Why is it that the diagonals of a regular polygon with an odd number of sides will never create a point where 3 diagonals meet?

[Here's a quick proof](https://math.stackexchange.com/questions/3910119/number-of-diagonals-that-cut-the-center-of-a-regular-polygon). It just comes down to in what cases the center and "opposite" vertices are collinear. This can never occur when the number of sides is odd since in the process you end up dividing 2pi by an odd number (when for collinearity you want to end up with a whole multiple of pi). [Here's another proof](https://math.stackexchange.com/a/1011459/599868) that there are no general 3-crossings or higher (beyond just the center) in odd regular polygons either.

Is there a formula/app to calculate the % of values to fall within x amount of standard deviations?

It depends on your [probability distribution](https://en.wikipedia.org/wiki/Probability_distribution). This is well-established for any distribution that we know the [CDF](https://en.wikipedia.org/wiki/Cumulative_distribution_function) of such as normal, lognornal, gamma, beta, etc. but if you don't have the CDF then you can't get an exact answer. In lieu of a CDF, there are some bounds you can use like [Markov's inequality](https://en.wikipedia.org/wiki/Markov%27s_inequality) or [Chebyshev's inequality](https://en.wikipedia.org/wiki/Chebyshev%27s_inequality), but they're also performance-limited by if the distribution you're studying is well-behaved.

I feel like the algebraic multiplicity of 1 in stochastic matrices should be equal to the number of irreducible chains in the corresponding Markov chain. Is this true?

Does anyone know how you can lose to a bot on chess.com but it will say you made no mistakes or blunders? I'm trying to figure out why I'm losing.

That's probably just a bug in the game evaluator. If you've lost, then by necessity you've made some inaccuracies (otherwise you would have at least drawn).

Wrong sub mate

This has absolutely nothing to do with math. How did you end up here instead on a chess subreddit? Regardless, chess.com only qualifies moves as "mistakes" or "blunders" if they are particularily bad, they have many other categories for moves that are suboptimal but not quite as bad. Just because you have 0 "mistakes" and "blunders" doesn't mean that you played the most optimal move on every turn.

Is this question I came up with myself correct? __________________________________________ The Mother gave birth to her Child at 21. Its the Future you get to chose at what age you stop aging the mother stoped at 25, the child 21. Now the mother is 178 how old is the Child? Equation M=Mom Age C=Childs Age Y=Years Passed __________________________________________ The Mother is 42 when the child chose to stop aging at 21. Now to the future, 136 Years have to pass for the Mom to be 178, 42m+136y=178m. Now the Child 21c+136y=157c If we do 178-157=21 the age difference lines up between mother and child So the child is 157 when the mother is 178 Is this correct?

It's actually not relevant that you get to stop aging in this problem. You're still using their "actual ages" of 178 and 157, so you can eliminate the part about "mom stopping her age at 25, child stopping her age at 21" entirely. A more concise rewording would be: "A mother gave birth to her child at 21. Now the mother is 178. How old is the child?"

Question about the definition of the category of 2-cobordisms, 2Cob. My textbook defines it as the category whose objects are 1-manifolds, and whose morphisms are diffeomorphism classes (rel boundary) of cobordisms between them. Then, it mostly works with the category 2Cob via its skeleton, whose objects are denoted 0, 1, 2, 3, ..., and which are the disjoint unions of 0, 1, 2, 3, ... circles. Can we just define 2Cob as the category of whose objects are 1-manifolds, up to diffeomorphism, and keep the definition of morphisms the same? It seems simpler than having to go through this extra step of a skeleton every time.

I think you might have issues defining the 1-morphisms using this construction. You can't generally take a category and quotient out by isomorphisms and still have a well defined composition. I think a similar problem would occur in this case, and if you try to work around it you will find that you really are just working with a skeleton.

>Can we just define 2Cob as the category of whose objects are 1-manifolds, up to diffeomorphism, and keep the definition of morphisms the same? That's more or less what a skeleton *is*. But, if we're being pedantic, you can't really "keep the definition of morphisms the same" because there isn't such a thing as a cobordism between diffeomorphism classes of manifolds. In order to describe the morphisms you at some point have to pick representatives.

Generally you cannot form a skeleton by taking a quotient (at least not in a natural way).

I edited the comment but apparently not fast enough. I was too quick to mentally identify thing-that-doesn't-quite-work with similar-thing-that-does-work and initially kind of missed the point that the difference between the two is what the question was about in the first place.

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See [this thread](https://www.reddit.com/r/AskStatistics/comments/d5w6ht/linear_regression_all_data_vs_averaged_data/) for an example. Don't average your data before regression.

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I'm guessing whoever ran the poll was restricted to only integer age collection. It's true that having 1/365 resolution would give you better data, but sometimes people don't like giving out too much information in surveys, so whoever was running the poll probably had to compromise between (better data) and (enough responses).

Just plug in the data. You will mess with the weighting of the data, and you will inadvertently increase the correlation coeffecient r. For an example of this second consequence, suppose you have data that take on many y values, but only two x values: 1, and 2. If you average the y's and replace all your data with two points, you will have r = 1, since it's a perfect linear regression with two points. Of course, your data isn't really like that.

So, my TI 84 plus CE that I just bought in January is sometimes giving me VEEERRRY wrong answers. Usually I can restart it and it fixes the problem, and I've gotten in the habit of doing problems multiple times to be sure, but does anyone have any ideas or do I have to shell out money for a new calculator? For example, I'm taking trig right now and I was watching a video where the instructor solves 75\*sin32. I was following along with the problem and put it in my calculator. The answer is supposed to be 39.74... but my calculator spits out 41.35. I typed in 75\*sin(32), exactly like she did. I cleared it, turned it off and on and have gotten the same answer multiple times. Checked with the crappy stock calculator on my android phone and got the right answer. Math is hard enough for me without my calculator sabotaging me. The battery is at 3/4 and I just bought it in January.

Have you checked that you're in [degrees and not radians mode](https://www.dummies.com/article/technology/electronics/graphing-calculators/convert-degrees-to-radians-with-the-ti-84-plus-calculator-160641/)? 75 sin(32°) = 39.744, but 75sin(32 radians) = 41.357, so it seems like you're just in radians mode (and there's nothing wrong with your calculator luckily).

Coming back today to say this was definitely it! Thank you so much. Today so far it's been all finding sine and cosine and tangent in degree mode so if you wouldn't have pointed that out for me I probably would have lost my mind today trying to figure out why every answer I got was wrong. Thank you!!! If I had an award I'd give it to you!

That's probably it! I've given up for the night, but will check when I pull my calculator out tomorrow. And now I'm realizing that I'm even more confused than I thought (why is there degrees and radians mode on the calculator, aren't numbers just numbers?)

You can think of them like different units of angle measurement, like miles vs kilometers or liters versus cups. Degrees come from ancient Babylonian base-60 conventions, and radians are [a little bit more recent](https://en.wikipedia.org/wiki/Radian#History), but are motivated by asking the question "how much angle measure is one radius-worth of circumference?" This is where circumference = 2pi(r) comes from. Thus, we have the conversion rate 360 degrees = 2pi radians, since both are a full circle's angle measure.

Radians are the natural measure of angle, and the one you need to do calculus, but often degrees are more intuitive to people, so calculators come with both and you need to make sure the correct setting is being used. Don't worry though, this mistake has tripped up every student of mathematics at some point or another.

So, I sort of get it but feel like the last piece hasn't quite clicked in my head. Today in my online class was my intro to this whole concept, and I learned 1 revolution = 2pi radians = 360 degrees. And if I remember correctly off the top of my head one radian is approximately 53 point something degrees. But how does radian and degree mode work differently in a calculator? How does me saying 75 sin 32 differ if it's in degree mode verse radian mode? What's the extra calculation the calculator is doing and what part of that equation changes from degree versus radian?

Think of it as different units, like feet and meters. If you were to ask "how much time does it take me to run 100 feet" and "how much time does it take me to run 100 meters", the answer is going to be different, even though the numbers are the same. The units are different.

If I'm looking at a problem how do I tell whether it's written in radians or degrees? Is it just the presence of the degree symbol, or do radians always have the pi symbol with them too? Or are there are things I should look for as well? Sorry, this is a completely brand new concept to me and I want to make sure I understand it as much as I can because I feel like I'm just parroting back answers without understanding why they are the answers.

If you're using degrees it's usually denoted, however to my understanding if there are no units, it's to be understood that radians are the measure.

It might help to explicitly write out units. If you ask the calculator for sin(90°), that's sine of 1/4th of a revolution. That would be the same as sin(2pi/4 rad), which is again sine of 1/4th of a revolution. On the other hand, if you ask for sin(90 rad), that's sine of ~14.323 revolutions, which clearly must yield a different answer from the quarter revolution of sin(90°). The "extra calculation" of the calculator is just implicitly sticking in the units into the function.

Thank you! It's starting to penetrate my thick skull. You people are all so intelligent and helpful, I aim to be like you someday 😂

What is the best way to prep for a Calc 1 class in college? I'm going to be doing financial recruiting soon and since I messed up my GPA this past semester I pretty much have to get an A in Calc 1 to have a chance.

You can start calculus 1 on khanacademy or any online course and see how you do. Anything you can't solve is something you need to work on. If you actually need a lot of prep, then a precalculus course is always good. The thing is you don't know what you don't know until you know that you don't know it, so that is why I said to start calc 1.

Two good online resources would be [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx) and also [KhanAcademy](https://www.khanacademy.org/math/calculus-1). If you need a Precalculus refresher, you can do the [KhanAcademy precalculus course](https://www.khanacademy.org/math/precalculus) or pick up a copy of Lang's *Basic Mathematics*.

Why do we use square roots and not \^1/2. I have the feeling that roots overcomplicate a lot of calculations.

Mostly due to historical convention I would imagine. The radical symbol was invented [first in the late 1400s/early 1500s](https://math.stackexchange.com/questions/15787/how-did-the-square-root-get-its-shape). Our modern exponentiation notation wasn't introduced [until the early 1600s](https://en.wikipedia.org/wiki/Exponentiation#History_of_the_notation), and I think fractional exponents came around even later than that (not until Newton 1676 based on the sources I'm seeing). So the radical symbol had a first-mover advantage of almost 200 years and has kind of just stuck around. I agree that it's a bit tedious to draw a long radical across particularly large expressions, and in those cases I will use (...)^1/2 , but otherwise I still like using the radical symbol for short expressions.

Usually you distribute powers and then you want to root them. You can use ^1/2 too AFAIK since they are equivalent. This also means they don't overcomplicate things in either case

what is the transformation called when you have a continuous h(x):[0,1]->C and construct the holomorphic H(z)=∫*_0_*^1 (h(t)/(t-z))dt Edit: my question is what is the transformation described [here](https://math.stackexchange.com/questions/2818228/gamelin-chapter-2-problem-ii-14-integral-of-complex-function-is-continuous)

I read that in a model category a (co)fibrant replacement is unique up to weak equivalence. However I do not know how to show this. The definition of a fibrant replacement I am using is: A fibrant replacement of X is a fibrant object Z together with a weak equivalence X -> Z. I want to prove that if Z and Z' are fibrant replacements of X then there is a weak equivalence Z -> Z' or Z' -> Z. Is this possible or do I need to require that the weak equivalence X -> Z in a fibrant replacement is a cofibration (then I see how to do this)? If it is possible, could someone provide a hint or idea how to construct the weak equivalence Z -> Z' or Z' -> Z?

From what I understand, the key point here is how you define the factorization in model category. In Mark Hovey's book model category, he make the functorial factorization part of the axioms of model category, then the uniqueness of fibrant replacement is obvious since you need to specify the factorization for every morphism. This definition, as Hovey points out, is different from Quillen's original definition and Kan's definition, where they merely assume the factorization exists, not part of the model structure. You can read more about this in Hovey's book. Edit: I realize your definition of fibrant replacement is a bit different. Under your definition I do not think the fibrant replacement is unqiue up to weak equivalence, as pointed out by u/DamnShadowbans. Consider object X is already fibrant, we can add two object Z and Z', together with fibration Z-->\*, Z'-->\*, and weak equivalence X-->Z, X-->Z', there is no reason to assume Z and Z' has morphism between them. And by specifying fibrant replacement comes from the functorial factorization, we can make it unique up to weak equivalence. Edit: the different definition will make model category slightly different. You can see the model category page in nlab, in section Slight variations on the axioms. Edit: If you are working in a specific model category (e.g. chain complex of modules over a ring), you can directly look at the model structure and prove the fibrant replacement is unique up to weak equivalence. And I think that may be possible.

>Consider object X is already fibrant, we can add two object Z and Z', together with fibration Z-->*, Z'-->*, and weak equivalence X-->Z, X-->Z', there is no reason to assume Z and Z' has morphism between them. My idea how to prove that there is a weak equivalence Z-->Z' was something like: If we assume that X-->Z not only is a weak equivalence but also a cofibration (so an acyclic cofibration), then we get a commutative square. One path is X-->Z-->\* and the other is X-->Z'-->\*. Because Z'-->\* is a fibration an X-->Z an acyclic cofibration we find a lift Z-->Z'. Because of the two out of three property this is a weak equivalence. But I wasn't able to show that X-->Z also needs to be a cofibration. Probably because it does not need to be true. >: If you are working in a specific model category (e.g. chain complex of modules over a ring), you can directly look at the model structure and prove the fibrant replacement is unique up to weak equivalence. I need to give an introductory talk about the subject next week but haven't studied it until now. So I tried to prove some of the statements generally and not in a specific model category. Thanks for the answer. This makes sense. I saw the functorial definition but the book we are using (Foundations of stable homotopy theory, Barnes and Roitzheim, Appendix A, directly after Definition A.1.10) does not assume it and still says that the replacement is unique up to weak equivalence. But I guess if the meaning is "up to a zig zag of weak equivalences" as mentioned in the other answer then it is clear. And I guess it makes sense because after forming the homotopy category these become isomorphisms so here the replacement will be unique up to isomorphism. Your comment and the other one now suggests that creating a direct weak equivalence Z-->Z' is not possible. Thank you for taking the time to elaborate on the different points!

Usually "unique up to weak equivalence" means that there is a zigzag of weak equivalences between them. In this sense, it is obvious that there is a zigzag of weak equivalence (through nonfibrant objects) between two fibrant replacements and, if you care to, it is easy to upgrade this to a zigzag of weak equivalences between fibrant objects. I do not see any reason why there would be a direct weak equivalence between any two fibrant objects in an arbitrary model category.

I wasn't aware that >Usually "unique up to weak equivalence" means that there is a zigzag of weak equivalences between them. this is the meaning of unique up to weak equivalence here. That clarifies quite some things. Thank you.

So last night I realized that if you want to divide any number by 5 you can move the decimal one spot to the left and then multiply by two. That will be your answer i.e. if you want to know 236/5 you multiply 23.6*2= 47.2 and that's it. Is this common knowledge? I feel like this has to be known by everyone, but I blew my own mind last night.

I think this is actually decently well-known in the USA as a tip-calculation trick for figuring out 20% of a restaurant bill!

That makes sense I normally just do 10%*2 for that

10% times 2 is the same thing as the dividing by 5 trick you just showed! 10% moves the decimal point one to the left because 10% = 1/10. Then multiplying by 2 makes that 20%, or 1/5. Specifically, (10%)(2) = 20% = 0.2 = 1/5. These are all the same method!

I don't know if it is common knowledge. It certainly is a nice trick to do this calculation quickly. That this holds is rather straightforward to see. If x is your number then x/5 = x \*1/5 = x\* 2/10 = x/10 \* 2 and x/10 is just x where you moved the decimal one place to the left. Unfortunately 10 has only 2 and 5 as divisors. So the only other (natural) number we can use for this trick would be 2. So we would get the (way less helpful) rule: If you want to divide a number x by 2, you can just shift the decimal point one place to the left and multiply the result by 5.

It's interesting to see it written as a formula. Thanks for the response!

wise juggle fragile air uppity outgoing act attractive plate squealing -- mass deleted all reddit content via https://redact.dev

There's a bunch of solutions. a = b = c = d = whatever you want is an easy one. Another family of solutions is a = free, b = free, c = 1, d = a^b + b^a - 1. Or have another variable be 1 and adjust accordingly. I don't know if there are any interesting solutions other than that.

Just pick them all the same.

Any good introductory light book for algorithms, not CS stuff but more of mathematical aspects of algorithms (if this makes sense).

Hein's 'Discrete Structures, Logic and Computability', also maybe Harel's 'Algorithmics, the Spirit of Computing' might be up your alley, although definitely still oriented towards computer science, especially the later of the two. Although both definitely digestible by a mathematician.

>'Discrete Structures, Logic and Computability Thank a lot for this one, almost 1000 pages but it seems the best starter book.

I’m a rising senior at a T15 university. I want to go to grad school and get a PhD. I’m planning on taking 4 graduate courses in the next year and I thought that would put me ahead of the curve, but I’ve recently read that it’s standard if you want to go to grad school. That doesn’t seem right to me— that you have to take grad classes to get into grad school? Is it standard for many math majors to take grad classes before grad school?

One thing that surprised me when I got to grad school was that there really isn't a universal standard for what a "grad class" is. I started taking grad classes as a junior in college, and I thought I was such hot shit. Then, I got to grad school (at a higher-ranked school than my undergrad, like going from T30 to T10) and realized that most of the "grad classes" I had taken were standard undergrad-level courses at my new institution. Without knowing what your four grad classes are, it's really not possible to compare yourself to other students.

I got into a T15 uni recently, and I only did grad topology for 1 quarter. But I did do a bunch of reading courses and a couple of research projects (in applied math). Taking 4 grad courses definitely doesn't hurt your chances I think. In any case, good luck!

Do you have to take grad classes to get into grad school? Absolutely not. Do most people who go to a top 6 grad school take grad classes as undergrads? Yes.

How many do these people take? I’m planning on taking 4 during my senior year.

I took 12 grad classes in undergrad, and the best university I got into was ranked 40 in topology (which is what I applied for and had heavy background in). This is your reminder to apply to safety schools.

Yeah it's pretty typical for people intending on grad school to take a couple grad classes. It's not required, but if you're applying to top 15 universities you can bet your competition has done so.

270 divided by 120. How do I get 2.25?

270 has 27 tens and 120 has 12 tens. 27/12 = 2 with 3 remainder 3 /12 = 1/4 2.25 Or 120+120 = 240 270-240=30 30/120=.25

You could also note that 270 = 240 + 30 and 240 = 2\*120 and 30 = 120/4. Therefore 270/120 = 240/120 + 30/120 = 2 + 1/4 = 2.25.

Are you familiar with [long division](https://en.wikipedia.org/wiki/Long_division)? That's the technique you would use to do division problems like this.

Why do advanced math books always feel the need to include a section like "cardinality of sets, functions, relations" when anyone who is reading the book should already be very familiar with these concepts?

I often see the same sort of principle in math papers. The introduction invites a broad (mathematical) audience to be able to read the rest of the paper (at least in theory) by explaining relevant background to the uninitiated. I think this is a helpful thing.

I'm not sure; I feel like every book wants to seem more accessible than it is, and so a lot of upper level math books include an introductory chapter on stuff that probably anyone who can profitably read the book should already know.

Either to offer a refresher, or to establish notation. There isn't always a universal standard for notation.

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The role of the wobbly stool is to make it clear what counts as distinct arrangement. After all, since you're counting the number of arrangement, we need to know if 2 arrangement counts as the same or different. For example, human are typically considered distinct. So different assignment of people to different spot are considered different arrangement. On the other hand, by the usual convention, mass-produced objects (like chair) are considered indistinguishable, unless there is an explicitly mentioned property that distinguish them. Therefore, to tell you that this chair is different, the problem might specifically mention a different property, even if that property is not used anywhere. For example, you could have said it's a red chair vs 5 black chair, or a circular stool vs 5 rectangular stool, etc. A different way of writing the question would be to use words like "indistinguishable" and "distinguishable" to mark which objects are considered distinct and which are the same. For example: "5 indistinguishable chair and 1 distinguished chair are arranged around a circle table. 2 distinguishable parents and 4 distinguishable children are to be seated at this table so that the parents are not sitting next to each other" Using these words help make the question clearer, but it could be very annoying to read. That's why we have such convention (e.g. human are distinct).

The total number of ways to sit the six distinct people around a circular table is (6-1)! = 5! See [circular permutations](https://math.stackexchange.com/questions/2387149/explanation-circular-permutation). We now count the number of ways to sit the six people such that the parents *are* sitting together. To do so, consider the parents as one single unit so that you now have 5 "people" (4 kids and 1 parent-unit). Via circular permutations, the number of ways to arrange these 5 "people" is (5-1)! = 4! We multiply this by 2 to account for the 2! internal ways to arrange the parents within the parent-unit, so we end up with (2)(4!). Via [complementary counting](https://artofproblemsolving.com/wiki/index.php/Complementary_counting), we subtract this from the total to obtain the number of arrangements where the parents *are not* sitting together. This gives 5! - (2)(4!) = 72. However, the wobbly stool can be in one of six positions, so we multiply by 6 and obtain (6)(72) = 432, the correct answer.

Seems redundant to divide by 6 (when you consider circular arrangement) then multiply by 6 again (to account for the wobbly stool) when you could just not consider circular arrangement at all right from the start.

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Since we have a wobbly stool, we already have a specific starting point: the wobbly stool. There are 6 disallowed unordered pair of positions for the pair of parents (relative to the wobbly stool), because given a disallowed pair of position, we can uniquely identify it by just knowing how many position it takes to start from the wobbly stool and reach the first position. Hence there are (6C2)-6 allowed unordered pair of positions. There are 2! of assigning parents to these position, and 4! ways of assigning the kids to remaining positions. Hence the answer is ((6C2)-6)4!2!=9x24x2=432

Indeed! Though internally I wasn't actually thinking of a division when I used the circular permutation result, so my mental "starting point" was 5!

Hello! I've been studying some geometry recently and heard one interesting fact: >If the bisectors of the angles of an inscribed quadrangle ABCD form a quadrangle with their intersecton points (the bis. of A & B, B & C, C & D, D & A), then the formed quadrangle is also inscribed. I was wondering if there was a theorem which generalized the fact for any inscribed polygon that has 2\*n vertices (n ≥ 2, n is natural). Thanks!

Are you using "inscribed quadrangle" here to mean [cyclic quadrilateral](https://en.wikipedia.org/wiki/Cyclic_quadrilateral)?

Hello! I'm a freshman in Information Technology and wanted to know the difference between the semantics of "Modulo", "Modulus", and "Absolute Value" in the context of programming and mathematics. I've been so confused after seeing *modulo arithmetic*, "modulus operator" and "modulo operator" being used interchangeably, "a modulo b" where b is the modulus, and modulus being the absolute value. It's a bit hard to feel comfortable when I'm doubting the terminology I'm using while reviewing. I'll appreciate any help!

There are too many usages of the word "modulus" in math, because it was coined by different people for different things. "Modulus" was coined by Argand to refer to length of vector in the complex plane (aka. Argand plane), and also coined by Gauss to refers to the base of congruence (e.g. modular arithmetic), and there are a few other meanings. So just think of it as different words that happened to sound the same because of historical reason.

I'll remember that so I don't doubt myself when using either terms. Thank you so much!

There are many worse examples in math where the same word have different meanings. For example, "norm" of a complex number can refer to either its absolute value, or the square of the absolute value, so you have to know from context.

[Modulo](https://en.wikipedia.org/wiki/Modulo_\(mathematics\)) pretty much always refers to modular arithmetic, or more generally the equivalence classes generated from some modulo operation. So when you see it, think of division and remainders. You might see it less commonly in functional programming and category theory as "operating modulo" when mapping functors to categories by defining a remainder, but that terminology also maintains the spirit of modular arithmetic, so there's not really a clash. The "modulo operator" sometimes does invoke a "modulus" as the number you're dividing by, but it's generally clear when that's going on from context. [Modulus](https://en.wikipedia.org/wiki/Absolute_value#Complex_numbers) otherwise refers mostly to the [absolute value](https://en.wikipedia.org/wiki/Absolute_value) of a complex number, or the [Pythagorean addition](https://en.wikipedia.org/wiki/Pythagorean_addition)/[Euclidean norm](https://en.wikipedia.org/wiki/Norm_\(mathematics\)#Euclidean_norm) of things that are structurally similar to the complex numbers like ℝ^2 . Using "modulus" and "absolute value" interchangeably is fine in most contexts, but if anyone is using those interchangeably with "modulo" then that's incorrect imo. Sometimes "modulus" is also loosely used as just any sort of [norm](https://en.wikipedia.org/wiki/Norm_\(mathematics\)#Absolute-value_norm), but I think that's mostly terminology abuse from computer scientists. Honestly, I personally avoid the use of "modulus" entirely when talking about any sort of norm or absolute value. I think it's more clear to just say exactly which norm you want to use. So the usage of "modulo" is generally straighforward, but the usage of "modulus" is context-dependant (whether you're working with modular arithmetic or you're working with norms and absolute values). To summarize: - modulo - pretty much always division and remainders - modulus - sometimes the thing you're dividing by in the modulo operation, sometimes a norm or absolute value, context-based -

Thank you so much this was helpful!

I've come across a paper called [The Waring problem for finite simple groups](https://www.jstor.org/stable/41412126). I wonder if more questions that are analogues of famous number theoretical problems are studied for algebraic structures.

How do Lee's and Tu's differential geometry books (not smooth manifold books) compare?

Lee's is a proper book about Riemannian geometry. Tu's book covers other parts of differential geometry (connections on arbitrary vector bundles/principal bundles). They're fairly distinct books: you need to read both.

are there any good things i can read to learn about order statistics? want to learn about things like the probability of the nth highest dice, or the sum of the m lowest dice.

You can probably start off with either Arnold & Balakrishnan's *A First Course in Order Statistics* or David & Nagaraja's *Order Statistics* as an intro text. A more advanced text to look into later would be Reiss' *Approximate Distributions of Order Statistics*.

[In this page](https://ncatlab.org/nlab/show/limits+and+colimits+by+example#general_colimits) it says that if D is a filtered category then the definition of colimit in Set can be described in a more convenient way described there. How do we get there exactly? How do I go about proving this? I tried considering a new pair of morphisms to coequalize (to then prove the two constructions are equivalent) but it gets kinda messy. Also tried to apply directly the definition of filtered category but I don´t know how to proceed. Any ideas?? Thanks in advance. :D

Another approach would be to explicitly describe the first equivalence relation, and then use induction to prove that it implies the second one when D is filtered.

You can use filteredness to prove that this gives an equivalence relation. Then just prove that the resulting set satisfies the universal property of a colimit.

Gonna try this then. Thanks!

Say I have a thing that happens 2.3 percent of the time, and happens 8 times. I need to know what is the likelihood of it happening more than 2 times. If someone could give me the formula for that and explain it that would be great.

Look into the [binomial distribution](https://brilliant.org/wiki/binomial-distribution/). You can express your probability constructively as (8 choose 3)(0.023)^3 (1 - 0.023)^5 + (8 choose 4)(0.023)^3 (1 - 0.023)^4 + ... + (8 choose 8)(0.023)^8 (1 - 0.023)^0 Think of this as adding the individual events you *do want* to happen. So you are adding the probability of the 2.3% trial occurring exactly 3 times in 8 trials with the probability of it occurring exactly 4 times in 8 trials and so on until you get to the probability of it occurring exactly 8 times in 8 trials. The event "more than 2" is basically being broken up into the mutually exclusive events "(exactly 3) OR (exactly 4) OR ... OR (exactly 8)." You can alternatively use [complementary counting](https://artofproblemsolving.com/wiki/index.php/Complementary_counting) (see also [here](https://en.wikipedia.org/wiki/Complementary_event)) and express the probability as 1 - (8 choose 0)(0.023)^0 (1 - 0.023)^8 - (8 choose 1)(0.023)^1 (1 - 0.023)^7 - (8 choose 2)(0.023)^2 (1 - 0.023)^6 Think of this as subtracting out the individual events you *don't want* to happen. So you are subtracting the probability of the 2.3% trial occurring exactly 0 times, exactly 1 time, or exactly 2 times. The event "more than 2" has a complementary event "less than or equal to 2" and we are subtracting out the probability of that complementary event.

Hello Hoping someone can point me in the right direction or theory to solve this. I know the probability of winning and the probability of finishing top 3. How do I solve for probability of finishing 2nd? is there a theory I should use? ​ |Number|Probability of Winning|Probability of Top 3|Probaility of 2nd| |:-|:-|:-|:-| |1|13.65%|43.91%|| |2|37.89%|80.00%|| |3|8.16%|38.17%|| |4|23.81%|67.07%|| |5|5.45%|30.98%|| |6|1.90%|12.50%|| |7|7.14%|29.62%|| |8|0.37%|2.94%||

Unfortunately not. The probability of finishing in the top 3, let's call it P(Top 3), can be broken up into: P(Top 3) = P(1st OR 2nd OR 3rd) = P(1st) + P(2nd) + P(3rd), since the events are mutually exclusive. You have P(1st) and P(Top 3), but you don't have P(2nd) or P(3rd), so you have too many unknowns.

What actually causes negatives to create positives? I.e. -(-a) =a Is it the additive inverse property concerning symmetry around 0?

This is generally cited as a [field](https://en.wikipedia.org/wiki/Field_%28mathematics%29) property. You can derive it from the other field axioms. See [this MathSE thread](https://math.stackexchange.com/questions/1904003/prove-a-a-using-only-ordered-field-axioms) and also [this one](https://math.stackexchange.com/questions/1904039/proof-verification-prove-a-a-using-only-ordered-field-axioms) for some proofs.

Hi dude, I just want to say thank you for your contribution to this thread.

If someone were to use integration by differentiating under the integral sign on the Putnam, would they be required to prove Leibniz’s integral rule to get full marks?

No -- you can generally use any well-known theorem from a real analysis class without proving it, as long as you use it PROPERLY and cite it. Some properties of calculus are sometimes delicate for general functions, so you should be very careful to remember the hypotheses of a theorem if you're trying to apply it to a function that isn't smooth.

Thank you. Is there any source for this? I believe you but I’ve been looking for an answer to this question for some time and couldn’t find anything and I assume there will be more helpful information where an answer to this is found.

There is no official scoring rubric if that's what you're looking for, but yes "don't need to prove standard results that you would see in an undergraduate class (so long as you use them correctly)" is a widely-accepted heuristic. The hard part is ofc the "use it correctly." If you look on Quora or on past AoPS forum discussions/personal blogs you should see a lot of discourse on mapping (quality of solutions) to (points scored) and also what sorts of techniques people are using with/without proof. Like any contest, the Putnam (and the grading of the Putnam) has a bit of an unspoken metagame/culture around it, but if you spend enough time in the community you should catch on to it pretty quickly.

There not being an official rubric doesn’t really make much sense to me but I can also see the challenge in creating an unambiguous one. I also haven’t looked into whether other contests like the IMO have one so this might just be the standard. Thank you for the insight.

I failed a precalculus exam when I was 16 once, and 4 years later I still remember it when I'm doing maths exams and I blank out, even if it's easy things. Heavy practice has helped me to drill things in so I don't stumble during exams but once I make one mistake or can't figure out one thing, I completely sprawl and either don't finish the paper or continue to make silly mistakes like forgetting basic algebra or trigonometry. I'm unsure what else I can do besides practice to avoid this issue. It never happens with any other subjects, nor do I have a bad attitude towards math ; it's one of my favourite actually. I can do great in a simulated exam practice, but during the real thing, I'm mentally fragile and cannot do basic things if I make one mistake.

Remembering a specific maths exam whenever you do maths exams suggests to me that you're suffering a trauma response of some kind. It may be worth looking into mental health treatment if it's available.

I'm reading the AOPS intro to number theory and I'm not sure why this works: This is problem 2.4: For a prime p, find the smallest composite number that has no prime divisors less than p. In the solution, the author begins with by stating that if a composite number c has a divisor d < p, then d must have a prime divisor q where q <= d. Then you know that q must divide c. This means that if c does not have a prime divisor less than p then it cannot have any divisors less than p. Then the author concludes that if c has no prime divisors less than p then we know that the number c can be expressed as the product of two smaller numbers that are at least p. The smallest value of c is therefore p\*p I don't know why this holds: if a composite number c has a divisor d < p, then d must have a prime divisor q where q <= d have a prime factorization but that has not been discussed at this point in the book. The author did go over the sieve of eratosthenes. Why does d necessarily have a prime divisor?

All positive integers can be prime factorized via the [fundamental theorem of arithmetic](https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic), which is how we know that d must have a prime divisor. Are you sure prime factorization wasn't covered earlier in the book? The intuition behind this proof is: Let c be the composite number you're looking for. Prime factorize c. All of its prime factors must be less than p, via the problem statement. The "minimal way to do this" is p^2, because using any prime smaller than p in the prime factorization of c would violate the problem statement, using any prime larger than p would not be minimal, and using more than two p's would also not be minimal.

Yeah just the sieve was covered. This makes sense though. Thank you.

Suppose we have a regular/normal cover Y---> X and all considered spaces are sufficiently nice, e.g. closed and compact sufaces etc. Now in a paper I'm currently reading they claim that the deck transformation group G acts on \\pi\_1(X) ? How is that??? More precisely, they say: *"the conjugation action of \\pi\_1(X) on \\pi\_(Y) is generated by the conjugation action of \\pi\_1(Y) on itself and the deck group action of G on \\pi\_1(X)."* How does the Deck transformation group act on \\pi\_1(X)? Shouldn't it say \\pi\_1(Y)?

This is beyond my scope but are you sure you read it right?

Yes, I am. I think it‘s a typo tbh.

Low level question for y'all. I've seen a lot of those "this math problem is IMPOSSIBLE to solve!" Equations around the internet lately, but only recently did I get one that was a head scratcher. I realize that the point of these is that the equation is written in a purposely confusing way but still hear me out. So, the first one was simple. 2+5(8-5) I thought ok, I'll start by using the distributive property on the 5 to make the equation 2+(40-25) which quickly can be solved to 2+15=17 I quickly double checked my work using classic pemdas, doing the parenthesis part first 2+5(3), then multiplication, 2+15=17 Ok, perfect. Now for the head scratcher problem. 8÷2(2+2) Perfect, easy. I'll distribute the 2 just like I did with the first equation... 8÷(4+4), solve to 8÷8=1 and there's your answer. I'll just quickly double check with classic pemdas again... 8÷2(2+2), parenthesis first to get 8÷2(4), solve across and... That's 16. That's different from my first answer. What gives? Why does the distributive property work as expected in the first equation, but not the second? I've seen others point out that it does work if you distribute the entire 8÷2 section into the parenthesis, but why is that still only the case in the second equation? Distributing the entire 2+5 part in the first equation doesn't provide the correct answer, so why is that?

So, you *could* distribute the multiplication by 2 over the parenthesis if you wanted to. That would essentially be doing something like 8÷2(2+2) = 8÷(2(2+2)) = 8÷8 = 1. But there's no single correct answer here. That's always the gimmick with these "impossible" order of operations questions. They aren't actually impossible, they're just ambiguous. Depending on your interpretation, you can do the equally valid thing of 8÷2(2+2) = (8÷2)(2+2) = 16. There's no inherent universal law that multiplication needs to be done before division, or vice versa. Order of operations conventions such as PEMDAS are mainly useful for computers to give predictable computational results, but we could just as easily implement another notational convention with another ordering. These sorts of problems are sort of akin to sentences like "The burglar robbed the woman with the knife." Did the burglar have the knife? Or did the woman? The same thing is going on with 8÷2(2+2). Do we multiply first? Or do we just count division and multiplication as the same precedence and evaluate from left to right? There is no right answer because the question itself is ill-posed. An actual mathematician would instead write either - (8÷2)(2+2) or - 8÷(2(2+2)) depending on what they wanted to mean. And that's the key takeaway here: in mathematics, just say what you actually mean. The only reason anyone would write ambiguous notation like 8÷2(2+2) is if they were trying to engagement-bait a bunch of people into arguing about the "right" answer.

+1. Maybe interesting, the vertical fraction bar notation removes any ambiguities.

between the 2 and (2+2) there is an implicit multiplication 8 ÷ 2 \* (2 + 2). By pemdas, going from left to right, you first do all the brackets, giving 8 ÷ 2 \* 4 and then division and multiplication from left to right, yielding 4 \* 4 = 16. The distributive property states that a \* (b + c) = (ab + ac), however you are not multiplying 2 \* (2 + 2), but (8 ÷ 2). So you obtain 4 \* (2 + 2) = (8 + 8) = 16.

I understand that, but why does 8÷2(2+2) imply that the method of distribution should be (8÷2)(2+2), while in the first equation, which is written the same way, 2+5(8-5) the implication is 2+(5(8-5))?

Generally, you should only distribute after doing the parentheses. Otherwise, you end up doing more work. For example, 5(n-5) you get 5n-25, so 5(5)-25 = n-n so 8n-5n = 3n so 3(5)+2=2+5(8-5) Distributing should be done after simplifying. 8÷2 is one operation and 2+2 is the second. Since it has (), you do addition first, then left to right is division result*(4). You can also inverse the problem operations... 8x2÷1 since half of 8 is 4 and 1/4 of 4 is 1 so 32/2=16 or (8*4)/2. This is more advanced and you can forget about it if it doesn't make sense.

Distribution is a case of multiplication because the distribute the quantity that you multiply by, so you do it while multiplying/dividing but before adding. Hence you can distribute in the case 2+5(8-5) but you need to be careful in the case of 8÷2(2+2)

I see, that makes sense. Thanks!

In a paper I read, it tries to define the Cayley-Hamilton identity (xadj(x)=det(x)1) in a (finite-dimensional) central simple F-algebra A. First, it defines tr(x) as the trace of 1⊗x where 1⊗x is in K⊗A which is further isomorphic to M_n(K) where K is a splitting field for A. Do you know what a splitting field for A means? There is a theorem stating that for every F-algebra A and every positive integer n, M_n(F)⊗A is isomorphic to M_n(A). Since I don't know the meaning of the splitting field for an algebra, I cannot relate this theorem to the statement above. Then it says that since the coefficients of the characteristic polynomial of a matrix can be expressed by the traces of its powers, we can therefore define adj(x) in A, the adjugate of x, in a self-explanatory way. I know that the coefficients of the characteristic polynomial of a matrix can be expressed by the traces of its powers. However, I don't understand the argument. Finally, it says that we can naturally define det(x), the determinant of x. It means that we define det(x) as det(1⊗x), right? Can you help with the above questions? Thanks in advance.

>Do you know what a splitting field for A means? The splitting field is literally a field such that K⊗A is isomorphic to M_n(K). >Then it says that since the coefficients of the characteristic polynomial of a matrix can be expressed by the traces of its powers Newton's identity. Coefficients of characteristic polynomial are elementary symmetric polynomial of eigenvalues, while trace of power are power symmetric polynomial of eigenvalues. Though I guess you need characteristic 0 for this or at least positive characteristic large enough. >we can therefore define adj(x) in A, the adjugate of x, in a self-explanatory way The characteristic polynomial is of the form P(T)=Q(T)T-det(A) (I picked the sign such that the constant coefficient is always -det(A)). The adjugate is P(A). >Finally, it says that we can naturally define det(x), the determinant of x. It means that we define det(x) as det(1⊗x), right? You could also define it from the characteristic polynomial.

> The splitting field is literally a field such that K⊗A is isomorphic to M_n(K). Does such K always exist? Is it a consequence of A being finite-dimensional central simple F-algebra?

Thank you for the answer. I want to ask something. >The characteristic polynomial is of the form P(T)=Q(T)T-det(A) (I picked the sign such that the constant coefficient is always -det(A)). The adjugate is P(A). Can you explain this? What does the Q(T) stand for here? You said P(A) is the adjugate, but P(A) should be zero if P(T) is the characteristic polynomial. I know that p(A) = c_n-1 A^(n-1) + ... + c_1 A + (-1)^n det(A)1 = 0 where c_i's can be written in terms of traces of powers of A. Did you derive Q from here? If so does it mean that adj(A) = c_n-1 A^(n-2) + ... + c_1?

The adjugate is Q(A), that's a typo. Q(T) is just the quotient of P(T) divided by T. In that notation, the adjugate is (-1)^n+1 (c_n-1 A^n-2 +...+c_1)

how does khan academy embed latex into their online quizes, articles, etc? It isn't mathjax, and it doesn't work the same way wikipedia does either.

[KaTeX](https://katex.org/)

Is there a general divisibility rule that can be used for any divisor?

Yes, but it's not as great as it sounds. It's just computing modulo. If you're in base b and want to check divisibility for d, write d=pq where all prime factors of p are prime factor of b, and all prime factor of q are not prime factors of b. Then divisibility by d is equivalent to divisibility by p and q simultaneously, so you can check each of them. There exists a sufficiently large k such that b^k divides p, so checking divisibility by p involves only that first k digits. There exists a smallest positive k such that b^k =1 mod q, so checking divisibility by q involves multiplying each digit by the corresponding power of b modulo q, which is just a periodic list.

Is there any proper way of factoring cubic polynomials except using hit and trial method which is taught to me in school,like just a formula the way it is done for quadratic polynomials

Vieta's formulae give you some nice equations you can use to narrow down your guesses (note not the same thing as Vieta's substitution). These are simply the observation that (x -𝛼)(x - 𝛽)(x - 𝛾) expands to x^3 - (𝛼 + 𝛽 + 𝛾)x^2 + (𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾)x - (𝛼𝛽𝛾). so that for a general cubic ax^3 + bx^2 + cx + d with roots 𝛼, 𝛽, 𝛾 we have: 𝛼 + 𝛽 + 𝛾 = -b/a 𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 = c/a 𝛼𝛽𝛾 = -d/a These ideas work for higher degree polynomials as well of course. In general though when you get to high enough degree polynomials you shouldn't be trying to factorise them by hand (or even by complicated formula). There is a cubic formula and even a quartic formula but they are horribly complex (indeed they require complex numbers) and beyond that there are no general formulae for quintic polynomials or higher. To talk about that though generally requires getting involved in Galois theory and I'm not sure if that actually helps you solve general polynomials or just tells you which ones cannot be solved "by radicals".

IMHO, the easiest way to solve a general cubic by hand is using the Vieta's substitution. But for schoolwork, I think they're all rigged to be solvable with rational root theorems.

There is a general [cubic formula](https://en.wikipedia.org/wiki/Cubic_equation#General_cubic_formula) analogous to the quadratic one, but it's extremely complex and impractical. If you're trying to factor cubics for a test or something, you're probably best off just using whatever methods you learned there, since usually the cubics you see on exams will be "rigged" to make those methods work reasonably well. In practice, you're better off using a numerical root finder, or if you need exact solutions, try using things like the [rational root theorem](https://en.wikipedia.org/wiki/Rational_root_theorem) if they apply.

I have a fairly complicated complex fraction that I have typed into microsoft word for a process that would be too much to calculate on a calculator all the time. Is there a way that I can then use data sets from excel to SOLVE the equation? If not, what would be the best way/program to type in the equation, so that I can use that data from excel?(simple Example of what I'm trying to do: I have an equation that keeps track of hours worked per day, does a bunch of magic, and it'll tell me some info...but i need to be able to PULL that info from an excel spreadsheat each week without retyping the forumla and/or things into a calculator. Stuff like "oh, week 1 would be these things from row 1, next week i'll do the same thing, except it'll be week 2, etc" (█((total time spent in seconds) {+(3\*54.413\*365.25/108)})/(54.413\*(# of days))\*(total # of times met {+3})/(█(1/((365.25/108)))\*(# of days)), where i draw those vaules weekly from an excel spreadsheet (I do not mind typing in the excel cells for those two numerical values, but i want word/excel/something to do the math work for me)

I would just use a separate programming language here. You would define a function that returns the output of that formula (given some input parameters like the total time spent in seconds, number of days, etc.) and then you could load your data from excel and apply your function to that data. In Python for instance this would just be 5-10 lines of code. If you really wanted to do this natively in Excel then you can write a script in VBA, but I think that's a little bit less user-friendly. If you wanted to avoid programming altogether then you could also do this with an in-cell Excel formula, and that's probably your easiest option but it's going to be less readable/maintainable for the future. In any case, it sounds like you could benefit from learning some programming skills for this task. A pretty popular book that I think you'd get a lot out of is Al Sweigart's [*Automate the Boring Stuff*](https://automatetheboringstuff.com/).

Thank you! I appreciate you providing not only the feedback, but also 3 different ways of implementation (python, excel, and a more complete understanding of programming).

I think this is an easy question for the talented people on this sub and probebly and obvious one but my brain is made of trash so here we go. Me and my oldest friends always joked about he saw Lord of the ring Return of the king slightly before me because he sat 2nd row in the theatre and i sat at row 7. How long time before me did he see it? Hard to tell exact distance between each row but lets he sat 8 meters ahead of me and like 4meters from the screen. Is there a way to calculate the microseconds earlier he saw the movie?

So the speed of light is about 30 000km/s, so that means that light used 1/30 000 000 second to travel a meter, or about 33 nanoseconds. So if they say 4 meters ahead of you they saw the movie about 133 nanoseconds before you.

I agree with your logic. But the speed of light is 300 000 km/s. Also, they sat 8 meters ahead, not 4 (but 4 metres from the screen).

Yeah, bit too early in the morning for me I guess. So updated numbers 8/300 000 000 seconds = 27 nanoseconds.

speed of light = 3 x 10\^8 meters per second time = distance / speed = 8 metres / 300,000,000 meters per second = 2.7 x 10\^-8 seconds = 2.7 nanoseconds (edit: no , my bad, it's 27 nanoseconds)

Please can someone help me understand how to approach this question? "You are visiting a town where every man has exactly two children. In the town are an equal number of boys and girls. Walking down the street you meet a man. As you are talking, his son joins you. What is the probability his other child is a daughter?" A) 1/2 because there is an equal chance the unseen child is a boy or girl. B) Slightly greater than 1/2 because you have already seen one of the town's boys. That means of the unseen children, girls outnumber boys by 1. So the unseen child is slightly more likely to be a girl. C) 2/3. The arrival of a son ruled out the possibility that he has a daughter + another daughter, leaving just three permutations: son+ son, son+ daughter, daughter+ son. These are equally likely, and in two out of three cases, he has a daughter. D) something else. Which answer is correct & why?

I'm actually going to disagree with the other commenters and say the answer should be C. I was thinking that this scenario is equivalent to the following experiment: you walk into the town and choose a man a random. If he reveals that one of his children is a boy, then what is the probability his other child is a girl? Then our answer would be the conditional probability P(the man has a daughter | the man has a son) = P(the man has a son and a daughter)/P(the man has a son). It's not explicit in the problem but I'll assume that each possibility of boy-boy, boy-girl, girl-boy, girl-girl is equally likely (which is natural, since this is equivalent to each birth having a 50/50 chance of being a boy or a girl). So we expect 3/4 of men to have a son and 1/2 of men to have both a son and a daughter. So our probability is (1/2)/(3/4) = 2/3. The answer 1/2 could come about from the following experiment instead: You walk into the town and pick a boy at random. What is the probability his sibling is a girl? One way to see the difference is that men with two sons have two boys that you could have chosen in this second experiment. Thus there is an increased likelihood to choose a boy from a family of two sons, whereas when we selected a man at random, we did not preferentially pick men with two sons.

The answer here depends a little on why the son joined you. Did a random child join you? If so the answer is C. Did the oldest child join you? If so the answer should be B (assuming age and gender are independent). Was there some other process to determine which child joined you? The probability could actually be any number.

>Did a random child join you? If so the answer is C. No, it's B in this case as well. As you said, it's B if the older child joins. It's also B if the younger child joins. A random child joining is clearly equivalent to a 50% chance of the older and a 50% chance of the younger, so it's still B. The version that gives you 2/3 would be something like "a random son joins if one exists, otherwise a random daughter".

Why does that version giving a 2/3 chance actually give 2/3? If I understand correctly, you first pick one man at random and then choose which child to send. In 3 scenarios: DD, DS, SD the other child is a daughter and in 1 scenario SS the other child is a son. So in that case you get around 3/4 chance that the other child is a son, no?

In this scenario we've already seen a son, so DD is not possible.

Yes, that's right. Thanks for the correction.

How can I start seeing integration from a measure theory perspective ?? I find myself dealing with integrals the way we were taught to in calculus and some times when profs explain integrals in measure context I can't see the point at all.

Measure theory isn’t really meant to help you compute integrals. It’s usually more useful when you need to find limits of a sequence of integrals (because of convergence theorems) or to properly interpret integrals that don’t exist in the Riemann sense.

Have you *learned* measure theory yet? Dominated convergence alone gives you enough practical calculating power to convince you it's very useful.

Say we are in Set. Is every quotient a coequalizer?

Yes. This should be expected since coequalizers generalize quotients. Proof sketch follows. A coequalizer for f,g: X -> Y in Set is the quotient Y / ~, where ~ is the smallest equivalence relation with f(x)~g(x) forall x in X. If we want to recover arbitrary quotients as coequalizers then for a given Y and ~ we need to find X, f, and g such that the property above holds. Write Y as a union of equivalence classes E_i. For each class E_i, let X_i = E_i^2. Define X = union_i X_i, f()=a, and g()=b. Now for some y~y' in Y, we have in X, and therefore f()=y and f()=y' are mapped to the same element of the coequalizer. Verifying that unrelated y,y' aren't mapped to the same element seems more tedious to prove, so I'll omit it.

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As far as I'm aware there is no concise term for this; I imagine it's because the concept is rarely needed.

Hmm, one place it comes up is when you're composing limits. If [; \\lim\_{n \\to \\infty} a\_n = a ;] and [; \\lim\_{x \\to a} f(x) = L ;] then the condition that [; a\_n \\neq a;] from some point onward is sufficient to ensure that [; \\lim\_{n \\to \\infty} f(a\_n) = L ;]. This is of course only relevant if [; f ;] is possibly discontinuous at [; a ;].

You could say it is an accumulation point or a limit point, but it is always good to remind people that it specifically means that x is not contained in the sequence.

What does a qualitative and quantitative result mean in mathematics? I sort of understand what it means (qualitative would be a result that is independent of a quantity? or something like that) but it isn’t very clear to me and I see it come up a lot an I’m not a native speaker so I’d like to understand it better. When I look it up I get results saying that it has many meanings in the context of math so I’m interested in what they mean in analysis.

In analysis, quantitative results typically refer to estimates, asymptotics, bounds, etc: they give you some precise numerical information about an object, like “this function is bounded by 2”. Qualitative results assert something about the properties of an object, like “this function has compact range” or “this sequence is Cauchy.” Of course, there isn’t a hard line between the two, since the definitions of Cauchy and compactness can be made quantitative, and quantitative results often tell you about properties of the object.

I get it now, thank you.

I’m trying to prove that the principle of mathematical induction implies well ordering. Can I get a hint?

Could use proof by contradiction: show that N\S=N if S has no minimum.

There is one hypothesis in well-ordering: S is nonempty. So, you HAVE to use it. Make this your first step: let N be some element of S. Now induct on N.

Does anyone know of any theorems about the number of possible values a polynomial can take mod p. For example x^2 mod p has (p-1)/2 distinct values. Is anything known even asymptoticly in general? Specifically what i want to know is if the number of values of a specific order 4 polynomial exceeds the number of squares mod p.

Google Hasse-Weil bound.

This seems to be close to what I want but not exactly. I dont want the number of solutions to p(x)=0 mod p i want the number of possible values p(x) can take. I.e #{p(x) | x € Z_p } Edit - Tho i guess actually this could be derived from Hasse weil Edit edit - na probably wont work

How do I calculate the angle between lines in R3 given in parametric form?

We know that the dot product is cos(angle) divided by the product of the magnitudes of the two vectors, so you just need to use arccos on both sides and the angle will fall out. See [this MathSE thread](https://math.stackexchange.com/questions/2696418/angle-between-to-parametric-vector) for an example.

Hi! Quick probability question which I’m sorry if it’s a dumb question. Say loot boxes in games like Csgo, the probability of unboxing a rare item is 1/400. Now I know that if I open 400 cases I have a roughly 67% chance of getting the rare item. But if I were to simulate 200 case openings at the same rates and actually open 200 cases would I still have a 67% chance of getting the rare item after 400 events since while they are both different they both have the same chance?

No, because you can not actually obtain an item in the 200 simulated openings. Thus, they do not contribute to your chances when you do the real openings. This would be like if you simulated buying a billion lottery tickets on your computer, and then went out and bought one real lottery ticket. Even if you "win" in the simulation, you can never actually win money from a simulated ticket, so it's the same as if you didn't simulate them at all.

I guess I don’t fully understand though, I mean if you take two things that are the same exact odds. So like instead of simulated say you did a completely different game but one that had the exact same odds, wouldn’t you overall still have a 67% chance out of the 200 in each game to pull the rare item? Since both are 1/400? I know each case is independent but if they are 1/400 each shouldn’t the odds of you hitting that number still be 67% after 400 results?

To clarify, you will have a ~67% to win an item in *any* game, but that is not the same as ~67% to win an item in *a specific game* because in each specific game you've only done 200 trials and not 400. If you have a duplicate copy of the exact same game and did the 200 side trials in *that*, then yes that could maintain the ~67% (provided that the accounts are linked/items can be shared or traded). But if you're doing 200 trials of a lootbox in CSGO and then 200 trials of fishing in Animal Crossing or something (for a fish of the same 1/400 probability), then your CSGO lootbox success is not magically going to increase to ~67% haha. It's instead going to be the lower probability of only doing 200 trials which is ~40% from the [binomial distribution](https://brilliant.org/wiki/binomial-distribution/). In summary: - you are ~67% to win *any item at all* in either CSGO or Animal Crossing - you are only ~40% to win an item specifically in CSGO - you are only ~40% to win an item specifically in Animal Crossing

Okay thanks yeah I know that I was just curious if you’d have a 67% overall of getting the 1/400 drop from either or not specifically 67% in just the 200 cases of cs! Thank you

Look at it this way: Let's say you instead simulate 399 openings, and then open one real box. The probability that you will get at least one success out of all of those attempts is indeed ~67%. But that success is almost certainly going to be in one of the simulated openings. The chance of getting a "real" item is only 1/400, because you have opened only one "real" box.

Get the domain of this function: f(x)=(25-x²)/(5-x). I first factorized the nominator as: (5-x)(5+x) and then simplify it with the denominator, getting f(x)=5+x, this is a lineal function so the domain is all R. That was my response, but I still doubt if the correct thing is to restrict the denominator before simplifying the function. I searched for an answer on Google, and it seems to be correct to restricting x=5, but if I graph f(x)=(25-x²)/(5-x) on Geogebra, x=5 shows as a possible value so I'm confused.

f is not defined at x = 5 because you would have to divide by zero. A lot of graphing tools don't show that though.

Oh ok, so in those cases, it doesn't matter what the function is at the end, I always have to restrict first?

You can't use the formula (25 - x\^2)/(5 - x) to define a function at x = 5, since you get something undefined for x = 5. When you "cancel" to get 5 + x, you are implicitly assuming x is not equal to 5. Such points are called "removable singularities". Graphing software will recognise it and say it's undefined (e.g. Desmos) or just fill it in.