - but better nish some
prob
Homework 9
Due Wednesday, October 19
Note : Though they are quite a few problems, most of them are easy. You
dont have any lthy computation. But please do this over the weekend, you
might get pressured if you postpone it for Wednesday. Also remember next