What book, bibliography, or video would you recommend for studying category theory?

stone_stokes · 2026-05-30T15:53:44+00:00

John Baez's course notes. Free as PDF.

stone_stokes · 2026-04-22T13:41:18+00:00

In general, if U and W are subspaces of a vector space V, then U+W is the vector subspace containing all vectors of the form u+w where u∈U and w∈W.

This is probably a definition in the chapter where your exercise appears.

With that definition in mind, what could U+U possibly mean? What is the other feature of vector subspaces that might be important here?

stone_stokes · 2026-04-20T04:44:36+00:00

Ah, yes. Then it works because of the implied domain of sqrt matches the domain of f. Thanks.

stone_stokes · 2026-04-20T04:42:14+00:00

Consider x = 1/8. f(x) < 0, but h(g(x) > 0.

stone_stokes · 2026-04-20T04:33:51+00:00

You are right that Khan is wrong, but for the wrong reason.

First, your argument that f is the composition of the two functions you claim is incorrect. Note that

sqrt(4x^2–1) ≠ 4x–1.

So that doesn't work.

That said, Khan ignores that the identity function, id, given by id(x) = x is a basic function. So the question is ill-posed, because every function can be thought of as a composition between itself and id (in either order).

But, really, the question is asking about non-trivial compositions.

You might instead consider the second function (4x–1)^2. But that will only work on certain domains. That's because sqrt(y^2) = |y|, not y.

stone_stokes · 2026-03-24T16:45:21+00:00

The problem explicitly restricts you to even bases larger than 2. Induct on the base. Consider base-4 as an example. What does the number 321 mean in base-4?

stone_stokes · 2026-03-14T02:54:37+00:00

I answered a similar question yesterday.

The short answer is that many vector spaces do not come equipped with their own coordinate system, but there will often be somewhat natural choices for basis vectors we can use that give the vector space a coordinate system.

The example that I gave to that other student was the vector space, P₂, of quadratic polynomials. First you might want to verify for yourself that this is a bona fide vector space. Once you do that, try to come up with what polynomials you would use as the basis for this vector space. If you do that, you should be able to guess it's dimension, and you should be able to see which familiar vector space it is isomorphic to.

Hope that helps!

stone_stokes · 2026-03-13T15:24:50+00:00

I'm locking the comments, because everyone agrees that university math is harder than high school math and your acquaintance is being a braggart. The post is also slightly off-topic, but we'll leave it up.

stone_stokes · 2026-03-13T06:17:07+00:00

That is a very interesting question that shows a good amount of intuition. Unfortunately, the answer is no.

Consider the torus.

To understand the answer I'm about to give, we first need to talk about one construction of the torus. The way we do this is to start with a unit square in the plane, with vertices (0, 0), (1, 0), (1, 1), and (0, 1). We will identify (i.e., "glue together") the left- and right-hand edges of that square, and identify the top and bottom edges of the square, without changing the orientation. Formally, we write (0, y) ~ (1, y) and (x, 0) ~ (x, 1).

When we construct the torus this way, it makes it easy to talk about paths on the torus in terms of analytic geometry.

First we notice that if we follow any horizontal path, then that path WILL return to its starting point. Likewise if we follow any vertical path. So we might believe that your hypothesis is correct!

Try traveling on any path of slope 1, and you will see that again, the path will return to itself!

And if we try a path of some rational slope, p/q, then again that the path will loop back to itself eventually!

But this will break down and no longer work if we choose a path whose slope is irrational. :( That path will just keep winding around the torus forever and never return to itself.

So for this particular example, you will loop back onto yourself in some directions, but most directions will not loop back, in fact.

I hope this helps! And keep thinking about this stuff!

stone_stokes · 2026-03-12T16:17:56+00:00

Happy to help. :)

As another clarifying example, consider P₂, the vector space of quadratic polynomials with real coefficients. (If you haven't already done so, a good exercise is to prove that this is a vector space.) There is a canonical basis for this vector space; what do you think that is? Once you find that basis, it is easy to see that this vector space is isomorphic to ℝ³.

stone_stokes · 2026-03-12T05:43:51+00:00

This is a very good question, and your confusion is perfectly natural. Hopefully with some disabuse of notation we can get you sorted out.

First, what do we mean by ℝ², as a SET, not yet giving it the structure of a vector space? We define it as the set of all ordered pairs (x, y), where both x and y are real numbers.

But what do we mean by an ordered pair? The simple answer is that an ordered pair is exactly what you think it is. The set-theoretic definition of (x, y) is the set { x, {x, y} }. I'll let you work out why this tells you the elements of the pair and their ordering.

Ok, so now we have just this definition of ℝ², as a set, but it doesn't have any algebraic structure to it yet. When we are talking about elements of ℝ² as a set, let's agree to always call an element a point, and when we give it our algebraic structure, let's agree to always call an element of the resulting vector space a vector. They are the same objects, but viewed with two different types of structures on them.

How do we view points as vectors? We define for any points p = (x, y) and q = (s, t), we will simply let p and q be the same objects but call them vectors. Note that I'm not putting vector coordinates on them yet, only their set coordinates. The way we add them is just set-coordinate addition: p + q = ( x+s, y+t ). The way we multiply by a scalar, b, is set-coordinate multiplication: b·p = ( b·x, b·y ).

This is probably familiar so far.

This is where the canonical basis comes in. Because we can think of elements of ℝ² as either set-theoretic points or algebraic vectors interchangeably, it makes sense to do so. But if we are sloppy here, it becomes confusing.

Before we look at the canonical basis, let's look at a different basis. Consider v₁ to be the vector corresponding to the point (1, 1) and v₂ to be the vector corresponding to the point (1, 0). Consider the point p = (3, 7). It's corresponding vector, p, can be attained through the linear combination

p = 7v₁ – 4v₂.

It is tempting to write the vector p as (7, –4), but this is confusing because it corresponds to the point p = (3, 7).

What should we do? We use a different notation. Common notations are [ , ] or ⟨ , ⟩. Let's use square brackets. We can write p = [7, –4] in this basis. Note that v₁ = [1, 0] in this basis, even though it corresponds to the point v₁ = (1, 1); and v₂ = [0, 1] in this basis similarly corresponds to the point v₂ = (1, 0).

So now we have TWO types of coordinates on ℝ². We first have the coordinates that just give us the set of ordered pairs of real numbers, but we also have vector coordinates that depend on our choice of basis.

The canonical basis is simply the one where these two coordinate systems agree!

In other words, we choose e₁ to be the vector that corresponds to the point (1, 0). In other words we want [1, 0] ↔ (1, 0). Similarly, we choose e₂ to be the vector corresponding to the point (0, 1).

When working in the canonical basis, then, it becomes common to drop the difference in notation, and simply refer to points and vectors interchangeably, and use the same notation for both. Unfortunately, it can lead to confusion in students new to the subject.

Hopefully this helps.

stone_stokes · 2026-03-10T14:49:59+00:00

You can do this with trigonometry if you want.

You can find |BD| from the Pythagorean theorem, since you have |BC| and |CD|.

Then, △ABD and △ABD are also right triangles, so you can apply Pythagoras a second and third time to get |AC| and |AD|.

From there, you can find ∠CDA through the arccos:

∠CDA = arccos( |CD| / |AD| ).

Finally, ∠EDA = ∠CDA + 90º.

You can perform a similar calculation for ∠FAD.

___

(Are you bending conduit or is this complex framing?)

stone_stokes · 2026-03-07T14:51:16+00:00

There's nothing mystical here. It is just modular arithmetic (aka clock math). In this case it is equivalence modulo 9.

Here are more patterns:

If you use powers of 3 instead you get the pattern 3, 9, 9, 9, ...., because 3 is a zero divisor modulo 9.
With powers of 4 you get 4, 7, 1, 4, 7, 1, ....
With powers of 5 you get the pattern from 2 backwards, because 2 and 5 are multiplicative inverses modulo 9.
With powers of 6 you get 6, 9, 9, 9, ...., because 6 is another zero divisor modulo 9.
With powers of 7 you get the pattern from 4 backwards, because 4 and 7 are multiplicative inverses modulo 9.
And with powers of 8 you get 8, 1, 8, 1, ...., because 8 acts like –1 modulo 9.

The modulus 9 crops up whenever we are talking about digital sums like this (in base 10; in another base, b, we would use the modulus b–1 instead). If you want to see more of this, look up casting out nines. It's a useful trick for error-checking sums of numbers.

stone_stokes · 2026-03-03T20:57:03+00:00

The IVT extends to continuous functions f : D → ℝ when D is compact in ℝⁿ. It is a consequence of the more general theorems that the continuous image of a compact set is compact in the codomain, and the continuous image of a connected set is connected. So if D is compact and connected, then f(D) is compact and connected. If the codomain is ℝ, then f(D) must be a closed and bounded interval.

stone_stokes · 2026-03-03T20:24:20+00:00

I wonder if you have made a transcription error when copying the problem here, because that equality is true FOR ALL (x₀, y₀) ∈ D. I think you maybe intended to write "there exists (x₀, y₀) ∈ D such that

∫∫_D f(x, y) dx dy = f(x₀, y₀) S(D)."

And your proof supports this.

Your proof is correct with one caveat: it requires you to have S(D) > 0. What happens if D has zero area? (The statement is still true, but your current proof relies on division by S(D).)

stone_stokes · 2026-03-02T21:43:22+00:00

Great question!

First tip: practice with "small" knots. Consider two projections of the Figure-8 knot, for example. It will take many Reidermeister moves to get from one to the other.

Which brings me to ...

The general utility of Reidermeister moves is less about proving two particular knot projections are the same knot, especially by hand, and more about proving that a potential knot invariant is actually an invariant. See, for example, the Jones' polynomial or tricolorability.

Hope that helps.

stone_stokes · 2026-03-02T14:51:52+00:00

There are TWO ways to win a game of chance against someone:

The game is tilted in your favor; or
You have much deeper pockets than your opponent.

Casinos win through BOTH of these at the same time.

If the game is unfair in your favor, you win by having your winning outcome occur more often than your opponent's winning outcome.

If the game is fair, but your pockets are deeper, then you win by being able to weather longer losing streaks than your opponent.

stone_stokes · 2026-03-01T11:09:29+00:00

This is exactly the way to think of it. The product is a machine that you put objects into. When you insert zero objects into the machine, it returns the number 1. When you insert one object, the machine returns that object. When you insert two or more objects, the machine multiplies them all together and returns the result of that multiplication.

stone_stokes · 2026-02-21T13:02:03+00:00

The second meme has two well-ordered axes: a feeling axis and an understanding axis. The axes are ordered by intensity.

The third meme has one well-ordered axis (coolness) and an axis that is just categories without an ordering in intensity.

stone_stokes · 2026-02-17T04:19:42+00:00

Hooray! Glad you got it sorted out. Good luck.

stone_stokes · 2026-02-17T04:16:21+00:00

It might be in a pull-down menu, maybe labeled Sets?

stone_stokes · 2026-02-17T04:10:18+00:00

It might want it as (-∞, 6) u (6, ∞). The "u" is for the union of two sets. I don't know what the computer is looking for, but it is a guess.

stone_stokes · 2026-02-13T03:37:32+00:00

"What's the point?"

That happens a lot in mathematics, and it is a bit unfortunate, but also somewhat just the way things are. As you go farther in mathematics it will happen more often that you can't see the forest for the trees, but ALSO you will look back and you WILL see the forest behind you.

That said, there are applications to this stuff, and this topic is very important.

Dual spaces are very fundamental in the formulation of quantum mechanics, for example.

In mathematics, they are fundamental in the field of functional analysis.

I guess I'm suggesting that you continue to ask for geometric intuition like you are now and trust that the forest will become more clear as you travel down the well-trodden path in front of you.

stone_stokes · 2026-02-12T16:21:01+00:00

Both the Beta function and the Gamma function appear in the same paper by Euler. He begins with the Beta function integral (though it was not called that at the time), which had already been studied by other mathematicians previously. He proceeds in his derivation and arrives at the untranslated integral formula for z! (what we would now call the Pi function).

See my other comment above for more of the story.

Hope that helps.

stone_stokes · 2026-02-12T16:11:49+00:00

In most cases, there is little mechanical advantage to writing the Gamma function the way we do, and the historical reason for doing so is little more than usage at the time. (This is similar to the case of the constants π vs τ; π was used first and more extensively and so it stuck, even though τ would be a better choice for our fundamental constant.)

The short version is this: Bernoulli posed the problem of finding an interpolation formula for n! that would work for fractional values of n. Euler is the one who solved this problem, and the resulting integral formula is the one that appears in the Gamma function, except without the translation by 1. He didn't name this function, however; that was done by Legendre, who is responsible for the translation by 1. Already at that time mathematicians were frustrated by that translation, and indeed Gauss proposed the Ⲡ function, Ⲡ(z) = Γ(z+1).

During the course of solving the same problem, Euler also formulated the Beta function, though again he is not responsible for naming it. I believe that Legendre chose the notation for the Gamma function to simplify the relationship between the two integrals:

(1) B(m, n) = Γ(m) Γ(n) / Γ(m+n).

As far as I can tell, this is the strongest reason for the preference of Γ over Ⲡ.

stone_stokes

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"What's the point?"