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Homework Help: Valid proof?

  1. Jul 17, 2006 #1
    valid proof? (Intro to Analysis)

    Hey all,

    Great forum here! I am working through "Elementary Analysis" by Kenneth Ross in preparation for an intro to analysis course I am taking in the fall. I was wondering if the following is a valid proof? Included are the axioms I used in the proof, but if you need the complete list I will post them. I'm still learning to use LaTeX so my appologies for not using it here.

    <= "is less than or equal to (or > for greater)"
    != "does not equal"

    Problem 3.4: Prove 0 < 1

    The following properties hold:
    M4. For each a != 0, there is an element a^(-1) such that aa^(1) = 1
    02. If a <= b and b <= a, then a = b.

    The following are consequences of an ordered field:
    (iv) 0 <= a^2 for all a

    Proof by contradiction: Suppose that 0>=1. By (iv), we also have that 0 <= 1^2 = 1. Therefore 0 <= 1 and 1 <= 0, and hence 0 = 1, by O2. But there is an element 1^(-1) = 1 such that (1)(1)^(-1) = 1, and 0 != 1, by M4. Thus 0 = 1 and 0 != 1, a contradiction. Therefore 0 >= 1 is false, and hence 0 < 1.

    I am not sure about using M4 to prove 0 != 1, I am pretty shaky on my logic when it comes to quantifiers.

    Thanks for any input!
    Last edited: Jul 18, 2006
  2. jcsd
  3. Jul 18, 2006 #2
    The only ring/field where you could have 1=0 is the singleton set {0}, ie: the ring/field whose only element is 0. So when talking about the reals or the rationals, it is taken as an axiom that 1 != 0.

    edit: Actually it is implied in the definition of a field that 1 != 0, so this is only true for rings (see my second post below). To arrive at a list of properties that defines a ring, just remove the following properties for a field
    - multiplication is commutative
    - existence of multiplicative identity
    - existence of multiplicative inverses for nonzero elements

    Here is where you should end your proof. You have just contradicted the axiom that 1!=0.

    Alternatively, it is not necessary to do a proof by contradiction, you could simply assert that 1 = (1^2) >= 0, and since 1 != 0 it must be that 1>0.
    Last edited: Jul 18, 2006
  4. Jul 18, 2006 #3
    Thanks. I guess that's where my confusion was, I wasn't sure if 0 != 1 was an axiom. It is not stated explicitly in the text I am using, but the author did present the axioms as properties of the rationals, so I guess 0 != 1 is assumed. Is it not possible to prove that 0 != 1 from the basic axioms of an ordered field?
  5. Jul 18, 2006 #4
    No, you can't deduce that 0 !=1 from the axioms of an ordered field. In fact it is stated in the defintion of a field that 1 != 0, and this is certainly true for ordered fields as well.

    In my first post, I stated the axioms for a ring (see edit). A ring with a multiplicative identity element which we call 1 is called a ring with unity.
    I showed you an example of a ring with unity in which 0 = 1
    Ie: The set {0} where addition and multiplication are defined by 0+0 = 0 and 0x0 = 0.
    Verify on your own that this set with the two operations defined above satisfies all the ring axioms. Notice in this case that 0 acts as both the additive and multiplicative identity.

    I also stated, that this is the only example of a ring where 0 = 1.
    To see this, note that for any ring K, we have
    0a = a0 = 0 , for all a in K, where 0 denotes the additive identity of K.
    (you should prove this on your own using only the ring axioms)
    Then you can deduce that for any ring with multiplicative identity 0, that
    a = 0 for all a in F, that is F = {0}.

    Since fields are rings with some added axioms, the above discussion must hold for fields also. But as I said, fields are defined such that 1 !=0 since the only possible candidate for a field with 1 = 0 would be {0}.

    I hope this is not too confusing. If so, it should relieve you to know, that this will be covered (in depth) if you take any course in abstract algebra, which covers topics in groups, rings, and fields.
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