Common Core math was the pinnacle of their evil genius, and low-point of our complacency... It disrupted the minds of an entire generation on how even the most basic things work.

Readin', Writin', & Rithmatic

How far did they get? Let's see.

1895 8th Grade Final Exam, Kansas:

More from Maths

$Spencer Greenberg \U0001f50d$

Ten components of being compassionate to yourself:

1. Self-friendship: treat yourself at least as well as you would treat a friend who is going through the same situation

[thread]

2. Worthiness: remember that you have value (as all people do), and that your happiness and preferences matter.

3. Self-talk: notice when you are saying something mean to yourself (aloud, or in your head), and rephrase it in a gentler, kinder, more considerate way.

4. Connectedness: reflect on the fact that, whatever challenge you are facing, there are others who are also facing a similar challenge. You are not alone in your pain.

5. Change: note that everything changes with time. If you are dealing with feelings that are difficult to stand at this moment, they will subside. This too shall pass. Things are usually not as important or permanent as they seem to us in the moment that we are focused on them.

6. Self-empathy: orient towards yourself with tenderness, understanding and compassion, especially when you are suffering. Take on the perspective that you would towards a child or loved one that was hurting.

Riuji
@weakconvergence

When I was starting out as a grad student, I remember being frustrated with analysis problems where it didn't use DIRECTLY the things we spent so much time proving in class.

As an example, let's look at a standard real analysis problem:

(1/23)

The problem looks like it can be done using the dominated convergence theorem.

This is an important theorem and is used in a lot of places outside a grad real analysis class. A student would go through a lot of buildup leading to this theorem.

(2/23)

Now, let's see whether the conditions for DCT are satisfied.

First, for fixed 00 as n->infinity.

So we have a.e. convergence of the integrand in [0,1].

To the student, this builds a stronger case to use DCT to solve the problem.

(3/23)

Finding an integrable dominating function is where the problem starts.

It would seem NATURAL (at least for me) to bound it with:

p.s. it feels "natural" to me because it is reminiscent of squeeze theorem problems, math students encounter in calculus.

(4/23)

However, 1/x^n is not integrable in (0,1) and hence cannot be (further) dominated by an integrable function.

So, what I would do is since I know it converges a.e. to 1, maybe it converges uniformly to 1?

This leads me into a BAD detour.

(5/23)

Algebra Etc.
@AlgebraFact

How to mentally compute the cube root of the cube of a two digit number.

Thread (1/10)

In base 10, the last digits of the cubes of digits are distinct.

If d = 0, 1, 4, 5, 6, or 9, then d^3 ends in d.

If d = 2, 3, 7, or 8, then d^3 ends in 10-d.

(2/10)

So if you're given the cube of an integer, you can figure out the last digit of the cube root.

For example, if 50653 is the cube of an integer, it's the cube of a number ending in 7.

(3/10)

Let n = 10a + b be a two-digit number, with a and b single digit numbers.

Then 10a <= n <= 10(a+1), and so n^3 is between (10a)^3 and (10(a+1))^3, i.e. between 1000 a^3 and 1000 (a+1)^3.

(4/10)

So suppose x is the cube of a two-digit number n = 10a + b.

Chop off the last three digits of x.

Then a, the first digit of n, is the largest number whose cube is no greater than what's left of x.

(5/10)

Michael Kinyon
@ProfKinyon

Loops and their multiplication groups

A thread in 15 parts

(0/15)

Recall that a quasigroup (Q,*) is a set Q with a binary operation * such that for each a,b in Q, the equations a*x=b and y*a=b have unique solutions x,y. Groups are quasigroups and this property is usually one of the first things proved in elementary group theory.

(1/15)

Note that we don't assume associativity of *!

A loop is a quasigroup with an identity element. The story of why they are called loops is an interesting one and may even be true, but I will save it for another day. I am going to focus on loops in this thread.

(2/15)

Natural examples of nonassociative loops:
- The nonzero octonions under multiplication
- The sphere S^7 under octonion multiplication
- I have discussed other examples

Rethinking Vector Addition
or
How I Learned to Stop Worrying and Love Nonassociativity

A thread in 29 tweets

(0/28)
— Michael Kinyon (@ProfKinyon) December 1, 2020

For each x in a loop Q, define the left & right translations L_x, R_x : Q->Q by L_x(y)=xy and R_x(y)=yx. These mappings are permutations of Q. The composition L_x L_y of two left translations is not necessarily a left translation because Q is not necessarily associative.

(4/15)

Gro-Tsen
@gro_tsen

OK, I may be guilty of a DoS attack attempt on mathematicians' brains here, so lest anyone waste too much precious brain time decoding this deliberately cryptic statement, let me do it for you. •1/15

\u2200x.\u2200y.((\u2200z.((z\u2208x) \u21d2 ((\u2200t.((t\u2208x) \u21d2 (t\u2208z) \u21d2 (t\u2208y)))) \u21d2 (z\u2208y))) \u21d2 (\u2200z.((z\u2208x) \u21d2 (z\u2208y))))
— Gro-Tsen (@gro_tsen) February 12, 2021

First, as some asked, it is to be parenthesized as: “∀x.∀y.((∀z.((z∈x) ⇒ (((∀t.((t∈x) ⇒ ((t∈z) ⇒ (t∈y))))) ⇒ (z∈y)))) ⇒ (∀z.((z∈x) ⇒ (z∈y))))” (the convention is that ‘⇒’ is right-associative: “P⇒Q⇒R” means “P⇒(Q⇒R)”), but this doesn't clarify much. •2/15

Maybe we can make it a tad less abstruse by using guarded quantifiers (“∀u∈x.(…)” stands for “∀u.((u∈x)⇒(…))”): it is then “∀x.∀y.((∀z∈x.(((∀t∈x.((t∈z) ⇒ (t∈y)))) ⇒ (z∈y))) ⇒ (∀z∈x.(z∈y)))”. •3/15

Maybe a tad clearer again by writing “P(u)” for “u∈y” and leaving out the quantifier on y, viꝫ: “∀x.((∀z∈x.(((∀t∈x.((t∈z) ⇒ P(t)))) ⇒ P(z))) ⇒ (∀z∈x.P(z)))” [✯]. Now it appears as an induction principle: namely, … •4/15

… “in order to prove P(z) for all z∈x, we can assume, when proving P(z), that P(t) is already known for all t∈z∩x” (n.b.: “(∀z.(Q(z)⇒P(z)))⇒(∀z.P(z))” can be read “in order to prove P(z) for all z, we can assume Q(z) known when proving P(z)”). •5/15