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Math Foundations from Scratch
Math Foundations from Scratch
Table of Contents
Intro
What is math
Start here
The natural numbers
Arithmetic with natural numbers
Laws of arithmetic 2
Subtraction and division
The Hindu-Arabic number system
Arithmetic with Hindu-Arabic notation
Division
Fractions
Arithmetic with fractions
Laws of arithmetic for fractions
Proofs
Nature of mathematical logic
Lectures on logic
Materials needed
2.1 The Peano Axioms
2.2 Addition
First proof
Second proof
Third proof
Fourth proof & Implication
Exercise 2.2.1
2.2. cont.
Exercises
2.3 Multiplication
Exercises
Set Theory
History of set theory
Appendix A: the basics of mathematical logic
Lectures on deduction
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Quiz: Logic
3.1 Fundamentals (Set Theory)
Quiz Sets
Exercises
3.2 Russell's Paradox
3.3 Functions
Quiz Functions
Exercises
Proof strategies
3.4 Images and inverse images
3.5 Cartesian products
3.6 Cardinality of sets
Integers and rational numbers
Introducing the integers
4.1 The integers
Exercises
Rational numbers
Visualizing the rational numbers
The deep structure of the rational numbers
4.2 The rationals
Exercises
Decimal numbers
Appendix B: the decimal system
4.3 Absolute value and exponentiation
Exercises
4.4 Gaps in the rational numbers
Real Numbers
What exactly is a limit
2D geometric interpretation of Cauchy sequences
Real numbers and Cauchy sequences I
Real numbers and Cauchy sequences II
5.1 Cauchy sequences
5.2 Equivalent Cauchy sequences
5.3 The construction of the real numbers
5.4 Ordering the reals
5.5 The least upper bound property
5.6 Real number exponents I
Series
7.1 Finite series
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7.3 Sums of non-negative numbers
Trigonometry crash course
Math circle
Discrete math lecture 1
Algebra chapter 1
Discrete Math Lecture 2
CMU Putnam Seminar
Lecture 1 Introduction
Lecture 2 Polynomials
Algebra 2 Polynomials
Intro
This is part IB of the AI Tripos to learn the basics of mathematical reasoning so deduction and
strategies for writing proofs. No background is required. We start with some typical linear
progression then it turns into a game learning through solving problems. There is a seperate linear
algebra workshop here.
Do 1% effort for 100 days - Prof Ryan O'Donnell.
What is math
As per Poh-Shen Loh it is one of the easiest subject you will ever take because there is little to
memorize, you can deduce whatever it is you don't know using mathematical reasoning. The
closest analogy would be lawyers, they deduce from statutes and case law new arguments. We will
be doing the same thing using methods of reasoning. Once you start to get familiar with this logic
then you can use it not only to manipulate models you build in math of real life things but in
everyday life your thinking will change to become more rigorous. In math the model studies you
while you study the model.
Terence Tao believes mathematics is complex, high-dimensional (as in moving in any direction,
like the depiction of an ancient deity with many arms) and evolving in unexpected and adaptive
ways.
Start here
We start from the very beginning counting sticks and work our way up to the CMU Putnam
Seminar.
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The natural numbers
Numbers are objects you build to act like the concept of a number. Here we begin to construct the
natural number object.
YouTube - Arithmetic and Geometry Math Foundations 1 *note these videos eventually get
better in quality
Arithmetic with natural numbers
YouTube - Arithmetic and Geometry Math Foundations 2
An interesting explanation of multiplication.
Laws for addition
n + m = m + n (Commutative)
Changing the order of the elements.
(k + n) + m = k + (m + n) (Associative)
Changing the grouping of the elements.
n + 1 = s(n) (Successor)
Laws for multiplication
nm = mn (Commutative)
(kn)m = k(nm) (Associative)
n ⋅ 1 = n (Identity)
Note the different notation for multiplication:
nm
n⋅m
n×m
Distributive laws
k(n + m) = kn + km
(k + n)m = km + nm
Exercise: Prove why the multiplication and distributive laws work. Since we have no idea what a
proof is yet, we can instead try going through some of our own examples.
Identity law: n ⋅ 1 = n.
We are creating n groups of 1. According to our definition of multiplication |||| ⋅ | would be 4
groups of | or ||||. This is still 4.
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Commutative law: nm = mn.
||| ⋅ || = || || || or 6. Rearranging this, || ⋅ ||| is ||| ||| which is again 6.
Associative law: (kn)m = k(nm).
If k = |, and n = ||, and m = ||| and parens expressions are evaluated first:
Left side: (kn)m. We first evaluate (kn) as || (2 groups of 1, the identity law), and 2m is
||| ||| (2 groups of 3) for a total of 6.
Right side: k(nm): In the parens nm is 2 groups of 3, so ||| |||. k(6) is the identity law,
one group of 6. Thus (kn)m = k(nm).
Distributive law: (k + n)m = km + nm.
If k = ||||, and n = ||, and m = |:
Left side: (|||| + ||) is 6, and (6) x 1 is six groups of 1 or 6 according to the identity law.
Right side: km + nm: 4 groups of one is 4, plus two groups of 1 is 2 (identity law).
Answer is again 6.
Laws of arithmetic 2
YouTube - Arithmetic and Geometry Math Foundations 3
Wildberger shows another interpretation of multiplication, and walks through the proofs of the laws
by rearranging according to the commutative law.
Subtraction and division
YouTube - Arithmetic and Geometry Math Foundations 4
Subtraction is the inverse of addition
If k + m = n then k = n - m
Note: n - m is only defined if n > m as we are still in the realm of the Naturals and negative
numbers/integers aren't yet defined.
+, -, are binary operations and can only do 2 operations at a time
Laws of Subtraction:
n - (m + k) = (n - m) - k
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n + (m - k) = (n + m) - k
n - (m - k) = (n - m) + k
Distributive laws of subtraction:
n(m - k) = nm - nk
(n - m)k = nk - mk
Note to remove parens: https://www.themathpage.com/alg/parentheses.htm we can drop the
grouping enforcement when using addition (associative law) but must distribute the negative sign,
as subtraction is not associative.
n - (m + k) = (n - m) - k
removing grouping: n - m - k = n - m - k
The Hindu-Arabic number system
YouTube - Arithmetic and Geometry Math Foundations 6
A simplified Roman numeral system is introduced to abstract our previous notation of one's so we
can manipulate bigger numbers.
Uses powers of 10
327 = 3(100) + 2(10) + 7(1)
Arithmetic with Hindu-Arabic notation
YouTube - Arithmetic and Geometry Math Foundations 7
Multiplication with the Hindu-Arabic notation:
Same laws as before:
(ab)c = a(bc)
a(b + c) = ab + ac
(a + b)(c + d) = ac + ad + bc + bd
A different way of hand multiplication is shown, taking advantage of the fact the notation is using
the distributive law.
Division
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YouTube - Arithmetic and Geometry Math Foundations 8
Division is repeated subtraction. Laws of Division:
1.
2.
3.
4.
k
m
k
m
+
⋅
k/m
n
k
m/n
n
m
n
l
=
=
=
=
k+n
m
kn
ml
k
mn
kn
m
Fractions
YouTube - Arithmetic and Geometry Math Foundations 9
We're still using only natural numbers.
Def: A fraction, is an ordered pair (m,n) of natural numbers also written m/n.
Fractions m/n and k/l are equal precisely when ml=nk.
Arithmetic with fractions
YouTube - Arithmetic and Geometry Math Foundations 10
We're still in the natural numbers, which he defined as beginning with 1 so we aren't working with
zeros yet (no division by zero corner cases). The prime notation (the ' character so c', a', b') used in
the proof represents another variable. What we are trying to prove is ab' = ba' and cd' = dc'. He
introduces some extra notation multiplying both sides by (dd') to help prove this. This is our first
experience with a proof strategy: multiply both sides of something linear by a scalar.
Laws of arithmetic for fractions
YouTube - Arithmetic and Geometry Math Foundations 11
Wildberger proves the distributive and commutative laws using fractions while still in the natural
numbers.
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Proofs
Now we will redo what we just learned rigorously in Terence Tao's Analysis I. The author explains
here why we need this rigorous stage. This book was originally supplementary notes for an honors
analysis class, and in the preface he mentions we will need some lectures to go with the material to
help us understand it. We will read only some of this text at first, then return to it later when we're
doing problem challenges, by that time you will be much more sophisticated through practice.
Nature of mathematical logic
Figure 1: What is a proof from 15-251
Axioms are laws without a proof, as there are no earlier laws from which we can deduce their
proofs from and they should be self-evident and simple enough that they can be fully understood
such as Euclid's parallel postulate that it is obvious there exists a concept of parallelism.
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Deduction can be described as if P then Q, assume P and prove Q which is part of the rules of
inference. There is a series of lectures given by Per Martin-Lof in 1983, which in clear language
anybody can understand, explains the meaning of logical constants and the justifications of the
rules of inference, and their history such as hypothesis a Greek word translated into Latin as
supposition, meaning an assumption. In math the contrapositive of statements is often used for
proofs such as 'if P then Q so not Q then not P' as these are logically equivalent statements.
Variables are used to express general laws, for example how do you express that every natural
number is equal to itself ie: 2 = 2 and 3 = 3. You introduce a free variable x = x and it's meaning
remains fixed inside this context. Often the prime symbol is used to denote two different variables
with the same letter such as b and b' called b and b prime.
The operation of replacing all occurences of one variable for a value is known as substitution such
as a + 0 = a and if a = 1, then substitute 1 + 0 = 1. There is also a substitution rule, where equivalent
terms or formulas can be substituted for one another and this becomes highly abstract interchanging
entire complex expressions. In books you will often see a succession of equal signs: (thing) =
(bigger thing) = (even more complex thing) meaning all 3 can be substituted for each other.
Math has data structures just like programming languages, and you can create new ones if you
want. There are sets like {3, 1, 5} in which order does not matter, and lists like (a1, a2, …., an)
which are ordered data. These are often interpreted as vectors or n-tuples like (x, y) coordinates. A
list (a, a, b) is different from (a, b) as order matters but set {a, a, b} is the same as {a, b}. Math is
object-oriented so theories describe classes of structures and the instance of those structures models
the theory/axioms.
That's it really: rules of inference, logical deduction, building models and manipulation of data
structures. Much of the above was taken from texts you can find on library genesis about
mathematical logic like Introduction to Metamathematics by Kleene if you want to see how
reduction is used to collect a bunch of theories and reduce them to axioms, and how similar it all is
to computer science with relational models, decidability, complexity theory and determining
semantics of a natural language.
Lectures on logic
Deductive systems - Lecture from 15-251
This lecture gives examples of an initial object, and deduction rules. You really just need to watch
up to 5:30, then skip to 13:35 to see his ATM example of deduction and how he deduces the proof
that the ATM can provide any postitive natural number denomination except for 1 and 3. The rest
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of the lecture is optional where he models deduction using a boolean circuit: given input X, Y or Z
these are passed through gates such as NOT, AND, OR to arrive at an output which is your deduced
formula.
Logic - Lecture from 15-251 CMU
Around 32:00 is the model for 'pretty much all mathematical reasoning' and logical quantifiers like
'for all' or 'there exists' and propositional connectives such as 'implies' or 'iff' are introduced. Most
of this is in Tao's appendix.
You only need to know these connectives:
AND ∧ (conjunction)
OR ∨ (disjunction)
NOT ¬
Implies ⟹ is only false when A is true and B is false
If and only if ⟺
These connectives can of course be used to define each other:
A
B is ¬ A ∨ B
implication is only true when either A is false OR B is true
Iff is (A ⟹ B) ∧ (B ⟹ A)
⟹
Implication seems confusing at first, it's modern language use relating to casuality is different from
the logician's use which is explained here and those Per Martin-Lof lectures above gives a complete
justification of the logical rules of implication.
Some more optional lectures on the history of logic from Aristotle to the present state:
Aristotle and deduction - Brief history of logic MF 251
Origins of modern logic
Stoics and other thinkers - Brief history of logic MF 252
Implication, disjunction and conjunction is discussed here and highly relevant to all
further chapters
Islamic/Medieval logic - Brief history of logic MF 253
Modal logic origins
Leibniz to Boole - Brief history of logic MF 254
From deduction to induction
Notation for relations
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Boole's book Analysis of Logic
De Morgan's laws
Materials needed
The 3rd edition of Analysis I by Terence Tao Buy it cheap from Amazon India/Abebooks, or
use library genesis (see Wikipedia page for their latest domain) or WorldCat to get it.
2.1 The Peano Axioms
Start with Analysis I chapter 2. We begin again with basic counting but rigorously defined.
The five Axioms
Axiom 2.1: 0 is a natural number
Axiom 2.2: If n is a natural number, then n++ is also a natural number
Axiom 2.3: 0 is not the successor of any natural number
Axiom 2.4: Different natural numbers must have different successors
Axiom 2.5: Let P(n) be any property pertaining to a natural number n. Suppose that
P(0) is true, and suppose that whenever P(n) is true, P(n++) is also true. Then P(n) is
true for every natural number n.
We make an assumption about a property P(n) being true meaning you do not
prove P(n) it is already assumed to be true, you prove P(0) and P(n++) are true,
and all three form the logical chain that also proves n+2, n+3, etc.
You can think of n++ being a unary successor operation to generate the natural numbers, while n +
1 will be defined later as a binary operation. The idea is no matter how many natural numbers you
have generated, n++ will generate one more.
Remark 2.1.10 the example where no natural number can be 'a half number', because of the
increment step n++. The base case of 0 (as per Axiom 2.5, and Axiom 2.1) fails if we are using
'half numbers' instead of natural numbers because if zero is a natural number, and any n++ is
another natural number, the only next successive number is 1 it can't be 0.5 or -3 or 3/4. The way
you should view these axioms about the natural numbers is:
1. 0 is an element of N
2. If x is an element of N, then so is x++
3. Nothing else is an element of N
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This is called an inductive definition: clause 1 is an application, clause 2 a succession of
applications, clause 3 is the extremal clause stating only steps 1 and 2 are instances of what N is.
P(n) is not a function it represents "n has the property P". Think about how the natural numbers are
generated with successors, those generated natural numbers all inherit properties of previous
generated numbers such as P(n): "any natural number n + 0 = n". If you prove that for 0, n and the
next successor n+1 that these all have property P(n), any new successor will inherit this property
too since they are derived in a chain from P(0) as per Axiom 2.1. You have already encountered
mathematical induction in elementary school when presented with a summing progression such as
'1+2+3+4…' the inductive step is 'and so on'.
2.2 Addition
The example for addition: 3 + 5 should be the same as (((5++)++)++) and adding 0 to 5 should
result in 5.
Definition 2.2.1 we can add n++ to m by defining (n++) + m to be (n + m)++. You can read this as
(thing)++ and anything in brackets is just a value to be determined, so you can consider it as a
single value to be incremented. Example (1 + 2 + 3)++ + 5 is (n++) + 5.
Try writing your own examples:
0+m=m
1 + m = ((0++) + m) or m++
2 + m = ((1++) + m) or ((m++)++)
3 + m = ((2++) + m) or (((m++)++)++)
4 + m = ((3++) + m) or ((((m++)++)++)++)
2 + 3 = ((3++)++) or (4++) or 5.
3 + 2 = (((2++)++)++) or ((3++)++) or (4++) or 5.
First proof
Let's go through his proof in Lemma 2.2.2. Notice first it is a written essay, and that's how all your
proofs should look. Second, he is using the exact same induction template from 2.1.11. Third he's
using a different variable to not confuse it with the already established definition of 0 + m = m
reminder these variables are just placeholders for value and if you see m twice or n twice it means
the same value.
Induction proofs are a logical chain:
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P(basis First link contains the property and is true, you prove this
P(n) Assume this second link contains the property, you do not prove this because you
assumed it's already true
P(n++) Prove successor is true, with ability to use the previous 2 like they are already
established definitions
How does the base case for n + 0 = n become 0 + 0 = 0, from the previous definition of 0 + m = m?
We are inducting on n so once we substitute in 0 for n we now have 0 + (value) = (value) which is
from the definition of addition: 0 + m = m. Since (value) is 0 it is 0 + 0 = 0.
Suppose that n + 0 = n (assumption). Then the successor (n++) + 0 = n++. Pointing to the definition
of addition where it defines (n++) + m = (n + m)++ we see it is in the same form as (n++) + 0, so if
we say something is equal to something else, we can replace them for each other: (n++) + 0
becomes (n + 0)++ as they are equal so interchangable.
Finally we are left with (n + 0)++ = n++. The value in the brackets (n + 0) is in the form of P(n) the
inductive hypothesis or assumption, since we have assumed n + 0 = n is true, we can again replace
n + 0 for n and arrive at the proof: (n)++ = n++. Remember to consider your assumption 'already
true' and a definition you can just point to while in the inductive proof. Why is induction used?
Because we are building up a structure. He defined natural numbers as being the successor of
another, which means an incremental layer upon layer that you can recursively take apart back to
the base case just like you can in computer programming. Property P(n) exists in base case 0, plus
1, plus another, so this property holds for all natural numbers derived from the Peano axioms.
The most important takeaway: no arithmetic was done here at all. We pointed at definitions, and
reasoned about them using replacement to exchange their forms until they were the same on both
sides of the equal sign. You are a logician now pattern matching forms and moving them around. If
thing = thing2, then thing2 = thing. At this early stage you could literally cut out the definitions and
assemble proofs by swapping out definitions for each other.
How can you check this is correct? Reverse the logic, if n + 0 does not equal n, then 0 + 0 does not
equal 0.
Second proof
Lemma 2.2.3 Here we are proving n + (m++) = (n + m)++, inducting on n. Since we are building a
logical chain of deductions we start at the first link, which is 0 and induct on variable n meaning we
replace n with 0 to get the form: 0 + (m++) = (0 + m)++. Looking at the right side, get out your
scissors and cut apart the definition of addition and replace (0 + m)++ with (m)++. So now we have
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0 + (m++) = m++. Since we know from the definition of addition that 0 + (value) = (value), you
can replace 0 + (m++) with m++ and now the base case is equal on both sides: m++ = m++
Next, we assume n + (m++) = (n + m)++.
We prove the successor: n++ + (m++) = ((n++) + m)++. Looking at the left side, refer to the
definition of addition and notice (n++) + m = (n + m)++ or (successor value) + (something) =
(value + something)successor, the left hand side is now (n + (m++))++. This form is similar to the
hypothesis of n + (m++) = (n + m)++ so replace the left side and now we have ((n + m)++)++. The
right side follows the same logic as the left. ((n++) + m)++ can be seen abstractly as (value)++ or
(successor + value)++, and the second is the definition of addition again so ((value +
value)successor)++
You can always insert concrete examples to see what is going on, ie for n+(m++) = (n+m)++ let
n=2, m=0 then 2+(0+1) = (2+0)+1.
Third proof
Proposition 2.2.4 (Addition is commutative) the base case is point-to definition of addition and
Lemma 2.2.2. We use the definition of addition, and lemma 2.2.2 to rewrite the successor of n to be
(n + m)++ = (m + n)++ and now it is in the form of the inductive hypothesis: (hypothesis
form)successor = (hypothesis form)successor so we can use our assumption that n + m = m + n to
commute m and n around.
Fourth proof & Implication
Proposition 2.2.6 we want to prove b = c. In the inductive step, he uses Axiom 2.4 to drop down
from the successor to the form: a + b = a + c which is the form of the inductive hypothesis, where
we claimed 'a + b = a + c implies b = c' so we can now imply b = c and have 'virtual cancellation
without subtraction'. Tao writes about implication in the appendix which you should read then
return to many times over as it comes up in later chapters.
Exercise 2.2.1
Prove 2.2.5 (Addition is associative) that (a + b) + c = a + (b + c)
We want to show that if a natural number is inside or outside brackets, it will have the same result
when added together. Induct on any variable (a, b or c) you want. Starting our logical chain at the
first link 0:
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Base case:
(a + b) + 0 = a + (b + 0) (substitute 0 for c)
looking at left side: (a + b) + 0
since b + 0 = b (lemma 2.2.2) then (value) + 0 = value
(a + b) + 0 = a + b (lemma 2.2.2)
looking at right side: a + (b + 0)
a + (b + 0) is a + (b) for the same lemma 2.2.2 reasons
now we have a + b = a + b and base case is done.
We selected c for the base case, but what if you choose a or b? Try it while looking at the definition
of addition. The next step in our logical chain is our inductive hypothesis(IH) that addition is
associative, and that (a + b) + c = a + (b + c). The last part of the logical chain is prove the next
link, the successor of our hypothesis, is true. There's a few ways to show this, I'll move stuff around
randomly and you can see where the proof is:
Inducting on c: (a + b) + (c++) = a + (b + (c++))
right side: a + (b + (c++))
a + (b + c)++ (lemma 2.2.3)
((a + b) + c)++ (by the IH)
((a + b)++) + c (def of addition)
((b + a)++) + c (commutative)
((b++)+ a) + c (def of addition)
c + ((b++) + a) (commutative)
c + (a + (b++)) (commutative)
left side: (a + b) + (c++)
(c++) + (a + b) (commutative)
(c + (a + b))++ (def of addition.. consider (a + b) to be (value)): Recall (n++) + m = (n
+ m)++ so this means (c++) + (a + b) is (c + (a + b))++ because (n++) + (value) = (n
+ value)++
(c + (a + b))++ is the same as ((a + b) + c)++ (commutative around the (a + b) bracket)
(a + (b + c))++ (by the IH)
((b + c) + a)++ (commutative)
etc., can keep moving these around
Try moving everything around until you understand exactly how commutativity and associativity
work. See if you can get it to a form where you can use the inductive hypothesis, replace and
permute them around some more until this starts to make sense.
2.2. cont.
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Lemma 2.2.10 has an exercise where induction is suggested. Always read the comments on Tao's
blog if you're stuck but this is the Peano Axiom 2.4 so not sure why it's even here, it's just restating
Axiom 2.4 and 2.3. Assume for contradiction that b++ = a and c++ = a so a positive number is the
successor of two different natural numbers. But by the Peano axioms this is not possible so c++
must equal b++. That is my good enough proof so we can later use this lemma to replace 'a positive
number' for b++.
Definition 2.2.11 n is greater or equal to m if n = m + some other natural number. If that number is
0, then the equality holds, if that number is 0++ then the inequality holds.
Another way to think about order of the natural numbers, is because they are inductively defined
and generated, they already have an order. You can think of m < n if m is generated before n.
Exercises
Try a few yourself, write a convincing argument to yourself like how Tao writes proofs. Click on
details button to see my scratchwork you do not have to do all the exercises.
Details
2.3 Multiplication
Proposition 2.3.4 follow along with definition 2.3.1 of multiplication where (n++) x m = (n x m) +
m also recall that n++ is n + 1: (n + 1)m so nm + m
left side:
a(b + (c++)) is a(b + c) + a because (m(n++)) = m(n) + m from def 2.3.1
a(b + (c++)) is a(value)++ or a(value) + a
right side:
ab + a(c++) is ab + ac + a
ab + ac + a is in the form of the inductive hypothesis so change it for a(b + c) + a
Remark 2.3.8 the contradiction is that ac = bc by corollary 2.3.7 when fed into the trichotomy of
order can only result in a = b. This is a scheme to produce a temporary notion of
division/cancellation.
Exercises
Scratchwork and notes
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Details
Set Theory
History of set theory
There is a lecture on the history of sets/functions, explaining ordinals, cardinals, various schools of
logic, and a second lecture about the history of set theory with Godel and Turing.
Appendix A: the basics of mathematical logic
This chapter on set theory in Analysis I relies on the fact that you've already read the appendix on
logic. Tao makes a point that this chapter is meant to be read and understood not studied and
memorized. Proposition A.2.6, proof by contradiction, we will learn later that when pi/2 (90deg)
the sine function is 1.
Exercise A.5.1
a) For every x++, and every y++, y*y = x or x is the square root of y. Is this true? Square root
of two is the obvious candidate for claiming this is a false statement since what y * y =
sqrt(2).
Lectures on deduction
Deduction - Lecture from 15/251 CMU
Optional lecture, you've already done some deduction from writing proofs, however this lecture
shows what is and what isn't deducible. The end of the lecture has overlap with the appendix and
the first logic lecture in this workshop.
Quiz: Logic
These multiple choice quizzes are from Tao's Analysis class, try the logic quiz. If you struggle
search YouTube for proofs by implication, or search 'discrete math inference rules' there are dozens
of tutorials.
3.1 Fundamentals (Set Theory)
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You may want to watch this lecture by Wildberger he visualizes sets.
Example 3.1.2 is similar to a nested list such as [list: 1, [list: 1, 2], 100] is length 3, even though
there is a nested list overall it still only contains 3 elements. Remark 3.1.3 is in the above historical
lectures on set theory. We are referred to Appendix A.7 Equality to read about the substitution
axiom which you should have already read.
Lemma 3.1.6 if set A is equal to set (empty), then if x is in A it must also be in set empty, a
contradiction, so A must be a non-empty set is how I interpret the lemma, which is a way to say
that if an empty set exists, there must also be a possibility of a non-empty set. Remark 3.1.8
Cardinality if you watched the Wildberger History of Mathematics lecture linked above assigns a
value.
Note the errata for Definition 3.1.4 from Tao's site: "Definition 3.1.4 has to be given the status of an
axiom (the axiom of extensionality) rather than a definition, changing all references to this
definition accordingly. This requires some changes to the text discussing this definition. In the
preceding paragraph, “define the notion of equality” will now be “seek to capture the notion of
equality”, and “formalize this as a definition” should be “formalize this as an axiom”. For the
paragraph after Example 3.1.5, delete the first two sentences, and remove the word “Thus” from the
third sentence. Exercise 3.1.1 is now trivial and can be deleted." So this is now Axiom 3.1.4, but
the definition remains the same.
Exercise 3.1.2 prove emptyset, {emptyset}, {{emptyset}}, and {emptyset{emptyset}} are all not
equal.
If ∅ = {∅ } then by Def 3.1.4 ∅ ∈ ∅ a contradiction to Axiom 3.2
Same reasoning as above for ∅ is not equal to {{∅ }} or {∅ {∅ }}
{∅ } is not equal to {{∅ }} because [list: 1] is not equal to [list: [list: 1]]. The first has element
1 in it, the second has element [list: 1].
Remark 3.1.21 clarifies the above exercise how {2} and 2 are distinct objects. is not in {1,2,3} but
2 is in {1,2,3}. The rest of this chapter is straight forward definitions of set operations.
Example 3.1.32 looks a lot like programming syntax. There seems to be an order of evaluation,
applying the axiom of specification first to create a new set, then applying the axiom of replacement
so working right to left in {n++ : n ∈ {3,5,9); n<6}
Quiz Sets
Try Tao's quiz. There are many Youtube lectures on basic theory.
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Exercises
You can skip these, even in the preface of this book Tao advocates to just read this chapter so I've
only done a few.
Details
3.2 Russell's Paradox
Wildberger has a lecture in his history of math series about this if you watch the lecture on set
theory. 3.9 Axiom of Regularity is worth reading to see some more examples of what is and what
isn't an element of a set otherwise you can skip this chapter.
3.3 Functions
Example 3.3.3 if you forgot range notation which Tao will teach us in later chapters. Remark 3.3.5
clears up some notation for functions notably a sequence subscript also denotes a function.
Example 3.3.9 asks why 3.3.7 asserts that all functions from {empty} set to X are equal, because
they have the same domain (nothing) and range.
Our first proof in this chapter is in Lemma 3.3.12, and if we can go by all the previous chapters, this
will be our template for the exercises. It also doesn't make sense, so I checked the errata page and
sure enough there is errata for this lemma depending what version of the book you're using.
Def 3.3.10 composition, you can think of that composition symbol as being a binary operator
(function) that accepts two functions as input and outputs a single function.
3.3.14 See this tutorial and notice Tao has defined this function by it's contrapositive, which you
can do as it is logically equivalent. Write your own examples for this definition. If two distinct
elements of A map to the same element of B, this is not injective. However if those two A elements
are equal to each other then it is injective:
A = {a, b, a} (function input)
B = {1, 2} (output)
f(a) = 1
f(b) = 2
f(a) = 1
but {a} = {a}, so f(a) = f(a) and this function is injective.
A = {b, c}
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B = {2}
f(c) = 2
f(b) = 2
but {c} doesn't equal {b} so this not injective
3.3.17 Onto functions, another tutorial and for Remark 3.3.24 here is a tutorial for bijective inverse
functions
Quiz Functions
Another multiple choice quiz from Tao's analysis class.
Exercises
All optional, do you get what a bijective/surjective/injective function is? Then you're good we will
do all these definitions again in a calculus course.
Details
Proof strategies
Watch this starting @ 36:00, from 41:00+ he shows how he solves problems step by step, trying
small cases to understand the problem better and poking at the problem enough to finally figure out
a proof.
3.4 Images and inverse images
This is only casual reading, skip the exercises and return to do them when these topics come up
again.
Chapter 3.4 of Tao's Analysis I. Example 3.4.5 makes more sense if you read Example 3.4.6 too, f1({0,1,4}) is the set of everything that could result in {0,1,4} if f(x) = x2, so {-2,-1,0,1,2}. The rest
of this chapter, if it doesn't make sense on first read search youtube ie: indexed sets.
3.5 Cartesian products
Again more casual reading, come back to do the exercises as needed.
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Another topic with plenty of youtube tutorials. Definition 3.5.7 notation for product is introduced, it
looks complicated but all it's doing is specifying an index (i is greater or equal to 1, and less than or
equal to n) and repeated cross product. The exercises some of the solutions are walked through on
youtube. The cartesian plane itself is a cartesian product of R x R, you have tuples (x, y) or (x, y, z)
so R x R x R.
3.6 Cardinality of sets
Skip the exercises like the last two chapters, we just want to be familiar with the material.
Reminder in the preface of this book, Tao himself advocates to skip the exercises as 'set theory is
not essential for analysis' but it will be essential for us to understand numerous other math topics, at
which time we can come back here.
A good tutorial. Tao is using #(A) notation which means finite set cardinality, while |A| cardinality
notation means infinite. There exists errata for the exercises specifically 3.6.8. The last exercise the
pigeonhole principle if set A has greater cardinality than B, then f(x) A -> B can't be injective,
since there would be two elements in A that point to a single element in B.
Integers and rational numbers
Introducing the integers
YouTube - Arithmetic and Geometry Math Foundations 12
Wildberger's definition is similar to Tao's, he walks through a proof of a-(b-c)=(a-b)+c
4.1 The integers
Chapter 4 of Analysis I by Tao. Definition 4.1.1 is exactly the definition from the above Wildberger
foundations video. Lemma 4.1.3 proof you can finish most of it yourself without reading the whole
proof. Lemma 4.1.5 Tao notes the same concerns Wildberger expressed why they both used the 'a
less b' style of definition instead of just using Lemma 4.1.5 as the definition of integers, because of
the many different cases that would be needed. Tao mentions this definition of the integers forms a
commutative ring because xy = yx.
Exercises
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You don't have to do the exercises, though proving equality as per A.7 for every object he creates is
worth trying. This is the poor scratchwork I did
Details
Rational numbers
YouTube - Arithmetic and Geometry Math Foundations 13
This definition is similar to Tao's in Analysis I. The inverse law, write your own example, the
rational number 4/1 inverse is 1/4 and 4/1 x 1/4 is 4/4 or 1. A rational number is (a-b)/(c-d)
according to it's definition but we write them as p/q and in some cases just p if q = 1.
Visualizing the rational numbers
YouTube - Arithmetic and Geometry Math Foundations 14
The rational number line sits on [0, 1], so you don't end up with a zero denominator which isn't
defined. A rational number is a line through the origin [0,0], and an integer lying on point [m,n].
Ford circles are an example of the massive complexity of the rational numbers, which are expanded
upon in the next lecture.
The deep structure of the rational numbers
YouTube - Real numbers and limits Math Foundations 95
This is about 35m long but worth the watch if you want to see what a convex combination is, learn
about rational intervals, and a more complete description of Ford Circles set up with Farey
sequences. This will help us understand what a sequence is later in Tao's book. The overall
takeaway of this lecture is that the smaller the interval you take on the rational line, the more
complex the numbers become with enormous denominators meaning it has a very deep structure.
The next videos, #96 and #97 about Stern-Brocot trees and wedges are highly recommended and
can also be found in the book Concrete Mathematics. In Lecture #97, he takes a visualization on the
plane of the Stern-Brocot tree and shows it also can be used as a geometric interpretation for the
rational numbers, it's kind of crazy when you see it. The 'fraction' 1/0 is also explained.
4.2 The rationals
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Tao's Analysis I chapter 4.2, definition 4.2.1: 3//4 = 6//8 = -3//-4 we know from Wildberger's
visualization of the rational numbers, these numbers all lay on the same line. Lemma 4.2.3, in the
proof of addition getting rid of +bb'cd is a simple case of subtracting it from both sides, cancelling
it out leaving ab'd2 = a'bd2 and he can divide both sides by d because d is known to be non-zero,
since it's in the denominator of a/b + c/d.
To recap we defined the natural number object, embedded them in the integers (a - b) to define
integers, then embedded integers inside the rational numbers (another constructed object) where the
arithmetic of the integers is consistent with the arithmetic of the rationals. The rationals are a field
because of the last identity in 4.2.4 the inverse algebra law. Tao's example proof is similar to the
proofs we've seen Wildberger do, convert the letters into integer // integer form, apply law,
commute the results together so that they are the same.
Exercises
These are pretty easy, try them. 4.2.6 in my copy must be a mistake 'show that if x, y, z are real
numbers…' that must be 'rational numbers' and sure enough checking Tao's errata page it is a
mistake. There's a mistake in exercise 4.2.1 as well depending which version you're using.
Details
Decimal numbers
The next chapter in Tao's Analysis I uses decimals to represent tiny positive quantities (epsilon) so
we should probably know what decimals are.
Wildberger has a basic introduction lecture using natural numbers that assumes you know nothing
about exponents and reviews the hindu-arabic notation, shows how to convert a decimal to a
fraction. The second lecture introduces rational number decimals and their visualization.
His definition is a decimal number is a rational number a/b, where b is a power of 10 (hence
deca/decimal), then shows various examples how you can manipulate rational numbers like 1/2 to
be 5/10, or 0.50, we also find out why 1/3 is not a decimal number. In binary, which is a base-2
system, the only prime factor is 2. If you load up any language interpreter and try and add 0.1 + 0.2
you will get surprising results, typically this number: https://0.30000000000000004.com/ because
of the way floating point is used to represent decimals.
At 36:00 arithmetic w/decimals is covered, showing how the rational number and decimal number
arithmetic methods are the same so you can just stay w/rational numbers if you want converting
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back to decimals at anytime.
Appendix B: the decimal system
From Tao's book Appendix B.1 The decimal representation of natural numbers his opinion is that
the decimal system is not essential to mathematics, computers have replaced it with binary/floating
point and 'we rarely use any numbers other than one-digit fractions in modern mathematical work'.
Of course to understand scientific notation we still need to know how decimals work and we just
did with those Wilberger lectures so this chapter is optional (we can't do it anyway before learning
the the real numbers). The single exercise in B.1 is interesting, 'the purpose is to demonstrate that
the procedure of long addition taught to you in elementary school is actually valid'.
4.3 Absolute value and exponentiation
Def 4.3.1 you apply that def to the rational number, so if x = -3, the absolute value is -x so -(-3)
which of course makes it a positive rational number, as seen in the example of 4.3.2 where the
absolute value of 3 - 5 = 2.
Def 4.3.4 that symbol is epsilon, meant to be a tiny quantity he defines as greater than 0. The
example 4.3.6 if two numbers are equal, and their difference is zero, then they are 'epsilon close' for
every epsilon since their difference is not bigger than any positive epsilon.
Proposition 4.3.7 recall that all these definitions are for d(x, y) or d(z, w) so x - y or z - w. His proof
he moves xz to one side, so it becomes negated (subtract from both sides). He then manipulates |az
+ bx + ab| according to previous definitions to get |a||z| etc., and then uses the various propositions
in 4.3.7 to rewrite them as eps|z| + delta|x| + eps(delta). This will make sense after we attempt to
prove the rest of the propositions.
Rest of this chapter covers exponents and their properties.
Exercises
n
Exercise 4.3.5 Prove that 2
≥ n
for all positive integers n using induction.
base case: n = 0
20 = 1 and since n = 0, 1 > 0.
but Tao specifically wrote 'positive integers' so base case must be 1
1
2 = 2 and since n = 1, 2 > 1
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n
assumption: 2 ≥ n for all positive integers n
successor: 2
≥ n + 1
left side:
2
is 2n*2 by def 4.3.9
2*2n is also 2n + 2n by prop 4.3.10 (a)
now we have 2n + 2n ≥ n + 1
since 2n ≥ n (pointing to our assumption), then 2n + 2n ≥ n + n
n is only positive, so n + n is ≥ n + 1
2n + 2n ≥ n + n ≥ n + 1
here we have standard transitivity proving 2
≥ n + 1
n+1
n+1
n+1
This of course falls apart if Tao changed it to 'for all integers' instead of positive integers, or if we
allowed our assumption 2n > n to have n = 0, then we could not assume n + n is ≥ n + 1. Search
around for proving inequalities on YouTube there are dozens of demonstrations of various
strategies.
4.4 Gaps in the rational numbers
The idea here is that certain things don't exist in the rational numbers object like the square root of
2, so there are 'gaps' where the square root of 2 should be.
Proposition 4.4.1 there is an integer n, and a rational number x, such that n <= x < n + 1 so
sandwiched between integers 2 and 3 are rational numbers such as 25/10 or 100/40 etc.
Proposition 4.4.3 there is always some rational you can squeeze in between two other rationals as
we've already seen in Wildberger lectures. 1/800 < 1/500 < 1/100 etc. If you want put in examples
for his generalized proof to better understand it like x = 2, y = 3 and z = (2 + 3)/2.
Proposition 4.4.4 famous proof by contradiction there does not exist a rational number x for which
x2 = 2 though Tao's is significantly longer than most and as per the preface this is for the reader's
benefit. He rewrites (p/q)2 = 2 into p2 = 2q2 by multiplying each side by the denominator, which is
q2 as p/q squared is p/q * p/q or p*p/q*q.
Proposition 4.4.5 is basically saying for any epsilon bigger than 0, there exists a rational number
that when squared it's less than two, and less than x + epsilon. Remember there is an infinite
amount of rational numbers between 0 and 2 like trillion/1000 billion = 1. 12 is 1, if epsilon is 1 so
that x + epsilon = 2, then (x + eps)2 is (1 + 1)(1 + 1) and bigger than 2. He asks why ne (e =
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epsilon) is non-negative for every natural number n in (ne)2 and the parens grouping ne and
squaring it obviously make it positive ie (-2)2.
Tao mentions Dedekind cuts will be avoided, and the theory of infinite decimals has problems so
also will be avoided. Wildberger has numerous lectures on both theories if you're interested the
most detailed ones are in his Famous Math Problems playlist. Wildberger makes the point that you
will only ever find the cherry picked example of the square root of 2 for Dedekind cuts in any
undergrad material because of the mind boggling complexity involved trying to partition a set for
pi2/6 and other irrational numbers.
The infinite descent exercise Tao has basically handed the solution to us, if you fix any sequence of
natural numbers then by their definition there are finitely many numbers from 0 to your fixed point,
so you can't choose infinitely many distinct natural numbers less than your fixed point so infinite
descent does not exist for the naturals because you'll run out of them. Integers you can infinitely
descend to negative infinity and of course rationals we know you can infinitely descend with bigger
denominators until you approach zero or descend to negative infinity.
Real Numbers
Real numbers are basically a convenient abstraction for doing pure math analysis that every popular
math text uses however our computers all use floating point arithmetic which is basically a stream
of data we grab from to take finer approximations. You could view the following chapters as
Cauchy sequences are just proving the real numbers exists instead of actually proving the real
number system.
What exactly is a limit
YouTube - Real numbers and limits Math Foundations 106
Watch the first 20 minutes of this lecture and you'll see what the notion of an epsilon displacement
is. Note that Tao will define all these things for us later, but at least you'll have an idea of what
epsilon of a sequence is. The rest of the lecture is excellent covering a different definition of limits
but optional. 3Blue1Brown has an epsilon-delta definition of a limit here.
2D geometric interpretation of Cauchy sequences
YouTube - Real numbers and limits Math Foundations 94
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Watch from 2:20 to 16:00, he shows the 2D geometric interpretation of Cauchy sequences, and how
they are interpreted as 1D on the real line, the idea of 'converging sequences'.
Real numbers and Cauchy sequences I
YouTube - Real numbers and limits Math Foundations 111
He explains the triangle inequality around 8mins, we just read about this in chapter 4.3 of Analysis
I (and is one of the exercises).
Real numbers and Cauchy sequences II
YouTube - Real numbers and limits Math Foundations 112
An excellent lecture why we even need these complex limit and real definitions in the first place,
showing how limits make many subjects in analysis much easier. Archimedes bracketed
measurement is covered when he tried to figure out the approximation of pi. We also get an
introduction to transcendental functions as a geometric series, something we will see in Analysis II
by Tao in detail. The next lecture in the series is optional but Wildberger presents an alternative
definition of real numbers, using Archimedean style of nested intervals where equality, algebra
laws, order etc are all trivial to prove.
5.1 Cauchy sequences
Wildberger has good lecture introducing what a sequence is here and more precisely here with
index notation.
Def 5.1.1 the notation outside the bracket is n = starting index, and the infinity symbol is how far
the index should go which is forever because we're now doing continuous mathematics.
Example 5.1.2 reminds us of Def 4.3.4 d(x,y) = |x - y| ≤ ϵ so if x and y are 2, then 2 - 2 = 0 and
every single epsilon is valid since x - y will always be less than epsilon.
The definition 5.1.3 of epsilon-closeness means two adjacent indexes like 1 and 2 in our above
array when subtracted are less than or equal to the epsilon we declare as being epsilon-close. 1.4 1.41 (absolute value). The next set of adjacent sequence indexes is 2 and 3, so 1.41 - 1.414.
Everytime we subtract two of the indexes their displacement or 'absolute value distance' should be
closer and closer meaning a smaller epsilon.
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Let's do an example:
a4 = 1.41 1.412 1.4143 1.41424
index 1 - index 2 = 0.1 (absolute value)
index 2 - index 3 = .004
index 3 - index 4 = .0002
This sequence is 0.1-close, after index 2 it's 0.004-close, after index 4 it's 0.0002-close.
Def 5.1.6 if you watched the Wildberger lectures on What Exactly is a Limit then you can see what
Tao is setting up here with epsilon-steadiness, there is some index in the sequence where the entries
past that index are epsilon-steady. Example 5.1.7 the sequence 10,0,0,0.. d(10,0) is 10, so this
sequence is not eps-steady for any epsilon less than 10, but if we choose N past 10, then all the
remaining sequence is 0,0,0.. and d(0,0) is 0 - 0 so as per Def 4.3.4 will always be less than any
epsilon so 'eventually eps-steady for every eps > 0'.
Def 5.1.8 if you saw the above Wildberger lectures on Cauchy sequences this is defining a
convergence so any sequence that eventually becomes epsilon-steady, is a Cauchy sequence.
Example 5.1.10 depending on your edition there's errata, in mine (3rd edition Springer from libgen)
the example makes no sense.
Prop 5.1.11 proof he cites the fact that there always exists some rational number you can squeeze in
between numbers and chooses 1/N noting that no beginner would come up with this proof and it is
merely staging in preparation of learning what a limit is. Later in Example 5.1.13 he notes Prop
4.4.1 can again be used to prove the infinite sequence 1, -2, 3, -4.. has no bound, I would guess
because of the 4.4.1 statement 'there exists a natural number N and rational x such that N > x ie: no
such thing as a rational that bounds all the natural numbers so an infinitely increasing and
decreasing sequence can't be bounded.
Exercise 5.1.1 since we declared it a Cauchy sequence, you can point to the definition of Cauchy
sequences being an object that eventually becomes epsilon-steady, so Lemma 5.1.14 proves the
finite part of Cauchy sequences up to epsilon-steady are bounded, and obviously epsilon-steady
means it's bounded since you can subtract two indices and they will always be less than epsilon if
this is a Cauchy sequence and epsilon-steady, or 1-steady, 2-steady, 1billion-steady. So both are
bounded are ergo all Cauchy sequences are bounded using his definition of episilon-steadiness.
There is a lecture here if you're interested more about the proofs used in this chapter.
5.2 Equivalent Cauchy sequences
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Prop 5.2.8 is the proof that 0.9999… = 1, and again prop 4.4.1 is brought out to claim we can
always choose any rational number we want to make 2/N <= epsilon. If you follow this proof:
(1 + 10-n ) - (1 - 10-n )
10-n + 10-n = 2 x 10-n
10-N is ≤ 1/N because of the chapter on exponents, if N = 3 then 10-3 is 1/3x3x3 which is
smaller than 1/3
Now choose an N that satisfies the requirement that 2/N ≤ ϵ and it will automatically be true
that it also satisfies 2 x 10-N ≤ ϵ
Prop 4.4.1: you can always choose some rational to shove in there that will satsify the
req
Only two exercises here, simple point to definition pretend you are a lawyer writing an argument
and convince an invisible jury through explanation why it's true an equivalent sequence of rationals
is a Cauchy sequence if and only if one of them is a Cauchy sequence.
5.3 The construction of the real numbers
Tao has one of the few books anywhere that actually defines the real numbers. Just like
understanding the constructions of rationals and integers helped us understand their operations
better, understanding the construction of the reals should help us better understand things like xe or
square roots.
Def 5.3.1 the formal limit of a Cauchy sequence, meaning the number a sequence is converging
towards, is a real number. Every rational number is also a real number, for example if we choose 5,
then a Cauchy sequence converging to 5 like 4.99999… is bounded by 5 and is a real number. The
notation is: x is a real number if x = the limit of a Cauchy sequence an, as the sequence goes from
index n to infinity so a1, a2, a3 …
The rest of the chapter covers the same definitions we've already seen except now we are working
with a different object, but since it's construction is entirely made up of rational numbers in a
sequence almost all the laws are the same. Division by zero not being allowed for real numbers is
explained using the sequence 0.1, 0.01, 0.001.. is actually converging towards 0, and it's inverse
1/0.1, 1/0.01, 1/0.001 is 10, 100, 1000… an infinite unbounded sequence therefore not a Cauchy
sequence, therefore not a real number, so division by zero is undefined.
The #1 takeaway for this chapter is the real numbers are all a limit. I am just casually reading the
proofs in this chapter, Tao has simplified it for us with his epsilon-close 4.3.7 propositions and his
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definition of equivalent Cauchy sequences. The proof of Lemma 5.3.6 was already explained by
Wildberger here if you're wondering what epsilon/2 means.
Looking at the exercises, I'm guessing to prove 5.3.5 we show the reciprocal of 1/n is unbounded so
we must be working with a sequence not bounded away from zero. In prop 5.1.11 Tao proved the
1/n sequence is a Cauchy sequence, and since all Cauchy sequences are bounded you can also show
it's bound is 0. There is worked out solutions for this chapter here but you will never need to know
how to prove any of these definitions.
5.4 Ordering the reals
For me it was interesting to go back to prior definitions and see how they work when cast into this
Cauchy series abstraction. Prop 5.4.8 if 5 > 4 then 1/5 < 1/4.
Prop 5.4.9 'non-negative reals' means 0 or positive, while positive reals do not include 0. Closed I
take to mean a set is closed iff the operation on any two elements of that set produces another
element of the same set, in other words adding two non-negative real numbers produces another
non-negative real. The explanation of the proof is in the comments here but Tao tells us we'll see
this more clearly in future chapters on closures.
Prop 5.4.14 we already know from previous proofs and Wildberger lectures that there is an infinite
amount of rational numbers you can squeeze in between two other numbers.
5.5 The least upper bound property
Example 5.5.7 the empty set doesn't have a least upper bound because of prior definitions where
you can always find a rational number (which are also contained in the real numbers) to act as an
upper bound to nothing like 1/trillion, 1/septillion, etc.
The proof for Theorem 5.5.9 is huge requiring going back to look at the Archimedean property
again and some exercises. Wildberger has broken down this proof showing the gist of what is going
on but the last part of the proof that the upper bound S is less than or equal to every upper bound of
E is easy to follow.
Remark 5.5.11 we get the answer to 'can you add negative infinity with positive infinity and get
zero' which is no.
The proof of prop 5.5.12 makes sense until you get to that inequality, wondering where he got x2 +
5e, which is the Archemedian property. This is basically saying x is the least upper bound, x plus
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some tiny displacement (epsilon) squared is less than or equal to the Archimedian property of x2 +
5(displacement) since if epsilon is really small, squaring it means it will be even smaller. He
derives a double contradiction that x2 is bigger than 2, and x2 is less than 2, so x2 must equal 2 and
we have a number that solves x2 = 2 which is not rational.
Exercise 5.5.1 you can type into youtube 'proof of infinum' and get a dozen tutorials or you can go
through Prop 5.5.8 and Theorem 5.5.9 to change everything to prove a least lower bound which
gives some more experience with the Archmedian property which all these proofs seem to rely on.
The rest of the exercises 'draw a picture' think of Wildberger with his within epsilon/2 picture in
earlier lectures.
5.6 Real number exponents I
Finally we learn square roots. Depending on your copy of Tao's book there's some minor errata here
to look up. Note in the proof of Lemma 5.6.8 he's using equivalency replacement, if y = x
then
we can replace them both for any occurences, such as (y ) = (x ) . Anytime in the future when
you see some complicated nested square roots you can go back to this chapter and understand them.
a
a
a
′
1/b
′
a
1/b
′
′
Exercises 5.6.2 and 5.6.3 are simple and worth doing, using replacement to manipulate exponent
laws. Exercise 5.6.3 that's the square root of x2, so x or using Lemma 5.6.6(g) or Lemma 5.6.9(b):
(x
)
is x
which is x1 or x.
2/1
1/2
2/1⋅1/2
Series
7.1 Finite series
Wildberger has a lecture on summation of arithmetic and geometric series to go with this chapter.
3
We are just manipulating indices here. Make an example series for yourself: ai = 11, 32, 53. ∑
means a2 + a3 or 32 + 53 as per Definition 7.1.1.
i=2
ai
In Lemma 7.1.4 the triangle inequality proof is here note how they get the base case of 0 by setting
the variable on top of the sigma less than the initial starting variable so all indexed summations are
zero (let b < a) like Tao's beginning of chapter examples with m - 2, m - 1.
Check the errata for this chapter, in my version (third edition (hardback)) there's a few problems.
Remark 7.1.9 the proof is straight forward when you flip back to Lemma 7.1.4 and note that
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because he already assumed x to be g(n + 1) he could not also assign x to h(n + 1), which is why he
needed to introduce h(j).
7.1.13 why is P(0) true? Because the assertion P(n) is true for any set X with n elements, so if it has
P(0) elements, it's sum is 0. Reminder to also re-read what a cartesian product is, the cartesian plane
itself is a cartesian product of every value of x with every value of y.
Proving all these finite sum identities are easy once you get one of them, all you need is 7.1.1 and
the recursive definition he uses to pull out values, manipulate and reindex the sum.
Try 7.1.4(d) without induction, expand the series, factor out the c, done:
∑
(ca ) = ca1 + ca2 + … + can
c(a1 + a2 … + an) by distributive law
rewrite as c(summation notation)
7.1.4(c) expand ai + bi, group them together:
(a1 + b1 +.. an + bn) = (a1 +.. an) + (b1 +.. bn)
rewrite in summation notation
7.1.4(a) let m = 1:
(a1 +.. an) + (an+1 +.. ap) = (a1 +.. ap)
n
i
i=1
7.3 Sums of non-negative numbers
7.3.7, do you get how he rewrote that sum to the geometric series?:
k
1
2
k
(2 )
k
q
−kq
2 (2
by Lemma 5.6.9(d)
k
k
k+k
) by exp laws x * x = x
) which is 2
by Lemma 5.6.9(b) (xq)r = xqr
the geometric series is the sum xn
k
n
(2
) is (thing) or x
)
k+(−kq)
(2
1−q
(2
k−qk
k
1−q
k
Just flip through the remainder of this chapter noting things you can always come back to if needed
to understand better, such as rearranging infinite sums.
Trigonometry crash course
Let's take a very short crash course to remember trig:
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Trigonometry Bootcamp - Brown University
Contains a review video and a solutions video
(Optional) Trig fundamentals by 3Blue1Brown that won't make sense unless you do
the Brown bootcamp
In case you forgot what π (pi) is (for us just an applied constant of 3.14… or 22/7 or 355/113)
Wildberger talks about the history of pi such as Archimedes interval for pi, Newton's formula, a
hexadecimal version of pi, his own formula.
Let's start with the crash course lecture.
If you forgot highschool this video @ time index 6:00 shows how sine(x) is the y-axis, and
cosine(x) is the x-axis, the whole thing is worth watching
Radians you could even make up your own notation and define it as DoublePI = 6.283… and
then 360deg is DouplePI, 90deg is 1/4DoublePI, 180deg is DoublePI/2, etc.
We really only need to know how sine/cosine functions work, all others are derived from
them
1
In the scaling of triangles example where
1
√2
⋅
√2
√2
is
√2
√ 2⋅√ 2
which is
√2
√4
or
√2
becomes
√2
2
they are rationalizing the denominator:
√2
2
Summary: There exists 3 triangles, which we should know their measurements, we scale them
down since this is a unit circle with maximum x and y values of 1. Then we superimpose the
triangles in the pi/2 (90deg) angle space on the unit circle in such a way that we have known values
of cos()/sin() and other trig functions, because we can derive those other functions from cosine and
sine.
Try the worksheet you can actually draw these out and count rotations if you want or use this radian
to angle picture from Wikipedia. The equations if one side is squared, then you need to take the
square root of both sides so it's (not-squared = value) instead of (squared thing = value-whensquared). If you're stuck watch the video that walks through all the solutions. How I solved these:
subtract 1 pi if it's bigger than 2pi, or a 1/2pi if smaller and see what you get, look at the 3 triangles
if they have a corresponding value for sin/cos. If not subtract pi/3, pi/6 etc.
If you hate this trigonomentry this guy reinvented it here with rational trigonometry and has a book
on library genesis where angles are removed and replaced with vectors, he even respecified
hyperbolic geometry to be completely rational it's crazy when you see it. If you want/need to
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practice Trigonometry, we will do so as this workshop progresses or search library genesis for Titu
Andreescu's 103 Trigonometry Problem a book for training the USA IMO team.
Math circle
Now we completely change from linear progression through textbooks to a solving olympiad style
problems. This is similar to a math circle essentially an informal group of all ages from professors
to student TAs to middle school students get together to group solve a problem and repeat.
Let's start with Poh-Shen Loh's combinatorics problem solving seminars and a Russian olympiadstyle book on basic algebra skills because you don't need a background more than what we've
already done for either.
Materials
Mathematics via problems: Part 1: Algebra by Arkadiy Skopenkov.
A masterclass in basic algebra everybody should know
"This book attempts to build a bridge (by showing there is no gap) between ordinary
high school exercises and the more sophisticated, intricate, and abstract concepts in
mathematics"
The author is a professor at MIPT and uses this book for an olympiad circle
21-228 Discrete Mathematics course (Sp 2021)
Lectures on his YouTube channel
Prof Loh is the current national coach of the US olympiad team who has won gold 4x
(and even wins the math competitions for former medalists) so surely his content will
be worth viewing.
Discrete math lecture 1
The first lecture begins ~14:30 mins. We start here so you can see how to investigate problems in
math. He's breaking down some elementary combinatorics problems to show the logic of solving
them. @37:40 why does the last digit have to be even? Because it's supposed to be an even 4-digit
number so ending in 0, 2, 4, 5, 8. The general form is crazy but he did it by writing down a table of
one digit numbers, two digits .. n + 1 digits to find the general pattern.
Assignment 1
Let's try the assignment. This discrete math course is unusual in that there is no hundreds of drill
problems to do in the book, just the challenges he hands out. I will attempt to work through them
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like he does, writing down little examples and ideas and seeing what happens:
1.) The first is a symmetric difference assignment which is defined here. Look under properties, the
answer is there 'from the property of the inverses in a Boolean group, it follows…'. We can
eliminate B and are left with:
(A △ C) △ (C △ A) and from that same article, we can eliminate C so are left with A △ A which
is just the empty set.
2.) We have a set bigger or equal to 3, and we need to know how many subsets contain exactly 2
permutations of 1, 2, 3 so {1, 2, 3, 5} or {1, 2, 3} does not count as it contains 3 permutations of 1,
2, 3 and the assignment wants only 2. This seems like a powerset problem, generating all possible
subsets. Google any free powerset online calculator and try the base case, set {1, 2, 3} it contains 3
sets with exactly 2 numbers from 1, 2, 3. Try the next bigger set {1, 2, 3, 4}, I count 6 occurences.
The set {1, 2, 3, 4, 5} has 12 occurences. I'm going to guess {1, 2, 3, 4, 5, 6} is 24 occurences, and
that {1, 2, 3, 4, 5, 6, 7} will have 48 occurences. The pattern is 3x2x2x2.. now how to figure out
what the n set is? If n = 4, it's 3x2, if n = 5, it's 3x2x2, if n = 6, it's 3x2x2x2. Writing out more
examples I see it's: Length 4 choose 1 of the 2's, length 5 choose 2, length 6 choose 3, 7 choose 4,
the choose n is the amount of 2's in 3x2x2x2.. and they differ by 3 each time. The solution seems to
be any set with length n, subtract 3 and that's how many 2s to multiply to get the answer. A set {1,
2, 3…. n} where n = 10, it's 3 times seven 2's or 3x2x2x2x2x2x2x2 or 384 subsets.
3.) This is exampled by Wildberger here where he shows how to count the number of paths on a
grid, it's Pascal's array or n choose k as detailed in his Famous Math Problems 5 lecture. There is
also a writeup here how to avoid paths on a grid and is the answer to this problem. First we can
visualize this with free online graphing software, either plotting y = 59 streets and x = 8 avenues or
reduce the scope, since we're only interested in the (59 - 34) and (8 - 5) avenues. Zoom to see all
the streets. If King Kong goes (59 - 34) streets and (8 - 5) avenues he has 25 streets + 3 avenues or
28 total moves in any direct path to the Empire state building then we have to figure out how to
subtract the paths that involve 42nd and 7th ave.
Algebra chapter 1
Reading Mathematics via problems book by Skopenkov. The contents of this book are similar to
undergrad courses like Math 135: Algebra for Honours Mathematics and if you search for the
course description they removed the sage on the stage approach of lecturing and students solely do
problems together in class, like a math circle. In the preface of Skopenkov's book we are informed
that you don't read this book linearly, you take sections that interest you solving the level 1 to 2
problems, then going back later and trying the level 3 and 4 problems as you will be 'at a new
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level'. Prof K in the software workshop recommends the same method that you iterate over material
you've already done multiple times each time returning to it at a higher level so you can see how
concepts there transfer over to concepts you're currently learning. The preface also says we are
doing this to learn both how a mathematician reasons their way through problems, and because 'the
lack of skill (manipulating algebraic expressions) amoung olypians often leads to ridiculous and
annoying mistakes'.
1.1.1(a) What are the rules of divisibility by 2, 4, 5, 10, 3, 9 and 11. This is an example of a great
problem you could figure out yourself during spare time using just a phone calculator. Investigate
first seeing if you can discover the rule, then look at author's proofs.
Page 2 are the proofs, he's using generalized hindu-arabic notation we already learned. First proof,
subtract from n the last digit, ie: 112 - 2 = 110. What does he mean by 'divides each term of the
sum' he means in hindu-arabic notation form, so 112 = 100 + 10 + 2 or 336 = 300 + 30 + 6 and 2
divides each term and the sum. The proof of divisibility by 4, any number n if you subtract the 10's
and the last digit, it will be divisible by 4 like 177 is 177 - 70 - 7 or 100. The 4|n notation is in the
preface, it means 4 divides n.
Exercise, prove divisibility of 5: The number n - a0 is divisible by 5. Suppose a0 is 5, then
rewrite his other example proofs.
Exercise, prove divisibility of 10: The number 10a1 is divisible by 10, could use the fact that
anything divides zero: 'suppose a0 is 0, or by definition a0 is < 10 so must be zero, then if a
number divides every term of the sum, it divides the sum'.
The rest of the proofs put in concrete values if you don't understand them like the proof of
divisibility by 3: 234 is (200 - 2) + (30 - 3) + (4 - 4) or 225 which is divisible by 3. 1224 is (1000 1) + (200 - 2) + (20 - 2) + (4 - 4) or 1215 which is also divisible by 3, and by 9. He inserts a
function f(n) to represent the algorithm of summing even/odd indexes of numbers divisible by 11.
1.1.1(b) Trying some examples, 111,111|(106m - 1). It will also divide repeating number of 1's that
are 106m digits long, so it's digit sum needs to be divisible by 6 ie 1012, 1018 etc. A huge number
consisting of all 1's who's digits sum to 1993 (a prime number, so sum of digits not divisible by
anything except itself) is not divisible by 111,111.
1.1.1(c) prove that a number consisting of 2001 1's is divisible by 37. We are only asked to prove
for 11111… not a general number or rule of division by 37. Trying examples, any repeating number
(111, 222, 333) who's digit sum is divisible by 3 is divisible by 37. Since 3|2001, it should be
divisible by 37. Further investigation reveals that 37|1110 and 37|11,100 and 37|111,000 etc. 3|2001
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is 667 so there are 667 'groups of 3' of the digits 111. If n = 111,111 then 37|n - (10a2 + 10a1 + a0)
and 37|(10a2 + 10a1 + a0).
Finish the rest of chapter 1, so 1.1.3, 1.1.4 and 1.1.5. We will return again later to the same
concepts over and over so you don't have to memorize anything.
1.1.3(a): 2|(a2 - a) input small examples, like 32 - 3 is 6, whatever you substitute for a ends up
even and is divisible by 2. A lot of Tao's rules are here, like not being able to cancel each side
if the cancelled number is 0.
1.1.3(f) if a = kb then ac = k(bc) which is still in the definition of division form: (thing) =
k(thing).
1.1.4 if a is divisible by 2 and 3, then a is obviously even, and since it is divisible by 3 as
well, 3 x (even) will be divisible by 6. He uses an interesting proof for these but if 17|a and
19|a then 17x19 = 323 and 323|a because 17|323 and 19|323 is how I read it.
A good first chapter covering division which Wildberger in his beginning math foundations lecture
claims is the one thing most students don't understand. Poh-Shen Loh also claims students usually
don't understand ratios/fractions/division, that's why we are doing this.
Discrete Math Lecture 2
CMU Discrete Mathematics lecture 2/3. This class covers some linear algebra, which I do in the
first assignment of 10-301 in the AI workshop and also here which also covers Eigenvalues/vectors.
You can also watch Wildberger's lectures on linear algebra to see how a 2x2 matrix is multiplied.
Around 43m if you're curious how to invert a matrix you can either type it into wolfram alpha and
see instant results or watch this other Wildberger lecture you take the determinant, factor it out and
multiply it by the 'adjoint' which is a matrix where every entry is it's determinant. Since this is all
1's we simply invert the minus/plus signs. This was a crazy lecture.
Last 10 minutes are about paths, which is the first homework we already did. Around 53mins notice
LRRLLRRL no matter how you permute those instructions, you get to the destination. If you did
LLLLRRRR you walk the edges of the square. If you did RRRLLLLR you end up in destination
again. End of lecture is the homework, total amount of paths minus top path multiplied by bottom
path. We figured this out ourselves. We will learn more linear algebra in the Putnam seminar, which
is next up.
CMU Putnam Seminar
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Another Prof Loh seminar we can do at the same time as the algebra and combinatorics problems,
this one follows the book Putnam and Beyond second edition you can find on library genesis what
has completely worked out solutions for all problems. Lectures are archived on YouTube and the
class page is here.
According to Stanford "The competition emphasizes ingenuity rather than knowledge, so freshmen
are not at much of a disadvantage compared to seniors".
Lecture 1 Introduction
Let's watch the Putnam Seminar 9/4 introduction.
At 13:31: the top 500 scores are only ~23 points out of 120 max possible points, the exam is that
hard, meaning you solved the 2 easiest problems and maybe had insight into a third but didn't solve
it. He explains having just 2 insights to a problem requires a mental search tree of 100 x 100
options repeatedly claiming this not an exaggeration, the Putnam is actually this hard since no 3rd
party materials are allowed during the exam though in 2021 it's not proctored and entirely online,
so all results will be unofficial.
@30:00 the first seminar starts, the pdf he pulls up is here. The very first problem, the 8x8 block
pattern he already has a talk about this that gives away the answer, you need 6 bits of memory.
Third problem @31:45 'integral coefficients' means they are integers. If you forgot what a
polynomial is Wildberger has a brief crash course describing their properties and graphing in this
linear algebra lecture. A root means an input to the polynomial so it evaluates to 0. We will of
course learn polynomials doing problems about polynomials as we go.
Now we go through ideas, poking at the problem to solve it like p(1) to sum all the coefficients, a
strategy to show all the integers are neither even nor odd so can't be rational. Notice how he's just
abstracting away using boxes ignoring all the values. If you forgot what modular arithmetic is
Wildberger also has a lecture for this in typical Wildberger style where he introduces the problem
it's trying to solve, the history and then at around 11:43 the rules of k congruent to l mod m. A
conceptual crash course is: consider an abstract system of the type (D, 0, successor) where D has
two distinct objects 0 and 1, and let 0++ = 1 and 1++ = 0. This is called residues modulo 2. The
natural numbers become this mod2 system when each natural number is replaced by it's remainder
after division by 2:
0/2 is remainder 0
1/2 is remainder 1
2/2 is remainder 0
3/2 is remainder 1
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4/2 is remainder 0
He cleared the fractions by multiplying the polynomial by cn which you can see here (rational root
theorem), which makes sense if (b/c)n then it's (bn/cn) ie (3/4)2 then 3/4 x 3/4 or 9/16. This
introduction is worth rewatching a few times to learn how this style of problem solving works.
His other site Expii Solve has similar kinds of challenges they start easy and get harder with links at
the bottom of each question where to find background material, you may want to try those too they
help you learn all kinds of topics such as spherical geometry, probability and even calculus. It's a
community written site so kind of like math overflow where anybody can write a tutorial for a
subject and these are upvoted if other people find them useful, which could be good or bad, a well
written Tao-like fundamentals post would likely not be upvoted compared to some bag of tricks
with a clickbait youtube embed.
Lecture 2 Polynomials
The only prereq for this lecture is Wildberger's crash course in calculus here. Note how one
function tells you how the other is changing and you can switch from velocity to acceleration to
position. You now know basic calculus.
Watching the second lecture on Polynomials. If you don't know what the constant e is, watch this.
The problem we are doing, try expanding the left hand side n = 3: $a1x1 + a2x2 + a3x3 is equal to 0
(has a real root, meaning not a complex number root) if the sum or the constants divided by the
exponents + 1, equals 0. Recall from Wildberger's calc crash course that a polynomial constant like
1 in: 3x2 + 1, the 1 is really 1x0 so our problem could have constant/0+1 as well. Once again notice
how he is brainstorming ideas and writing them all down.
What is an 'antiderivative'? First let's learn what an integral is.
Integral of Xn (a) - FMP 10
An optional lecture, Wildberger breaks down exactly what an integral for a polynomial is. The
derivative for xn is nx
and the integral for xn is a /n + 1 or x4 = x5/5. @6:48 he begins
concrete examples starting at n = 0. If you watch part 2 he'll show different interpretations of the
integral like Lebesgue integrals.
n−1
n+1
Fundamental theorem of calculus - Essence of Calculus
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More integration and what an antiderivative is, which is f' = f meaning a function who's derivative
is equal to the original function f. If f = 3x2 then the antiderivative is x3.
Back to the Poh-Shen Loh polynomial seminar. Do you now get why the antiderivative of a generic
polynomial is how he wrote it? Take the antiderivative's derivative and the n+1 both cancel, you're
left with a x +. . . +a . Idea #4, if you watch 3Blue1Brown's lecture on epsilon-delta definition of
Limits, then this proof that all polynomials are continuous functions will make sense. A continuous
function just means it has no abrupt changes to it's output/input ratio, ie: it it's geometric form is a
continuous line with no breaks in the graph. MVT solution you can google, MIT's 18.01 has
lectures and slides showing the geometric interpretation but Poh-Shen Loh gives a clear enough
explanation. He explains he tried to get advanced placement when he was a student by just
guessing his way through a differential calc class by using polynomials/power series to solve
everything which actually worked.
n
n
0
He moves on to another question, a classic result for derivatives. "We're doing these problems to
do a very fast review of all concepts in math" which is exactly what we want. Here we even see
the fundamental theorem of algebra. He mentions complex number derivatives, Wildberger has a
whole series on this with his dihedron algebra infinitesimal theory.
Very end of this lecture… it's the Russian algebra book chapter on divisibility we did! 'If I have a
'not multiple of 5', plus a multiple of 5, it's still not a multiple of 5'.
Algebra 2 Polynomials
I was shaky on the Putnam polynomial roots seminar, so let's see what the Russian olympiad book
can teach us about finding roots for polynomials. Reading Chapter 3 Polynomials and complex
numbers of the book Mathematics via Problems by Skopenkov.
3.1.1 and 3.1.2 we already know, since a rational number is two integers that produce a quotient.
3.1.3 is interesting in that I've never seen it before, that if a(p) = 0 then p divides a0. Example: A(x)
= 2x + 4 and A(-2) = 0, since it divides a0 which is 4. This was taught to me a different, more
confusing way. Make an example for (c): A(x) = 2x - 1 then A(1/2) = 1 - 1 = 0. Pick arbitrary k: 13.
A(13) = 2(13) - 1 = 25 and this should be divisible by (1 - 13(2)) or -25, which it is.
3.1.4 because this is a Russian text, we are saved from cringe pi memes being used here. You don't
have to hand calculate these, search for a trig unit circle diagram on google images. Remember
cos() is the x value, sin() is the y value. 3.1.4(a) is 1/2 so rational, b) is square root of 3 over 2 so
obviously not rational, the rest you can type into some online trig calculator and see they are
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infinite decimals, or see their phi definitions and phi if you search contains the square root of 5, an
irrational number.
3.1.5 you can enter all of these into desmos and compare the graphs. Many of these were covered in
the 3blue1brown lockdown math lectures on trig which I already went through in the calculus
workshop. Just enter them and think about how you would prove this, that's good enough.
3.1.6 these are Chebyshev Polynomials and looking at section 3 Some properties of the polynomials
Tn, and the identities in 3.1.5, you can guess the answer for example b), if you input cos(x), and get
back cos(nx) then the polynomial is one of the Chebyshev polynomials but what is it for any n? The
hint is 'use induction' or n + 1, and of course there is a helpful tutorial from on YouTube completely
solving this problem if you're interested it's typical 'this is now in the form of an identity, so we will
swap it out' type substitution we are already used to, and solving recurrence relations. The whole
video is worth watching, seeing him substitute recurrence relations if you didn't get it from the
compsci workshop. This guy is pretty good, if you search for his name Prof Sankaran Viswanath
you get a lot of good lectures in other topics. Prof Viswanath runs an abstract algebra course using
Artin/Dummit & Foote texts, and the Algebra II course is full of category theory like functors.
3.1.7 we didn't do the section on mod yet, but you can also guess and check using an online trig
calculator. None of this is what we want though, we wanted to solve polynomials which is in
chapter 3.2. TODO
Home
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