# Math Foundations from Scratch

## Intro

This is a (ongoing) workshop to learn the math needed for various prereqs in university AI intro courses, and is Part IB of the AI Tripos. No background is assumed. The idea here is that you do this workshop until you can figure out the courses in the AI workshop and then can either continue doing both or switch, up to you. We start by writing proofs of basic arithmetic with the natural numbers and work our way up. Proof writing is a skill you practice like every other skill, if you persist and practice it enough eventually you pick up the skill. I watch a lecture one day, next day read some of a chapter, mainly 1 - 2 hrs a day sometimes less or more on the weekends but I try and do it everyday.

*Do 1% effort for 100 days* - Prof Ryan O'Donnell

### Exercises/psets

If you can't figure out an exercise write it down in a notebook called 'Things I Don't Know' and every once in a while as you progress through more of the material, review your notebook of unsolved problems and see if you can now solve them. For proofs write out full sentences, it helps you remember the material and is the proper way you write a proof anyway. Don't just do scratch work and think you have the answer you'd be surprised how stuck you get when you try explaining it to yourself in the form of a written proof, it means you don't truly understand and need to either review the material again, or put that proof in your 'book of things I don't know' and go back to it later.

If you write all your own examples everytime you come to a definition, you probably can easily do many of the exercises. Write them like you would example tests in the software workshop, find all the corner cases and see if the definition or lemma holds. Remember why we write examples in that workshop, it's to learn more about the problem domain since there may be subtle things that escaped our understanding. It's the same deal here.

## Start here

We start with some 10 minute lectures on basic arithmetic, from the point of view of a mathematician.

### The natural numbers

Presented here is a definition of a mathematical object called a natural number, as a string or sequence of strokes, which are successors of each other. ie: the successor of a number, is another number.

- 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)

- n + m = m + n (Commutative)
- Laws for multiplication
- nm = mn (Commutative)
- (kn)m = k(nm) (Associative)
- n \(\cdot\) 1 = n (Identity)

- Note the different notation for multiplication:
- nm
- n \(\cdot\) m
- n \(\times\) 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 \(\cdot\) 1 = n.

- We are creating n groups of 1. According to our definition of multiplication |||| \(\cdot\) | would be 4 groups of | or ||||. This is still 4.

Commutative law: nm = mn.

- ||| \(\cdot\) || = || || || or 6. Rearranging this, || \(\cdot\) ||| 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
- 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

- YouTube - Arithmetic and Geometry Math Foundations 8

Division is repeated subtraction. Laws of Division:

- 1. \(\frac{k}{m}\) + \(\frac{n}{m}\) = \(\frac{k + n}{m}\)

- 2. \(\frac{k}{m}\) \(\cdot\) \(\frac{n}{l}\) = \(\frac{kn}{ml}\)

- 3. \(\frac{k/m}{n}\) = \(\frac{k}{mn}\)

- 4. \(\frac{k}{m/n}\) = \(\frac{kn}{m}\)

### Fractions

- YouTube - Arithmetic and Geometry Math Foundations 9

We're still using 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). He defines the operations of addition and multiplication with fractions using the laws of division we already learned.

Note how he sets up the proof. The 'prime' notation (the tick, so c', a', b') represents another variable. You could rewrite this 'Suppose a/b = y/z so that a/z = b/y, and c/d = something/another, so that c/another = something/d does it then follow …'.

What we are trying to prove is ab' = ba' and cd' = dc'. He introduces some extra notation, multiplying both sides by (dd') to help do this.

### 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.

## 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.

### Materials needed

- The 3rd edition of
*Analysis I*from WorldCat or LibGen (or buy it, it's an inexpensive book if you get the international version from abebooks or Amazon India) - The book
*How to Think Like a Mathematician*by Kevin Houston. Most of this won't make sense until we do a few chapter's out of Tao's book, we will read it at the same time. - The optional book
*An Infinite Descent into Pure Mathematics*by Clive Newstead which provides another view of the same material if you're confused by something, look it up here (or search YouTube for a tutorial)

#### How to read math

See chapter 2 *Reading mathematics* and chapter 12 *Examples and counterexamples* and Kevin Houston's book. The way I read is go through the chapter and attempt to understand with my own examples, don't worry about memorizing anything when you do the exercises you will flip back and forth to every definition in the chapter repeatedly as you write the proof, this is where you start to remember the material and why exercises exist.

### 2.1 The Peano Axioms

We begin like we did with Wildberger, 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), you prove P(0) and P(n + 1) are true, and both these form a logical implication meaning that if they are both true, then your initial assumption of P(n) being true holds.

Remark 2.1.10 and the template in Proposition 2.1.11 is probably the clearest definition of proof by induction I've seen. Note the example where no natural number can be 'a half number', because of the increment step n + 1. 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:

- 0 is an element of N
- if x is an element of N, then so is x + 1.
- nothing else is an element of N

### Induction

Let's understand Axiom 2.5 by watching a lecture from MIT on proofs though you will end up writing so many induction style proofs on the natural numbers in Tao's book this will become second nature.

- 1. Induction lecture
- From MIT's 6.042j
*Mathematics for Computer Scientists*

- From MIT's 6.042j
- Read chapter 24
*Techniques of proof IV: Induction*in Kevin Houston's book

### Back to 2.1

Proposition 2.1.16 (Recursive definitions) in the book *Analysis I*. Tao uses a function, it's definition can be read in chapter 3.3. Here \(a_0\) is function notation so \(a_0\) is \(f_0(0)\), \(a_1\) is \(f_1(1)\). This proposition is using a new function (a rule that maps from one input to one output) every time, I'm guessing to avoid sequences where they could repeat ie 0, 1, 1, 2, 3..

Let's do an example.

- \(a_0\) becomes 0 (for c = 0)
- \(a_{0++}\) becomes \(f_1(a_1)\) or 1
- \(a_{1++}\) becomes \(f_2(a_2)\) or 2

The proof notes Axiom 2.3 protects the value \(a_0\) from being redefined by one of the functions. His base case for this proof is \({a_0}\) (Axiom 2.3) and he is assuming P(n) which is that the procedure will always give a single value to \({a_n}\). Axiom 2.4 is the inductive step, that this procedure will also give a single value to \(a_{m++}\). All of the Axioms are used to prove this. If this didn't make sense it will when we come back here in chapter 3.5 for an exercise.

#### Aside: Induction and Recursion

Induction and recursion are the same concept. As an example, the specification of a program function twoExp(n) that when given a value n, returns \(2^n\). The examples you write for this program are as follows:

- twoExp(0) = 1
- twoExp(1) = 2
- twoExp(2) = 4
- twoExp(3) = 8
- twoExp(4) = 16
- twoExp(5) = 32

You notice the pattern: the result of the next value, is 2x the previous value. twoExp(5) is 2 * twoExp(4). You write an algorithm using the above examples: if input = 0, then return 1, otherwise return 2 * twoExp(n-1). You make an assumption, that the input will always be non-negative. Walking through the program execution, twoExp(3) evaluates:

- 2 * twoExp(3-1)
- 2 * (2 * twoExp(2-1))
- 2 * (2 * (2 * twoExp(1-1)))
- 2 * (2 * (2 * (1))) = \(2^3\)

To prove this by induction on n, we try the base case P(0). If n = 0 then twoExp(0) evaluates to 1, which is \(2^0\). What is P(n)? According to our recursive definition, n = 2*P where the value of P = twoExp(n-1) or \(2^{n-1}\). Our inductive step is n, since our algorithm is inductively defined n-1, and we wish to prove the successor n + 1, which is n. 2*P = \(2*2^{n-1}\) or \(2^n\).

Note that we didn't have to do anything extra to prove this recursive function. That's because an inductive proof is a recursive function, and a recursive function is an inductive proof.

### 2.2 Addition

Let's understand his example for addition. 3 + 5 should be the same as (((5++)++)++) and adding 0 to 5 should result in 5.

Definition 2.2.1 notice he has a base case (0 + m is m), the hypotenuse "Suppose we inductively defined how to add n to m" which we saw at the beginning of 2.2 where 3 + 5 is defined as incrementing five three times, finally the inductive step, we can add n++ to m by defining (n++) + m to be (n + m)++.

Try writing your own examples as advised in the book *How to Think Like a Mathematician*

- 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.

Lemma 2.2.2 notice first how he is using the exact same induction template from 2.1.11, you will eventually get very familiar with this template. He's using prior established definitions to set up the proofs, notably definition 2.2.1 where 0 + m = m is defined. He assumed that n + 0 = n. Next the inductive step, where (n++) + 0 = n++. No arithmetic was done here, he pointed to the definition of addition from 2.2.1 that (n++) + 0 is (n + 0)++ or n++, so we have n++ = n++ and the proof is done. He used a different variable to not confuse it with the previously established definition of 0 + m = m, since we are now going the other way with n + 0 = n on our way to prove a + b = b + a.

Lemma 2.2.3 Prove: n + (m++) = (n + m)++. We are inducting on n, and keeping m fixed which means we apply the inductive step to n++ instead of m++. Replace all instances of n with 0 for the base case and show the inductive hypothesis holds that 0 plus the increment of some other number, equals that number. Try picking apart this proof yourself to follow it:

- base case:
- 0 + (m++) = (0 + m)++
- left side is m++ because of definition 2.2.1
- right side is m++ because of definition 2.2.1

- inductive hypotenuse: 'Suppose n + m++ = (n + m)++'
- inductive step:
- n++ + (m++) = (n++ + m)++
- left side is ((n + m)++)++ because of 2.2.1
- right side is ((n + m)++)++ because of 2.2.1

Corollary: Tao asks why n++ = n + 1 is a consequence from the two proved lemmas. Compare lemma 2.2.3 *For any natural numbers n and m, n + (m++) = (n + m)++* to definition 2.2.1 where we were able to show m++ = m + 1. Compare lemma 2.2.2 to the base case in definition 2.2.1

Proposition 2.2.4 (Addition is commutative) the base case is straight forward. 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, so we can use our hypothesis of n + m = m + n to show that (m + n)++ is the same as (n + m)++. Proposition 2.2.6 is similar.

#### Exercise 2.2.1

Recall Exercises/psets if you can't figure out something, write it down somewhere and move on. Routinely revisit to see if you can now figure out the proof. Note this is all scratchwork:

Proposition 2.2.5 (Addition is associative) what we want to prove is (a + b) + c = a + (b + c) Obtain the inductive template from the beginning of the chapter and try this yourself, making sure you write out full sentences exactly like Tao does for his proofs. Here is all the scratch work I do before writing a proof:

**Base Case**: What is the base case? We want to show that even if a natural number is inside or outside brackets, it will have the same result when added together. Let's pick c or a, as on each side of the equality they are either inside the bracket, or outside it. This will use substitution, which is in the appendix. If we say x = y, then every occurrence of y we can substitute for x:

(a + b) + 0 = a + (b + 0) (substitute 0 for c)

- left side: (a + b) + 0 since b + 0 = b (lemma 2.2.2) then we have:
- a + (b + 0) (lemma 2.2.2)
- base case: a + (b + 0) = a + (b + 0)

- or choose a: right side: 0 + (b + c) = b + c by definition of addition
- lef side: (0 + b) + c by definition of addition
- another base case: 0 + (b + c) = 0 + (b + c)

**IH**:
The inductive hypothesis we assume addition is associative, and that (a + b) + c = a + (b + c). If any steps of your proof are either in form (a + b) + c or a + (b + c) you can substitute them for each other by the IH. For example: x + y + a + (b + c). Note a + (b + c) is the form of the IH, so can change this to x + y + (a + b) + c

**Inductive step**: There's a few ways to show this, I'll move stuff around randomly and you can see where the proof is:

(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 one value
**Recall (n++) + m = (n + m)++ ergo (n++) + (m + b) is (n + (m + b))++**

- ((a + b) + c)++ (commutative)
- (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, substitute and permute them around some more until this starts to make sense. Remember to think as (n + m) as one value, so (n++) + (n + m) is (n + (n + m))++ and (n + (n + m)) can be rewritten according to commutative property ((n + m) + n).

- The proof, write full sentences:
- We induct on
*a*keeping*b*and*c*fixed. We first consider the base case a = 0 and we have to prove (0 + b) + c = 0 + (b + c). By the definition of addition (0 + b) + c = b + c, and 0 + (b + c) = b + c, thus both sides are equal to b + c and are equal to each other. Suppose inductively that addition is associative and (a + b) + c = a + (b + c), now we have to prove that ((a++) + b) + c = (a++) + (b + c) to close the induction. By the definition of addition, the left hand side is equal to ((a + b)++) + c which is also equal to ((a + b) + c)++ by the definition of addition. By the inductive hypothesis, ((a + b) + c)++ is equal to (a + (b + c))++. The right-hand side is equal to (a + (b + c))++ by the definition of addition and now both sides are equal to each other, and we have closed the induction.

- We induct on

### 2.2. cont.

Proposition 2.2.6 (Cancellation Law) uses Axiom 2.4 which states '*Different natural numbers must have different successors; i.e., if n, m are natural numbers and n = m, then n++ = m++. Equivalently, if n++ = m++, then we must have n = m*'. So (a + b)++ = (a + c)++ is = a + b = a + c by the line *if n++ = m++, then n = m* in axiom 2.4. We dropped down from the successor by using axiom 2.4 and can now use our inductive hypothesis of the cancellation law for a to show b = c and close the induction. .

Lemma 2.2.10 has an exercise where induction is suggested but this is actually much easier to prove using contradiction (See Kevin Houston book). Always read the comments on Tao's blog if you're stuck: "You can modify the statement of the exercise to an equivalent statement which makes sense for a=0. (The statement will probably look something like “For any natural number a, either a=0, or else there exists exactly one…”.)" If you see this stackexchange answer they use the peano axioms of 2.2 existence and 2.4 uniqueness. We can also try the chapter on contradiction proofs in Kevin Houston's book or the appendix of *Analysis*: Suppose for the sake of contradiction that that there exists two different natural numbers with the same successor. Then a = m++ and b = m++. But then by Axiom 2.4 we have a = b, a contradiction.

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.

#### Exercises

Proposition 2.2.12 / Exercise 2.2.3: Prove the basic properties of order for natural numbers. Here I will show how you can 'reverse' a lemma or other proved proposition, none of this is a properly written proof only notes on moving through steps to understand the problem then after write a proof in the style Tao has been teaching us. We just want to point to definitions and lemmas here not use induction:

**Order is reflexive**- Reflexive property is that any natural number is equal to itself.
- We want to show order is reflexive: a \(\ge\) a
- a = a + n (def 2.2.11 of ordering of natural numbers)
- From lemma 2.2.2 we know that a + 0 = a and so a = a and is reflexive
- By definition 2.2.11 we also have a \(\ge\) a
**Transitive order property**- a \(\ge\) b and b \(\ge\) c, then a \(\ge\) c
- If a = b + 0, b = c + 0, and a = c + 0, then we have lemma 2.2.2 and transitivity (equality) holds with a = b, b = c and a = c.
- Substitution:
- a = b + n
- b = c + n
- b is equal to (c + n) so let's put that in place of b in a = b + n:

- a = (c + n) + n
- a = c + (n++)
- By lemma 2.2.10 we have a = c + m since b++ will always equal some nonzero natural number.
- By definition 2.2.11 a = c + m is a \(\ge\) c + m

Proof attempt, though we could write our own def or lemma that a + b = a natural number it's largely implied by everything we've read so far: By the definition of Ordering of the natural numbers, if a = b + m and b = c + n, then a = c + m + n. By associativity a = c + (m + n) and since if a = b when m or n are 0, then a = c when m and n are 0. By substitution d = m + n, and since m and n are natural numbers, d is a natural number by the Peano axioms and 2.1.3 (m + n)++ = d++, and by Axiom 2.4 (m + n) = d so we have a = c + d as desired by the definition of Ordering of the natural numbers.

**Anti-symmetric order**- If a \(\ge\) b and b \(\ge\) a then a = b
- See the entry for Antisymmetric relation
- If these two inequalites hold then they must be equal
- We need to show a = b:
- a = b + m by 2.2.11
- b = a + n by 2.2.11
- Substitute b for (a + n):
- a = (a + n) + m
- a = a + (n + m)
- a + 0 = a + (n + m) ('reversing' lemma 2.2.2)
- 0 = n + m (cancellation law)
- 0 = 0 + 0 (corollary 2.2.9)
- If n = 0, and m = 0 by above, then a = b + 0 which is a = b
**d) Addition preserves order**- a \(\ge\) b iff a + c \(\ge\) b + c
- a + c = b + n + c by 2.2.11
- a = b + n by cancellation law and a \(\ge\) b by 2.2.11
- Another method: a + c \(\ge\) b + c
- a + c = b + n + c by 2.2.11
- a + c = b + c + n (commutative law)
- (a + c) = (b + c) + n (associative law)
- This is nat number = nat number + another number or by 2.2.11 a \(\ge\) b

**e) a < b iff a++ \(\le\) b**- b = (a++) + n (def ordering of natural numbers)
- b = (a + n)++ (def of addition)
- successor of a number cannot be itself, so b does not equal a and by 2.2.11 a < b
- We skipped steps, Tao is more thorough so let's try it again:

- b = a + n and n does not equal zero (2.2.11)
- Since n must be a positive natural number by 2.2.7 and 2.2.11:
- b = a + (n++)
- def of a positive natural lemma 2.2.10

- b = a + (n + 1) (corollary from 2.2.2/2.2.3)
- b = a + (1 + n ) (commutative law)
- b = (a + 1) + n (associative law, grouping doesn't matter for addition)
- b = (a++) + n by previous corollary of n++ is n + 1
- a++ \(\le\) b since by 2.2.11

**f) a < b iff b = a + d for some positive number d**- For the sake of contradiction, assume a \(\le\) b in the a = b case for some positive number d. Then a + 0 = a + d by lemma 2.2.2, and 0 = d by cancellation law. This is a contradiction that d must be a positive number so a cannot equal b and a \(\lt\) b

Proposition 2.2.13 fill in the gaps:.

- When a = 0, we have 0 \(\le\) b for all b (why?)
- 2.2.1 Addition of natural numbers, 0 + m = m so 0 + b = b and by 2.2.11 0 \(\le\) b

- If a > b then a++ > b (why?)
- a = b + d (as per 2.2.12, d must be positive)
- c++ = d (lemma 2.2.10)
- a++ = b + (c++) (substitution)
- a++ > b by def of ordering
- or not so rigorous way:
- a++ = (b + c)++ (def of addition, (b + c)++ must be positive
- a = b + c (Axiom 2.4)
- a > b (2.2.12 d is positive)

- If a = b, then a++ > b (why?)
- If a = b, then a++ = b++ by Axiom 2.4
- a++ = b + 1 (corollary 2.2.2/2.2.3)
- this is a++ > b by 2.2.11

Proposition 2.2.14 (Strong induction) From MIT this lecture on strong induction, the intuition here is Q(m) if all the dominoes are already knocked down and there's one left standing, then it too will be knocked down where 'weak induction' intuition is P(n) if you knock down one domino, they will all fall. If P(n) is true because P(n++) implies P(n) true, then for all m \(\le\) P(n), Q(m) is true.

First look up vacuous truth or see the appendix on mathematical reasoning where Tao defines it.

- Let Q(n) be the property that P(m) is true for all m\(_0\) \(\le\) m < n for n \(\ge\) m\(_0\).
- Q(0) is true since:
- m\(_0\) \(\le\) m < 0 is vacuously true as stated by Tao in the exercise hint
- 2.2.14 proposition of for all n \(\ge\) m\(_0\) is true 0 \(\ge\) 0\(_0\)

- Show that Q(n++) is true:
- P(m) m\(_0\) \(\le\) m \(\lt\) n \(\lt\) n++ is true (2.2.11)
- P(m) m\(_0\) \(\le\) m \(\lt\) n++ is true for all n \(\ge\) m\(_0\)

Try an example.

- P(n) is the property that n + 0 = n
- Q(0) is 0 + 0 = 0 and P(0)
- Q(1) is 1 + 0 = 1 and P(1), P(0)
- Q(2) is 2 + 0 = 2 and P(2), P(1), P(0)

Exercise 2.2.6: Your proof could roughly be 'Suppose P(n++) is true such that all P(m) is true for m \(\le\) n++. Then per Axiom 2.5 P(n) is true, so P(m) is true for all m \(\le\) n' since Tao hinted at 'backwards induction'.

### 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
- right side:
- ab + a(c++) = ab + ac + a
- ab + ac + a is in the form of the inductive hypothesis so 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

**Exercise 2.3.1**
n x m = m x n:

- base case: 0 x m = m x 0
- left side: 0 x m = 0 by definition 2.3.1

For the right side, m x 0 = 0, write your own lemma that mimics 2.2.2 but for multiplication, roughly as so:

Lemma X: For any natural number n, n x 0 = 0:

- base case: 0 x 0 = 0
- IH: n x 0 = 0
- Inductive step: (n++) x 0 = 0
- (n x 0) + 0 = 0 (def 2.3.1)
- Remove the + 0 (lemma 2.2.2)
- Now it is in the form of the IH: n x 0 = 0

Lemma Y: m x (n++) is m + (n x m), rewriting Lemma 2.2.3 but for multiplication

- base case: Our lemma X and def 2.3.1
- IH: m x (n++) = m + (n x m)
- Inductive step: m x (n++)++ = m + (n++ x m)
- Left:
- m + (n++ x m) by IH
- m + (n x m) + m by def 2.3.1

- Right:
- m + (n x m) + m by def 2.3.1

Now we have to prove m x n = n x m

- base case: def 2.3.1 and lemma X
- IH: m x n = n x m
- m x n++ = n++ x m
- Lemma Y = def 2.3.1

**Exercise 2.3.2**

The hint is to prove the second statement first, if n and m are both positive, then nm is also positive. You could use Proposition 2.3.6 as a template for this:

- If n > 0, and m > 0, then nm > 0
- We have nm = 0 + d for some positive d
- Since d is positive, n and m must be positive (or else 0 = 0 + d, a contradiction)

Or you could use Lemma 2.2.10, n x m = (b++) x m and then mimic 2.3.6 to determine n x m is positive if n and m are positive. Note that (b++) x m expands to the form (b x m) + b, and since you know a x b + a is the successor form of multiplication it can't be zero.

Lemma 2.3.3: n x m = 0 iff n = 0 or m = 0. You could try a contradiction using the nm > 0 proof we already wrote, so it must be n = 0 or m = 0 if nm = 0 or a contrapositive of Tao's statement: see the appendix on mathematical reasoning or look up contrapositive in Houston's book. If Tao states something, and then proves it, then there is a contrapositive statement we can use to prove another fact.

**Exercise 2.3.3**
A rewrite of 2.2.5: For any natural numbers a, b c, we have (a x b) x c = a x (b x c). You could induct on b, convert to the 2.3.1 form, and rearrange whenever you move something around and it is in the form of the inductive hypotenuse.

Tao also suggests using the distributive law:

- Inductive Step: (a x b++) x c = a x (b++ x c)
- (a(b++)c) = (a(b++)c) by distributed law
- (ac(b++)) = (ac(b++)) by commutative law
- (ac(b + 1)) = (ac(b + 1)) by corollary n++ = n + 1
- acb + ac = acb + ac by distributed law again
- cancellation law the two +ac:
- acb = acb or (ab)c = a(bc)
- or permute any way you want like we did for 2.2.5

**Exercise 2.3.4**

Prove the identity.

- (a + b)(a + b) = (a + b)a + (a + b)b by distributive law/exponentiation
- aa + ba + ab + bb by distributive law
- a\(^2\) + ba + ab + b\(^2\) by def 2.3.11/Examples 2.3.12
- ba + ab is ba + ba by commutative law
- ba + ba is 2ba by def 2.3.1 where 2 x m = 0 + m + m
- (a + b)\(^2\) = a\(^2\) + 2ab + b\(^2\)

**Exercise 2.3.5**

Prove proposition 2.3.9 by fixing q and inducting on n for n = mq + r and 0 \(\le\) r \(\lt\) q

For the base case, if 0 = mq + r, then m and r must be zero because this still satisfies the inequality where q is greater than r and remember we fixed q, so it can't be zero. For the inductive step, n + 1 = mq + r + 1 (adding + 1 to both sides as per previous lectures on induction), we have two cases, r + 1 < q or r + 1 \(\le\) q. If r + 1 < q then we have the inductive hypothesis and there's nothing more to prove since 0 \(\le\) r + 1 \(\lt\) q. Remember the order of multiplication/addition, this is still (number) = (number) plus n in the form of (n + 1) = (mq) + (r + 1).

If r + 1 \(\le\) q, or (r = q), then we have n + 1 = mq + q, and from the definition of multiplication/distributive law this is n + 1 = (m + 1)q or m++(q) + 0 and 0 still satisfies the inequality of r < q.

## Geometry

Before we learn some algebra and modern trigonometry, we should at least see some geometry. These lectures follow the John Stillwell book *Mathematics and it's History* where Wildberger slices various shapes with planes so you can see why we study things like a parabola or hyperbola. This is only casual viewing.

### Euclidean

- Pythagoras' Theorem 1 and Pythaoras' Theorem 2
- A geometric proof of original Pythagoras theorem
- A proof by contradiction why the square root of 2 is irrational
- Part B shows classical cos/sin/tan trigonometry using chords

- Greek Geometry 1 and Greek Geometry 2
- Euclid's book the Elements
- Part B covers different planes, ellipse/conic geometry

### Analytic

This is the geometry we will be doing shortly that allows for computational Euclidean geometry. We don't have a definition of polynomials yet but you can probably understand most of this lecture anyway.

- Analytic Geometry
- Cartesian coordinate plane explained
- Part B covers periodic continued fractions method to estimate irrational numbers

### Projective

Abstract geometry, we will need to use projections later when we study linear algebra and project into affine spaces.

- 3D Projective Plane
- Shows you different perspectives when looking at objects
- Lay the x,y cartesian plane flat, add a z dimension like a flag pole, add another flat cartesian plane in z, project lines from x,y into z

### Non-Euclidean

Even more abstract geometry, construct a new plane by slicing a sphere and studying a hyperbola. Hyperbolic geometry seems better suited for deep learning as it has 'more capacity'. Maybe we will end up back here projecting our data into a hyperbolic plane.

- Non-Euclidean/Hyperbolic Geometry
- Multiple slicing planes/perspectives

### Differential

Geometry of space curves, osculating plane

- Differential Geometry
- Abstractions on top of abstractions
- There's even more abstract geometry

## Some Algebra

### Crash course in solving equations & polynomials

Flip through sections 1.3.1, 1.3.2 and 1.3.3 *Review, Redo, Renew* from this draft book used in CMU's 15-151/21-28 course. The author has a list of errata on his page. You will get a crash course in some quick arithmetic, solving linear equations and polynomials. There's a few interesting methods here you may not know about like how to variables from equations. Wildberger covers much of this material too for example Problem 1.3.8, an infinite nest is also in a continuous fractions lecture from the *History of Mathematics* playlist: \(x = \sqrt{2 + x}\)

If you're interested there is a summary of Algebra here that was originally made for non-english speaking university students.

### Lectures

Mostly 10 minute lectures on algebra basics, such as how Pascal's array/Khayyam triangle/Yang Hui's triangle encodes the binomial theorem, this series sets up understanding algebra and later calculus and combinatorics.

- Introduction to Algebra
- Wildberger notes there is a subtle change in variables as you get more sophisticated with algebra

- Baby Algebra
- The basics of solving equations

- Quadratic equations I and part II
- Completing the square
- The quadratic formula

- How to find a square root
- Manual square root algorithms

- Number patterns I
- Table of differences, summation notation introduced

- Number patterns II
- Table of differences geometric interpretation

- Euler and Pentagonal numbers
- Recursive partition function

- Algebraic identities
- Common identities

- The Binomial theorem
- Coefficients are related to paths on a two dimensional array

- Binomial coefficients
- N choose K intro to combinatorics

- Trinomial theorem
- 3D Pascal's tetrah

#### A different modelling of Polynomials

Wildberger has a unique encoding of polynomials in the rest of his math foundations playlist by abstracting coefficients into columns and rows. Pascal's array is then used, and things like factoring or polynomial division is surprisingly trivial to do. Later in the same playlist he teaches some algebraic calculus using this polynumber encoding and taking derivatives and doing integration is again trivial arithmetic. They could teach this to primary school kids. If you have time, I highly recommend watching the rest of the playlist.

### Lectures about pi

Before we do trig, let's learn what \(\pi\) is. Wildberger talks about the history of pi such as Archimedes interval for pi, Newton's formula, a hexadecimal version of pi, his own formula. The end of this lecture he claims pi is some kind of complex object we still don't understand. At this stage we only need to know it's a constant, or the applied approximation of 22/7 or 355/113.

## Trigonometry Boot Camp

Let's take the Brown Univeristy trig boot camp. There is a crash course lecture. I assume you know the basic gist of what an angle is which will be better defined when we take calculus.

- Radians you could even make up your own notation and define it as dblpi = 6.283… and then 360deg is dblpi, 90deg is 1/4dblpi, 180deg is dblpi/2, etc.
- We really only need to know sine and cosine functions, all others are derived from them
- In the scaling of triangles example where \(\frac{1}{\sqrt 2}\) becomes \(\frac{\sqrt 2}{2}\) they are rationalizing the denominator:
- \(\frac{1}{\sqrt 2}\cdot\frac{\sqrt 2}{\sqrt 2}\) is \(\frac{\sqrt 2}{\sqrt 2 \cdot \sqrt 2}\) which is \(\frac{\sqrt 2}{\sqrt 4}\) or \(\frac{\sqrt 2}{2}\)

- You can download the slides to keep a copy of the scaled triangles
- The reflections think about subtracting 1/6 from 1/1, you get 5/6 so 5pi/6 is pi (180 deg) minus 1pi/6, and a negative direction means we are moving clockwise so move to pi, subtract pi/6 in other direction.
- The end of the lecture we finally discover why sin() and cos() are wave functions, since it's a function that moves around the unit circle. Tan() is the slope 'rise over run'

Summary: There exists 3 triangles, which we should know their measurements, and 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. We reflect these triangles into the other areas of the circle so we can get x/y values there as well by rotating around the circle and then adding up all the angles to indicate how much we rotated counter-clockwise (positive) and clockwise (negative). A reflection is demonstrated here so if f(x) = sin(x), cos(x) and tan(x) then y = -(fx).

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-when-squared). If you're stuck watch the video that walks through all the solutions then do the worksheet again until you can finish it without needing to look at the video walkthru.

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're wondering why we are using radians instead of degrees it is explained here. Wildberger's rational trigonometry uses a turn, so pi/2 is 1/4, pi is 1/2, and 2pi is 1. Angles are replaced by 'the spread of vectors'.

### Some Calculus

Now that we know trig functions, watch this basics of calculus crash course which was designed for non-english speaking students to learn math notation (with Chinese subtitles). Note in the last few examples how the derivative works d = \(nx^{n-1}\) so in the examples the derivative of \(y = \frac{-g}{2}t^2+v_0t+y_0\) is (\(2\cdot\frac{-g}{2}t^{2-1}+ 1\cdot v_0t^{1-1}\)) or \(-gt + v_0\) you take the exponent power, multiply the coefficient of the base, and subtract the power, so the derivative of \(-gt^1 + v_0\) is (\(1\cdot-gt^0 + v_0\)) which is -g, as we'll see later any terms without a variable (in this example t) cancels out or as Wildberger mentioned are so small so they are discarded.

Summary: if you know the acceleration, then you integrate to find the velocity, then integrate again to find the position. If you know the position, you take a derivative to find it's velocity, and take another derivative to find the acceleration. One function tells you how another function is changing. These are also higher order functions we learned in CS019, where a function produces another function.

Try reading the first chapter from Gilbert Strang's calculus and see if you understand it. His book was designed for self-learning, so that first chapter assumes little background and builds up this intuition of one function that tells you about another. There is also a lecture for this chapter and a self-study guide that acts as a recitation.

## Set Theory

### History of set theory

### 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 and have an idea what implies means or disjunction. Now that we have done some proofs ourselves, and the basics of trigonometry we can finally understand some of the examples he uses in this appendix. Tao makes a point that this chapter is meant to be read and understood not studied and memorized, and thankfully he is not bombarding us with piles of confusing logic notation that you normally see in other 'how to do proofs' books. At the same time you read this appendix, read *How to Think Like a Mathematician* by K. Houston it will clear up any cryptic definitions like how implication works.

### A.2 Implication

A.2 Implication seems confusing on first read, it's modern language use relating to casuality is different from the logician's use which is explained here. The only permutation where implication is false is if p implies q and p is true, but q is false. See the Kevin Houston book for more details.

An optional short series on the history of various schools of logic that led to our 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 discussed

- 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
- Boole's book Analysis of Logic
- De Morgan's laws

#### Using Boole's Algebra w/implications

Another optional lecture from a short series on Boole's Algebra which is different than Boolean Algebra you've already seen (truth tables). For example at around 24:29, he shows how you can verify Modus Ponens (p implies q) with some very simple algebra instead of deducing symbols.

### Appendix A cont.

Proposition A.2.3, confusing the hypotenuse "Suppose that…" by assuming the conclusion. We always assume the hypotenuse, and prove the conclusion. Proposition A.2.6, proof by contradiction, we know that when pi/2 (90deg) the sine function is 1, because remember it is the y axis of the unit circle. Tao also makes a note here to not use confusing logical notation in proofs.

Remark A.5.1 is in the above history of logic lectures about Aristotle logic.

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).
- Think about the rest, using the K. Houston book as you go to find a contrapositive

A.6 the point of the examples is to make a point about arguments that use outer variables depending on inner variables being false. We can come back later and reread this section when we learn more concepts like limits, indeed we will likely have to reread this entire appendix many times over while writing future proofs.

#### Quiz: Logic

These multiple choice quizzes are from Tao's Analysis class, try the logic quiz. If you struggle read the other two resources from Materials needed relevant chapters on logic, search YouTube for proofs by implication, or search 'discrete math inference rules' there are dozens of tutorials. Take the quiz again.

### 3.1 Fundamentals (Set Theory)

We are back to reading Tao's *Analysis I* and still in the realm of the natural numbers. 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. Firstly, 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 \(\emptyset\) = {\(\emptyset\)} then by Def 3.1.4 \(\emptyset\) \(\in \emptyset\) a contradiction to Axiom 3.2
- Same reasoning as above for \(\emptyset\) is not equal to {{\(\emptyset\)}} or {\(\emptyset\){\(\emptyset\)}}
- {\(\emptyset\)} is not equal to {{\(\emptyset\)}} because [list: 1] is not equal to [list: [list: 1]]. The first has element 1 in it, the second has element [list: 1]. More precisely if {\(\emptyset\)} = {{\(\emptyset\)}} then by Def 3.1.4 every element of {\(\emptyset\)} (it's only element is \(\emptyset\)) must be in {{\(\emptyset\)}} (it's only element is {\(\emptyset\)}) and vice-versa, so claiming {\(\emptyset\)} \(\in\) {\(\emptyset\)} which contradicts Axiom 3.3, where y \(\in\) {a} iff y = a. In this case y must = \(\emptyset\) to be in {\(\emptyset\)}.

Remark 3.1.21 clarifies the above exercise how {2} 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, better than the usual treatment of venn diagrams. Remark 3.1.26 distinct is defined as *there are elements of one set which are not elements of the other* so \(\emptyset\) and \(\emptyset\) are not distinct because neither can have elements. In Proposition 3.1.28 notice how any operation on two sets results in a new set, like a higher order filter function over a list results in a new list.

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 \(\in\) {3,5,9); n<6}

#### Quiz Sets

Try Tao's quiz. If you totally failed this try secondary reading material of the other two books in Materials needed. Watch Youtube lectures and take the quiz again.

#### Exercises

- 3.1.3 prove the remaining claims in 3.1.13:
- If a and b are objects, then {a, b} = {a} U {b}:
- By (new axiom) 3.1.4 we need to show every element x in {a, b} is in {a} U {b}. Suppose that x is an element of {a, b} then by axiom 3.3 x = a or x = b. By Axiom 3.4 this means at least one of the following statements is true: x is in {a} or x is in {b}. We divide into cases. If x is in a, then by axiom 3.4 it is in {a} U {b}, similarly if x is in b, it is also in {a} U {b}. Thus in both cases we see every element of {a, b} is in {a} U {b}.

- The next claim, prove the union operation is commutative:
- Follows the same proof he did for the union operation is associative. Suppose x is an element of A \(\cup\) B. By def 3.1.4 and axiom 3.4 pairwise union then x \(\in\) A or x \(\in\) B. By disjuntion this is also x \(\in\) B or x \(\in\) A. So x is in B \(\cup\) A and x is in A \(\cup\) B.

- Last claim, prove A \(\cup\) A is equal to A \(\cup\) \(\emptyset\) is equal to \(\emptyset\) \(\cup\) A is equal to A. Suppose x is an element in A \(\cup\) A, then x is in A or x is in A, so by disjunction (we can pick either, but there's only one choice) x is in A. If x is in \(\emptyset\) \(\cup\) A, x is in \(\emptyset\) or x is in A, so x is in the union of A and \(\emptyset\) and since the emptyset is empty x is in A.

- 3.1.4 prove claims in Proposition 3.1.18
- You can write both these proofs exactly like his 3.1.18 proof for transitivity. Suppose A \(\subset\) B and B \(\subset\) A. To prove that A = B we have to prove that every element A is an element of B, and every element B is an element of A by def 3.1.4. Let us pick an arbitrary element x \(\in\) A. Then, since A \(\subset\) B, x must also be an element of B. But since B \(\subset\) A, then every element of A is an element of B and every element B is an element of A. By 3.1.4 they are equal.

- 3.1.15 Show how union, intersection and subset are logically equivalent:
- Look in the Houston book in the appendix, see the advice for how to prove that two statements are equivalent. For this exercise if A is a subset of B, and the union of A and B = B, then this implies the intersection of A and B will be A, which in turn implies A is a subset of B.
- We will show that any two set operations implies another. If x is in A \(\cup\) B, then by definition of union x is in A or B, so by disjunction x is in B. If x is in A, since A is a subset of B then x is again in B. Thus x \(\in\) A \(\cup\) B if and only if x \(\in\) B and by definition of union, and axiom of equality 3.1.4, A \(\cup\) B = B. Since x is in B, if x is in A \(\cap\) B it must also be in A, so x is in A and B, which is the definition of intersection, and if A \(\cap\) B = A, and if A \(\cup\) B = B, then every element of A is in B so A \(\subset\) B.

- 3.1.16 Prove proposition 3.1.28
- de Morgan's laws a proof exists for them everywhere including youtube.
- Rough proof strategy: what you want to do is get your arbitrary element into the form of the definitions for union, intersection and difference sets.

- 3.1.8 You can google the absorption laws (disjunction(OR) absorbs conjuction(AND)) then try your own proof
- 3.1.9 If A intersection B = emptyset, then they are disjoint (have no elements in common). This is the definition of 3.1.27 difference of sets, where x is in A but not in B, so A = x - b and B = x - a if the union of A and B = X.
- 3.1.10 All I can think of is the emptyset, since emptyset union emptyset is still emptyset, and emptyset intersection emptyset is also disjoint.
- 3.1.11 There is a solution here, but suppose we have a statement P(x, y) pertaining to x and y such that for each x in A, there is at most one y for which P(x, y) is true. Then by the axiom of replacement there exists a set {y: y = f(x) for some x in A}, then y = f(x) for some x in A, and y is in the set of A precisely when P(x) is true, which is the definition of the axiom of specification. Not a complete proof but something like that is the strategy.

#### 3.2 Russell's Paradox

Wildberger covered some of this already, 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.

### 3.3 Functions

Example 3.3.3 if you forgot range notation. 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.

3.3.14 See this tutorial. 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 elements are equal to each other then it is injective:

- A = {a, b, a}
- B = {1, 2}
- f(a) = 1
- f(b) = 2
- f(a) = 1
- but f(a) = f(a), so a = a and this function is injective.

- f(a) = 1
- f(b) = 1
- f(a) = 1
- but f(a) doesn't equal f(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. If you totally failed this try secondary reading material of the other two books in Materials needed. Watch Youtube lectures you can find. Try the quiz again.

#### Exercises

A all of these proofs have a youtube tutorial or stackexchange answer, these are common proofs first year students do.

- Ex 3.3.1 was deleted in errata (See Tao's site)
- Ex 3.3.2 The proof for this is in the book https://infinitedescent.xyz/.
- Ex 3.3.3 The empty function is always injective since it's 'vacuously true' that f(x) = f(x'). It is never surjective since there are no elements in the emptyset that fit the surjective definition of every element in Y is a value of f(x) of f(x) = y, and thus never bijective, since a function must be both injective and surjective to be bijective.
- Ex 3.3.4 The tilde hat thinghy is just notation for two different fuctions, like f and f'. Cancellation law means we're trying to prove two different functions are the same so g = g' and f = f'. If f and g are injective, then g(f(x)) = g(f'(x)) and f(x) = f'(x), and by one-to-one definition of injective functions f = f' (since f is g's input). Surjective is similar, g(y) = g'f(x)) = g(f(x)) = g'(y) so g = g'. Does g need to be injective for this proof to always hold? Try corner case examples from the text, like if g = n
^{2}, then g(f(-1)) and g(f'(1)) have the same output, so two different inputs to g with same output means f(x) can't equal f(x') as per definition of one-to-one functions so we can't always derive f = f'. - Ex 3.3.5 If f is injective in g(f(x)), then g doesn't need to be injective. Consider an example A = {1}, B = {2, 3}, and C = {4}. f(1) = g(2) = 4 so f is still injective. The difference in 3.3.4 is the wording 'show that these two input functions are always equal if g in injective' and the second question 'Is this still true if g is not injective?' which means a single case where if g is not injective, and we get two non distinct inputs that equal the same outputg, the whole proof is false whereas here we are asked 'must g always be injective' so just needa single case of not needing to be injective.
- Ex 3.3.6 Is described here see Remark 3.3.24 if you forgot what an inverse function is. Note how the inverse of f and f cancel each other out.
- Ex 3.3.7 Similar to above and his association proof
- Ex 3.3.8 todo

### 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. This is what I usually do:

- re-read the definitions of every object mentioned, re-read that chapter's definition of equality.
- look at any proofs he gave in the chapter for style/strategy
- make a guess and update the exercises section here so I don't forget it later (usually wrong guesses)
- look at the appendix of
*How to Think Like a Mathematician*and see if anything applies to the strategy I guessed - get notepad like he does and write down summary of definitions/lemmas I think will be needed so I can stare at them while thinking
- reread the exercise carefully, note the exact task asked for (only started doing this after multiple failure, see last step)
- try a bunch of small examples
- play around with brackets and input, for example if n + (m + n) then consider it m + (one object) or g(f(x)) is g(object) so it's input is a whole function, and notice that in one-to-one function definition if x = x' then this is f = f' as instead of x, it's f() that is input.
- try examples from the text, like Example 3.3.15, and my own corner case examples
- try a contrapositive example if the exercise is 'show that if A then B' which usually leads to a solution
- if I still don't get it find youtube tutorials and look at other books, like Infinite Descent text
- all else fails look up the answer on math stackexchange, which sometimes is incomprehensible or wrong
- if there is an answer, try to rewrite it in full sentences to see how tripped up I get or not
- negate my proof see if still holds, do what he did for 'adversary proofing'
- if still no idea ask a question on stackexchange, being extremely careful to reproduce the exercise, at which point, I've usually discovered the problem, 9 out of 10 times when I can't prove something it's because I misread the exercise and was hopelessly lost in proving something that wasn't asked

## Linear Algebra

We're doing *Analysis I* and Linear Algebra in parallel. These sets of lectures cover Linear Algebra from a geometric approach so later when we work through Tao's Linear Algebra notes and take vector calculus we can understand what is happening.. Reasons I decided to do this course is it is similar to the other workshop where we are learning difficult computer science abstractions using Pyret: all the complexity is stripped away so you can just focus on building insight into how all this works. The foundations of this geometry are simple and from Pappus of Alexandria's law of parallel lines.

### Introduction to Linear Algebra

- Introduction - Wild Linear Algebra A1

Welcome to 320 AD in ancient Alexandria you're learning linear algebra with just parallel lines. Around 4000 years ago, the people of Babylon knew how to solve a simple 2x2 system of linear equations with two unknowns and around 200 BC the book *Nine Chapters of the Mathematical Art* displayed how to solve a 3x3 system of equations so linear algebra in some form or another has been around a very long time.

He constructs the affine plane from scratch. The vector (8/3, 10,3) you need to full screen to better see exactly where they are. From B, you move right 2 lines, and 2/3 so \(2 + \frac{2}{3}\) which is 8/3. From there you move up 1/3, and 3 so \(3 + \frac{1}{3}\) which is 10/3. Reasons for using the affine plane are you do not impose anything on it artificially like the origin (0,0) in the Cartesian plane, you can do projections easily, and it's construction is simple, since only parallel lines.

The biggest a takeaway here is what vector direction is, what a basis vector is and how you can combine them to form larger vectors, and how to translate basis vectors to suit different perspectives. The exercises at the end the delta symbol will return in the lecture on area and the determinant.

In this optional related lecture he shows how ancient people made a ruler using just parallel line geometry, which leads to modelling the rational continuum.

### Vector Arithmetic

- Geometry with Vectors - Wild Linear Algebra A2

All this vector arithmetic will come up later in other courses, at least now you will be somewhat familiar with it. The proof for diagonals of a parallelogram bisect each other, note how quickly this can become a pile of symbols.

- let lambda = mu in: lambda = 1 - mu
- if lambda = mu, then moving mu over to the other side means 2lambda
- 2lambda = 1
- Divide each side by 2 to isolate lambda
- lambda = mu = 1/2

@38:00 or so solving linear eq with 2 unknown variables we already covered in the first lecture or review Crash course in solving equations & polynomials.

todo exercises