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Set of ALL numbers?

  1. Jan 15, 2012 #1
    If I want to denote the set of ALL numbers, is the set of complex numbers fitting? a + bi, where b is 0, allows for reals, and where a is 0, allows for imaginary.
  2. jcsd
  3. Jan 15, 2012 #2


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    What are you looking for? The set of all complex numbers includes all reals and all pure imaginary.
  4. Jan 15, 2012 #3


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    Define number. Seriously, all you have done is written down the definition of the complex numbers. There are numbers which are outside the set of complex numbers.
  5. Jan 15, 2012 #4

    Char. Limit

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    That's not sufficient. You've neglected to mention quaternions, octonions, vectors, tensors, the like.
  6. Jan 16, 2012 #5
    If you consider transfinite numbers as actual numbers, then there is no set of all numbers. The collection will be too large. It will be a proper class.
  7. Jan 17, 2012 #6
    The problem is that '(the collection of) all numbers' doesn't actually mean anything - what we have here is a bunch of different interpretations of this, all of which are equally correct (and equally incorrect).

    If you are looking for the set of all numbers which are the solutions of equations involving arithmetic operations (including exponentiation) on the integers, you want the set of complex numbers as in the original question.

    I think that in all ordinary usage, the word 'number' implies a scalar quantity, so Char. Limit's post misses the mark for me.

    Similarly, transfinites such as [itex]\aleph_0[/itex] do not fit most people's definition of a number since they do not behave the same way as all finite numbers do (e.g. [itex]x + 1 > x[/itex] is not true when [itex]x =\aleph_0[/itex]).

    Working from the other end, the whole numbers can be considered to be 'all numbers' if you infer that a number is a number of things.

    Many people would add 0 to that to get the set of integers, so we can now have 'no things'.

    But that leaves the simple equation of [itex]x = 1 - 2[/itex] without a solution, so the negative numbers are surely numbers too.

    Next come the rationals as the solutions to [itex]x = a \div b[/itex].

    The next one is a bit more of a leap because most of the time we accept that the four arithmetical operations of addition, subtraction, multiplication and division are all we need to deal with the 'real' world. But the length of the diagonal of a square with sides [itex]1m[/itex] long is [itex]\sqrt 2 m[/itex], and [itex]\sqrt 2[/itex] cannot be expressed in terms of rational numbers so I want irrational numbers in my set of all numbers. Now if I mark out a circle with radius [itex]1m[/itex] it will have a circumference of [itex]2 \pi m[/itex] so we need to add the transcendentals too and now we have now got to the reals. If we think of numbers as points on a line that extends infinitely in two directions, each real number is represented by a point on the line and vice versa. So for me, this is the set of 'all numbers'.

    But what about [itex]\sqrt {-1}[/itex]? Taking square roots is not an unreasonable thing to do with a number - the diagonal of a square is an obvious example. So perhaps the number that is the solution to the equation [itex]x = \sqrt {-1}[/itex] which we call [itex]i[/itex] belongs in the set of all numbers too - we are then going to end up admitting all the complex numbers to the 'set of all numbers' and have the answer proposed in the original post.

    For me, 'all numbers' are represented by points on a line so this is a step too far - you cannot have a square with sides of length [itex]-1{m}[/itex], so the concept of its diagonal is meaningless (or if you like, imaginary) in this context, so I'll stick with the reals. But I'll admit that the complex plane is quite tempting - after all, the rules of arithmetic hold true for all complex numbers, so they certainly behave like the rest of the numbers - they just don't (all) represent the distance between two points like all the real (and 'real') numbers do.

    One question with many answers!
    Last edited: Jan 17, 2012
  8. Jan 17, 2012 #7


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    The transcendentals are defined as real numbers that are not algebraic.
    You can't take the algebraic numbers, and union it with the set "real numbers not algebraic" and claim "we now got to the reals".

    Edit: nothing you post argues we should use reals over p-adics.
    Last edited: Jan 17, 2012
  9. Jan 18, 2012 #8
    Perhaps my ninth paragraph could be improved (I erroneously imply that the irrationals do not include [itex]\pi[/itex]):

    The next one is a bit more of a leap because most of the time we accept that the four arithmetical operations of addition, subtraction, multiplication and division are all we need to deal with the 'real' world. But the length of the diagonal of a square with sides [itex]1m[/itex] long is [itex]\sqrt 2 m[/itex], and [itex]\sqrt 2[/itex] cannot be expressed in terms of rational numbers. Now if I mark out a circle with radius [itex]1m[/itex] it will have a circumference of [itex]2 \pi m[/itex] and again [itex]\pi[/itex] is not a rational number so we need to add irrational numbers too and now we have now got to the reals.

    I don't think that anyone could argue that the union of the rationals and the irrationals is not the reals.

    Nothing I post argues that you should use anything. My point was that 'all numbers' has no universal meaning: for me it implies the reals, for someone with a better knowledge of number theory than mine it may mean something else.

    However, the mapping of the reals to the points on a line is the clincher for me - the p-adics cannot do this.
  10. Jan 18, 2012 #9


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    You realize that there are one-to-one mappings from Rn onto R, right? So all n-tuples of real numbers can be identified with points on a line too. Granted these maps do not preserve algebraic structure, but you never indicated that this was an important point to you.

    If the fact that R can be identified with points on a line in a way that preserves algebraic structure is what the real clincher is, then R is no more special than Q or R∪{_∞}∪{-∞} or the hyperreal numbers for that matter.
  11. Jan 18, 2012 #10
    I did rule out n-tuples as not fitting any definition of 'number' that is satisfactory to me (although I did acnowledge at the end that number pairs are a difficulty because a + ib certainly behaves like a number).

    Er, yes it is, Q cannot do that. However if cardinality equal to that of the continuum is all I am interested in, I may as well pick [ 0 , 1 ] as my set of all numbers! Clearly I also want my set to be a complete metric space and R is (I believe) the simplest set that satisfies both of those conditions.
  12. Jan 18, 2012 #11


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    No. The embedding QR identifies points of Q with the line in a way the preserves algebraic structure. If you want completeness, then you need to state that separately.
  13. Jan 18, 2012 #12
    Yes you are right. My original statement was "each real number is represented by a point on the line and vice versa", but I ommitted the reverse mapping in my later comment.
  14. Jan 20, 2012 #13
    What a ridiculous post.
  15. Jan 20, 2012 #14
    What do you mean by "a point on the line".

    Do you mean "a real number"?
  16. Jan 20, 2012 #15
    I'm not exactly sure whether you are asking one question or two.

    If it is one question, no I do not mean 'a real number' because then my statement would indeed have been a ridiculous tautology.

    If it is two questions, (i) I am not sure if I can add to the clarity of 'a point on the line'. I am not trying to construct a formal argument, I am trying to explain what the phrase 'all numbers' means to me. And (ii) yes I do mean a real number.

    I do appreciate the element of tautology, essentially I am saying that the elements of a continuum map to the elements of a continuum. This is straying from my point however which is the sufficiency (and necessity) of the reals to satisfy my concept of 'all numbers'. I am perfectly willing to accept that you find that ridiculous.
  17. Jan 20, 2012 #16

    aside from the choice of words and phrases of MrAnchovy,

    you are not disagreeing with that the set of algebraic numbers

    unioned with the set of transcendental numbers equals the

    set of Real numbers?
  18. Jan 20, 2012 #17


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    The algebraic numbers unioned with transcendentals is equal to the set of all reals.
    That is not in contention.

    MrAnchovy's used the terms "we got the reals". The connotation is that there's this magical process that let's you go from the algebraics -> real by adding on "numbers which are not algebraic".

    That's not how it works. We define the rationals, then use that to construct the reals (using Dedekind or Cauchy). After that we consider the algebraic and transcendentals.

    This isn't just theoretical. I was taught in highschool "the real numbers are numbers like square root 2 and transcendental numbers like pi". As if that explains anything. :uhh:
  19. Jan 20, 2012 #18
    To be fair, you can define an irrational number as a number whose decimal expansion is non repeating. That is: you can treat real numbers as decimal expansions. That is a third way to construct the reals and it's probably the most honest way. But it's terribly tedious!!!!
  20. Jan 20, 2012 #19


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    but...but...to be fair, if you are going to start with decimals, as one's basis for understanding numbers, you don't even have the full set of rationals to begin with, you just have the ring of all terminating decimals (that is, we might just as well regard 1/3 as the Cauchy sequence:


    mirroring in theory, what we do in practice).

    is there even a well-established name for this ring?

    it seems to me this ring is dense in the reals (isn't this obvious?), and that it's completion gives us the reals. the fact that some integers aren't units in this ring (4 is, but 3 isn't) is, um, inconvenient, but if you really want to use decimals as your jumping-off point, there is a piper to be paid.

    by the way, your definition needs a bit of work. for example, 1/12 isn't a "repeating decimal" (it's "eventually repeating"), but it is hardly irrational.
  21. Jan 20, 2012 #20
    I don't really see why this must be true.
  22. Jan 21, 2012 #21
    I don't need to construct all the reals in order to know that there are some numbers that are not rational, in the same way that I do not need to construct all the integers to know that 'one, two, many' is not sufficient to comprise 'all numbers'. It is enough (for me) to know that an infinity of numbers that are not algebraic exist to necessitate the extension of my set. Everything that I have been calling a number can be represented by a point on a line; I know that there are some (infinitely many) numbers (i.e. points on the line) that are not algebraic, so I simply extend my set to all points on the line and call it 'all numbers'.

    Of course I could have taken a different approach and extended my set using a different method: I might arrive at the p-adics and decide that is enough, but sooner or later I am going to want to go further and require Cauchy sequences or I might use the simpler method of decimal expansion (thanks micromass - and there is no reason to be restricted to base 10 so I don't think Devono's objection is a problem).

    So all of these methods (plus those of Dedekind and of others) arrive at the same conclusion: no (proper) subset of the reals is sufficient to merit in my eyes the label 'all numbers'.

    Me too, but I don't think I was ready for Dedekind Cuts at high school.
    Last edited: Jan 21, 2012
  23. Jan 21, 2012 #22


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    How do you know that it needs to be an uncountable set? From what I'm reading constructable reals is sufficient for your purpose.

    ...apart from the "complete metric space" bit. :tongue:

    It's more elementary, but not necessarily simpler. You need to worry about how carrying works for multiplication, and deal with the 0.9...=1 problem directly.
  24. Jan 21, 2012 #23
    Was my question not simple enough as it is? If you can't "add to the clarity of a point on the line", then your post is meaningless, because I don't know what you mean by a point on the line.
  25. Jan 21, 2012 #24
    I don't - I can see that I need numbers that are in the continuum but are not rational, so I am 'innocently' grabbing the whole of the rest of the continuum. The facts that this is uncountably many numbers, and that a subset of these numbers are algebraic numbers are things that I only learn when I start investigating my collection.

    Yes, elementary (or perhaps intuitive) is a better description than simlple - nothing is ever simple :)
  26. Jan 21, 2012 #25


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    This depends on what you mean. If you assume a priori that certain numbers like √2 and π exist then you can show that there are some numbers that are not rational. But formally you need to construct them in the first place before you can really talk about them at all.

    Granted there are constructions of things like √2 that do not yield all of the real numbers. But if you want to talk about the entire collection of real algebraic numbers or transcendental numbers, then you really do need all of the reals.

    Why do you assume that a line is best modeled by R? Non-standard models of R ought to be perfectly good candidates too.

    How are decimal expansions more simple or natural? Mathematicians rarely use the fact that the real numbers have decimal representations. The important properties that most mathematicians use are better captured in the constructions via Dedekind cuts or equivalence classes of Cauchy sequences.
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