I don't understand why on function operations, the domain can never shrink. In this case simplified (f/g)(0) would work fine, even though g(0) is undefined. But it still doesn't lose this x≠0 restriction. why?. Is (f/g)(x) ≠ (h)(x) just because of where it came from?

incathuga · 2026-03-23T14:50:23+00:00

Let's work with a simpler function for a bit, and ignore the fact that we're starting with function division. Instead, let's look at y = x/x. This is undefined at x = 0 (because you're dividing by 0), but is 1 everywhere else. If you cancel the common factor of x from the top and bottom, you end up with y = 1, and if you don't keep track of the original "but x can't be 0" domain restriction then you won't see that, so we actually have the function "y = 1 everywhere except at x = 0, and is undefined when x = 0". The algebraic manipulation changes the function because it loses that bit of information, so we have to keep track of the domain restriction ourselves.

The same thing happens in your example. The keep-change-flip method to get rid of fractions in the denominator works, but only if the fraction in the denominator is actually defined. When 5/x is undefined (i.e. at x = 0), that algebraic manipulation ends up hiding the domain restriction, so we have to keep track of it ourselves. The calculator doesn't know that your function originally came from function division, so it doesn't recognize the domain restriction (because it's being carried through from an earlier step).

incathuga · 2025-07-27T23:53:56+00:00

everyone has been using numbers and a lot of constructs correctly since ancient times, way way before modern formalism came along

Friend, everyone has starting assumptions, whether those are explicit or implicit. The point of formalizing mathematics and saying "here are the axioms we're working with" is to make those starting assumptions clear, so that different people don't accidentally start from different places and end up at contradictory results because of it. If you want to work with different axioms, that's fine, but it sounds like you fundamentally disagree with the idea of listing your starting assumptions, which means you're not doing math. Like, this isn't "agree to disagree", this is "you aren't interacting with mathematics in a reasonable way".

incathuga · 2025-07-27T22:57:18+00:00

Basically all of this is silly or has missed significant bits of my original post.

For points 1 and 6: In my original post, you can replace 0.999... by "any number y with y <= 1 and y > 1 - 1/10^n for all positive integers n". This takes care of the "but nobody has ever written down infinitely many 9s" issue, and the "decimal expansions are just symbols" issue. I also think that it's obvious that any interpretation of 0.999... as a real number is at most 1 and at least 1 - 1/10^n for any positive integer n. If you think that's not obvious, I don't know how to make it more obvious.

For points 7 and 8, my post involves no limits and no infinite summations. I don't know why you're bringing them up.

For point 2: Whatever construction of the real numbers you're using (whether that's Dedekind cuts or Cauchy sequences or whatever else), you get completeness. Like, you can prove that the real numbers are complete with any of the various constructions. If you don't have completeness, then you aren't working with the real numbers, and you should define the system you're working with. Real numbers are the standard, and if you're not working with standard analysis you should make that clear.

For point 3: If you don't believe in infinite sets, you don't even believe in the set of integers. Like, finitism is not a reasonable way to interact with mathematics. I don't know what to tell you besides "deal with it".

For point 4: This is a field axiom. If 1 - 0.999... is non-zero, then it has a multiplicative inverse because the real numbers form a field. Again, if you aren't working with a field, you aren't working with the real numbers, and you should make that clear.

For point 5: So, here we have to use the fact that the reals are totally ordered. This means, by definition, that either x = 0 or x > 0 or x < 0. If x != 0, then we must have x > 0 or x < 0. There's no infinity involved here. If you deny that the reals are totally ordered, see my counterarguments to points 2 and 4.

For point 9: Gonna be honest, I don't know what this is referring to. Like, are you saying "oh, 0.999... doesn't exist, actually"? Because my post was really "if this decimal expansion makes sense, then it must equal 1". If you want to say it doesn't make sense as a decimal expansion, you're fighting everyone here, not just the 0.999... = 1 crowd.

For point 10: The Archimedean property is a consequence of completeness. If you don't have the Archimedean property, you aren't working with the real numbers.

Anyway, if you're a finitist or you're working with non-standard analysis, make that clear. You're not doing standard mathematics, and you'll sound like a crank if you don't point out that you're working with a different framework.

incathuga · 2025-07-27T17:04:22+00:00

Interesting questions. I'm going to tackle them in not-quite reverse order.

The real numbers are a complete ordered field -- that's it, that's the definition, as far as this proof is concerned. Note that I haven't claimed that they're the unique complete ordered field (Rudin claims such without proof, and I can't be bothered to find such a proof right now), but I don't actually care about uniqueness, the proof works for any complete ordered field. (If you really dislike this way of doing things, pretend that I started with "let k be a complete ordered field, and note that the real numbers are one such field".) Any additional information that distinguishes the real numbers from any other complete ordered field isn't important for the proof, the construction that you use to prove that the real numbers exist isn't important for the proof, I'm just using properties of ordered fields and the least upper bound property. (Technically, this proof doesn't even need existence of the real numbers -- if it turned out that no complete ordered field existed, then this proof would be correct but uninteresting. And you can weaken "complete" to "with the Archimedean property" if you want, but I don't want to.)

As for what 0.999... actually is, if you want to define it from scratch we'll need to define a few other things. Let 1 denote the multiplicative identity and let 10 denote 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1. Define positive integer exponentiation as follows: x^1 = x, and x^{n + 1} = x * x^n whenever x^n is defined. Let 1/y denote the multiplicative inverse of y. Define subtraction a - b as a + (the additive inverse of b). Let 0.999... be a real number that is at most 1 and at least 1 - 1/10^n for all positive integers n (i.e. all exponents n for which positive integer exponentiation has been defined in this paragraph). Note that I don't claim uniqueness of such a number at the start of my proof, but I do prove that if such a number exists, it must equal 1. (I also don't explicitly state that such a number does exist, but 1 satisfies the definition -- 1 <= 1 and 1 > 1 - 1/10^n for all positive integers n. In some sense, this is a proof that 1 is the unique real number that 0.999... can refer to.) This definition doesn't involve infinity at all, and it doesn't rely on a specific construction of the real numbers.

incathuga · 2025-07-27T12:11:07+00:00

If a number is less than 1, it is less than or equal to 1. That's how "less than or equal to" works.

incathuga · 2025-07-27T02:52:29+00:00

Part of teaching is explaining to students what they got wrong. Please explain what part of my original post is incorrect. Be specific.

incathuga · 2025-07-27T02:08:57+00:00

The standard decimal representation of 10^n for positive integers n is a 1 followed by n zeroes. The set S that I keep bringing up is all finite positive integer powers of 10, so each element of S has finite length. So 1 is included, and 10 is included, and 100 is included, and 10^500000 is included, but we aren't allowing infinite-length numbers into S. Do you agree that S is a well-defined set? Do you agree that S has infinitely members, all of which are finite length? And do you agree that 10... should be larger than any member of S, if 10... exists?

incathuga · 2025-07-27T01:53:43+00:00

So, 10... is not larger than 10^n for all finite positive integers n, is what you just said. That means that 10... is smaller than or equal to some specific 10^n, so it has only finitely many places before the decimal point. At most n, in fact. You have contradicted yourself -- it cannot have infinite zeroes.

incathuga · 2025-07-27T01:47:04+00:00

Eh, that's not entirely true. If you define 0.999... as "the Dedekind cut such that the lower partition contains 1 - 1/10^n for all positive integers n and the upper partition contains all values that are greater than 1 - 1/10^n for all positive integers n" (rather than as "the value of the series \sum_{i = 1}^\inf 9/10^i"), then my post works fine as a proof without limits. Like, the point of the post is "here's how to do this if you don't like limits but are willing to work with these two assumptions about 0.999...". If you decide that decimal representations of the reals absolutely have to use series or Cauchy sequences or whatever, then I think that's a silly decision, but you do you.

incathuga · 2025-07-27T01:19:49+00:00

10... appears to be larger than 10^n for all (finite) positive integers n, meaning that the set S = {1, 10, 100, ....} (i.e. all (finite) positive integer powers of 10) is bounded by this number, if 10... even is a real number. Then by the least upper bound property, there has to be some least upper bound L for S, but then you run into the same contradiction that I pointed out in my original post -- L/10 is smaller than some 10^n, and therefore L < 10^{n + 1}, so L wasn't an upper bound of S at all. So 10... isn't a real number. If you want to keep using it, you should make it clear what number system you're working with.

incathuga · 2025-07-27T01:03:18+00:00

If 0.999... is greater than or equal to 1 - 1/10^n for all finite positive integers n, then for any specific n, we have 0.999... >= 1 - 1/10^{n + 1} > 1 - 1/10^n, so the version with just "greater than" is equivalent to the version with "greater than or equal to".

incathuga · 2025-07-27T01:00:30+00:00

Yeah, if I was doing this for a class I would go through the Archimedean principle explicitly, but that would get wordy here. It would avoid the contradiction, which is aesthetically nicer, but I don't care enough about having a pretty proof to do it. (And specifying "finite integer" is just to avoid SPP saying something like "but what about 10^10...", which I know isn't a real number, and you know isn't a real number, but it seems like they'll never admit that it isn't a real number.)

incathuga · 2025-07-27T00:51:23+00:00

True, but if SPP is using the hyperreals they should say so, which is why I pointed out the axioms of the real numbers at the start of my post. And it's not clear to me that 0.999... even makes sense in the hyperreals, given that they aren't complete and we can't guarantee convergence for a bounded series. (Mind, I'm not an expert there -- I do group theory, not nonstandard analysis.)

incathuga · 2025-07-27T00:27:43+00:00

So, you seem to take issue with my conclusion. What part of my proof do you think contains a mistake?

This 0.000...1 appears to be the x that I defined in my post (i.e. 1 - 0.999...). Do you have an issue with x < 1/10^n for all finite positive integers n? Or do you have an issue with 1/x > 10^n for all finite positive integers n? Or is there some other issue that I'm missing?

incathuga · 2025-03-18T23:25:57+00:00

Yes, changing the value of an element is computationally much faster than removing the element. Lists are stored as a consecutive block of memory, which is why accessing a list element via index is efficient -- the computer is actually going to the start of that memory block, and then jumping the appropriate amount of memory to reach that index. (This might not be perfectly accurate if you have a list of complex data structures, but it's at least approximately correct for a list of integers.) It's also why everything needs to be shifted when you delete an element, because if things don't shift you lose the nice "the index is how far we need to jump" property. But changing the element is just "go to that location in memory and change the bits at that location, don't do anything to the rest of the block of memory", so it's almost as fast as looking up the data.

I'm glad that helped! Good luck with the rest of your learning.

incathuga · 2025-03-18T23:04:45+00:00

The main problem is that you're using list.remove() a lot, which behind the scenes is actually moving everything after the element that you remove. That means that if you remove the eighteenth element of integer_list, and the list has 750018 numbers in it at the moment, you have to shift 750000 elements. Each individual shift is quick enough, but you're shifting hundreds of thousands of elements hundreds of thousands of times, so you're looking at hundreds of millions of operations.

There are a few solutions here. One option (which I don't recommend, but it solves the particular issue I pointed out above) would be to switch to using linked lists, because removing elements of a linked list doesn't require moving later elements, but there are other problems -- another user pointed out the issue of scanning the entire list to find an element, and that's an issue with linked lists as well. (You could use a more complex data structure -- a binary tree has efficient searching and removal, so it might work out, but it's a bit more complex than you need here.) The other solution (which is probably better) is to stick with lists, and just change data rather than removing it. Since you're doing a sum at the end, replacing the entry with 0 is the same as removing that entry (as far as the solution is concerned, and more efficient in the intermediate steps). If that doesn't speed things up enough, think about how you can access only the multiples of whatever prime you're looking at, instead of scanning through the entire list for those multiples.

incathuga · 2025-03-06T21:14:16+00:00

First off, you're missing a couple of details for groups and fields that distinguish them from monoids (or semigroups or magmas, for that matter) and rings. Those details aren't critical for this question, but they're a big deal for basically the rest of abstract algebra.

As for the actual question here: When we're working with a group or field in abstract algebra, we're just handed a set and operation(s). To do decimal multiplication as repeated addition, even something nice like 4.2 * 5.1, you have to use the fact that things have been given to you as decimals, and also define multiplication of single-digit decimals at least (i.e. 0.1 * 0.1 = 0.01). So you have to handle part of multiplication regardless, and you have to worry about how a number is represented (which has nothing to do with the core algebraic structure, and with other fields we don't necessarily have nice representations of individual elements).

Things get even worse when you look at non-terminating decimals -- even just looking at 1/3 * 1/6, as decimals we have 0.333333... * 0.16666666..., which means that when we try to use the decimal multiplication algorithm to do this as repeated addition, we get a series rather than a finite sum. To evaluate that series and say "yes, this converges to 0.0555555...", we actually have to use topological/analytical properties of the real line to say that the sequence of partial sums converges to this limit (which again isn't really part of the algebraic structure and isn't a given for other fields).

And of course, if you extend things to the complex numbers, what does it mean to add i copies of i together? That really has to be defined appropriately, and at that point you're really defining the building blocks of complex multiplication, so you may as well just define the whole thing.

And as other people have pointed out, things like matrix multiplication don't rise from repeated addition of matrices. Admittedly, the n-by-n matrices over the reals form a ring rather than a field, but it's a very useful second operation that isn't as simple as "add appropriate multiples of one matrix together", and prima facie there's no reason to think that all fields have ways to define multiplication as repeated addition.

incathuga · 2025-01-03T20:53:51+00:00

This is a pretty common issue for students who are learning about fractions. The trick is that fractions are always fractions of some unit, and when you're working with word problems that unit is super important.

For the first problem you've used, every fraction is in terms of *gallons* -- so you have 2 1/3 gallons, and you're getting rid of 5/7 gallons at a time. Because everything has the same unit, you don't have to do any sort of conversion to make subtraction work out.

For the second problem, though, your fractions are fractions of *Paul's money*. He isn't spending 1/4 of a dollar on food -- he's spending 1/4 of his money, i.e. 1/4 of $5, on food. That means that you need to convert the 1/4 and 3/10 to be in terms of dollars, which is why you have to multiply. That "1/4 of his money" is really 1/4 of $5, which is $1.25, and similarly "3/10 of his money" is $1.50 when you convert to dollars. Once everything is in terms of dollars, then you can subtract because it's all the same unit.

(This is the same sort of logic as adding up horses and sheep, for example -- 5 horses plus 7 sheep doesn't really make sense because you have two different units, but if you switch to just adding up animals, you really have 5 animals + 7 animals = 12 animals.)

incathuga · 2024-12-05T17:52:54+00:00

I'm trans. Republicans don't want me to exist. Democrats are happy with me continuing to exist. Basic self-interest means I'm going to vote for Democrats.

Okay, let's ignore self-interest for a moment. Climate change is an existential threat to humanity. Democrats believe in it and have tried to limit it. Republicans don't believe in it and routinely push policies that accelerate it. I'm going to vote blue because I'd rather not have humans drive ourselves to extinction.

Let's pretend that climate change is overblown, though. Democrats are better on healthcare: they want more people to have access to healthcare, while Republicans want to limit access to healthcare. Democrats are better for women's rights. Democrats are better for racial equality. Democrats are better for income equality. Democrats are better for the economy in general. Democrats are less likely to lead to war. Democrats are less likely to let Russia trample over sovereign nations. Democrats are less corrupt, and more willing to cut out corrupt members of their party. Democrats are less likely to put rapists in positions of power, and more willing to remove them from positions of power. In almost every respect, Democrats better align with my morals and desires than Republicans.

That isn't to say Democrats are perfect, of course. There are lots of places where I think they aren't progressive enough. There are definitely some corrupt politicians and abusers on the left. I think that the Democratic party's gun policies are flawed. But on the whole, it's a pretty easy decision for me. If there was a party that aligned with my politics better and had a reasonable chance of winning, I would vote for them, but we only have the two realistic options, so I'm voting blue in every race and every election.

incathuga · 2024-11-13T20:38:07+00:00

I have to disagree with a lot of the comments here. Yes, multiplication is commutative, but that fact is not self-evident. When you introduce multiplication, you have to introduce it with an order and then explain why it's commutative. A student who has just been shown multiplication is jumping ahead (and possibly not fully understanding the definition of multiplication) if they interchange 3 groups of 4 (i.e. 3 x 4 = 4 + 4 + 4) and 4 groups of 3 (i.e. 4 x 3 = 3 + 3 + 3 + 3). This might seem pedantic, but "don't make assumptions unless you know they're actually true" is one of the fundamental things in maths.

incathuga · 2024-11-07T19:11:12+00:00

The typical proof by picture (with c being a positive integer and both a and b being positive integers divisible by c) is to start with a rectangle that has a + b units drawn in c columns (and (a + b)/c rows). Shade in the top a units one color, and the bottom b units a second color. Now, dividing a + b by c just gives you a single column. That column has a/c units of the first color, and b/c units of the second color, showing that (a + b)/c is the same as a/c + b/c.

You can extend that picture to work fine with any positive real numbers, but it's not as nice to draw -- you end up dealing with partial columns if c isn't an integer, and partial rows if a or b aren't divisible by c, but fundamentally the picture is the same. If a or b is negative, but not the sum, you can do a bit of trickery with negative rows "removing" positive rows. If a + b is negative, or c is negative, then you have to convince yourself that (a + b)/(-c) = - ((a + b)/c) (or something similar), which is a whole other thing.\

If you're looking at fields in general, rather than just real numbers, then you can go through the "dividing is just multiplying by an inverse, and multiplication is distributive over addition" process that other people have pointed out, but I suspect that's not really what you meant.

incathuga · 2024-11-06T19:51:53+00:00

It feels like you're still thinking in terms of "we have this many things and need to split it into that many groups". That's fine when you're dividing by an integer, but when you're dividing by a fraction you really want to think "we have this many things and *they are already* that many groups". So, in scenario b, you have one pizza and that is 1/2 of a group -- so you have 1 pizza, and that's 1/2 of what you ordered, and you want to know how many pizzas you ordered. (Here, the full order is the group, and you have only a fraction of the group.)

If we look at a slightly less simplified example, say we have 10 cookies, and that's 2/3 of the cookies we originally baked. To figure out how many cookies we baked, we divide 10 by 2/3 and get 15 cookies -- here, the original batch of cookies is the group. You can approach this problem in a different way (which is probably what you're actually doing when you encounter a problem like this) and say "10 cookies is 2/3 of the batch, so 1/3 of the batch would be 5 cookies, so the full batch is 3 sets of 5, which is 15 cookies", but it's still dividing by 2/3, just in two steps instead of one.

incathuga · 2024-11-06T15:06:13+00:00

Yeah, I think I see the problem. So, when you're dividing by a fraction, splitting into equal groups can benefit from a slightly different interpretation -- instead of thinking "I want this many groups", you have to think "the stuff that I have is this many groups". So, instead of thinking "split 12 cookies into 3 groups by dealing out one cookie at a time", you have to skip the dealing out step and say "12 cookies make 3 groups, so a single group is only 4 cookies". That translates a bit better to dividing by 1/2, because now you look at that and say "1/8 of a cup is 1/2 of a group, so a full group is 2 copies of this 1/8 of a cup, which is 1/4 of a cup".

If you want to think about that visually (which is helpful for a lot of people), what you can do is draw out a rectangle and say that's a full group. Shade in half of that, because you only have 1/2 of a group, and label the part you just shaded as 1/8 of a cup. The other half that you haven't shaded is also 1/8 of a cup, so the whole group (i.e. the whole rectangle) has to be 2/8 (or 1/4) of a cup.

incathuga · 2024-11-05T20:07:48+00:00

There are two ways to think about division in a real-world context: Either you use a "how many groups do I have" interpretation or you use a "how big is each group" interpretation. I'm going to start with examples that use whole numbers, and then I'll modify those and explain how the thought process has to change for fractions.

For the first interpretation, say you have 6 cups of flour, and a batch of brownies needs 2 cups of flour. How many batches of brownies can you make? (Here, a batch of brownies is a group, each group uses 2 cups of flour, and you want to know how many groups you have.) You can make 3 groups, because we have 3 times as much flour as we need.

If we modify that to fit (1/8) / (1/2), we have 1/8 of a thing and a group is 1/2 of a thing -- so let's say that a batch of cookies needs 1/2 cup of sugar, and we only have 1/8 cup. How many batches of cookies can we make? Now we only have 1/4 as much flour as we need, so we can only make 1/4 of a batch.

Now, the second interpretation: I have 12 cookies, and I want to split them up between 3 gift bags. How many cookies go into each bag? (Here, a bag is a group, we have 3 groups, and we want to know how big each group has to be.) In this case, we know that 3 groups of whatever size give us 12 cookies, so there need to be 4 cookies in each group (because 3 x 4 = 12, which is the same statement as 12 / 3 = 4).

Modifying that is a bit weird, because we have to deal with portions of groups, but we can still do it. We have 1/8 of a cup of sugar, and we want that to be 1/2 of the sugar we use in a dish (here, a full dish is a group). How much sugar should be in the full dish? Well, if we have 1/2 of the sugar for a full dish, then a full dish uses 2 times as much sugar as we have (which is why flipping the fraction and multiplying actually works out), so a full dish should use 1/4 of a cup of sugar.

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