Mathematics Exploration
Modeling Ponzi
Scheme
Name : Mugdha Gupta
Session : May 2015
Subject : Math Higher Level
Candidate Number : 002730 0046
Supervisor : Mr. Ken Adler
School : Dubai International Academy
Candidate number: 002730 0046
Mugdha Gupta
Mo
Lecture 2 Tuesday, September 15
Have you done these things?
1. Read the course outline
2. Picked up the textbook and started reading
(reading schedule on course website)
3. Visited the course website: LEARN
4. Submitted Assignment 0 to Crowdmark
1 / 13
Cr
Math 135: Lecture 2: Statements, (In)equations and RWD
Using the Cosine Law for supplementary angles \AP B and \AP C, and then clearing denominators and simplifying gives dad + man = bmb + cnc as required.
Proof D
The Cosine Law on 4AP B tells us that c2
Math 135: Lecture 3: Truth Tables, Logic and Implication
Example 3.18. Prove: if x is an integer and 22x is an odd integer, then 2
2x
Proof: Assume that x is an integer and 22x is an odd integer. Then 2x
would not be an integer).
Also, 2x < 1 (otherwise 2
Math 135: Lecture 3: Truth Tables, Logic and Implication
Denition
3.1
(Compound
Statement).
A
compound
statement
is
Example 3.2 (Compound Statement). The compound statement (8 is a positive integer) and
(5 > 2) has components
A:
B:
The compound statement
Math 135: Lecture 2: Statements, (In)equations and RWD
Denition 2.1 (Statement). A statement is a sentence that has a denite state of being
either
Example 2.2. Identify each statement as being either TRUE or FALSE.
1. 3+1=4.
2. 1 2.
3. x2 2 = 0 has two di
Lecture 3: Truth Tables, Logic and Implication
Math 135: Sections 018, 019
Math 135: Sections 018, 019
Lecture 3: Truth Tables, Logic and Implication
Oce Hours
Oce Hours are times for Math 135 students to seek help from me
in MC 6244.
Monday 3:30 pm  5:0
Math 135: Lecture 4: More Implication and Divisibility
Denition
4.1
(Prime).
An
integer
Question: Is 1 a prime number? Why or why not?
Answer:
Discover
Example 4.2. Prove that if (p, p + 2, p + 4) are all primes, then p = 3.
Recognizing Implications
Examp
Lecture 4: More Implication and Divisibility
Math 135: Sections 018, 019
Math 135: Sections 018, 019
Lecture 4: More Implication and Divisibility
Clicker Setup
To set clicker for this room, hold the power button until the two
letters ash. Then press the c
Math 135: Lecture 5: Introduction to Sets
Recall Divides:
If a and b are integers,
such that
we say that b divides a (ba) if
Proposition 5.1 (Divisibility of Integer Combinations (DIC). If a, b and c are integers
where
and x and y are integers, then
Exam
Lecture 6: Subsets, Set Equality, Converse and I
Math 135: Sections 018, 019
Math 135: Sections 018, 019
Lecture 6: Subsets, Set Equality, Converse and I
Denition 6.1 (Subset)
A set S is called a subset of a set T , and written
Math 135: Sections 018, 019
Math 135: Lecture 6: Subsets, Set Equality, Converse and I
Denition 6.1 (Subset). A set S is called a subset of a set T , and written
Symbolically, we write
S T means
.
Denition 6.2 (Proper
a set T , and written
Subset). A set S is called a proper subset
MATH 135, Fall 2015: Lecture 40
Polynomials
Notation
When we write F, we will mean one of Q, R, C or Zp .
F stands for eld. (Every nonzero element has a multiplicative
inverse in a eld.)
Question
Why is Z not a eld?
Polynomials
Denitions
An expression o
MATH 135, Fall 2015: Lecture 16
Midterm Info
It is important that every student read this entire announcement
very carefully.
1. The midterm exam is on Monday, Oct 19, 2015 from 7:00
p.m. to 8:50 p.m.
2. You must write the exam in your assigned room unles
MATH 135, Fall 2015: Lecture 27
iClickers
Clicker Question #1
I enjoy trying to discover and write MATH 135 proofs.
A) Strongly disagree
B) Disagree
C) Neither agree nor disagree
D) Agree
E) Strongly agree
iClickers
Clicker Question #2
When I have diculti
Lecture 11 Tuesday, September 30
From last class:
1. Contrapositives
(B) = (A) is the contrapositive of A = B.
The two statements are logically equivalent. (Why?)
Writing down the statement and the contrapositive gives you two
chances at a direct proof:
A
Lecture 2 Tuesday, September 15
Stewarts Theorem
Let ABC be a triangle with AB = c, AC = b and BC = a.
If P is a point on BC with BP = m, PC = n and AP = d,
then dad + man = bmb + cnc.
Stewarts Theorem
Let ABC be a triangle with AB = c, AC = b and BC = a.
Lecture 9 Monday, September 28
Universal Quantier (for all)
For every x in the set S, P(x) is true.
To prove:
Consider a representative object, say x, from the set S and show that
P(x) is true.
This representative must be a placeholder, not a specic eleme
MATH 135, Fall 2015
Section 026, Lecture 12
Friday 02 October 2015
Division Algorithm
If a and b are integers and b > 0, then there exist unique integers q and r such that
a = qb + r and 0 r < b.
Examples
(a, b) = (85, 11): 85 = 7(11) + 8
(a, b) = (2015,
MATH 135, Fall 2015
Section 026, Lecture 04
Friday 18 September 2015
Logical equivalence (continued)
Prove that A (B C) (A B) (A C).
Proof
We construct a truth table:
A
T
T
T
T
F
F
F
F
B
T
T
F
F
T
T
F
F
C
T
F
T
F
T
F
T
F
Conditional Statements
Read Sectio
MATH 135, Fall 2015
Section 026, Lecture 08
Friday 25 September 2015
Example
Prove that S = T if and only if S T = S T .
Proof (continued)
Suppose S T = S T .
We need to show S = T .
To do this, we show S T and T S.
Without loss of generality, we show tha
MATH 135, Fall 2015
Section 023, Lecture 32
Monday 09 November 2015
Example
Solve n12 5 (mod 55).
We dont have a good method of doing this now.
(And why is this in a section on systems of congruences?)
Splitting the Modulus (SM)
Let p and q be coprime pos
MATH 135, Fall 2015: Lecture 03
Proposition
For all x R, we have x +
1
2.
x
Proof?
Since (x 1)2 0 for all x R, then
x 2 2x + 1 0
x 2 + 1 2x
1
x+ 2
x
as required.
Proposition
If a = b, then 2 = 1.
Proof?
a=b
a2 = ab
a2 b 2 = ab b 2
(a + b)(a b) = b(a b)
a
MATH 135, Fall 2015: Lecture 05
Divisibility
Proposition (Bounds By Divisibility (BBD)
Let a and b be integers. If a  b and b = 0, then a b.
Things we can deduce
Divisibility
Proposition (Bounds By Divisibility (BBD)
Let a and b be integers. If a  b
MATH 135, Fall 2015: Lecture 25
LDE Example
Example
Alyssa has a lot of mail to send. She wishes to spend exactly $100
buying 49cent and 53cent stamps. In how many ways can she do
this?
LDE Example
Solution
Let x be the number of 49cent stamps that she
MATH 135, Fall 2015: Lecture 13
Functions
Denitions
Let S and T be two sets.
A function f : S T is onetoone (or injective) if and only if for
every x1 S and for every x2 S, f (x1 ) = f (x2 ) implies x1 = x2 .
A function f : S T is onto (or surjective) i
MATH 135, Fall 2015: Lecture 02
Proposition
A group of six people attend a party. It must be the case that
either three people are all friends or all not friends.
Proof
We label the people A, B, C , D, E , F .
We use green lines to connect pairs that are
Math 135: Lecture 7: Universal and Existential Quantiers
Open Sentence
A statement like x < 9 has no truth value until a value of x is specied.
Let P (x) denote the statement x < 9.
Then P (0) is
but P (10) is
Concept 7.1 (Open Sentence). We call P (x) an
Lecture 7: Universal and Existential Quantiers
Math 135: Sections 018, 019
Math 135: Sections 018, 019
Lecture 7: Universal and Existential Quantiers
Open Sentence
A statement like x < 9 has no truth value until a value of x is
specied.
Math 135: Sections
Chapter 3. Discovering Proofs
Practice (Course Note)
1. Prove the following statement. Let abc Z. If ac  bc and c 0, then a  b.
2. Prove the following statement. Let x be an integer. If 2  (x2  1), then 4 (x2 1).
3. Consider the following statement:
MATH 135 Fall 2013
Practice Problems: Congruence (continued)
Solutions to selected problems
1. [24]1 = [42] in Z53 . (Since 53 is prime, any nonzero class in Z53 is invertible.) There are
several ways of finding [24]1 :
Method 1. Solve the Diophantine eq
MATH 135 Fall 2013
Practice Problems: Divisibility
Solutions to selected problems
1. This is false. Counterexample: take x = 0 and y to be any irrational number.
0
= a/b for some integers a and b, with b 0.
2
2
Squaring both sides, x = a /b , which is rat

odeCaK Wm
H. Ame
5: NOW
7* >'+m~e
M
s
(4
we H + h: 10H
"
arsomeerk'sv'ue.
e metk+ )Lav +kq+ 7 HM,
3 [Ca I 59w. ? EM and 7 2 1 4M 5 62<
_+[_L_*
9 if (91/ +9"
3
lr'M'lM 1