12 ms·

What's the difference between 0/0 and 1/0?

194 points by korethr 7y ago

saagarjha 7y ago
> Last year a new Math Stack Exchange user asked What's the difference between 0/0 and 1/0? I wrote an answer I thought was pretty good, but the question was downvoted and deleted as “not about mathematics”. Never change, Stack Exchange.
- kouh 7y ago
  There is something inextricably human about this sort of anecdote though, like modern reenactments of the drowning of Hippasus by the Pythagoreans after bringing up irrational numbers.
  - edoo 7y ago
    He showed that the square root of 2 was an irrational.
- jandrese 7y ago
  It's more hostile than Wikipedia to new users. Ask a tough question and there's a decent chance of getting the dreaded 1 downvote and lock.
  - btrettel 7y ago
    This hasn't been my experience. I've asked questions that in retrospect were fairly stupid and I've only been treated with respect.
    - starpilot 7y ago
      how does one shot web
    - jandrese 7y ago
      My experience is more like this: https://serverfault.com/questions/966765/private-network-dns-configuration https://serverfault.com/questions/966765/private-network-dns... Closed due to "off-topic". With a comment that basically says "I don't like your problem, try to be more normal."
      - ciupicri 7y ago
        You basically had a problem related to a DNS server, so you thought that Server Fault was the place to ask for help. In reality it's more related to workstations, i.e. try asking on Super User or Unix & Linux Stack Exchange. I suggest having a look at dnsmasq [1]: > -S, --local, --server=[/[<domain>]/[domain/]][<ipaddr>[#<port>][@<source-ip>|<interface>[#<port>]] > Specify IP address of upstream servers directly. Setting this flag does not suppress reading of /etc/resolv.conf, use --no-resolv to do that. If one or more optional domains are given, that server is used only for those domains and they are queried only using the specified server. This is intended for private nameservers: if you have a nameserver on your network which deals with names of the form xxx.internal.thekelleys.org.uk at 192.168.1.1 then giving the flag --server=/internal.thekelleys.org.uk/192.168.1.1 will send all queries for internal machines to that nameserver, everything else will go to the servers in /etc/resolv.conf. [1]: http://www.thekelleys.org.uk/dnsmasq/docs/dnsmasq-man.html http://www.thekelleys.org.uk/dnsmasq/docs/dnsmasq-man.html
        jandrese 7y ago
        Note that the closing message didn't suggest better places to ask though. And your advice is exactly what I was looking for. Thank you.
      - deleted 7y ago
        [deleted]
    - ineedasername 7y ago
      Yeah, browsing a few if my own old questions always makes me think, "seriously? Was I that bad at Google-fu?" Though generally it boiled down to having a problem, and not knowing enough to even ask Google the right questions.
  - wuliwong 7y ago
    I found wikipedia's interface to be pretty unfriendly for me as a new user. Stackoverflow was a lot easier for me to get started with. Maybe that lower barrier to entry leads to more random posts from people that are "low quality?" Side note: I find a similarity between SO and HN. Both can feel pretty hostile and over-the-top with penalizing you for not saying things "the right way." But as much as it does annoy me sometimes, these are two of the most useful websites for my purposes.
    - human20190310 7y ago
      Stack Overflow is worse because the questioners are approaching it in the vulnerable state of "not knowing". This makes the smack-downs worse, because the user already had a problem, and now they have haters as well.
      - wuliwong 7y ago
        Maybe worse for you. My experience certainly contradicts your claim. And I suspect that your claim is anecdotal as is mine. The difference is I am only speaking about my experience where you are attempting to make a sweeping judgement based on your experience.
        human20190310 7y ago
        Stack Overflow is great for me; I was empathizing with someone in the position of receiving criticism instead of answers.
- gowld 7y ago
  https://math.stackexchange.com/questions/26445/division-by-0 https://math.stackexchange.com/questions/26445/division-by-0 has a long discussion of this problem.
- JdeBP 7y ago
  It's Mathematics Stack Exchange, not Stack Overflow.
  - saagarjha 7y ago
    Fixed, thanks.
  - RyJones 7y ago
    It's pretty much the same across all of the properties, though. I quit using them; they're run poorly and, especially in system administration or programming areas, the quality is uselessly low
    - inopinatus 7y ago
      Years ago it was the first place I looked - but now, Stack Overflow has a currency problem. Many programming answers are obsolete, the languages or frameworks to which they relate having evolved, and this is reinforced by aggressive moderation policy and toxic attitudes. Not sure if the curators are unaware or simply don't care as long as they're still getting fresh questions for whatever's shiny and new. In practice, nonetheless, it is now the last place I look, not the first, and also the last place I contribute. There are plenty of other forums for relevant and current answers, and without dripping poison.
      - na85 7y ago
        >There are plenty of other forums for relevant and current answers, and without dripping poison. Would you mind listing a few examples?
        inopinatus 7y ago
        They tend to be language or framework specific. e.g. for Ruby/Rails questions I haunt the GoRails slack and a couple of Ruby/Rails specific subreddits. In quite a few cases the official dev forums for a particular vendor or technology are actually very active, and I think that's because forums just got a lot easier to implement. For example I get a lot back from the Apple Developer forums (not the public discussion community, which is hot garbage) and Vue.js's own forum. A fair chunk of what I previously used Stack Overflow for (finding code snippets to learn from) has been replaced by the ease of browsing public repositories on Github.
- felipellrocha 7y ago
  I just now assume that whenever I post there, the first few comments will be people berating me out for going out of my comfortable zone to ask/answer a question.
  - BMorearty 7y ago
    For a second there I assumed you were talking about HN.
    - kbenson 7y ago
      Eh, as long as you take a bit of care to make the question come across as genuine and not as snark or feigning lack of knowledge to draw out a comment to be hostile to (which people here are often primed to interpret things as), I think you'll generally either get a useful response and upvotes or no response. Working around other people's hangups is a real pain, but that's basically what a large portion of participating in a pseudo-anonymous internet forum is about, unfortunately.
      - ineedasername 7y ago
        Yeah, if someone says here, "do you have any examples to share?" then it can be hard to distinguish between an honest request for details vs. a passive aggressive way of saying "I think you're an idiot who can't back up their claims"
  - sockpuppet999 7y ago
    I've felt the same way here many times. Although I'm trying to keep it under control now. I understand the mods motivation a little bit better
- dreamcompiler 7y ago
  Almost every useful answer I've ever found on SE or SO has been marked down with some snarky judgmental horseshit meta-comment. At this point, if I don't see such markdowns, I assume the answer isn't very useful until proven otherwise.
  - ultrarunner 7y ago
    I wonder if this is a repeatable phenomenon wherein a universally-useful resource is more likely to be accessed by a wider gamut of people, thus essentially guaranteeing a judgmental response. That would mean those snarky responses really are a mark of a good and useful question.
    - wallace_f 7y ago
      After the world solved the scarcity problem with respect to everyone's ability to survive a normal, healthy, dignified lifespan, most of the economy continued to shift towards conspicuous consumption. If you look honestly at humanity, the motivations of the average person are selfish and unenlightened. An internet mod enjoys authority and an elevated sense of self worth by putting other people down. There really are some people genuinely enthused by virtues such as intellectual discovery and shared dignity, but they're in short supply and likely to be outnumbered in the population clamoring to become internet mods. I've had a few posts removed from a SO site after I corrected someone who happened to be a mod there. And the Innocence Project fights to give innocent people their lives back because some prosecutors don't want a scratch on their resume. I think it would be better if we acknowledged these realities more-often rather than pretending all the evil people in the world died in Nazi Germany, or wherever, and they weren't just normal people. I think this is also why I always retreat into nerdy or intellectual endeavors. Not because I am particularly intelligent, but because the real world is so ugly.
      - ineedasername 7y ago
        I agree with everything you said, except that first part about solving the scarcity problem. I'd need you to elaborate there, because it doesn't seem like that's been done in many parts of the world.
- TrevorJ 7y ago
  The worst part about it is, all these deleted/locked questions rank really high on google so they pollute the search results.
  - deleted 7y ago
    [deleted]
LoSboccacc 7y ago
as most of these discussion they go from "these number are impossible!" to "but wait! there's algebra!" without really providing a framing context of what's going on. this is what's going when the article say "slides smoothly in toward 2" https://www.google.com/search?q=(x^2%2B2x)/x https://www.google.com/search?q=(x^2%2B2x)/x and this is another common example, which "slides" toward a different number (which is why when you see the division alone is "undefined" and only acquires value in context) https://www.google.com/search?q=sin(x)/x https://www.google.com/search?q=sin(x)/x but beyond the "just apply limits" party trick it's the implication however that's the most interesting part, and I'll use the words from here[1]: > Things that appear to be zero may be nonzero in a different dimension (just like i might appear to be 0 to us, but isn’t) [1] https://betterexplained.com/articles/why-do-we-need-limits-and-infinitesimals/ https://betterexplained.com/articles/why-do-we-need-limits-a...
lucasgonze 7y ago
I'm inspired by how clean his reasoning is. Something basic but fuzzy which we refer to all the time - "divide by zero is undefined" - actually contains meaning! This makes me happy.
ggm 7y ago
The one thing his article didn't say for the arithmetically challenged person (me) is very much about 1/0 - he certainly talked about anything/0 and function()/0 but dived off why 1/0 and 0/0 have subtle difference. It's a good read but he missed his clear mission. One extra paragraph?
- mannykannot 7y ago
  The middle section is the author's answer: they are different because they fail to yield a definite answer in different ways and for different reasons. You will find some alternative ways of looking at it in the comments here.
marknadal 7y ago
The main argument I've seen against division by zero is because: "Then 2 + 2 could equal 7!" If you go through the examples of how they derive that, you can actually fix it quite easily by assigning each 0/0 pair to different variables. This prevents the underhand commutative rule violation. Then all that you are left with is linear algebra. Whether that equation is solvable depends upon how many constraints you have. But it at least doesn't cause math to violate itself.
- throwawaymath 7y ago
  That doesn't work. You're changing the setting of the problem from scalars to vectors in order to avoid breaking the uniqueness properties of fields and rings. The solution set for a system of linear equations is not unique (if it is at all) in the same sense that e.g. additive and multiplicative identity elements are unique.
- maxk42 7y ago
  My layman's interpretation: Taking the author's argument that a/b = the number x, for which x * b = a, it's easy to draw this out to it's logical conclusion: For numerators (that is: a) other than 0, you get impossible equations. That is: 1 / 0 = "the number which, when multiplied by 0 is 1." In equation form: x * 0 = 1 Well no such number exists. Therefore x can be said to be "no number." But for the case where a and b are both zero, the equation becomes: x * 0 = 0. And in this case, the answer is any number, since any number multiplied by zero = 0. So we can think of 0 / 0 as "any number" and non-0 / 0 as "no number". That means assigning each instance of 0/0 to a variable is a very good solution. Because we use a letter to represent an unknown number in basic algebra, but in linear algebra we also use a letter to represent a value which could be any number. As in the equation for a line: y=mx+b, where x and y can both take on any value.
- mjd 7y ago
  I wrote a rather detailed article on this exact topic. You might enjoy it. https://math.stackexchange.com/a/125208/25554 https://math.stackexchange.com/a/125208/25554
tim333 7y ago
Well, in javascript 0/0 == NaN while 1/0 == Infinity. An interesting thing is they have the value -0 so you can set a=-0 and 1/a returns -Infinity.
- strainer 7y ago
  This is not only javascripts behavior, it is the IEEE standard for floating-point arithmetic which most modern hardware and software strictly adheres to.
- dylan604 7y ago
  This should be part of an update to the Destroy All Software's Wat video: https://www.destroyallsoftware.com/talks/wat https://www.destroyallsoftware.com/talks/wat
  - garmaine 7y ago
    Except that is defined, standard IEEE 754 behavior and is true on nearly all programming languages.
- stOneskull 7y ago
  -0 is my favorite number
ikeboy 7y ago
>When f(0)≠0, we can state decisively that there is no such Q This is false. This relies on an assumption not stated - that f is continuous. For a simple counterexample: f(x) = 1, x=0 f(x) = x, x!=0 g(x) = x Limit of the ratio as x approaches 0 is 1.
TomMckenny 7y ago
>I wrote an answer I thought was pretty good, but the question was downvoted and deleted as “not about mathematics”. Why post on some else's site? Especially quality work that took real effort. I'm guilty of it too (although nothing so cool is this article). We complain about the web becoming dangerously centralized compared to the past. Much of this is because of free high grade labor that adds value to an already dominant organization.
- jobigoud 7y ago
  The usual answer to that is "discoverability". You can either post your content to someone else's website and have it read/viewed/watched, or post it where virtually no one will see it so you might as well not post it.
  - TomMckenny 7y ago
    Yes but discoverability to what end? In one case the reader then thinks and says "I learned this on Stack Exchange" rather than a possibility of "I learned this from Mark". In one case an organization that is much complained about becomes even more of a necessity and so has less incentive to fix itself. In another the reverse happens. In hyperbole: in one case you may as well not post it, in the other case you may as well have not written it.
8note 7y ago
my calculus answer is that usually you see 0/0 when looking for limits 1/0 definitely has no real value 0/0 might have one, but you've framed the question poorly -- if you rewrite the problem in a different way, you might find the value of 0/0 for your case eg, by applying L'Hôpital's rule
ddxxdd 7y ago
Someone needs to copy this poster's simplistic style, and use it to explain to me why Laplaces equation of electrostatic potential is solved with Bessel functions times an exponential times a sine/cosine wave (edit: in cylindrical coordinates).
- codethief 7y ago
  I suppose you've already looked at: https://en.wikipedia.org/wiki/Cylindrical_harmonics https://en.wikipedia.org/wiki/Cylindrical_harmonics ?
  - ddxxdd 7y ago
    No, but that page has all the info that Jackson's Classical E&M textbook already explains. I was being slightly facetious, but any average person will tune out a complete derivation after 15 integral signs, 50 terms, and 100 factors unless there is a pleasing narrative woven throughout the discussion.
    - codethief 7y ago
      Haha, I feel you!
- mjd 7y ago
  I haven't done electrostatic potential yet, but I have been blogging for 13 years. Maybe you would enjoy one of my other articles. I like this one: https://blog.plover.com/math/60-degree-angles.html https://blog.plover.com/math/60-degree-angles.html
- whatshisface 7y ago
  Here's the answer: an exponential times a sine/cosine wave is what you already have in rectangular coordinates. The Bessel functions show up because they are the cylindrical coordinate equivalent of sine, cosine and exponential. "What does that mean," you say? It is quite simple, as sine, cosine, the exponential function and all Bessel functions are defined as the solutions to specific "fundamental" differential equations. The basic thread between solving Laplace's equations in any coordinate system is, you separate variables and break down the equations in other ways until you have factored them into a set of standard ("prime?" I don't know if that's a good word, I'm not a mathematician) DEs that somebody like Bessel has tabulated the numerical solution of.
tzs 7y ago
Well, 1/0 - 0/0 = (1-0)/0 = 1/0, so the difference is 1/0. :-)
- aijony 7y ago
  or -1/0
  - ngcc_hk 7y ago
    Or whatever/0 For the article what it try to do is using limit to see whether the difference converge to a finite number diff(x) = one(x)/zero’(0) - zero(x)/zero’’(0) Where one and all zero function approach 1 and 0, smooth etc Find diff(x) when lim x-> 0
codethief 7y ago
The author's explanation is based on how the rationals are defined explicitly but this leaves the question open whether the rationals couldn't be extended somehow, so that the inverse of 0 is suddenly defined in this extension. More generally, couldn't there be other fields where 0 has a multiplicative inverse? Of course the answer to both questions is negative but it's the consequence of a much more general fact: In any field the additive zero element can never have a multiplicate inverse. (If it did, the field's other algebraic operations would suddenly lead to contradictions.) And in this sense, there is also no difference between the expressions 1/0 = 1 · 0⁻¹ and 0/0 = 0 · 0⁻¹ because 0⁻¹ simply doesn't exist. I've always found this explanation much more illuminating. No matter what you do, you can never have 0⁻¹!
- sunstone 7y ago
  Yes this seems like the right answer, both results are undefined so they are the same but not equal.
- daveFNbuck 7y ago
  You can have a multiplicative inverse of 0 if 0 = 1. This results in a field with only a single element.
  - gizmo686 7y ago
    Unless your definition of field include the non-triviality axiom that 0 != 1.
  - Certhas 7y ago
    Not to be confused with the field with one element :P https://en.wikipedia.org/wiki/Field_with_one_element https://en.wikipedia.org/wiki/Field_with_one_element
  - throwawaymath 7y ago
    The definition of a field explicitly requires the additive and multiplicative identities to be distinct elements. You can't induce a field structure on a singleton set. You can construct a trivial ring, but the trivial ring does not comprise a trivial field because it fails to be an integral domain. To be fair this is mathematical pedantry when we're talking about trivial objects. But in principle it's meaningful because every field you can construct will lose critical properties if you allow one element to be both the additive and multiplicative identity. Unfortunately authors don't always make it clear that you need both existence and uniqueness when they state the field axioms.
graycat 7y ago
To answer the question, move to the extended real numbers as in, say, H. Royden, Real Analysis. In short, the extensions work and at times are convenient but work less well than the rules of the real numbers; the extensions are a bit tricky and have to be careful.
pugworthy 7y ago
In a kind of both joking way, and in a serious as a programmer way, either way is the same - and not exceptional. I mean beyond that it will raise an exception. Either way, you divided by zero.
gizmo686 7y ago
>But to really understand the difference you probably need to use the calculus approach I am probably showing my bias as an algebraist here, but there is a purely algabraic way of seeing the difference between 0/0 and 1/0, which is (in my view) more intuitive. The standard way of constructing the field of fractions is to start with the integers (or your choice of base ring), and define an equivalence relation on the set: Q = { (x,y) | x,y are integers, y!=0 } Given by: (a,b) ~ (c,d) iff ad = bc Intuitively, the pair (a,b) is the fraction a/b, and the equivalence relationship is defining that: a/b = c/d is equivalent to ad =bc (assume b,d are non-zero). From that you can define multiplication and division in the expected way and verify that the resulting structure is a field that behaves as you would expect the rationals to behave. One thing to notice in the above view is that the prohibition on dividing by 0 is entirely artificial. By removing 6 characters from the above definition, we arrive at the following structure: W = { (x,y) | x,y are integers } (a,b) ~ (c,d) iff ad = b Let [W] be the set of equivalence classes on W, and [Q] be the set of equivalence classes on Q. If we define addition and mulitplication on [W] and [Q] in the "obvious" way, then [Q] behaves identically to the field of rational numbers (indeed, this is a common construction. Further, [W] contains [Q] as a subset, so we can view [W] as an extension to the rational numbers. We can verify that [W] contains exactly two elements not present in [Q], which I will denote: ⊥ = [(0,0)] = 0/0 ∞ = [(1,0)] = 1/0 = a/0 for a!=0 and a!=⊥ The question now becomes, is "division" meaningful on [W]. If we keep the same definition of multiplication as with [Q], it is clear that we cannot define an inverse operation [0]. However, we can define a unary operation called the reciprical, given by: /[(x,y)] = [(y,x)] And define a division operation as multiplication by the reciprical. So: [(a,b)] div [(x,y)] = [(a,b)] * (/[(x,y)]) = [(a,b)] * [(y,x)] = [(ay,bx)] And note that, on the subset [Q], this operation is identical to our standard notion of division. With this, we have extended [Q] into a structure, [W], for which division by 0 is defined. Further, by doing so, we added exactly 2 elements: 0/0 and 1/0. So we can see that 0/0 and 1/0 are distinct (in this structure) in a way that 1/0 and 2/0 are not. (as 1/0 ~ 2/0 in the same way that 1/1 ~ 2/2) In fact, the behaviour of ⊥ ∞ should not be suprising to this forum. ⊥ behaves essentially like NaN in that it "absorbs" the other number in all operations. So: x + ⊥ = ⊥ and x * ⊥ = ⊥. ∞ behaves mostly like you would expects: 0 * ∞ = 0, x * ∞ = ∞ | x != 0 and x != ⊥ x + ∞ = 0 | x != ⊥ The main "weirdness" here is that ∞=-∞ In fact, this [W] I have been describing is an established structure refered to as a Wheel [1], and can be constructed over the real or complex numbers just as easily (it can also be viewed as an extension of the more well known Riemann sphere by adding ⊥) [0] I will note that the lack of division by 0 means that we cannot define an inverse operation for multiplication in [Q] either. [1] https://en.wikipedia.org/wiki/Wheel_theory https://en.wikipedia.org/wiki/Wheel_theory
cairo_x 7y ago
Nothing has an infinite number of nothings in it. One has zero nothings in it...? It also has an infinite number of somethings in it. And each something has another infinity of divisible somethings within that. Therefore anything that isn't nothing (in the universe of abstraction which doesn't exist in reality, but whatever) is infinite. OBVIOUSLY. NOTE: Hasn't read the article and is mathematically challenged.
Rhinobird 7y ago
For x⋅0=a (i.e. 1/0) no number fits. It's like a database search that finds nothing. It's a null. A point that is not in the set. For x⋅0=0 (i.e. 0/0) any number fits. Is there such a thing as an anti-null? In the infinite set, 0/0 is infinite. If a/0 isn't any number if you exclude 0, then a/0 should be zero. Tada! ...of course zero wasn't in the set, which makes it a null
meuk 7y ago
This can be explained clearly in two sentences: Division is the inverse of multiplication (a * b / b = a / b * b = a). However, multiplication by 0 is not injective (we have x * 0 = 0 for every x), so multiplication by zero can't have an inverse operation, and both 0/0 and 1/0 are undefined.
- EForEndeavour 7y ago
  This explanation correctly explains why 1/0 and 0/0 are not numbers, but doesn't answer the question of what makes 1/0 and 0/0 different. And they aren't both undefined. 1/0 is undefined because no value of x satisfies 1/0 = x. 0/0 is indeterminate because any value of x satisfies 0/0 = x.
  - throwawaymath 7y ago
    0/0 is only indeterminate if your axioms allow or require it to be. Strictly speaking you can define 0/0 to mean something, but you will lose a lot of useful properties along the way. The only time 0/0 is not indeterminate is when you're dealing with something exotic and obscure like an algebraic wheel. It's correct to say 0/0 is undefined because in the vast majority of cases where someone doesn't explicitly call out the algebraic setting, they're working with a field.
bahhh 7y ago
George Boole proposed something similar in The Laws of Thought: he considered a definition of c = a/b where a = c AND b (even if he didn't write it like that). Thus, 0/0 was an indeterminate value, and 1/0 was seen as an impossibility. Boole's example: if m means "is mortal" and h means "is human", then saying "there is no immortal humans" is h AND NOT m = 0, and Boole explains that from it you could deduce: m = h + 0/0 and not h meaning "mortals are humans and something indeterminate which is not human" here it's still George Boole trying to cram normal arithmetics into logic, but still.
yuriko 7y ago
I'm definitely joking, but $ node > 1.0/0.0 Infinity > 0.0/0.0 NaN
jchook 7y ago
I read a relevant article about how the semantics of division by Zero ultimately rest on arbitrary decisions of the mathematics community. Trying to dig this up I found these: - https://www.1dividedby0.com/ https://www.1dividedby0.com/ - https://www.hillelwayne.com/post/divide-by-zero/ https://www.hillelwayne.com/post/divide-by-zero/
askvictor 7y ago
Also fun: 0^0 (not an emoji; 0 to the power of 0). Windows Calculator gives an answer of 1; Excel throws a number error, while Android hedges its bets (shows "Undefined, or 1").
murkle 7y ago
Extended number line is one way to formalise this https://en.m.wikipedia.org/wiki/Extended_real_number_line https://en.m.wikipedia.org/wiki/Extended_real_number_line

194 points by korethr 7y ago

> Last year a new Math Stack Exchange user asked What's the difference between 0/0 and 1/0? I wrote an answer I thought was pretty good, but the question was downvoted and deleted as “not about mathematics”. Never change, Stack Exchange.

kouh 7y ago
There is something inextricably human about this sort of anecdote though, like modern reenactments of the drowning of Hippasus by the Pythagoreans after bringing up irrational numbers.
- edoo 7y ago
  He showed that the square root of 2 was an irrational.
jandrese 7y ago
It's more hostile than Wikipedia to new users. Ask a tough question and there's a decent chance of getting the dreaded 1 downvote and lock.
- btrettel 7y ago
  This hasn't been my experience. I've asked questions that in retrospect were fairly stupid and I've only been treated with respect.
  - starpilot 7y ago
    how does one shot web
  - jandrese 7y ago
    My experience is more like this: https://serverfault.com/questions/966765/private-network-dns-configuration https://serverfault.com/questions/966765/private-network-dns... Closed due to "off-topic". With a comment that basically says "I don't like your problem, try to be more normal."
    - ciupicri 7y ago
      You basically had a problem related to a DNS server, so you thought that Server Fault was the place to ask for help. In reality it's more related to workstations, i.e. try asking on Super User or Unix & Linux Stack Exchange. I suggest having a look at dnsmasq [1]: > -S, --local, --server=[/[<domain>]/[domain/]][<ipaddr>[#<port>][@<source-ip>|<interface>[#<port>]] > Specify IP address of upstream servers directly. Setting this flag does not suppress reading of /etc/resolv.conf, use --no-resolv to do that. If one or more optional domains are given, that server is used only for those domains and they are queried only using the specified server. This is intended for private nameservers: if you have a nameserver on your network which deals with names of the form xxx.internal.thekelleys.org.uk at 192.168.1.1 then giving the flag --server=/internal.thekelleys.org.uk/192.168.1.1 will send all queries for internal machines to that nameserver, everything else will go to the servers in /etc/resolv.conf. [1]: http://www.thekelleys.org.uk/dnsmasq/docs/dnsmasq-man.html http://www.thekelleys.org.uk/dnsmasq/docs/dnsmasq-man.html
      - jandrese 7y ago
        Note that the closing message didn't suggest better places to ask though. And your advice is exactly what I was looking for. Thank you.
    - deleted 7y ago
      [deleted]
  - ineedasername 7y ago
    Yeah, browsing a few if my own old questions always makes me think, "seriously? Was I that bad at Google-fu?" Though generally it boiled down to having a problem, and not knowing enough to even ask Google the right questions.
- wuliwong 7y ago
  I found wikipedia's interface to be pretty unfriendly for me as a new user. Stackoverflow was a lot easier for me to get started with. Maybe that lower barrier to entry leads to more random posts from people that are "low quality?" Side note: I find a similarity between SO and HN. Both can feel pretty hostile and over-the-top with penalizing you for not saying things "the right way." But as much as it does annoy me sometimes, these are two of the most useful websites for my purposes.
  - human20190310 7y ago
    Stack Overflow is worse because the questioners are approaching it in the vulnerable state of "not knowing". This makes the smack-downs worse, because the user already had a problem, and now they have haters as well.
    - wuliwong 7y ago
      Maybe worse for you. My experience certainly contradicts your claim. And I suspect that your claim is anecdotal as is mine. The difference is I am only speaking about my experience where you are attempting to make a sweeping judgement based on your experience.
      - human20190310 7y ago
        Stack Overflow is great for me; I was empathizing with someone in the position of receiving criticism instead of answers.
gowld 7y ago
https://math.stackexchange.com/questions/26445/division-by-0 https://math.stackexchange.com/questions/26445/division-by-0 has a long discussion of this problem.
JdeBP 7y ago
It's Mathematics Stack Exchange, not Stack Overflow.
- saagarjha 7y ago
  Fixed, thanks.
- RyJones 7y ago
  It's pretty much the same across all of the properties, though. I quit using them; they're run poorly and, especially in system administration or programming areas, the quality is uselessly low
  - inopinatus 7y ago
    Years ago it was the first place I looked - but now, Stack Overflow has a currency problem. Many programming answers are obsolete, the languages or frameworks to which they relate having evolved, and this is reinforced by aggressive moderation policy and toxic attitudes. Not sure if the curators are unaware or simply don't care as long as they're still getting fresh questions for whatever's shiny and new. In practice, nonetheless, it is now the last place I look, not the first, and also the last place I contribute. There are plenty of other forums for relevant and current answers, and without dripping poison.
    - na85 7y ago
      >There are plenty of other forums for relevant and current answers, and without dripping poison. Would you mind listing a few examples?
      - inopinatus 7y ago
        They tend to be language or framework specific. e.g. for Ruby/Rails questions I haunt the GoRails slack and a couple of Ruby/Rails specific subreddits. In quite a few cases the official dev forums for a particular vendor or technology are actually very active, and I think that's because forums just got a lot easier to implement. For example I get a lot back from the Apple Developer forums (not the public discussion community, which is hot garbage) and Vue.js's own forum. A fair chunk of what I previously used Stack Overflow for (finding code snippets to learn from) has been replaced by the ease of browsing public repositories on Github.
felipellrocha 7y ago
I just now assume that whenever I post there, the first few comments will be people berating me out for going out of my comfortable zone to ask/answer a question.
- BMorearty 7y ago
  For a second there I assumed you were talking about HN.
  - kbenson 7y ago
    Eh, as long as you take a bit of care to make the question come across as genuine and not as snark or feigning lack of knowledge to draw out a comment to be hostile to (which people here are often primed to interpret things as), I think you'll generally either get a useful response and upvotes or no response. Working around other people's hangups is a real pain, but that's basically what a large portion of participating in a pseudo-anonymous internet forum is about, unfortunately.
    - ineedasername 7y ago
      Yeah, if someone says here, "do you have any examples to share?" then it can be hard to distinguish between an honest request for details vs. a passive aggressive way of saying "I think you're an idiot who can't back up their claims"
- sockpuppet999 7y ago
  I've felt the same way here many times. Although I'm trying to keep it under control now. I understand the mods motivation a little bit better
dreamcompiler 7y ago
Almost every useful answer I've ever found on SE or SO has been marked down with some snarky judgmental horseshit meta-comment. At this point, if I don't see such markdowns, I assume the answer isn't very useful until proven otherwise.
- ultrarunner 7y ago
  I wonder if this is a repeatable phenomenon wherein a universally-useful resource is more likely to be accessed by a wider gamut of people, thus essentially guaranteeing a judgmental response. That would mean those snarky responses really are a mark of a good and useful question.
  - wallace_f 7y ago
    After the world solved the scarcity problem with respect to everyone's ability to survive a normal, healthy, dignified lifespan, most of the economy continued to shift towards conspicuous consumption. If you look honestly at humanity, the motivations of the average person are selfish and unenlightened. An internet mod enjoys authority and an elevated sense of self worth by putting other people down. There really are some people genuinely enthused by virtues such as intellectual discovery and shared dignity, but they're in short supply and likely to be outnumbered in the population clamoring to become internet mods. I've had a few posts removed from a SO site after I corrected someone who happened to be a mod there. And the Innocence Project fights to give innocent people their lives back because some prosecutors don't want a scratch on their resume. I think it would be better if we acknowledged these realities more-often rather than pretending all the evil people in the world died in Nazi Germany, or wherever, and they weren't just normal people. I think this is also why I always retreat into nerdy or intellectual endeavors. Not because I am particularly intelligent, but because the real world is so ugly.
    - ineedasername 7y ago
      I agree with everything you said, except that first part about solving the scarcity problem. I'd need you to elaborate there, because it doesn't seem like that's been done in many parts of the world.
TrevorJ 7y ago
The worst part about it is, all these deleted/locked questions rank really high on google so they pollute the search results.
- deleted 7y ago
  [deleted]

LoSboccacc 7y ago

as most of these discussion they go from "these number are impossible!" to "but wait! there's algebra!" without really providing a framing context of what's going on. this is what's going when the article say "slides smoothly in toward 2" https://www.google.com/search?q=(x^2%2B2x)/x https://www.google.com/search?q=(x^2%2B2x)/x and this is another common example, which "slides" toward a different number (which is why when you see the division alone is "undefined" and only acquires value in context) https://www.google.com/search?q=sin(x)/x https://www.google.com/search?q=sin(x)/x but beyond the "just apply limits" party trick it's the implication however that's the most interesting part, and I'll use the words from here[1]: > Things that appear to be zero may be nonzero in a different dimension (just like i might appear to be 0 to us, but isn’t) [1] https://betterexplained.com/articles/why-do-we-need-limits-and-infinitesimals/ https://betterexplained.com/articles/why-do-we-need-limits-a...

lucasgonze 7y ago

I'm inspired by how clean his reasoning is. Something basic but fuzzy which we refer to all the time - "divide by zero is undefined" - actually contains meaning! This makes me happy.

ggm 7y ago

The one thing his article didn't say for the arithmetically challenged person (me) is very much about 1/0 - he certainly talked about anything/0 and function()/0 but dived off why 1/0 and 0/0 have subtle difference. It's a good read but he missed his clear mission. One extra paragraph?

mannykannot 7y ago
The middle section is the author's answer: they are different because they fail to yield a definite answer in different ways and for different reasons. You will find some alternative ways of looking at it in the comments here.

marknadal 7y ago

The main argument I've seen against division by zero is because: "Then 2 + 2 could equal 7!" If you go through the examples of how they derive that, you can actually fix it quite easily by assigning each 0/0 pair to different variables. This prevents the underhand commutative rule violation. Then all that you are left with is linear algebra. Whether that equation is solvable depends upon how many constraints you have. But it at least doesn't cause math to violate itself.

throwawaymath 7y ago
That doesn't work. You're changing the setting of the problem from scalars to vectors in order to avoid breaking the uniqueness properties of fields and rings. The solution set for a system of linear equations is not unique (if it is at all) in the same sense that e.g. additive and multiplicative identity elements are unique.
maxk42 7y ago
My layman's interpretation: Taking the author's argument that a/b = the number x, for which x * b = a, it's easy to draw this out to it's logical conclusion: For numerators (that is: a) other than 0, you get impossible equations. That is: 1 / 0 = "the number which, when multiplied by 0 is 1." In equation form: x * 0 = 1 Well no such number exists. Therefore x can be said to be "no number." But for the case where a and b are both zero, the equation becomes: x * 0 = 0. And in this case, the answer is any number, since any number multiplied by zero = 0. So we can think of 0 / 0 as "any number" and non-0 / 0 as "no number". That means assigning each instance of 0/0 to a variable is a very good solution. Because we use a letter to represent an unknown number in basic algebra, but in linear algebra we also use a letter to represent a value which could be any number. As in the equation for a line: y=mx+b, where x and y can both take on any value.
mjd 7y ago
I wrote a rather detailed article on this exact topic. You might enjoy it. https://math.stackexchange.com/a/125208/25554 https://math.stackexchange.com/a/125208/25554

tim333 7y ago

Well, in javascript 0/0 == NaN while 1/0 == Infinity. An interesting thing is they have the value -0 so you can set a=-0 and 1/a returns -Infinity.

strainer 7y ago
This is not only javascripts behavior, it is the IEEE standard for floating-point arithmetic which most modern hardware and software strictly adheres to.
dylan604 7y ago
This should be part of an update to the Destroy All Software's Wat video: https://www.destroyallsoftware.com/talks/wat https://www.destroyallsoftware.com/talks/wat
- garmaine 7y ago
  Except that is defined, standard IEEE 754 behavior and is true on nearly all programming languages.
stOneskull 7y ago
-0 is my favorite number

ikeboy 7y ago

>When f(0)≠0, we can state decisively that there is no such Q This is false. This relies on an assumption not stated - that f is continuous. For a simple counterexample: f(x) = 1, x=0 f(x) = x, x!=0 g(x) = x Limit of the ratio as x approaches 0 is 1.

TomMckenny 7y ago

>I wrote an answer I thought was pretty good, but the question was downvoted and deleted as “not about mathematics”. Why post on some else's site? Especially quality work that took real effort. I'm guilty of it too (although nothing so cool is this article). We complain about the web becoming dangerously centralized compared to the past. Much of this is because of free high grade labor that adds value to an already dominant organization.

jobigoud 7y ago
The usual answer to that is "discoverability". You can either post your content to someone else's website and have it read/viewed/watched, or post it where virtually no one will see it so you might as well not post it.
- TomMckenny 7y ago
  Yes but discoverability to what end? In one case the reader then thinks and says "I learned this on Stack Exchange" rather than a possibility of "I learned this from Mark". In one case an organization that is much complained about becomes even more of a necessity and so has less incentive to fix itself. In another the reverse happens. In hyperbole: in one case you may as well not post it, in the other case you may as well have not written it.

8note 7y ago

my calculus answer is that usually you see 0/0 when looking for limits 1/0 definitely has no real value 0/0 might have one, but you've framed the question poorly -- if you rewrite the problem in a different way, you might find the value of 0/0 for your case eg, by applying L'Hôpital's rule

ddxxdd 7y ago

Someone needs to copy this poster's simplistic style, and use it to explain to me why Laplaces equation of electrostatic potential is solved with Bessel functions times an exponential times a sine/cosine wave (edit: in cylindrical coordinates).

codethief 7y ago
I suppose you've already looked at: https://en.wikipedia.org/wiki/Cylindrical_harmonics https://en.wikipedia.org/wiki/Cylindrical_harmonics ?
- ddxxdd 7y ago
  No, but that page has all the info that Jackson's Classical E&M textbook already explains. I was being slightly facetious, but any average person will tune out a complete derivation after 15 integral signs, 50 terms, and 100 factors unless there is a pleasing narrative woven throughout the discussion.
  - codethief 7y ago
    Haha, I feel you!
mjd 7y ago
I haven't done electrostatic potential yet, but I have been blogging for 13 years. Maybe you would enjoy one of my other articles. I like this one: https://blog.plover.com/math/60-degree-angles.html https://blog.plover.com/math/60-degree-angles.html
whatshisface 7y ago
Here's the answer: an exponential times a sine/cosine wave is what you already have in rectangular coordinates. The Bessel functions show up because they are the cylindrical coordinate equivalent of sine, cosine and exponential. "What does that mean," you say? It is quite simple, as sine, cosine, the exponential function and all Bessel functions are defined as the solutions to specific "fundamental" differential equations. The basic thread between solving Laplace's equations in any coordinate system is, you separate variables and break down the equations in other ways until you have factored them into a set of standard ("prime?" I don't know if that's a good word, I'm not a mathematician) DEs that somebody like Bessel has tabulated the numerical solution of.

tzs 7y ago

Well, 1/0 - 0/0 = (1-0)/0 = 1/0, so the difference is 1/0. :-)

aijony 7y ago
or -1/0
- ngcc_hk 7y ago
  Or whatever/0 For the article what it try to do is using limit to see whether the difference converge to a finite number diff(x) = one(x)/zero’(0) - zero(x)/zero’’(0) Where one and all zero function approach 1 and 0, smooth etc Find diff(x) when lim x-> 0

codethief 7y ago

The author's explanation is based on how the rationals are defined explicitly but this leaves the question open whether the rationals couldn't be extended somehow, so that the inverse of 0 is suddenly defined in this extension. More generally, couldn't there be other fields where 0 has a multiplicative inverse? Of course the answer to both questions is negative but it's the consequence of a much more general fact: In any field the additive zero element can never have a multiplicate inverse. (If it did, the field's other algebraic operations would suddenly lead to contradictions.) And in this sense, there is also no difference between the expressions 1/0 = 1 · 0⁻¹ and 0/0 = 0 · 0⁻¹ because 0⁻¹ simply doesn't exist. I've always found this explanation much more illuminating. No matter what you do, you can never have 0⁻¹!

sunstone 7y ago
Yes this seems like the right answer, both results are undefined so they are the same but not equal.
daveFNbuck 7y ago
You can have a multiplicative inverse of 0 if 0 = 1. This results in a field with only a single element.
- gizmo686 7y ago
  Unless your definition of field include the non-triviality axiom that 0 != 1.
- Certhas 7y ago
  Not to be confused with the field with one element :P https://en.wikipedia.org/wiki/Field_with_one_element https://en.wikipedia.org/wiki/Field_with_one_element
- throwawaymath 7y ago
  The definition of a field explicitly requires the additive and multiplicative identities to be distinct elements. You can't induce a field structure on a singleton set. You can construct a trivial ring, but the trivial ring does not comprise a trivial field because it fails to be an integral domain. To be fair this is mathematical pedantry when we're talking about trivial objects. But in principle it's meaningful because every field you can construct will lose critical properties if you allow one element to be both the additive and multiplicative identity. Unfortunately authors don't always make it clear that you need both existence and uniqueness when they state the field axioms.

graycat 7y ago

To answer the question, move to the extended real numbers as in, say, H. Royden, Real Analysis. In short, the extensions work and at times are convenient but work less well than the rules of the real numbers; the extensions are a bit tricky and have to be careful.

pugworthy 7y ago

In a kind of both joking way, and in a serious as a programmer way, either way is the same - and not exceptional. I mean beyond that it will raise an exception. Either way, you divided by zero.

gizmo686 7y ago

>But to really understand the difference you probably need to use the calculus approach I am probably showing my bias as an algebraist here, but there is a purely algabraic way of seeing the difference between 0/0 and 1/0, which is (in my view) more intuitive. The standard way of constructing the field of fractions is to start with the integers (or your choice of base ring), and define an equivalence relation on the set: Q = { (x,y) | x,y are integers, y!=0 } Given by: (a,b) ~ (c,d) iff ad = bc Intuitively, the pair (a,b) is the fraction a/b, and the equivalence relationship is defining that: a/b = c/d is equivalent to ad =bc (assume b,d are non-zero). From that you can define multiplication and division in the expected way and verify that the resulting structure is a field that behaves as you would expect the rationals to behave. One thing to notice in the above view is that the prohibition on dividing by 0 is entirely artificial. By removing 6 characters from the above definition, we arrive at the following structure: W = { (x,y) | x,y are integers } (a,b) ~ (c,d) iff ad = b Let [W] be the set of equivalence classes on W, and [Q] be the set of equivalence classes on Q. If we define addition and mulitplication on [W] and [Q] in the "obvious" way, then [Q] behaves identically to the field of rational numbers (indeed, this is a common construction. Further, [W] contains [Q] as a subset, so we can view [W] as an extension to the rational numbers. We can verify that [W] contains exactly two elements not present in [Q], which I will denote: ⊥ = [(0,0)] = 0/0 ∞ = [(1,0)] = 1/0 = a/0 for a!=0 and a!=⊥ The question now becomes, is "division" meaningful on [W]. If we keep the same definition of multiplication as with [Q], it is clear that we cannot define an inverse operation [0]. However, we can define a unary operation called the reciprical, given by: /[(x,y)] = [(y,x)] And define a division operation as multiplication by the reciprical. So: [(a,b)] div [(x,y)] = [(a,b)] * (/[(x,y)]) = [(a,b)] * [(y,x)] = [(ay,bx)] And note that, on the subset [Q], this operation is identical to our standard notion of division. With this, we have extended [Q] into a structure, [W], for which division by 0 is defined. Further, by doing so, we added exactly 2 elements: 0/0 and 1/0. So we can see that 0/0 and 1/0 are distinct (in this structure) in a way that 1/0 and 2/0 are not. (as 1/0 ~ 2/0 in the same way that 1/1 ~ 2/2) In fact, the behaviour of ⊥ ∞ should not be suprising to this forum. ⊥ behaves essentially like NaN in that it "absorbs" the other number in all operations. So: x + ⊥ = ⊥ and x * ⊥ = ⊥. ∞ behaves mostly like you would expects: 0 * ∞ = 0, x * ∞ = ∞ | x != 0 and x != ⊥ x + ∞ = 0 | x != ⊥ The main "weirdness" here is that ∞=-∞ In fact, this [W] I have been describing is an established structure refered to as a Wheel [1], and can be constructed over the real or complex numbers just as easily (it can also be viewed as an extension of the more well known Riemann sphere by adding ⊥) [0] I will note that the lack of division by 0 means that we cannot define an inverse operation for multiplication in [Q] either. [1] https://en.wikipedia.org/wiki/Wheel_theory https://en.wikipedia.org/wiki/Wheel_theory

cairo_x 7y ago

Nothing has an infinite number of nothings in it. One has zero nothings in it...? It also has an infinite number of somethings in it. And each something has another infinity of divisible somethings within that. Therefore anything that isn't nothing (in the universe of abstraction which doesn't exist in reality, but whatever) is infinite. OBVIOUSLY. NOTE: Hasn't read the article and is mathematically challenged.

Rhinobird 7y ago

For x⋅0=a (i.e. 1/0) no number fits. It's like a database search that finds nothing. It's a null. A point that is not in the set. For x⋅0=0 (i.e. 0/0) any number fits. Is there such a thing as an anti-null? In the infinite set, 0/0 is infinite. If a/0 isn't any number if you exclude 0, then a/0 should be zero. Tada! ...of course zero wasn't in the set, which makes it a null

meuk 7y ago

This can be explained clearly in two sentences: Division is the inverse of multiplication (a * b / b = a / b * b = a). However, multiplication by 0 is not injective (we have x * 0 = 0 for every x), so multiplication by zero can't have an inverse operation, and both 0/0 and 1/0 are undefined.

EForEndeavour 7y ago
This explanation correctly explains why 1/0 and 0/0 are not numbers, but doesn't answer the question of what makes 1/0 and 0/0 different. And they aren't both undefined. 1/0 is undefined because no value of x satisfies 1/0 = x. 0/0 is indeterminate because any value of x satisfies 0/0 = x.
- throwawaymath 7y ago
  0/0 is only indeterminate if your axioms allow or require it to be. Strictly speaking you can define 0/0 to mean something, but you will lose a lot of useful properties along the way. The only time 0/0 is not indeterminate is when you're dealing with something exotic and obscure like an algebraic wheel. It's correct to say 0/0 is undefined because in the vast majority of cases where someone doesn't explicitly call out the algebraic setting, they're working with a field.

bahhh 7y ago

George Boole proposed something similar in The Laws of Thought: he considered a definition of c = a/b where a = c AND b (even if he didn't write it like that). Thus, 0/0 was an indeterminate value, and 1/0 was seen as an impossibility. Boole's example: if m means "is mortal" and h means "is human", then saying "there is no immortal humans" is h AND NOT m = 0, and Boole explains that from it you could deduce: m = h + 0/0 and not h meaning "mortals are humans and something indeterminate which is not human" here it's still George Boole trying to cram normal arithmetics into logic, but still.

yuriko 7y ago

I'm definitely joking, but $ node > 1.0/0.0 Infinity > 0.0/0.0 NaN

jchook 7y ago

I read a relevant article about how the semantics of division by Zero ultimately rest on arbitrary decisions of the mathematics community. Trying to dig this up I found these: - https://www.1dividedby0.com/ https://www.1dividedby0.com/ - https://www.hillelwayne.com/post/divide-by-zero/ https://www.hillelwayne.com/post/divide-by-zero/

askvictor 7y ago

Also fun: 0^0 (not an emoji; 0 to the power of 0). Windows Calculator gives an answer of 1; Excel throws a number error, while Android hedges its bets (shows "Undefined, or 1").

murkle 7y ago

Extended number line is one way to formalise this https://en.m.wikipedia.org/wiki/Extended_real_number_line https://en.m.wikipedia.org/wiki/Extended_real_number_line