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Thread: Tetration of reals

  1. #1 Tetration of reals 
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    I was thinking about tetration of real numbers and was wondering how other operators were extended to the reals.

    I'm assuming that addition is defined on the reals by definition of the reals. In other words, 1/2 is the number such that 1/2 + 1/2 = 1, etc. (If I'm wrong, how's this done?)

    If addition is defined on the reals, then multiplication can be defined as:
    x*a = x + x*(a-1)
    x*1 = x
    x*0 = 0

    How is this definition extended to the reals? For example, how can x*1/2 be defined using only addition? (Edit: And limits, I guess...)

    For exponentiation the definition is:
    x^a = x * x^(a-1)
    x^1 = x
    x^0 = 1

    Similarly for tetration:
    x^^a = x ^ (x ^^ (a-1))
    x^^1 = x
    x^^0 = 1

    So how was x^(1/2) defined? According to Wikipedia, nothing tried so far has been able to define x^^(1/2) in a satisfactory way.

    Obviously, the limit lim(e->0), x^^(n+e) = x^^(n-e) should hold, but what other properties should a good solution satisfy? (Wikipedia lists monotonically increasing and continuous. Is that limit the same as continuous?)


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  3. #2  
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    The steps to go from the integers to the reals has a few more steps then you have allowed for here. For starters you need to construct the rationals from the integers (the standard field of quotients construction - if you want i can describe it) and then use equivalence classes on those rationals to complete the field of rationals so that it does not have any holes.

    One way of doing this construction is to use dedekind cuts, or equivalence classes of convergent sequences on the rationals. Then addition in the reals is just addition of the appropriate equivalence classes, ditto for multiplication (which answers your first question in general).

    To get to general exponents you need to define the natural logarithm (log) first (which needs integration) and thus obtain the natural exponential function e<sup>x</sup>. a<sup>b</sup> is then defined to be e<sup>log(a) x</sup>.

    I dont see why the extension of tetration should be continuous though, the natural extension of another fast growing function, the factorial is not continuous on the entire real line.


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  4. #3  
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    It is continuous on the positive reals though.

    So you're saying that the definition of multiplication on the reals is defined first on the rationals, then on the reals?

    I'll try to walk through what I think I know (correct me if I'm wrong):

    Define a/b as being equivalent to c/d if there is an n s.t. n*a = c and n*b = d
    Define a/b + c/d as (a*d + b*c)/(b*d) and a/b * c/d as (a*c)/(b*d)
    Define n = n/1
    Then 1/n is the multiplicative inverse of n since 1/n * n = 1/n * n/1 = (n*1)/(n*1) == 1/1 = 1 (where == is 'is equivalent to')
    Where * is defined as above (with extensions to cover negatives)

    So this defines the rationals and addition and multiplication on the rationals. (I'm sure I missed a definition in there somewhere though...)

    Then these definitions can be extended to reals with limits? Then the step from multiplication to exponentiation requires integrals?
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  5. #4  
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    If you are constructing the reals from some smaller set then the standard way would be to start with the naturals and slowly extend until you arrive at the reals. Its the easiest way to see what a "real number" is.

    The standard equivalence class on Z<sup>2</sup> to define the rationals is to define the pair (a, b) ~ (c, d) iff a*d = b*c. Your equivalence class does not work as is (i dont think it is symmetric).

    You can extend to the reals in a few ways, the equivalence class of convergent (i.e. cauchy) rational sequences uses limits while dedekind cuts do not use them explicitly. To obtain general exponents though you need the integral.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  6. #5  
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    Well... I can accept that multiplication is just by definition on the reals (somewhat), but is there any way to define the reals that doesn't involve the multiplicative inverse?

    In any case, my question about the definition of exponentiation still stands. Just because it requires an integral doesn't mean I'm not interested in how it works. Can you describe the process of extending the exponential function to the reals? I know that the definition of x^(1/n) can be derived from the identity (x^a)^b = x^(a*b), but is that enough? If so, does anyone know any similar identities for tetration?
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  7. #6  
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    Multiplication is not by definition on the reals - and if we are talking about the reals as a complete ordered field then you cant get away from the multiplicative inverse (it comes part and parcel of the field axioms). The construction does not use multiplicative inverses though - im not sure what you are trying to get at.

    Okay, first define the logarithm Log(t) as the integral of 1/x from 1 to t. This has all the properties that you would expect for a logarithm and as it is monotonically increasing it has an inverse (which we call e<sup>x</sup>). General exponents are defined to be a<sup>b</sup> = e<sup>Log(a) b</sup> Thats all that there is to it really.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    You don't need log to define real exponents. You can define rational exponents first then extend it to the reals with continuity (the rationals are dense, this extension will be unique). You can then define log's as the inverse of these exponential functions.

    It's also not necessary to use an integral to define log, the approach above is one way. You can define the exponential e^x before any of the above though, using power series for example. Then take log to be it's inverse.


    When defining the rationals on ordered pairs of integers (a,b), you only take pairs with non-zero b (I'm sure you knew this, just to make it explicit).
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    Quote Originally Posted by shmoe
    You don't need log to define real exponents. You can define rational exponents first then extend it to the reals with continuity (the rationals are dense, this extension will be unique). You can then define log's as the inverse of these exponential functions.

    It's also not necessary to use an integral to define log, the approach above is one way. You can define the exponential e^x before any of the above though, using power series for example. Then take log to be it's inverse.


    When defining the rationals on ordered pairs of integers (a,b), you only take pairs with non-zero b (I'm sure you knew this, just to make it explicit).
    True - but the machinary to define exponents using the denseness of the rationals takes a few more steps then just using the riemann integral. First you have to prove that the rationals are dense if the reals are defined by equivalence classes of rational sequences (ok - its not difficult but i have no idea what magi's background with topology and real analysis is). Then you have to prove continuity on the rationals before you can extend to the reals - and i think that is a pain as the rationals are not complete. Using a power series leaves open the question if what you have called e<sup>x</sup> really is the exponential - checking the properties of the log i think is easier.

    I think the logarithm route is the easiest, but there are many ways of skinning this cat.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  10. #9  
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    Well, I've taken both courses at college level, so I'm fine with it. What I really want to know, regardless of complexity, is whether or not there is a way to define exponentiation/logarithms on reals using only addition, mutliplication, limits and statements such as 'The number such that ...'. Since you can define integrals with those statements, integrals would be ok too, although I'd like to find a way to avoid them, since the next step involves a lot of very difficult functions.

    One other question. For exponents (x^a)^b = (x^b)^a but for tetration this isn't at all true. Does it make sense to allow (x^^a)^^(1/a) = (x^^(1/a))^^a = x? The equation (x^^a)^(x^^b) = (x^^(b+1))^(x^^(a-1)) holds, but isn't a lot of help.
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  11. #10  
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    Quote Originally Posted by river_rat
    True - but the machinary to define exponents using the denseness of the rationals takes a few more steps then just using the riemann integral.
    I would doubt that. Defining the integral and proving increasing functions are integrable, etc. takes a fair bit of work.

    Quote Originally Posted by river_rat
    First you have to prove that the rationals are dense if the reals are defined by equivalence classes of rational sequences (ok - its not difficult but i have no idea what magi's background with topology and real analysis is).
    This is one of the first things you do starting from the axioms for the reals after proving all the basic algebraic properties. The archimedian property is usually the first non-trivial result, followed closely by showing that between any two reals there is rational.

    If you've just gone through a construction of the reals from the rationals (from whichever method) and it's not immediately obvious that the rationals are dense, I think it would be fair to say you didn't understand the construction!

    Also, if you've proven the existence of a decimal represantion for any real then you know the rationals are dense.

    Quote Originally Posted by river_rat
    Then you have to prove continuity on the rationals before you can extend to the reals - and i think that is a pain as the rationals are not complete.
    Rationals not being complete is irrelevant for this. Once you have the usual exponent property of a^(r+s)=(a^r)*(a^s), continuity in whatever form you want is not a problem. Start with a>1, a<1 is handled by looking at 1/a. a^r is an increasing function of r from the exponent property and knowing a^r>1 for any rational r>0 (this is straightforward to prove). From the exponent property it's enough to show continuity at r=0, for this increasing tells us it's enough to show a^(1/n)->1 as n->infinity (from which it follows a^(-1/n)->1 as well). This limit exists as it's a decreasing sequence, so it converges to some L>=1. a^(2/n) then converges to L^2 (product of limit is the limit of the product), since a^{1/n} is a subsequence of a^(2/n), they converge to the same limit, so L=L^2 and L=1.

    You can then define a^x as the limit of a^(r_n) for any sequence of rationals r_n converging to x. The above shows this is independant of your choice of sequence r_n (consider the difference of any two such sequences, it converges to 0)

    Some machinery is needed, but it's all stuff that you have long before integrals in any treatment of analysis I've ever seen. Basic supremum or limit properties necessarily come before defining integrals.

    Quote Originally Posted by river_rat
    Using a power series leaves open the question if what you have called e<sup>x</sup> really is the exponential - checking the properties of the log i think is easier.
    proving exp(x+y)=exp(x)*exp(y) from power series is not difficult either, nor are any other relevant properties.
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  12. #11  
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    I would doubt that. Defining the integral and proving increasing functions are integrable, etc. takes a fair bit of work.
    Defining the Riemann integral does not take that much work (about as much as setting up the machinary to prove power series make sense) and proving that continuous functions have riemann integrals only uses that the closed bounded subsets of the real line are compact (Heine-Borel) which is not too difficult to prove either (about as hard as proving that you can do the rearrangments of the power series for e<sup>x</sup> without messing up the convergence). To prove uniqueness is worse as you need to show term by term differentiation holds and then that the resulting DE has a unique solution - i cant think of a quicker way?

    This is one of the first things you do starting from the axioms for the reals after proving all the basic algebraic properties. The archimedian property is usually the first non-trivial result, followed closely by showing that between any two reals there is rational. If you've just gone through a construction of the reals from the rationals (from whichever method) and it's not immediately obvious that the rationals are dense, I think it would be fair to say you didn't understand the construction!
    Yep, so denseness is shown after the fact that the reals have no infinitesimals - and this is fairly obvious. But being fairly obvious is not the same as being easy to prove (just ask Jordan). Though in this case the result does not take a brain haemorrhage to get.

    Also, if you've proven the existence of a decimal represantion for any real then you know the rationals are dense.
    Yep, but nowhere have we proven the existence of base-n expansions - and that does not fall out from the definitions and also takes some work.

    Rationals not being complete is irrelevant for this. Once you have the usual exponent property of a^(r+s)=(a^r)*(a^s), continuity in whatever form you want is not a problem. Start with a>1, a<1 is handled by looking at 1/a. a^r is an increasing function of r from the exponent property and knowing a^r>1 for any rational r>0 (this is straightforward to prove). From the exponent property it's enough to show continuity at r=0, for this increasing tells us it's enough to show a^(1/n)->1 as n->infinity (from which it follows a^(-1/n)->1 as well). This limit exists as it's a decreasing sequence, so it converges to some L>=1. a^(2/n) then converges to L^2 (product of limit is the limit of the product), since a^{1/n} is a subsequence of a^(2/n), they converge to the same limit, so L=L^2 and L=1.
    Personally i think proving 1/t continuous is easier - but easier is quite subjective. To get the exponent property is not a two step problem, as mentioned before and you have used all the properties of the reals here that are needed to obtain the heine-Borel theorem as well (namely completeness).

    You can then define a^x as the limit of a^(r_n) for any sequence of rationals r_n converging to x. The above shows this is independant of your choice of sequence r_n (consider the difference of any two such sequences, it converges to 0)
    Any continous function on a dense subset extends continuously to its closure - nothing new there. Doesnt this characterise metric completions?

    Some machinery is needed, but it's all stuff that you have long before integrals in any treatment of analysis I've ever seen. Basic supremum or limit properties necessarily come before defining integrals.
    True, but deep analysis is not really needed here - and power series are usually handled properly after the introduction of the integral. I dont see how extending exponents via the densness of the reals is somehow simplier then using the integral to define logs. It may be "cool" but once you have gone through all the work, you could have proven that all continuous functions are riemann integrable which is a whole lot "cooler"

    proving exp(x+y)=exp(x)*exp(y) from power series is not difficult either, nor are any other relevant properties.
    You can prove the existence of a function f(xy) = f(x) + f(y), f(0) = 1; to prove that this is unique (before all the hassle of rearranging series and convergence of power series) is not easy, and thats what i am refering to about proving that this power series is the exponential. I think the integral has the same "uniquesness" issue though.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  13. #12  
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    Quote Originally Posted by MagiMaster
    Well, I've taken both courses at college level, so I'm fine with it. What I really want to know, regardless of complexity, is whether or not there is a way to define exponentiation/logarithms on reals using only addition, mutliplication, limits and statements such as 'The number such that ...'. Since you can define integrals with those statements, integrals would be ok too, although I'd like to find a way to avoid them, since the next step involves a lot of very difficult functions.
    If you can prove your binary operation f:Q<sup>2</sup> -> R continuous on the rationals then you have a unique extension to the reals (and you bypass the need for integrals, though if you can prove that you can use integrals anyway). I have no idea if tetrations are continuous in both arguments though (i doubt it, else you would not have this problem)
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    As always, a proof to the contrary is still a proof, but I don't think it'll be any easier than finding the function...

    Actually though, I was recently thinking about what (x^^(p/q)) might be. Since (x^^p)^^(1/q) won't generally equal (x^^(1/q))^^p, I'm having a little bit of trouble with this. If we assume (x^^(1/n))^^n = (x^^n)^^(1/n) = x, then it's easy to define x^^(1/n). (x^^(1/2) would be log(x)/W(log(x)) where W is Lambert's W function. Finding formulas for 1/3, etc. isn't so easy though.)

    On a related note, if f(x)=log(x)/W(log(x))=x^^(1/2), does the equation log<sub>a</sub>(f(a))+log<sub>a</sub>(log<sub>a</sub>(f(a)))=0 make sense (as far as the meaning of x^^y goes)? (Of course, that's just an observation. I can't prove that it's true yet.)
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    Quote Originally Posted by river_rat
    Yep, so denseness is shown after the fact that the reals have no infinitesimals - and this is fairly obvious. But being fairly obvious is not the same as being easy to prove (just ask Jordan). Though in this case the result does not take a brain haemorrhage to get.
    If you haven't proven the archimedian property before you go on to deal with integrals you should be shot.

    Anyways, you were talking about constructing the reals from the rationals, if this construction didn't make denses obvious....

    Quote Originally Posted by river_rat
    Yep, but nowhere have we proven the existence of base-n expansions - and that does not fall out from the definitions and also takes some work.
    I know "we" hadn't, we hadn't done anything structured here at all... it's just another thing that's often proven early. It does fall right from the definitions with very little difficulty though. if x>0 is real, take n1 to be the largest integer <=x. Take n2 the largest integer where n1+n2/10<=x and so on. A decimal n1.(n2)(n3)... is defined to be sup{n1,n1+n2/10,n1+n2/10+n3/10^2,...} and you're done.

    Quote Originally Posted by river_rat
    Any continous function on a dense subset extends continuously to its closure - nothing new there. Doesnt this characterise metric completions?
    Not sure in what sense you mean, but I think no for any interpretation I can think of. Any continuous function on Q extends continuously to it's closure in Q itself. You should have something like a metric space is dense in it's smallest (in some sense) completetion. Here you will be able to extend to the whole space.

    Anyways, you can avoid worrying about considering limits of a^{r_n} entirely and whether a^r is continuous on the rationals if you don't bother with sequences. Just take a^x to be the sup of {a^r|r<=x and r rational}. Increasing for rationals is needed to show this is consistant with your earlier definition of a^x when x was rational and that this sup does exist.

    I dont see how extending exponents via the densness of the reals is somehow simplier then using the integral to define logs. It may be "cool" but once you have gone through all the work, you could have proven that all continuous functions are riemann integrable which is a whole lot "cooler"
    First of all, it's a more usefull approach for the initial question here, trying to define ^^ for rational things then extending is the 'obvious' way to proceed (though not likely to succeed, it's been tried by many people!). Secondly- you should look through all the details required, it's a common approach in analysis texts and my half hearted explanations are sure to not give the entire picture. Baby Rudin goes this way (partly in exercises iirc) as does Johnsonbaugh and Pfaffenberger with all the details to see. You're just using basic theory that will exist before you ever hit integration.

    Oh, and the reason I mentioned defining exp(x) by a power series then define log as it's inverse was just a reaction to your statement "To get to general exponents you need to define the natural logarithm (log) first (which needs integration)", not because I thought power series would be a quick route to real exponentiation. I'm sure you knew you didn't need an integral to define log, I just wanted to clarify and give an example this was the case.
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    Ok. I may have come up with a way to define ^^ on rationals (and in the limit, irrationals).

    First define (x^^(1/n)) as the number y s.t. y^^n = x
    Then using the recurrence x^^n = x^(x^^(n-1)), you can decompose any rational using it's continued fraction expansion. (Any rational has a finite continued fraction. Using limits, you can do this for irrationals too.)

    So some quick calculations (assuming I did it right) has 4^^(3/5) ~= 2.2266

    4^^(3/5) = b => 4 = b^(b^^(2/3))
    b^^(2/3) = c => b = c^(c^^(1/2))
    c^^(1/2) = d => c = d^d

    MatLab can solve the equation c = d^d, and then b = c^(c^^(1/2)), so then just plugging in numbers for b, 2.2266 gets pretty close.

    Now... This definition isn't actually very good for calculating the values of things like x^^phi (even though phi = [1;1,1,1,1,1,...]) or even x^^(1/3). Also, it's hard to tell if this definition of x^^y is monotonic in y, much less continuous.

    How could I go about trying to show that this is monotonic, continuous, etc. and then finding a power series for it?

    Edit: Oh yeah. I haven't checked yet, but (4^^(3/5))^^(5/3) should equal (4^^(5/3))^^(3/5), both of which should be 4. I guess this follows from the definition really, but I haven't gotten to any kind of proof yet...
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    If you haven't proven the archimedian property before you go on to deal with integrals you should be shot. Anyways, you were talking about constructing the reals from the rationals, if this construction didn't make denses obvious....
    Well Dedekind cuts do not make denseness entirely obvious (perhaps because they ignore the metric topology on Q and merely "complete" the partial ordering?) and they are a legit. construction method. If you are doing the standard metric completion then yes, proving that the rationals are dense is easy (i never denied that) but not every construction makes it that easy. I found a cute construction of the reals starting from the group of integers which is quite cute as well, and finding the rationals in that construction seems painful!

    I know "we" hadn't, we hadn't done anything structured here at all... it's just another thing that's often proven early. It does fall right from the definitions with very little difficulty though. if x>0 is real, take n1 to be the largest integer <=x. Take n2 the largest integer where n1+n2/10<=x and so on. A decimal n1.(n2)(n3)... is defined to be sup{n1,n1+n2/10,n1+n2/10+n3/10^2,...} and you're done.
    What i meant is that we have not even mentioned that up until then, Magi wanted a simple construction and i gave him the one i think is the simpliest. Put it this way, its a construction that a final year high school student (or first year engineer! I should know, i tutor the buggers!) would get without any hassle - though i do agree that using the denseness is slicker.

    Not sure in what sense you mean, but I think no for any interpretation I can think of. Any continuous function on Q extends continuously to it's closure in Q itself. You should have something like a metric space is dense in it's smallest (in some sense) completetion. Here you will be able to extend to the whole space.
    I went and looked it up, a space M is the completion of a metric space N if for every uniformly continuous function f : N -> (a complete metric space) there exists a unique uniformly continuous extension of f to M. Dont you just love universal properties?

    Anyways, you can avoid worrying about considering limits of a^{r_n} entirely and whether a^r is continuous on the rationals if you don't bother with sequences. Just take a^x to be the sup of {a^r|r<=x and r rational}. Increasing for rationals is needed to show this is consistant with your earlier definition of a^x when x was rational and that this sup does exist.
    Im being lazy here (as i dont want to check) but doesnt this formulation make proving the exponential property a bit more painful?

    First of all, it's a more usefull approach for the initial question here, trying to define ^^ for rational things then extending is the 'obvious' way to proceed (though not likely to succeed, it's been tried by many people!). Secondly- you should look through all the details required, it's a common approach in analysis texts and my half hearted explanations are sure to not give the entire picture. Baby Rudin goes this way (partly in exercises iirc) as does Johnsonbaugh and Pfaffenberger with all the details to see. You're just using basic theory that will exist before you ever hit integration.
    I was just using the fist method i was ever shown when i was still a first year undergrad. Perhaps defining it on the rationals first is entirely the wrong way to go? Another way of obtaining the exponential is to look at the limit of (1 + x/n)<sup>n</sup> as n -> infinity. Perhaps a radical "rethink" like that is required.

    Oh, and for a list of possible constructions try http://en.wikipedia.org/wiki/Definit...ntial_function
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    Quote Originally Posted by river_rat
    Well Dedekind cuts do not make denseness entirely obvious (perhaps because they ignore the metric topology on Q and merely "complete" the partial ordering?) and they are a legit. construction method.
    They do make it obvious though. The real numbers are all defined in a way to be the supremum of the set of rationals in the cut once you've identified these rationals with their corresponding real numbers.

    Quote Originally Posted by river_rat
    I found a cute construction of the reals starting from the group of integers which is quite cute as well, and finding the rationals in that construction seems painful!
    I did specify constructions from the rationals for a reason, to avoid something exotic. I have yet to see a construction from the rationals that doesn't make dense obvious if you understood the construction.

    Quote Originally Posted by river_rat
    What i meant is that we have not even mentioned that up until then, Magi wanted a simple construction and i gave him the one i think is the simpliest. Put it this way, its a construction that a final year high school student (or first year engineer! I should know, i tutor the buggers!) would get without any hassle - though i do agree that using the denseness is slicker.
    Another reason I brought decimals- every high school student is familiar with them, even if we hadn't mentioned them yet. Depending on where you are 'starting' from, you'll be willing to accept different things. A full blown treatment is ideal to go through of course.

    Quote Originally Posted by river_rat
    I went and looked it up, a space M is the completion of a metric space N if for every uniformly continuous function f : N -> (a complete metric space) there exists a unique uniformly continuous extension of f to M. Dont you just love universal properties?
    uniformly continuous, that makes more sense.

    Quote Originally Posted by river_rat
    Im being lazy here (as i dont want to check) but doesnt this formulation make proving the exponential property a bit more painful?
    Nope. You prove it algebraically for the rationals first, and it follows for arbitrary real exponents using properties of sup.

    Quote Originally Posted by river_rat
    I was just using the fist method i was ever shown when i was still a first year undergrad. Perhaps defining it on the rationals first is entirely the wrong way to go? Another way of obtaining the exponential is to look at the limit of (1 + x/n)<sup>n</sup> as n -> infinity. Perhaps a radical "rethink" like that is required.
    The log approach is common in first year calculus classes, where they don't usually bother with supremums and the like. Defining exponentials as that limit or as a power series are the more usual approaches in an analysis book, and of course proving they are equivalent.
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    They do make it obvious though. The real numbers are all defined in a way to be the supremum of the set of rationals in the cut once you've identified these rationals with their corresponding real numbers.
    mmm, I agree obvious - but not as easy (and let me be specific here, im talking to a first year audience) to "prove" (or if you want justify in 3 lines in a forum post to an audience which obviously has limited mathematical skill).

    I did specify constructions from the rationals for a reason, to avoid something exotic. I have yet to see a construction from the rationals that doesn't make dense obvious if you understood the construction.
    Even the ultrafilter version?

    Another reason I brought decimals- every high school student is familiar with them, even if we hadn't mentioned them yet. Depending on where you are 'starting' from, you'll be willing to accept different things. A full blown treatment is ideal to go through of course.
    Cant fault that
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by river_rat
    mmm, I agree obvious - but not as easy (and let me be specific here, im talking to a first year audience) to "prove" (or if you want justify in 3 lines in a forum post to an audience which obviously has limited mathematical skill).
    their level of skill is irrelevant- if they understood the dedekind cut construction then they will understand the rationals are dense in the reals. I'm not making any claims about someone who didn't understand the construction.

    If you want to 'prove' it to your average bloke, then "decimals" would be the start. If you are going to assume they can handle integrals, then a safe bet is they can handle decimals as well.

    Quote Originally Posted by river_rat
    Even the ultrafilter version?
    I've only looked at this long ago, so I could be off. In this approach, the reals get defined as ultrafilters on the rationals correct? The rationals get identified with their corresponding principal ultrafilters. The topology is generated by open, bounded intervals of rationals, the corresponding open set in your 'reals' is all ultrafilters that contain this rational interval. therefore every open interval contains rational numbers, since the principal ultrafilter for a rational number contains every rational interval containing it.

    Or maybe you mean as a quotient of the finite hypperrationals? The reals in this case are just going to be equivalence classes of sequences of rationals (under your ultrafilter on N that defined the hyperrationals and the ideal of infintessimals you modded out by). Take any sequence representing the real number and the corresponding sequence of rationals will converge to it. I'm not too familiar with non-standad analysis, but I suspect this construction of the reals is just the same as the cauchy sequence version, it seems modding out by the maximal ideal of infintessimals is going to be only superficially different from not having bothered with the infintessimal approach at all. I'm not positive though.
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    I've only looked at this long ago, so I could be off. In this approach, the reals get defined as ultrafilters on the rationals correct? The rationals get identified with their corresponding principal ultrafilters. The topology is generated by open, bounded intervals of rationals, the corresponding open set in your 'reals' is all ultrafilters that contain this rational interval. therefore every open interval contains rational numbers, since the principal ultrafilter for a rational number contains every rational interval containing it.
    Arent you construction the Stone-Cech compactification of the rationals here?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Quote Originally Posted by river_rat
    Arent you construction the Stone-Cech compactification of the rationals here?
    An important detail I didn't mention before, you take the ultrafilters to contain only closed, bounded, and non-empty intervals of rationals. This isn't the Stone-Cech compactification, you only get something locally compact.

    Is this not what you had in mind by an "ultrafilter version"?
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    It was looking similar to one of the constructions i know for discrete spaces - though i cant remember the general ultrafilter version.

    I was thinking more of the second version you gave (i.e. use an ultraproduct on Q though on a second look i see your point)
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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