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Of course you can. Consider the sequence of rational numbers in (0,1). Now consider the sequence of rational numbers in (0,1]. The latter has had an extra term
by nilkn 10y ago
Of course you can. Consider the sequence of rational numbers in (0,1). Now consider the sequence of rational numbers in (0,1]. The latter has had an extra term appended.
(This is of course flawed, but I think it illustrates that the question isn't completely trivial. It requires us to carefully distinguish between the notion of an ordered set and a sequence and even then we'll have to deal with the fact that the rationals can be made into a sequence, but not with the same ordering.)
- lisper 10y agoNo. There is no such thing as the sequence of rational numbers in an interval. The rationals have a total order, and they can be enumerated, but the total order of rationals does not define an enumeration of the rationals and hence the rationals are not a sequence. If you want to dispute this, tell me: what is the next rational after 1/2?
- nilkn 10y agoI took some liberties with the word "the", but I'm pretty sure that you know that it's easy to construct an ordered sequence of all rational numbers using any of a number of standard diagonalization arguments. I'll skip the details you already know. Eventually, yes, you're going to be able to poke a hole in my argument. It is definitely flawed. I don't know how long it'd take us to get there, but it doesn't really matter. But we're already at the point where this cannot be considered "basic", and that is my true point here. Attempting to demonstrate that 0.999... = 1 while meticulously avoiding any rigorous definition of what 0.999... means is not very easy and will require you to fend off all sorts of potential jabs from various directions. It's much easier to just talk about infinite sums and be done with it.
- lisper 10y ago> It's much easier to just talk about infinite sums and be done with it. Indeed. But it's a lot less fun :-)