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Apostol's Field Axioms

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data

    I've been steadily working through Apostol's Calculus, Volume 1 and trying to prove all of the theorems and solve all of the exercises, but I skipped a few theorems and exercises in the Field Axiom section (I 3.2 & I 3.3). The lack of continuity irks me, so I sometimes flip back to that section and reattempt the problems, only to find myself in the same predicament. I'll post a few of the problems in hope that some one can shed some light on the situation. :)

    [itex](a/b) + (c/d) = (ad + bc)/(bd)[/itex] if [itex]b \neq 0[/itex] and [itex]d \neq 0[/itex]

    [itex]-(a + b) = - a - b[/itex]

    2. Relevant equations

    Apostol says that we can only use Axioms 1-6 and Theorems I.1 through I.4 for the first problem. This includes commutativity, associativity, distributivity, existence of identity elements, existence of negatives, existence of reciprocals as axioms, the cancellation law, and the possibility of subtraction which states that there is a unique real number [itex]x[/itex] such that [itex]a + x = b[/itex] for real numbers [itex]a[/itex] and [itex]b[/itex].

    The second problem can use the Axioms and all of the theorems.


    3. The attempt at a solution

    I proved the first equality using Theorem I.7 (the possibility of division), which is not on the included list, and some shoddy applications of associativity. But I cannot seem to solve it only using the given axioms and theorems.

    There exists a unique [itex]x[/itex] such that [itex](bd)x = ad + bc[/itex] and it is denoted as [itex](ad + bc)/(bd)[/itex]. Then I replace [itex]x[/itex] with [itex](a/b) + (c/d)[/itex] and it comes out to [itex](ad + bc)[/itex], so it must be equal to [itex](ad + bc)/(bd)[/itex], as asserted.

    To prove the second equality I noticed that [itex]-(a + b)[/itex] is the negative of [itex](a + b)[/itex]. I also assumed that the negative of a real number is unique because of the possibility of subtraction, but this might be a formally weak argument. Since we know that there is a unique real number [itex]x[/itex] such that [itex]a + x = b[/itex] for real numbers [itex]a[/itex] and [itex]b[/itex], then if we are given a real number [itex]a[/itex] and [itex]0[/itex], there is a unique number [itex]x[/itex] (denoted as the negative of [itex]a[/itex]) such that [itex]a + x = 0[/itex]. If this is not sufficient or ideal, please let me know.

    So then we replace [itex]x[/itex] with [itex](-a -b)[/itex] and show that [itex](a + b) + (-a -b) = 0[/itex]. It ends up working out, but I am slightly concerned that my applications of the associativity axiom were not concrete. The associativity axiom states that [itex]x + (y + z) = (x + y) + z[/itex], so we can take [itex](a + b) = x[/itex], [itex]-a = y[/itex] and [itex]-b = z[/itex] and then everything works out. Is that logically sound?

    I thought it was until I read the next exercise which reads [itex](a - b) + (b - c) = a -c[/itex], which seems to follow from the same argument. So perhaps I cannot assume the above application of associativity for that problem.

    Any help is greatly appreciated. Thanks!
     
  2. jcsd
  3. Feb 26, 2012 #2

    Deveno

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    first of all, what is a/b? i am assuming what is meant is (a)(1/b) = ab-1.

    i would start off like so:

    (a/b) + (c/d) = ab-1 + cd-1 (by definition)

    = (ab-1) + cd-1)(1) (multiplicative identity)

    = (ab-1)(1) + (cd-1)(1) (distributive law)

    = (ab-1)(dd-1) + (cd-1)(bb-1) (definition of (multiplicative) inverse)

    (or using "slash" notation:

    = (a/b)(d/d) + (c/d)(b/b)).

    i prefer to use the form ab-1 instead of a/b, because it makes where to use the associative laws and commutative laws (of multiplication) more apparent.

    you will need one more lemma (or sub-proof):

    (1/x)(1/y) = 1/(xy)

    that is x-1y-1 = (xy)-1

    in order to finish.

    your use of the associative law in the second part isn't quite right:

    if x = a+b, y = -a, and z = -b, then the associative law says this:

    (a + b) + (-a + -b) = ((a+b) + (-a)) + (-b).

    now it's true the LHS is the same as (a + b) + (-a - b),

    but it's not immediate that the RHS is automatically 0. you need to use the commutative law to show that:

    a + b = b + a, so (a + b) + (-a) = (b + a) + (-a),

    and then use the associative law again to show that:

    (b + a) + (-a) = b + (a + (-a)),

    and then use the definition of -a to show that (a + (-a)) = 0

    to conclude that (a + b) + (-a) = b + 0 = b (i used the identity law for addition in there somewhere).

    and you're not "home-free" yet....there's still more steps involved.
     
  4. Feb 26, 2012 #3
    Thank you for your response. :smile:

    Unfortunately, your proof for the first inequality runs into the same problem mine did - it uses a lemma outside of the permitted axioms/theorems.

    I'll post most of my proof of the second equality for clarification. I didn't list the intermediate steps initially because I was more concerned with my application of the associative law in the fourth line.

    By Axiom 5 and the definition of negatives, we know that [itex]-(a+b)[/itex] is the negative of [itex](a + b)[/itex] and [itex](a + b) + - (a + b) = 0[/itex]. It follows from Theorem I.2 (there exists a unique real number [itex]x[/itex] such that [itex]a + x = b[/itex] for real numbers [itex]a[/itex] and [itex]b[/itex]) that the negative is unique. It suffices to show that [itex](a + b) + (-a -b) = 0[/itex].

    [tex]
    \begin{align*}
    (a + b) + (-a -b) &= (a + b) + [-a +(-b)] \\
    &= (a + b) + [(-b) - a] \\
    &= (a + b) + [(-b) + (-a)] \\
    &= [(a + b) + (-b)] + (-a) \\
    &= [a + [(b + (-b)]] + (-a) \\
    &= (a + 0) + (-a) \\
    &= a + (-a) = 0
    \end{align*}
    [/tex]
     
  5. Feb 27, 2012 #4

    Deveno

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    i'm not sure which "axiom" you're talking about, is it:

    1/(xy) = (1/x)(1/y)?

    one doesn't need "uniqueness" of an inverse to show that, just "existence":

    (xy)[(1/x)(1/y)] = x(y[(1/x)(1/y)] (associativity)

    = x(y[(1/y)(1/x)] (commutativity)

    = x([y(1/y)](1/x)] (associativity)

    = x((1)(1/x)) (using y(1/y) = 1, since 1/y is "a" inverse for y)

    = x(1/x) (multiplicative identity property of 1)

    = 1 (since 1/x is "a" inverse for x).

    thus 1/(xy) is "one" inverse for xy (it might be, that there are others, but at the very least we know for sure that (xy)(1/(xy)) = 1 if x(1/x) = 1, and y(1/y) = 1).

    what is going on is this:

    the actual process of computing a/b + c/d goes like this:

    1. a/b + c/d = (a/b)(1) + (c/d)(1) =
    2. (a/b)(d/d) + (c/d)(b/b) =
    3. (ad/(bd)) + (bc/(bd)) =
    4. (ad + bc)/(bd)

    1-->2 is obvious
    2-->3 uses commutativity and associativity of multiplication

    (which is why writing 1/x = x-1 makes it clearer)

    3-->4 uses the distributive law

    for the above to work, we just need the existence of a-1 and b-1, we don't need uniqueness (the possibility of division).

    of course, uniqueness is easy to prove given existence:

    suppose ab = ba = 1, and ca = ac = 1:

    then c = c(1) = c(ab) = (ca)b = (1)b = b.
     
  6. Feb 27, 2012 #5
    I apologize for not being clear.

    The first equality is a theorem from Calculus, Volume 1 by Apostol. The proof is left as an exercise, but we are restricted to the field axioms and the first four theorems of the section. I'll list them all.

    Axioms:
    • Commutativity
    • Associativity
    • Distributivity
    • Existence of Identity Elements
    • Existence of Negatives
    • Existence of Reciprocals

    Theroems:
    • If ab = ac, then b = c.
    • There exists a unique x such that a + x = b.
    • b - a = b + (-a).
    • -(-a) = a.

    I was able to prove the equality using another theorem from the section, but I have not found a way to prove it under the conditions of the exercise. Your method most certainly works, but it also does not satisfy those conditions.
     
  7. Feb 28, 2012 #6

    Deveno

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    you'll have to be a bit clearer by what you (that is Apostol) means by "existence of reciprocals".

    usually this means: for every field element x ≠ 0, there is a field element y with:

    xy = 1.

    the element y is then denoted x-1, or 1/x.

    none of this mentions anything about whether of not such a y is unique (although it turns out it has to be). you want to prove a certain sum in F, namely:

    a/b + c/d

    is equal to another element in F, namely (ad+bc)/(bd).

    well, for that to be true, first of all, all of a/b, c/d and (ad+bc)/(bd) have to be in F. so, b,d and bd all have to be non-zero (because 1/x's existence is not assured when x = 0).

    i assume that a/b is DEFINED as (a)(1/b), where 1/b is DEFINED to be "some" element of F with b(1/b) = 1.

    i don't see anyway around using the fact (it IS a fact, even if you have to prove it) that (1/b)(1/d) = 1/(bd).

    personally, i don't like the notation "1/x". and here is WHY:

    you can have a field with just 3 elements: {0,1,2}, where one defines:

    1+1+1 = 0.

    in this field 2-1 = 2, although it makes one uncomfortable to think of writing:

    1/2 = 2 (although "cross-multiplying" gives you 1 = 4, which makes more sense, because

    4 = 1+1+1+1 = 1+(1+1+1) = 1+0 = 1, by definition).

    that is, there are "other fields" that "don't look like rational numbers", and our intuition about how division works can lead us astray.


    now, if the definition of a/b is such that a/b = c/d iff ad = bc (this is a very old definition, dating back to eudoxus), you are reduced to proving:

    (bd)[(a/b) + (c/d)] = ad + bc

    but you're still going to need to prove that y(x/y) = x (which means showing that x/1 = x, so you have something to "cross-multiply"). that is, to make the notation a/b work, you need to know when you can "cancel things".
     
  8. Feb 28, 2012 #7
    For every real number [itex]x[/itex] there is a real number [itex]y[/itex] such that [itex]xy = 1[/itex]. Later he defines this [itex]y[/itex] as [itex]x^{-1}[/itex].

    I am not refuting the accuracy of your proof - it's correct; however, it does not satisfy the conditions of the exercise. We're only allowed to use the axioms and theorems that I previously listed. The fact that the reciprocal of a product is the product of reciprocals is left as an exercise after this theorem, so we cannot use it.

    I proved it using another theorem, which states that there is a unique real number [itex]x[/itex] such that [itex]ax = b[/itex] for real numbers [itex]a[/itex] and [itex]b[/itex]. It also works, but the theorem is not included on the list of permitted theorems. I've been attempting to prove the theorem given the conditions of the exercise, but I have had no success.
     
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