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If a is not equal to zero then b/a=b*a^-1

  1. Jun 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove: if a[itex]\neq[/itex]0 then b/a=b*a^-1

    I don't know if my proof is sufficient with this one but I cannot think of another way. Also I am only supposed to use the basic axioms (e.g. commutative, distributive, existence of reciprocal...)

    2. Relevant equations
    1/a=a^-1 (I am not supposed to use this)
    Distributive law


    3. The attempt at a solution

    b/a=b*(1/a)
    =b*(a^-1)
    =b*a^-1

    I am not allowed to use the (1/a)=a^-1 thing, but I can't think of how to tackle this one any other way. Could someone at least get me pointed in the right direction? I am confused because the property seems too obvious hahaha.
     
  2. jcsd
  3. Jun 16, 2013 #2

    Dick

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    It is obvious, hahaha. And I don't think you can 'prove' it. It's a definition. As far as I know, b/a is defined by b*a^(-1). If you have an alternate definition of b/a then you might be able to prove it. What might that definition be?
     
    Last edited: Jun 16, 2013
  4. Jun 16, 2013 #3
    I don't see any alternate definitions of a/b. I only have these six axioms:

    AXIOM 1. COMMUTATIVE LAWS. X +y =y + X, xy = yx.
    AXIOM 2. ASSOCIATIVE LAWS. x + (y + z) = (x + y) + z, x(yz) = (xy)z. AXIOM 3. DISTRIBUTIVE LAW. x(y + z) = xy + xz.
    AXIOM 4. EXISTENCE OF IDENTITY ELEMENTS. There exist two distinct real numbers, which we denote by0and 1,such that for ecery real x we have x+0=x and1’ x =x.
    AXIOM 5. EXISTENCE OF NEGATIVES. For every real number x there is a real number y such that x + y = 0.
    AXIOM 6. EXISTENCE OF RECIPROCALS. For every real number x # 0 there is a real number y such that xy = 1.

    and these 4 theorems:

    THEOREM 1.1. CANCELLATION LAW FOR ADDITION. Zf a + b = a + c, then b = c. (In particular, this shows that the number 0 of Axiom 4 is unique.)
    THEOREM 1.2. POSSIBILITY OF SUBTRACTION. Given a and b, there is exactly one x such thata+x=6. Thisxisdenotedbyb-a.Inparticular,0-aiswrittensimply -aand is called the negative of a.
    THEOREM 1.3. b - a=b+(-a).
    THEOREM 1.4. -(-a) = a.
     
  5. Jun 16, 2013 #4
    If you can't directly use [itex]\dfrac{1}{a} = a^{-1}[/itex], then can you use [itex]a \cdot a^{-1} = 1[/itex]? If not, then what exactly is [itex]a^{-1}[/itex]? How is it defined?

    Edit: I don't see either [itex]\dfrac{1}{a}[/itex] or [itex]a^{-1}[/itex] defined in the axioms you posted. I believe it's safe to assume [itex]\dfrac{1}{a}[/itex] is defined as the multiplicative inverse of [itex]a[/itex]. If [itex]a^{-1}[/itex] isn't defined in the same way (from which the desired proof would follow from definition), then how exactly is it defined?
     
    Last edited: Jun 16, 2013
  6. Jun 16, 2013 #5

    Dick

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    There is no definition in there of what b/a means or what a^(-1) means, though I assume the latter is that a^(-1) is the y in the expression that is supposed to exist in ay=1 in AXIOM 6. There is no way to prove anything if your axioms don't tell you what the notation means in terms of the axioms. See the problem?
     
    Last edited: Jun 16, 2013
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