Every. Single. Answer. To this question seems to miss the point. A function is a subset of a Cartesian product. It consists of three data: (1) the domain, (2) the codomain, and (3) a selection of ordered pairs in the Cartesian product of the domain and codomain which satisfy certain properties (that is, a "mapping" which pairs every element in the domain to exactly one element of the codomain).
The formula given in the question could define many functions, depending on what the domain and codomain are supposed to be.
@XanderHenderson this is a typical approach used in many calculus texts (especially old ones). When a function is given by a formula in some intro calculus text, the domain is assumed to be the maximal subset of $\mathbb {R} $ for which the formula makes sense in the exactly the given form.
The modern approach is to specify the domain explicitly instead of asking users to guess it by the formula. In fact the idea that functions need a formula (at least a convenient one) had to be ditched somewhere during analysis of Fourier series.
@ParamanandSingh I know---I teach these kinds of classes, and it always frustrates me how they are treated. Thomas is at least fairly explicit about this at the start (he defines the "natural domain" of a function as the maximal domain on which the formula is defined).
But in the modern formality, a function needs a domain. If instructors were explicit about this, questions like the one cited above wouldn't need to be asked.
@XanderHenderson Luckily, Arturo already enjoyed that question:
Which is what you started with. What you need to understand is that your assertion that the disjunctive "loses information, namely that the object must belong to one of them" is quite simply utter nonsense. If you don't get that, you've actually learned nothing here. — Arturo Magidin7 hours ago
@XanderHenderson I and Jech and many other logicians disagree with your (2), and I think you also miss the point. The core concept of a "function" has nothing to do with set theory or real numbers. If you work in a typical set theory such as ZFC, then you would encode a function as a certain kind of set (which still does not encode codomain contrary to your statement), but that encoding is just an encoding. In a different foundational system you might use a different encoding.
However, all we need to know for this particular question is that functions need a domain (regardless of encoding), and we all agree on that.
I personally found the terminology "removable discontinuity" (that is how I learnt it) more useful than the "modern" point of view, namely that the function is not discontinous at a point where it is not defined. And I never had a problem with the assumption that the domain is meant to be the set of all real numbers $x$ for which $f(x)$ is a valid expression , unless explicitely stated otherwise.
Is this a duplicate? It sure does answer the question, but I wonder if it's too much of a generalization of the original question to be considered a proper duplicate.
@Peter But you are still specifying the domain: it is the largest set of numbers on which the function is defined. The question being asked there is fundamentally about whether or not $x=2$ is in the domain of that function---it is not. There is a removable singularity there (not a discontinuity---a function can't be discontinuous at a point where it isn't defined).
@Peter The question is stated wrong by the OP, actually. It isn't an if and only if , it's an only if. That is, if $x_n$ is prime, then $n$ must either be $0$ or of the form $2^k$.
It's just like : if $2^n-1$ is prime, then $n$ is prime. The reverse is not true (e.g. $n=11$).
@MartinR: unfortunately I don't see the read more button. It may be because I see the page on a mobile device. Once they enable full responsive design on profile pages it should work fine. Till then I have to just avoid reading any bio on profile pages.
:59151344 Sorry, @Xander, it was not the linked post that has habitual answerers of PSQ (see immediately below). I'll flag, and be in touch when I encounter one or both of them again.
$\displaystyle\frac{(x + ny)}{(n+1)} = y + \frac{(x-y)}{(n+1)}$
I dont understand how one can go about from the first term to the second. Can somebody please explain the steps?
Best regards
$\displaystyle\frac{(x + ny)}{(n+1)} = y + \frac{(x-y)}{(n+1)}$
I dont understand how one can go about from the first term to the second. Can somebody please explain the steps?
Best regards
