Math Foundations from Scratch

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.

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

• 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 \(\cdot\) 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.

• 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.

• 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

• 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)

• 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.

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

• 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.

• 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.

• YouTube - Arithmetic and Geometry Math Foundations 11

Wildberger proves the distributive and commutative laws using fractions while still in the natural numbers.

• 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)

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

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
• Read chapter 24 Techniques of proof IV: Induction in Kevin Houston's book

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.

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.

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.

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.

• 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

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

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

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

Geometry of space curves, osculating plane

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

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.

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

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.

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.

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.

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 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

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.

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}

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

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

• 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.

• 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