21 Nov 14
Originally posted by SoothfastThe symbols came out fine on my system, I'm reading this on Firefox. I have to admit to some surprise that completeness is required as an axiom for the reals - whereas that they are dense follows from a simple argument. Rigour in maths wasn't emphasised particularly in my maths for physicists undergraduate course, although I've got to admit they had the maths for physicists lectures at nine AM on a Friday morning the day after the campus disco, which didn't make them easy to attend. By the time I was a graduate student it started to matter, but I had to teach myself this stuff so it's a little patchy.
There are many notions of "completeness" in mathematics. The two most common ones, in my experience, are:
1) Completeness of a totally ordered set S.
2) Completeness of a metric space (X,d).
The set of complex numbers, while complete in the second sense (with respect to the usual Euclidean metric), cannot be complete in the first sense because it ...[text shortened]... r bound that is an integer. By definition this means the set of integers Z is complete.[/quote]
An example is your formal proof that the set of integers is complete, given the ordered set definition of completeness a subset of the integers with an upper bound must have a largest element (otherwise it's not bounded) and that element is a least upper bound; this is enough to convince me, but doesn't fulfil the rigour requirements of mathematicians (I assume or you'd just have said what I just did) which always leaves me wondering how I know if a proof I come out with is rigorous or not.
21 Nov 14
Originally posted by DeepThoughtOf what do you disagree? That his proof is not a real proof? Then refine it and publish it. I'm happy with this proof, nothing wrong with this.
I disagree, if someone knows, or thinks of, another proof then it might be fun to see it. There are over 370 proofs of Pythagoras' theorem from the other Euclidean axioms. Although it is neater to use it as an axiom and drop the parallel lines axiom.
You could easily use anything already proven as an axiom. Like noone need to prove that 1+1=2, people use it as an axiom. Because most of us wouldn't understand a stringent mathematical proof that 1+1=2 anyway.
21 Nov 14
Originally posted by FabianFnasNo, that it's the last word. If someone thinks of a way of proving that 0.999··· = 1 which is different to the ones so far presented it would be fun to see it.
Of what do you disagree? That his proof is not a real proof? Then refine it and publish it. I'm happy with this proof, nothing wrong with this.
You could easily use anything already proven as an axiom. Like noone need to prove that 1+1=2, people use it as an axiom. Because most of us wouldn't understand a stringent mathematical proof that 1+1=2 anyway.
I think in things like Peano arithmetic it's not so much proved it's just that 1 has a successor (1 + 1) and the symbol for it is 2.
21 Nov 14
1 edit
Originally posted by DeepThoughtYes, to say a set of integers with an upper bound must have a largest element (a.k.a. maximal element) is a statement that, technically, would require some kind of proof. There is a somewhat related statement known as the Well-Ordering Principle that says every non-empty set of positive integers contains a least element. Seems dead obvious, but, as discussed at
The symbols came out fine on my system, I'm reading this on Firefox. I have to admit to some surprise that completeness is required as an axiom for the reals - whereas that they are dense follows from a simple argument. Rigour in maths wasn't emphasised particularly in my maths for physicists undergraduate course, although I've got to admit they had th ...[text shortened]... did) which always leaves me wondering how I know if a proof I come out with is rigorous or not.
http://en.wikipedia.org/wiki/Well-ordering_principle
it's a tricky matter. It can be proved if you take the Principle of Mathematical Induction as axiomatic, or else you can take it as an axiom and use it to prove the Principle of Mathematical Induction. It depends on how you go about constructing your mathematical universe -- you have to start with some axioms no matter what. I'm not particularly well-versed in the rigorous foundations of mathematics (the kind of stuff Bertrand Russell & Co. spent a lifetime delving into), however. Using set theory to generate a 1000-page proof that 1+1=2 is just not my thing, especially when you witness a Gödel come along and knock it all down in half a page of logic symbols!
As an aside, I tested out the symbols in my prior post using Chrome, Firefox, and Safari, and everything looks good. I'm finding the website http://www.fileformat.info useful for this.
21 Nov 14
Originally posted by DeepThoughtConcerning your last sentence, I'd have to say that what constitutes a "rigorous" proof does depend on context. Any analyst would likely say my proof that the set of integers is complete is rigorous. But perhaps a Peano or Russell would take exception with my claim that if m and n are integers and n>m, then n ≥ m+1. Somewhere near the bedrock of mathematics that claim is a big deal, but not in the working lives of most mathematicians.
An example is your formal proof that the set of integers is complete, given the ordered set definition of completeness a subset of the integers with an upper bound must have a largest element (otherwise it's not bounded) and that element is a least upper bound; this is enough to convince me, but doesn't fulfil the rigour requirements of mathematicians (I ...[text shortened]... did) which always leaves me wondering how I know if a proof I come out with is rigorous or not.