$\{0,1,2,3,i,2i,1+i, 2+i, 3+i, 1+2i, 2+2i, -1+i, -1+2i\} + (3+2i)$, is this an exhaustive list of (distinct?) elements of the gaussian integers quotient ring $\mathbb Z[i]/(3+2i)$?
we have 13 elements because of isomorphism with $\mathbb Z/(13)$
monoidal: there basically aren't any, as above ("the background vibe is that any construction remotely like this with measurable sets as ingredients will be measurable"). one of the vibes of basic measure theory is that you don't escape the world of measurable stuff with piecewise constructions, countable sums, and limit operations.
monoidal: but there's no useful way of formulating that theorem outside of some pretty exotic stuff (e.g. models of set theory or logic where everything is measruable), or a long series of extremely routine verifications that things are, indeed, measurable.
@leslietownes I'd like to bring a legal matter which affects all Stack Exchange to your attention. I'm not asking you for professional legal advice, but perhaps you can shed some light on this matter as someone who understands legalese and copyright law. ;)
Recently, the Stack Exchange CEO expressed a desire to get financial compensation for the SE data used to train ChatGPT & other LLMs. From stackoverflow.blog/2023/05/31/…
> LLMs are trained off Stack Overflow data, which our massive community has contributed to for nearly 15 years. We should be compensated for that data so we can continue to invest back in our community.
Now, it doesn't seem right that SE can sell our data & waive our Creative Commons right for proper attribution in exchange for cash. However, some members claim that our contributions are dual licensed, see meta.stackexchange.com/q/388760 & opensource.stackexchange.com/q/5663 Personally, I think that's baloney.
The issue was brought to a head when it was discovered a few days ago that SE Inc had silently cancelled the regular data dumps a few months ago, see meta.stackexchange.com/questions/389922/…
So, I'd like to know if the dual license thing is BS. It would be fantastic if someone like yourself could clarify these issues. But I also understand if you'd rather not comment.
@shintuku When using SageMathCell on a phone, the windows may be a bit narrow. Here's a bookmarklet that lets you adjust the width: javascript:(()=>{let%20w=prompt('Width?','130%');if(w)jQuery('.sagecell').css('width',w);})()
Speaking of quadratic fields, a few days ago there was a discussion about $a+b\sqrt2$ being dense, with $a,b$ integers. It got me thinking about algorithms for finding such integer pairs so that $a+b\sqrt2\approx x$ for a given real $x$. I wrote Sage code that uses the continued fraction of $\sqrt2$ in a greedy algorithm to approach $x$, but I'm wondering if there's a better / more efficient algorithm. Here's one example I found: $\pi\approx 70113791\sqrt2 - 99155871$
@PM2Ring i don't think there is a clear answer. haven't read details but it looks like some version of the terms say both "you agree that A" and "you agree that B", where A explicitly imposes attribution obligations on SE when certain things happen, while B does not. even if this did create a 'dual license' (far from clear), there'd be a question of what SE is free to do if the obligations imposed are inconsistent with one another (also far from clear).
"far from clear" meaning a court could come out differentlyon the issue. forum dependence always exists with TOS type "agreements", and it only gets worse when terms are ambiguous or even contradictory. the result may turn on what a forum allows (e.g. what evidence is citable, who has which burdens of proof). SE has many arguments but are probably also kicking themselves. if this were litigated in 100 forums, they would not have a 100% win rate, maybe not even a 50% win rate.
US-specific detail: if a bad guy breaches the terms of a copyright license, under some circumstances you can only sue in the US for breach of contract, not copyright infringement. litigants often want to sue under US copyright because the remedies are often better, but sometimes this isn't available (and "licensee agrees to provide attribution, then doesn't" is generally one of those times). a lot of internet lawyering ignores this or is unaware of it (maybe because this issue is US-specific)
@leslietownes Thanks. I really appreciate it. The relevant sections in the TOS have changed slightly over the years, but SE Inc have never responded on the meta site(s) with a clear statement on this dual license stuff. :(
I guess it's even more complicated, from a legal POV, because (AFAIK) the courts haven't decided on the copyright issues surrounding LLM training. Do we class the trained net as a lossy compressed copy of the training data? Or is the trained net like a human reader, who may coincidentally repeat material that it's read...
yeah, it's not an easy fit for copyright as it has developed in most countries.
"databases" broadly speaking (e.g. compilations of information that might be collected from a range of sources) presented, and still present, similar difficulties. a lot of the doctrines around copyright generally don't fit it very well, so you get these patchworks of protections, some via copyright, some not.
it doesn't help that some of the stuff that LLMs put a focus on (e.g. rights of attribution, e.g. controls that 'stick' to the author of something and maybe their descendants but nobody else) already presents pretty broad points of disagreement among 'big' IP regimes even in the cases of like paintings and plays and stuff.
even pre-chatgpt there were some off the shelf "AI-lite" computer assistants for generating patent applications. ticking at least some of the boxes of the US requirements is fairly robotic, and it's not unreasonable to ask a computer to give a first draft of something (assuming that the starting input is what we would recognize as something patentable, and not an AI hallucination)
I was having trouble with a weird piece of thought. Are the two statements "A set $K\subset \Bbb R$ is compact in $\Bbb R$ if for every sequence in $K$, their exists a convergent subsequence in K, that converges to a point in $K$." and "A set $K\subset \Bbb R$ is compact in $\Bbb R$ iff for every sequence in $K$, their exists a convergent subsequence in K, that converges to a point in $K$." ---same ?
when was dealing with us berkeley administration in my first week in the usa, i had to repeat myself 5x times on a regular basis. very frustrating
@ThomasFinley i suppose when its a definition it is more of a linguistic equivalence where as in a tautology it is a logical equivalence if you see what i mean
@SouravGhosh Thanks, and I think: if the former statement is taken as a definition, then the two statements are equivalent. Am I correct in my understanding of the scenario ?
@copper.hat Danica McKellar says there's no conflict between being a mathematician & being sexy. But you need the right personality to carry it off successfully.
@copper.hat Pardon me, but till now, I have no idea what a "linguistic equivalence " or a "tautology" is . But this is the reasoning, I came up with i.e: if the former statement is taken as a definition, then the two statements are equivalent.
And I want to know, whether my reasoning is correct or not?
@ThomasFinley yes. what i mean by linguistic equivalence is that the definition is just a shorthand for the rhs. a logical equivalence is where you prove that the lhs implies the rhs and vica versa
@copper.hat Hmm, that might be true, but please dont create further suspense, it gets frustrating sometimes. (Like in this case, it was for me). But the last "yes" in your comment, was truly intended towards me, right ????
given $e_1,e_2$ are integers, one can show that $e_1^4 + 2e_2^2 = e_1^2 -2e_2$ holds $\mod 4$, is there a situation when this holds in more general rings of algebraic integers? Say we take $\mathbb{Z}[D] \subset \mathbb{Q}(d)$, where $D = \sqrt{d}$ if $D = 2,3 \mod 4$ and $D = \frac{1 + \sqrt{d}}{2}$ if $d = 1$ $\mod 4$. I would guess we can only expect it to hold if $2$ is prime in $\mathbb{Z}[D]$
does anyone know if there are concrete cases of $D$ for which this holds?
Sandy wrote or arranged a lot of good songs, but Who Knows Where The Time Goes is generally considered to be her greatest, written when she was only 18.
yeah, i would assume it is implied even if it's not stated and expect most people don't even notice it.
if there's some reason you don't want comments (e.g. the paper is already published and is being put on the arxiv for some archival reason, e.g. because a more recent paper refers to it) it might be helpful to add that.
relatedly, while many people often forget to add publication data when it becomes available (maybe some don't know that it's possible to do this), it's really helpful to do that.
of course you can. but if the person has submitted the DOI already, you can also just click.
and sometimes the web is littered with a plurality of junky versions, which makes searching for the 'right' published version more than just title + search + click.
perhaps. i just remember encountering a lot of preprints via google because the arxiv is pretty well indexed, and sometimes journal sites are not (e.g. paywalls might prevent indexing of material that appears in a paper's contents but not its title). and often having that moment of "i wonder if this is the most recent version of this, or just the last version they happened to put on the arxiv."
and making me go back to google to find that out is cruel.
there shoudl also i guess be some way of more explicitly flagging "by the way, in fact, i never published this, this really is the last of it." i don't know how to do that in arxiv metadata.
Is there sometimes a minor (mathematical) typo in a published paper? My college told me once that he found a published paper in applied mathematics that contains some minor typos (but not serious).
peer reviewing for minor typos is difficult because if you "know" what they're saying, you mentally correct for it and don't notice. same as regular language typos.
sometimes the publisher introduces their own typos... they just ruined the bibliography of paper (and I didnt think to proofread their version of the bibliography.. because come on..), eg "HF Münzner" is cited as "FM Hans", papers with two authors are reduced to one and taht guy just got his first name put into his last name ....
outside of math and other fields where the math might be a real focus of attention, you could expect the typo rate in equations to be significantly higher than the typo rate in other stuff, because reviewers are likely not bothering to check equations at all.
if the authors are honest and a proposition is incorrect (not just a typo) then they will issue an erratum or retract the paper (if the nothing can be saved)
as one example, in some technical fields people might put equations in submissions to the patent office. those are virtually never checked and often are not even typeset correctly, even if the initial submission was.
but mathematics doesnt usually function in a way that a small error somewhere gives rise to a major flaw. happens in differential geometry because everything they do is a computation
but more often than not if it walks like a duck, talks like a duck, and enough people think it is a duck, its a duck (replace duck by "a valid paper/result/theorem")
enflo's original work on the invariant subspace problem was notoriously hard to understand and for a long time people weren't sure if he had produced an example.
oh, yeah, well i'm not sure how widely it is known. the wikipedia page is understandably very neutral about it.
at least one other person, not as prominent as enflo, suggested for a while that his example (published in the 1980s) was the first example, i.e. implied that enflo had not gotten there.
i am not naming names because i do not know how long or how adamantly he continued to do that.
anyway, it's maybe more than a few parallels here with louis de branges, who was famous for solving the bieberbach conjecture (although when he announced his proof of that, many did not believe it, and it maybe took the work of others for people to be convinced), who in extremely advanced age repeatedly claimed to have solved the riemann hypothesis, or some modified version of it.
i.e. "very wild claim, but coming from someone who not only previously made a wild claim, but was actually right, even if it took a while for everyone to figure that out"
anyway, when i first heard of enflo's preprint i thought "i wonder if the proof involves some extremely complicated constructions by induction" and surprise! it does.
but that is not reason to count it out. if i were still in academia i might spend time trying to figure it out, haha
@leslietownes I am sorry. I don't understand. Why would it take a decade? Isn't the paper just 13 pages long. I mean I guess it'll just take probably a week to verify the paper. Do you mean it uses some advanced technique?
jakob: i think the arxiv metadata allows for something just like "Other URL" as indeed many published papers do not have doi's.
@PNDas it looks elementary but extremely complicated to me. and you absolutely can put a whole lot of difficult work into 13 pages, if you skip verifying technical details (or are so deep into your own personal view of something that you don't notice how others would need to see a verification).
i spent some time with one of the de branges RH proofs and it was very slow going even though i was familiar with a lot of his toolkit. i didn't find any fatal errors, but he used a lot of language in nonstandard ways and it took a ton of effort just to figure out what he was saying. then he came out with an "updated version" of it and i thought, oh, f--- this.
I've been searching for paper, hard to find on google scholar, no DOI, I finally found it on researchgate but it's hidden behind a paywall. Nothing on illegal sites either. I have to reinvent the wheel at this point
@PNDas "a decade" was also a joking but very specific reference to enflo's earlier work, where approximately a decade elapsed between his announcement/sketch of a result and its publication in final form. he did spend at least some of that decade trying to make it simpler and explain it to other people and it still took that long.
@Jakobian Some papers are cited all the time but are impossible to get your hands on, you might have to make a request to your librarian who will somehow magic a physical copy of the journal it was published in into existence
There was a question which asked:"Justify whether the set of rational numbers satisfy Cantor's nested interval property ?" . I know, that Cantor's nested interval property does not hold for rational numbers. But my answer will be: "While proving, Cantor's nested intervals property for real nos, we essentially used Axiom of Completeness(AoC), but AoC is never valid for the set of rational numbers, so we can't really prove Cantors Nested Property for rationals." But I feel that my answer if not wrong, is incomplete. This is because, maybe AoC can't be utilised to prove Cantor's Nested Interva…
But this is where the problem comes in. My idea for finding a counterexample was to show that there exists a Nested Sequence of closed intervals that has $\sqrt 2$ as the element of its intersection. But $\sqrt 2$ is not in Q, a contradiction. But I failed to produce any such examples of a nested sequence of closed intervals. I need some help in here...
if you want something explicit see above, I'd just say there exist rationals $p_n$ and $q_n$ with $p_n < \sqrt{2} < q_n$ and $\lim p_n = \lim q_n = \sqrt{2}$
Say for contradiction we assume that the NIP is valid in Q. Also, we know that there exists an increasing sequence , $p_n$ which converges to $\sqrt 2$ (for eg: 1,1.4,1.414,1.4142,... is a sequence which is increasing and converges to $\sqrt 2$ ) and there exists a sequence $q_n$ converging to $\sqrt 2$ and is decreasing (for eg:2,1.456,...,1,1.4,1.414,1.4142,...). Now, we construct intervals, as $I_1=[p_1,q_1], I_2=[p_2,q_2],I_3=[p_3,q_3]$ and so on, then,
we note that $....\subset I_4\subset I_3\subset I_2\subset I_1$ is a nested sequence of closed intervals, with $\bigcap I_n=\sqrt 2\notin \Bbb Q$ and so, $\bigcap I_n$ is empty in $\Bbb Q,$ a contradiction. This, proves, NIP doesn't necessarily hold in $\Bbb Q.$- Can this be considered a perfect answer to the question I mentioned, @Jakobian ?
@Jakobian This idea was inspired from this comment of urs..
@Jakobian That's the point of problem solving. But to be specific, my concern is, that I asserted that $\exists $ a montonically decreasing sequence convergent to $\sqrt 2$ by intution, however, I found no such examples of this scenario, taking into account that the example I wrote in my 2nd last comment, is erroneous.
@Jakobian In fact, you asserted the same. Is it enough to assert it just because this seems pretty obvious, or one has to prove this rigorously ?
most people already encountered how to obtain such sequence/seen a proof and verified it/assumed it's true
for anyone experienced it'll be just a fact you can use
if you never seen such fact and you want to convince yourself it's true, you can prove that $\frac{\lfloor \sqrt{2}n\rfloor + 1}{n}$ is such sequence of rational numbers
obvious just means "yeah it's true, boring, let's move on"
pretty much
or "yeah it's true I've seen it before, I don't have to verify it"
when you learn math, you probably will want to assume a lot because there's only so much a human brain can handle in a short time
The principal branch is $z \mapsto e^{{1 \over 2} Log (1-z)}$, defined on $\mathbb{C} \setminus [1,\infty)$. Note the change in the real axis part of the definition. — copper.hatFeb 3, 2019 at 4:01
@copper.hat: Am I right in saying that $\arg(1-z)$ in the case of the principal branch is in $(-\pi, \pi)$?
@Jakobian this was the thing, I wanted to say, in a roundabout way. The thing that went on my mind is:"Ok, such a monotonically decreasing convergent sequence converging to $\sqrt 2$ seems to exist obviously. C'mon, just look at the number line, where there is $\sqrt 2$, it will be 100% possible to find such a sequence. Now, let's use this fact such a monotonically decreasing seqn converging to $\sqrt 2$ exists to prove the real thing, i.e NIP is not necessarily, valid in Q."
I was precisely, asking: Under these thought processes in my mind, should I move on just like that, without proving that such a monotonically decreasing sequence converging to $\sqrt 2$ exists or should I try to prove it ?(considering that: I am fully convinced about the existence of such a sequence)
@Jakobian what's your opinion , then ? (If you consider sharing it with me, which will be indeed, much helpful for me. )
I found out a popular question on real analysis is this:" Let $X$ be any set. Then show that Card X < Card P(X)". But isn't this the thing, Cantor's Theorem implies ? For my version of Cantor’s Theorem is: "Given any set A, there does not exist a function f : A → P(A) that is onto. " Doesn't this directly imply that Card A< Card P(A), considering the fact that there exists an injective mapping between them. So, will writing the proof of Cantor's Theorem suffice as an answer to this question?
The third follows from $\lambda^\kappa = 2^\kappa$ for $2\leq \lambda \leq 2^\kappa$ and $\kappa\geq \aleph_0$, which you can show like $2^\kappa\leq \lambda^\kappa \leq (2^\kappa)^\kappa = 2^{\kappa^2} = 2^\kappa$
@ThomasFinley Show that if $a_i$ is a monotonically increasing sequence converging on $\sqrt2$ then $2/a_i$ is a monotonically decreasing sequence converging on $\sqrt2$.
Hi all, i'm planning on writing a paper regarding the Fukushima nuclear incident. More specifically, I am looking at how long it takes for the nuclides to decay such that it becomes safe for humans to be around.
This is my plan so far
However, one problem with this is that it lacks complexity.
Does anyone have any suggestions of higher level math I could use?
@Ajay This topic may be more suitable for the Physics chat room chat.stackexchange.com/rooms/71/the-h-bar But consider what happens when a radioisotope doesn't directly decay into a stable isotope but instead decays into another radioisotope.
I don't mean this specific case but approach you should have in general, should you try to verify things or maybe not, how should I know if it's even important for you to verify those things, maybe you just got an assignment from your class and it won't be anyhow relevant in your future to be diligent and check all of those
@PM2Ring It's not so much about the radioactivity part. It's more about the probability distributions and models. I'm looking for something more advanced. I'm doing a math research paper.
you should probably do what your teacher tells you to do, unless you want to verify everything I guess, and even if you want to become a mathematician, verifying everything is not always a good thing either I guess, and should probably be more like verify some things, assume others
it boils down to how important is it to you, how important is it for your teachers etc.
> Consider $\mathbb{R}^2$ with the Euclidean norm (or any other $\ell^p$-norm). If $f:\mathbb{R}\to\mathbb{R}$ is continuous, then $$A=\{(x_1,x_2)\in \mathbb{R}^2 : x_2<f(x_1)\}$$ is open in $\mathbb{R}^2$.
Any hints on how to approach showing this? I need to show that for every $x\in A$ there is some $r>0$ such that the open ball $B_r(x)$ is contained in $A$, but I just do not know where to start.
@sunny You could consider for a fixed $x$ the set $A(x):= \{(t,x)\in \mathbb R^2: x<f(t)\}$. $\bigcup_x A(x)= A$. If you show $A(x)$ to be open for every x then arbitrary union of open sets is open.
@Ajay Sure. Calculating the proportions of all the members of a decay chain is harder than calculating the proportions of something that decays directly to a stable isotope. Especially if the isotope can decay in multiple ways.
Also, as a rule of thumb, activity is roughly inversely proportional to half-life, so nuclides that have a short half tend to emit more energetic particles. So if an isotope with a long half-life has isotopes in its decay chain with a shorter half life, then that stuff can become more radioactive over time. You may like to calculate that activity. However, finding the energies of the emitted particles can be a little harder than finding half-lives, which you can just get from Wikipedia.
@Ajay Eg, if you start with 100 g of radium-226 (halflife ~1600 years), what will you have 100 years later? Maybe take a look at en.wikipedia.org/wiki/Decay_chain
Show that $(a, b]$ and $[c, d)$ have the same cardinality, where $a, b, c,d$ are real numbers and $a <b, c<d.$ This is a well-known classical question in real analysis, if I am not wrong. The thing is, these problems are new to me. I could prove that any two closed intervals in R have the same cardinality, but till now, I haven't been able to solve this problem (above). But I have a feeling that all real intervals be it open intervals or closed intervals all have the same cardinality.
Now, I have two questions to ask: How can I proceed to prove that $(a, b]$ and $[c, d)$ have the same cardinality?
(Till now, I have been trying to construct a bijection in between them, but neither of my attempts went well.) My second question, is: Whether my general opinion about equal cardinalities of all possible intervals is correct or not? If it it's true, how should I proceed.
@Ajay Sorry, maybe I said something misleading. We can get a rough idea of how radioactive an isotope is from nuclear models. But I wasn't suggesting that you try to do those calculations. All I was suggesting is that you find the emission data for your Fukushima isotopes & their daughter nuclei (from standard references) and use that to calculate the types & intensities of the alpha, beta, and gamma rays that your material emits over time.
hey everyone, i need to show that S, which is the set of all finite bit strings, is countable. i tried using the bernstein-schroder theorem by doing the following:
1. Injection from N to S - each number has a unique representation in the binary system.
2. Injection from S to N - We first prepend each binary string with "1", so that for example "0001" becomes "10001". we do this so "1" and "0001" are not mapped to the same element in N.
Are these two mappings injections?
forgot to mention then we convert each prepended bit string to its decimal representation, which essentially is the mapping
@Koro what you said, seems much obvious. But if we think abt cardinal numbers, then the prblm asks us to show, that excluding fintely many elements leaves the cardinal number unchanged- Which is something , I feel not eligible to try considering my course. Also, it would be very helpful if you consider telling how will it help with my concerns?
Hi guys! This was a question concerning Derived sets.
The two questions are very trivial, for eg: The derived set in 1st case, will be $\{1/2^n,1/3^m,1/5^p ,\sum (1/2^n+1/3^m),\sum (1/5^p+1/3^m),\sum(1/2^n+1/5^p),0\}$ and in case of the second problem, it will be all real nos.
But the thing, is, do I need to prove that there does not exist any other limit points for both the cases, apart from them ? Or will these listing suffice ?what is your general opinion ?
@ThomasFinley you said in your message that you have proved that two closed intervals have the same cardinality. So [a,b] and [c,d] (where a<b. c<d, a,b,c,d are real nos.) have the same cardinality. Now, by what I asked you to prove: removing a from [a,b] will not change cardinality, i.e., [a,b] and (a,b] have the same cardinality.
(this is because [a,b], a<b is also an infinite set.)
@TedShifrin $\{1/2^n\}$ is by it's own right a convergent sequence converging to $0$, so any of it's (sub)sequences converges to $0$ as well. So, $0$ is the only limit point.
@TedShifrin ok, but how? I mean, in this case we had only one limit point, but in that case, it has more than one. I don't see how the exact same argument applies there directly.
Are you suggesting, we find out all possible convergent sequences that are in that given set? And then, apply this argument?
@TedShifrin Sorry but I am not sure what ur talking about, $m,n,p\in\Bbb N$ so, there isn't any definite maximum value of $m,n,p$ . (But, surely u never meant this,right?)
@TedShifrin Ah, you meant the number of $\frac{1}{2},\frac{1}{3},\frac{1}{5}$ appearing in each term of a subsequence? If not, then I surely do smell some miscommunication here.
@TedShifrin whatever, you call it, it's just that: one just needs to state how he/she is getting each limit points of the set given, i.e by first fixing each power and varying the other two to infinity, then, by fixing two powers out (m,n,p) and varying the left power (to infinity) and finally, varying all 3 of the powers to infinity- This way/method/procedure/reflex action, will suffice as a proof to the thing. Is it so ?
@XanderHenderson Just to be clear, when I say “Islam is worse” I’m talking about all of it, not merely the Quran. You should hear some of the stuff their clergy can say openly without anyone batting an eye. I’ve heard it personally and it took all of my mental strength to resist punching that idiot and breaking his skull
@TedShifrin By natural human intuition, I strongly feel yes. But I have no words to prove that it exhausts all possibilities, and it might be so that no has a "true" proof for the fact, that going like this way, I mentioned exhausts, all possibilities (?) (It's probably a kind of affirmation we get from within)
@TedShifrin Now, studying that requires some more concept of sequences in higher dimensions. But yes, I worked on some sequences, $(a_k)$ where each term, say $a_n$ is an ordered tuple of the form $(m,n,p)$ but I don't understand, what sort of conclusion you insist upon deriving from there ?
@TedShifrin I haven’t even heard of it, there’s no way it’s the ‘best car ever built’ there’s too many candidates for that position, not to mention, you have to define what you mean by “best”
@shintuku hmm...how can I check if my function that maps a number $\mathbb{N}$ to its binary equivalent is injective if I don't have an explicit formula $f(n) = ... n \in \mathbb{N}$?
@TedShifrin Yes, (now this is starting to get funny) the thing, is, I could infer that. But then? Sorry, if I sounded ambiguously, I mean what relation are you insisting on hatching (from either of these) ?
@TedShifrin Ah, um, ok. Yes, we might put it in somewhat this way. But just to clarify things at hand, concerning the question stated like that, will it suffice, if I only, write out the set of limit points just like how I wrote in the OP as an "answer"? Do you suggest writing out all these things we did in our answer to that particular question?
@THE_CRANIUM maybe more formally: you show the function is injective on domain {0,1,2}, then suppose it is injective on {0,1,...,n} and show it is injective on {0,1,...,n+1}
@THE_CRANIUM this will give you a set of functions $f_1, \cdots, f_n, \cdots, $ on domains $\{0\}, \cdots, \{0, \dots, n\}, \cdots$, you take the set of all of these functions and take its union, for which you need to prove that union of injective functions agreeing on common domains is an injective function
this will complete a formal proof of the injectivity of your desired function
@SineoftheTime I just ranted on that question about the sloppiness of confusing an element of $\Bbb Z_n$ (which is an equivalence class) with an integer. The business with associates is annoying. Saying $\gcd(a,b)\sim D$ rather than $\gcd(a,b)=d$ is annoying, but it allows the $\pm1$. Why do you think there are other correct answers?
