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An ordered field in which N is bounded.

  1. Sep 26, 2005 #1
    I have absolutely no clue how to start here.

    Let F be the set of expressions of the form a = sum from i in Z of a-sub-i*x*i, where each a-sub-i is an element of R and {i < 0 : a-sub-i does not equal 0) is finite. (X is a formal symbol, not a number). An element a belonging to F is positive if the least indexed nonzero coefficient ak (what does this mean!) in the expression for a is positive. The sum oif a in F and b in F is the element c in F defined by c-sub-i=a-sub-i + b-sub-i for i in Z. the product for a in F and b in F is the element c in F defined by c-sub-j = sum from i in Z of a-sub-i * b-sub-j-i for j in Z.

    a) prove that the sum and product of the two elements of F is an element of F.

    b) WE have defined addition, multiplication, and order on F. Prove that with these operations, F is an ordered field.
    c) Interpret each real number a as the element a in F with a-sub-i = 0 for all i, except a-sub-0 = alpha; this interprets R as a subjset of F. PRove that N is a bounded set in F. COnclude that F does not satisfy the Archimedean property.

    I'm a freshman in College and this problem is giving me nightmares. any help at all would be appreciated.
  2. jcsd
  3. Sep 26, 2005 #2


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    It sounds like you're saying that the elements of F are formally written as:

    \sum_{n = -\infty}^{\infty} a_n x^n

    where there are finitely many nonzero [itex]a_i[/itex] with i < 0.

    Incidentally, these things are called formal Laurent series over R, and F is usually denoted as R((x)).

    "The least indexed nonzero coefficient" is exactly what it says. :smile: Among all the nonzero coefficients, you are to look at the one with the smallest index.

    Anyways, what have you done on any of the problems?
  4. Sep 26, 2005 #3
    Do you think you could break the problem down into less formal language for me? That might help me get anywhere!
  5. Sep 26, 2005 #4
    to reiterate: i honestly have no clue where to start and i'm not even sure i understand the problem correctly.

    to give you a clue as to where the class is mathematically: it's a class for people who might want to be math majors (or minors), so pretty much everyone in it is a freshman in one of their years of calculus (mostly second year). after talking with many students in the class, i can safely say that no one seems to know exactly what to do -- although problem might be less scary than it seems.

    am i going to have to see that it satisfies the axioms (like the Distributive Law and that x*1 = 1 etc.?). thank you so much.
  6. Sep 26, 2005 #5

    Tom Mattson

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    Let's see if I can help.

    That's just a sum over [itex]i[/itex], and the domain of [itex]i[/itex] is all integers.

    The reason for this is that there has to be an [itex]a_i[/itex] with a lowest index. If there were infinitely many nonzero [itex]a_i[/itex] with [itex]i<0[/itex] then your set would have no elements, because you would have to go infinitely far back to find the nonzero [itex]a_i[/itex] with lowest [itex]i[/itex].
  7. Sep 26, 2005 #6
    how does [itex]i<0[/itex] relate to [itex]a_i[/itex]. like [itex]a_i[/itex] seems to me to be just a constant, with the [itex]_i[/itex] part just differenting it from other [itex]a[/itex]s. when they say that [itex]i<0[/itex] do they mean that [itex]a_i <0[/itex]?

    and am i correct in saying that all of the elements of F are [itex](a_-nx^-n,...,a_0x^0,a_1x,a_2x^2,...,a_nx^n)[/itex]? and the [itex]a[/itex]s can assume any values in [itex]R[/itex] except that there can't be infinitely many [itex]a_i[/itex]s less than zero for [itex]i < 0[/itex] to ensure that we know where to look for [itex]a_x[/itex] to see where it is positive?

    is a) true because [itex]c_i[/itex] is of the form [itex]a_ix^i + a_jx^j[/itex] which is in the original [tex]
    \sum_{n = -\infty}^{\infty} a_n x^n

    [/tex]. the reasoning (using [tex]a*b=c[/tex] and [itex]x^n * x^a=x^d[/itex] for [itex]d=a+n[/itex]?
    Last edited: Sep 26, 2005
  8. Sep 26, 2005 #7
    yes? no? just let me preemptively thank you all for all the help you've given me thusfar and for all the help you might choose to give me in the future. you guys rock!
  9. Sep 26, 2005 #8

    Tom Mattson

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    The negative values of [itex]i[/itex] have to be explicitly mentioned because your problem talks about the "least indexed nonzero coefficient". That part in quotes refers to the [itex]a_i[/itex] with the lowest value of the index [itex]i[/itex]. Well, if the index simply took on values of, say, the natural numbers (something with a lower bound) then we wouldn't have to do anything special about that phrase. But since the index can be any integer, there must only be finitely many of the [itex]a_i[/itex] that are nonzero, or else there will be no least indexed nonzero [itex]a_i[/itex].

    [itex]a_i[/itex] is not just a constant, but a sequence of them.

    No, it means just what it says: The index is less than zero. [itex]a_i[/itex] itself need not be less than zero.

    An example would be [itex]a_{-3}=12.315[/itex]. The index is negative, but the constant itself need not be.

    There can't be infinitely many [itex]a_i[/itex]'s that are nonzero (not just negative) for [itex]i<0[/itex], and the reason is what I said before: If there are infinitely many with [itex]i<0[/itex], then there will be no "least indexed coefficient".

    I think that you have to to part a by cases: One in which the lowest index of each element of [itex]F[/itex] is the same, and one in which they are different. Your proof is going to make use of the fact that the [itex]\mathbb{R}^+[/itex] is closed under addition.
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