Activity for Tim Pederick
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #288113 |
Even the mathematical part of the question is really about a legal principle—the “balance of probabilities”—much like [another of the asker’s questions](https://math.codidact.com/posts/288101). I doubt there are “official terms” (OP’s request) in mathematics, but there may be in law.
To my underst... (more) |
— | 11 months ago |
| Comment | Post #288101 |
Agreed: this is about what facts a court can establish, not about the calculation of a probability. Mathematically, **yes**, Friend changed Ungrateful’s chance of winning the lottery from very low (the probability that *that* ticket was the winner, multiplied by the probability that Ungrateful would ... (more) |
— | 11 months ago |
| Edit | Post #288090 |
Post edited: Tweak wording. |
— | 11 months ago |
| Comment | Post #288090 |
The scenario described is a game played with dice to introduce an element of chance. If that’s not a type of game you’re familiar with, it’s not going to be helpful. (I chose it because it let me stick with the idea of “avoiding injury”, while also having well-defined probabilities.)
Rather than m... (more) |
— | 11 months ago |
| Edit | Post #288090 |
Post edited: Added a second analogy. |
— | 11 months ago |
| Comment | Post #288087 |
Of course it’s not *sufficient*. My point was only that the question lies somewhere in the intersection of maths and language. It raises some doubt (in contrast to simply saying “Not a math question”), and I chose to give OP the benefit of that doubt.
Other parallels:
① Verizon’s [conflation of... (more) |
— | 12 months ago |
| Edit | Post #288090 | Initial revision | — | 12 months ago |
| Answer | — |
A: 25% probability that there was a chance of avoiding injury $\quad$ vs. $\quad$ 25% chance of avoiding injury One is a chance of doing something. The other is a chance of having a chance of doing something. Here are some analogies that may or may not help, depending on your familiarity with the kinds of situations described. Analogy 1: A tabletop roleplaying game (RPG). Your character is sneakin... (more) |
— | 12 months ago |
| Comment | Post #288087 |
I disagree. It strikes me as akin to meta-statements like $A\to B$ versus $\left(A\to B\right)\to C$. I think it would *also* be on topic as an English usage question (no need to impute learner status to OP), but that doesn’t invalidate its presence here. (more) |
— | 12 months ago |
| Comment | Post #288081 |
Yes, it does clarify. I’m now more certain that I was reading it correctly: what’s needed here is to draw a distinction between “winning twice” (= “winning Monday’s *and* Tuesday’s draws”) and “winning a second time” (= “winning Tuesday’s draw, given that you won Monday’s”).
This is just one examp... (more) |
— | 12 months ago |
| Comment | Post #288080 |
① Yes, it is impossible, and I believe that’s Peter’s point. It is intuitively obvious that winning zero times is **not** independent of winning once. So it makes sense that winning *twice* wouldn’t be independent either, or indeed that winning $n$ times isn’t independent of winning $m$ times.
② B... (more) |
— | 12 months ago |
| Edit | Post #288082 | Initial revision | — | 12 months ago |
| Answer | — |
A: How can school children intuit why over 100, D is larger? But under 100, D% is larger? (I’m going to do this post in pounds so that I don’t have to escape dollar signs everywhere. But the currency doesn’t matter.) “Per cent” is, literally, “for each hundred”. Imagine making a literal pile of money from the cost of the item, and splitting it up into stacks of £100. So if the item cos... (more) |
— | 12 months ago |
| Edit | Post #288081 | Initial revision | — | 12 months ago |
| Answer | — |
A: How to intuit P(win the same lottery twice) $= p^{2}$ vs. P(win the same lottery twice | you won the lottery once) $= p$? I may be misreading your post, but it seems like there is some confusion between these two things: 1. Winning the lottery twice 2. Winning the lottery a second time All of the remarks about the probability being the same are relevant to the second situation. Your calculations, however, address... (more) |
— | 12 months ago |
| Edit | Post #287438 |
Post edited: Added missing word. |
— | over 1 year ago |
| Edit | Post #287438 | Initial revision | — | over 1 year ago |
| Answer | — |
A: Given two angles of a triangle, finding an angle formed by a median This is a synthetic geometry approach to the problem. It turned out more involved than I expected, and yet I haven’t found anything that could be omitted! I don’t say there isn’t a shorter way to do it, though, probably using some theorems that I haven’t thought of. (Maybe circle inversion? Seems lik... (more) |
— | over 1 year ago |
| Edit | Post #287431 | Initial revision | — | over 1 year ago |
| Question | — |
Is posting multiple answers encouraged, and under what circumstances? Users are able to post multiple answers to a question. But is this encouraged, or discouraged? Does the answer change depending on the circumstances? For the sake of a concrete example: Say there’s a geometry question that could be approached in different ways: with synthetic geometry, with trigon... (more) |
— | over 1 year ago |
| Edit | Post #286729 | Initial revision | — | almost 2 years ago |
| Answer | — |
A: Isn't "any, some, or all" redundant? Why not write just "any"? Language is ambiguous. Mathematics should not be. We can either state our own definitions at the outset, or we can use wording that (hopefully) rules out any of the ambiguous cases. “Any” may mean “at least one” ($n\ge 1$) or “exactly one” ($n=1$). “Some” generally implies both “more than one” and... (more) |
— | almost 2 years ago |
| Edit | Post #286034 |
Post edited: Fixed my mistake regarding rotations about axes: my assumption doesn’t hold in 4D! |
— | almost 2 years ago |
| Comment | Post #286034 |
“In quaternion algebra, $1$ is not identified with the vector $(1,0,0)$.” I think I’m starting to get it. Or rather, I think I *understand* it, but I still don’t *get* it. My lack of formal quaternion education is probably betraying me here. Why doesn’t the real unit 1 correspond to the 4D vector $(1... (more) |
— | almost 2 years ago |
| Comment | Post #286034 |
@#53398 “But where are you getting triplets in 2D?” From the trivial $(x,y)\mapsto (x,y,0)$.
“…modulo sign (the negation is not orthogonal)…” Ah, see, I was counting the negations, so your three I would’ve called six. But I still don’t see how we get that many! In two dimensions, there’s one (pair... (more) |
— | almost 2 years ago |
| Comment | Post #286034 |
‘I was implying orthogonal when I said it produces 3….’ You’re still beyond me here. What three roots are they? I only count two new ones, plus the two that we had in 2D: $(0,\pm 1,0)$ and $(0,0,\pm 1)$.
‘…an "axial vector" and a "position vector".’ Ah, now we’re back on ground I know—or once knew... (more) |
— | about 2 years ago |
| Comment | Post #286034 |
@#53398 Ah, I see my mistake now. I hadn’t even considered that the normal to a plane would no longer be unique in 4D. But just as there’s only one line orthogonal to a line in 2D, but infinitely many (a whole plane of them) in 3D, it makes sense that the same would happen to the normal(s) to a plane... (more) |
— | about 2 years ago |
| Comment | Post #286034 |
@#53398 And to a couple of your specific points that I didn’t have room for in the above…
“Moving to 3D immediately produces 3 square roots of -1.” Does it? I had thought it produced *infinitely many* roots lying on the circle $x=0,y^2+z^2=1$. And I worried that I’d glossed over that part too quic... (more) |
— | about 2 years ago |
| Comment | Post #286034 |
Let me start by saying that I came at this as someone who learnt complex numbers “properly”, but is entirely self-taught in quaternions out of sheer amateur interest. So I’m far from being an expert, and intended simply to share with the asker a mental model that worked for me.
With that proviso, ... (more) |
— | about 2 years ago |
| Edit | Post #286037 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Why 1. multiply the number of independent options? 2. add the number of exclusive options? First, a note: while the question is tagged “probability”, the quoted text talks about the “number of options”, which is equally applicable to counting and combinatorial problems. In what follows, I’ll talk about probability, but this too is equally applicable to counting. I don’t much like th... (more) |
— | about 2 years ago |
| Edit | Post #286034 | Initial revision | — | about 2 years ago |
| Answer | — |
A: How $ijk=\sqrt{1}$? First, there’s a couple of algebraic errors in your question: $\sqrt{a} \sqrt{b} = \sqrt{ab}$ is only true for real numbers, not complex numbers or quaternions $x^2 = y$ does not necessarily imply that $x = \sqrt{y}$ Keep in mind that the radical sign $\surd$ means the principal root. Even whe... (more) |
— | about 2 years ago |
| Suggested Edit | Post #285984 |
Suggested edit: Made equation in title match the one being asked about. Added an extra word for good grammar, but more importantly to meet the length requirement. (more) |
declined | about 2 years ago |
| Edit | Post #285757 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Missing a solution: Are A and B always equal if A-B=0 To directly answer the question in the title: yes, $a-b=0$ always implies $a=b$. (Well, okay, I won’t say there isn’t some abstract algebra, somewhere, where it doesn’t hold, and where $a-b+b=a$ isn’t necessarily true. But in our ordinary, everyday algebra, it’s true.) (more) |
— | about 2 years ago |
| Edit | Post #285731 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Out of 4 people, why does ways to choose a 2-person committee overcount by 2 the ways to divide the 4 into 2 teams of 2? The situations are different. In (a), two distinct groups of two are formed, “committee” and “not committee”. But in (b), we instead form two non-distinct “team” groups of two each. In fairness, the non-distinctness of the teams in (b) is not explicit in the question. You could certainly argue tha... (more) |
— | about 2 years ago |
| Comment | Post #285723 |
Indeed, but as I read it, the author is saying it’s obvious *from the values of $m$ and $n$*, before ever calculating that the side length exceeds $\sqrt{16}$. (more) |
— | about 2 years ago |
| Edit | Post #285723 | Initial revision | — | about 2 years ago |
| Answer | — |
A: How can you "easily see that such squares [of side length $\sqrt{13}$ and $\sqrt{18}$] will not fit into the [4 × 4] grid"? The key point here is what $m$ and $n$ represent. Why is the side length of each square $\sqrt{m^2+n^2}$? Why do they have to be less than 4? And why does the author say that swapping $m$ and $n$ gives an identical case? Because $m$ and $n$ are the legs of the right triangles that surround the squ... (more) |
— | about 2 years ago |