10
\$\begingroup\$

You must be wondering, why is this a so important subject? Well, take the example where a challenge asks you to take a "natural number" as input. What's a natural number? A non-negative integer, or a positive integer? The same would happen with Fibonacci numbers, do we include the 0 or not?

Another example would be for a challenge asking you to handle a case where you have to do 0⁰. Does that equal 1 or 0?

In both cases, and generally in a case where a mathematical concept has many accepted definitions and the question doesn't specify which one to use, one would either

  • wonder what definition to follow, or
  • perceive it as "too obvious" and use the definition they see themselves fit.

In the first case, one can simply leave a comment asking the original poster what definition to use. In the second case, they will use a particular definition to make an answer, and then there are many possibilities, so that the definition used can be:

  • free for the answerer to choose,
  • the one the original poster intended, or
  • different from the intended definition.

I emphasize on the last case, since there can be the one answer with the wrong definition, or a swarm of answers with wrong definitions, the latter being more difficult to face, so more of a problem.

To avoid such cases of ambiguity, what should be done about it?

P.S. This discussion actually arose from a CMC!

| |
\$\endgroup\$
  • \$\begingroup\$ Another (less ambiguous) example could be for people who include 1 in the primes \$\endgroup\$ – Stephen Jun 27 '17 at 14:14
  • \$\begingroup\$ @StephenS I thought of that but...they do? I don't think so, the only definition I could find is "a natural number greater than 1 that only has divisiors 1 and itself". \$\endgroup\$ – Erik the Outgolfer Jun 27 '17 at 14:15
  • \$\begingroup\$ I don't think they are anyone who's studied it, but it's natural to assume, before studying prime theory or whatever, that 1 is prime since it doesn't have any factors (it's only not prime by definition and because that's how mathematicians use primes) \$\endgroup\$ – Stephen Jun 27 '17 at 14:17
  • \$\begingroup\$ @StephenS I think that's another issue though. \$\endgroup\$ – Erik the Outgolfer Jun 27 '17 at 14:18
  • 1
    \$\begingroup\$ AFAIK, it's just as accepted by mathematicians that 0^0=1 and 1 is not prime \$\endgroup\$ – Stephen Jun 27 '17 at 14:21
  • 3
    \$\begingroup\$ @StephenS Mathematically, 0^0 is as undefined as 0/0. \$\endgroup\$ – Martin Ender Jun 27 '17 at 14:21
  • \$\begingroup\$ @MartinEnder Well, the latter could've been just , hadn't been defined as undefined... \$\endgroup\$ – Erik the Outgolfer Jun 27 '17 at 14:24
  • \$\begingroup\$ @EriktheOutgolfer 0/0 = , 0/-0 = -∞, so it's undefined \$\endgroup\$ – Stephen Jun 27 '17 at 14:24
  • 2
    \$\begingroup\$ @EriktheOutgolfer Or it could have been defined as 0 (because it's 0/x) or as 1 (because it's x/x), which is why it's undefined (same for 0^0 which could be 0 or 1, depending on whether it's 0^x or x^0). \$\endgroup\$ – Martin Ender Jun 27 '17 at 14:25
  • \$\begingroup\$ @MartinEnder I think x/x where x≠0 is defined as 1 since x=x*1...for 0, it can even be 0*∞...wait, division is borked, since it could be 0*x=0 for any x...pretty much that's the definition of undefined. \$\endgroup\$ – Erik the Outgolfer Jun 27 '17 at 14:27
  • 2
    \$\begingroup\$ @MartinEnder, both of those are contextually defined. In practice you never see 0/0: you see f(x)/g(x) where lim x->a: f(x) = g(x) = 0, and then you evaluate f(a)/g(a) using l'Hôpital's rule; similarly you never see 0^0 but f(x)^g(x). Ask a combinatorialist what 0^0 evaluates to and they'll say 1 because in any combinatorial context in which it occurs that is the appropriate limit. \$\endgroup\$ – Peter Taylor Jun 27 '17 at 15:38
25
\$\begingroup\$

The challenge poster must specify the intended definition

If such a mathematical ambiguity is spotted, leave a comment and vote/flag to close the challenge as "unclear what you're asking".

The reason why? Well, this is the close reason one should use if they think a question isn't clear enough for them to answer. The closure will prevent such possible swarms of answers being posted, provided the challenge gets closed quickly enough.

So far so good, but one shouldn't need to deal with such a problem if they post their challenge in the Sandbox first, another reason sandboxing is encouraged.

| |
\$\endgroup\$
  • 3
    \$\begingroup\$ I think ask first, VTC later is an important principle. Voting to close should be a last resort. \$\endgroup\$ – isaacg Jun 28 '17 at 3:43
  • 1
    \$\begingroup\$ @isaacg VTC isn't the same as downvote, and in this case a fast closure is way better imo...if you disagree you can downvote or post your own answer. ;) \$\endgroup\$ – Erik the Outgolfer Jun 28 '17 at 8:38
  • 7
    \$\begingroup\$ @isaacg, votes to reopen also exist. If there's a definite ambiguity then the question needs to be closed before there are conflicting answers such that one of them will have to be deleted no matter how the ambiguity is resolved. \$\endgroup\$ – Peter Taylor Jun 28 '17 at 9:59
  • \$\begingroup\$ @PeterTaylor And it'll be just one only if you're lucky enough...what if they're, say, 15 or 20? \$\endgroup\$ – Erik the Outgolfer Jun 28 '17 at 10:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .