@NathanMerrill 1. yes. 2. by presenting a useful error message and requiring an explicit specification of the interface to be used (by cast or similar). 3. see 1. 4. see 2. 5. in that case you could consider going with the more specific overload.
I mean, the other side is that you can determine all available methods at compile time, ensure that there will never be two methods with the same priority at compile time, and then at run-time, choose the method to call
so, for the above example, if you had a T that happened to also implement A, then it would call the A function
because T doesn't actually declare any interfaces, so it has a lower priority than A
What kind of points are we talking about here? Real numbers? If so, then I don't arguments of proportionality apply. Or do you just mean integers along the line? If so, then yes.
because there's an obvious one-to-one mapping between the integers and the even integers: if I map each integer N to the integer 2N, then I've assigned each integer to a different even integer. so there can't be more integers than even integers.
not really. you can still find an injective function that maps each point in the first line to exactly one point in the second line, without leaving out any points in the longer line
if your line only contains a single point, it contains exactly one point (not an unknown finite number of points, just one). if contains at least two points, it contains an infinite number of points, because you can always add the mean of two points to the set.
there is no way to have anything in between
(not with the usual definitions of real numbers and lines anyway)
infinitesimals are not actual objects you can treat like that. they are limits.
in particular, an infinitesimal length dl is the limit of ever smaller finite lengths delta-l. but each of those lengths that tend towards the limit have an infinite number of points in them but an ever smaller length that tends to zero.
actually I think what I said isn't even quite right yet. because for infinitesimals you only take the limit after performing arithmetic on your delta-l.
which is why you can't treat them like any other finite length.
no. We talk about approaching a length of 0, and despite the fact that the length of the line gets smaller, the number of points between those two endpoints are the same
The coastline paradox is the counterintuitive observation that the coastline of a landmass does not have a well-defined length. This results from the fractal-like properties of coastlines. The first recorded observation of this phenomenon was by Lewis Fry Richardson and it was expanded by Benoit Mandelbrot.
More concretely, the length of the coastline depends on the method used to measure it. Since a landmass has features at all scales, from hundreds of kilometers in size to tiny fractions of a millimeter and below, there is no obvious size of the smallest feature that should be measured around...
depending how you measure the coastline, you can get lengths that are absolutely massive
differentials/infinitesimals are very subtle concepts. the way they are defined, they are just a shorthand notation for inserting a value delta-x, and then after performing some arithmetic (usually a differentiation for derivatives or a sum for integrals) we take the limit of delta-x going to zero.
when you write df/dx, there is no infinitely short line dx involved. this is a convenient notation for lim(delta-x -> 0) (f(x + delta-x) - f(x)) / delta-x ... in this equation there is no infinitely short line. there is a finite line delta-x (which contains infinitely many points), and we're looking at a sequence of such expressions with ever-shortening delta-x.
an infinitely short line doesn't really enter into it. that's a simplification that works as a first explanation for infinitesimals, but it doesn't really hold up to any formal properties, because that's not how it's defined.
@MartinEnder well, I never meant that dx to be infinite small length. I was considering a infinitesimal length on our line of length l , to be dl
That dx is a infinitesimal change in x I suppose
But now I conclude something like this : As point is something like infinity packed into finite (line) , hence though it may seem to be proportional , proportionality don't hold. Right?
But wait. @NathanMerrill if dl is a point, then proportionality must hold.
Hence dl can't be a point
It must be a infinitesimal lengthed line with infinite points
my point is that dx (or dl) is neither a point nor a line. it's an entirely different concept, that you can in some contexts treat like a very short line but in practice it obeys completely different rules.
I'm not even sure whether the cardinality of it is well-defined, but if I had to guess then I'd say, yes it contains infinitely many points and has length zero.
Because removing elements from a collection , we must achieve another collection, and at some point we are bound to get a finite lengthed collection. Aren't we?
But we just proved that such collection don't exist
@MadhuchhandaMandal no, why would we? start from the natural numbers. repeatedly remove the smallest integer from the set. you'll never end up with a finite collection. there are always infinitely many values left.
there is no smooth transition from finite to infinite
of course it is. the problem isn't that it doesn't contain those 50 points. the problem is that there are no 50 adjacent points. what are the 50 smallest numbers greater than 1? the question doesn't make sense because there is no smallest number greater than one. whenever you talk about a contiguous range on the real line, you've immediately got infinitely many points.
if you talk about a finite number of points, there will always be other points between then.
you can remove 50 points from a line, but they won't be "next to each other", so this is no different from taking the points 1, 3 and 5 out of the line from 0 to 6. what you end up with is not a line, because it has gaps.
if by "collection" you mean set then yes that's right. but when we start to talk about lines then we are talking about special sets with geometrical meaning.
Ok. So you say line is not a normal set of points. (I was told so from the first day of my birth..) Ok then I think there is a long way for me to go before I understand that special set.
no, a line is just a set of points (although since our objects are points, clearly there's some geometrical relationship between them). however, not every set of points is a line. so if you take a line...and then you add or remove points, you get a different set. and there's no reason a priori why that new set should still be a line.
that is what I meant by "special sets": they are (normal) sets with specific properties (in this case, the property that they contain all the points between two endpoints... if a set does not have this property, it's not a line).
I'm heading out. might be able to talk again later, but I really recommend you try reading up on infinity, cardinality and the real line a bit. I'm sure there are some good introductory examples online somewhere.
But it seems me to be counterintuitive to think that there cannot be any last to last element in a collection. It might be undermined. But I think it must exist.
*undetermined
By the way thank you very much for your patience and interest
@MadhuchhandaMandal no it's not undetermined, it really does not exist. assume there is a largest number x < 1. now consider y = (x+1)/2. we have x < y < 1/2 which is a direct contradiction to "x is the largest number less than 1". so our initial assumption that such an element exists must be wrong.
this is really no different from saying that there is no largest number (in general).
if there was a largest number x < 1, then 1/(1-x) would be "the largest number".
@El'endiaStarman I just got a book that you might enjoy too! It is the magic of mc escher (there are various editions). It is a large collection of his prints but it also contains sketches where you can see how he developed some of his works. There is not too much text (a few anectodes and quotations).
Might make a great gift (for someone else or for youself=)