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Proof using the Axioms, Analysis

  1. Aug 31, 2009 #1
    1) For any x > 0 and 0 ≤ h < 1 we have (x + h)^2 ≤ x^2 + h(2x + 1).
    2) For any x > 0 and p > 0 with x^2 < p there exists y > x with y^2 < p.

    Prove the following statements (only using the axioms for the real numbers). At each step say which axiom you use.

    The problems is that my professor expects everyone to know how to do this from previous class; however my former professor never explained anything like this.

    I don't want an answer but I am hoping someone could guide me through proving this.
    So, could anyone help me get started, I really have no idea where to begin.
     
  2. jcsd
  3. Aug 31, 2009 #2

    Hurkyl

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    What exactly are you having trouble with? Understanding why the statements are true? Or trying to translate an argument into a formal one? Or something else?
     
  4. Aug 31, 2009 #3

    Mark44

    Staff: Mentor

    For the first one, the key is that if 0 ≤ h < 1, h2 < h.

    For the second one, draw a number line with x, y, and p on it.
     
  5. Aug 31, 2009 #4
    I suppose I am having trouble with both, in my previous class "Introduction to Mathematical Reasoning" I have never proved anything using axioms, most of the "proving" done in class was by induction and contradiction.
    I know what the axioms are but I don't know how to apply them, in this case I have no idea where to begin.
     
  6. Aug 31, 2009 #5

    Hurkyl

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    Then worry about "why are they true" first.

    These are elementary arithmetic statements. Forget about axioms and issues of formal proof for a moment -- just do arithmetic, and argue why these should be true statements.
     
  7. Aug 31, 2009 #6
    I think I should concentrate on one problem at a time.
    So:

    (x + h)^2 ≤ x^2 + h(2x + 1)
    x^2 + 2hx +h^2 ≤ x^2 + 2hx +h^2
    h^2 ≤ h

    or by example, Let x=1, h=0
    (1 + 0)^2 ≤ 1^2 + 0(2(1) + 1)
    1 = 1

    or, Let x=1, h=0.5
    (1 + 0.5)^2 ≤ 1^2 + 0.5(2(1) + 1)
    2.25 < 2.5

    Therefore, h^2 ≤ h is true.

    Where would I go from here?

    P.S. I did not expect such a quick response, thanks a lot.
     
  8. Aug 31, 2009 #7

    Mark44

    Staff: Mentor

    Given that 0 ≤ h < 1, h2 < h, so
    (x + h)2 = x2 + 2xh + h2
    < x2 + 2xh + h = x2 + h(2x + 1) QED


    You can omit the examples above. No amount of examples can prove anything.
    You weren't required to prove the statement above, but you need to use it to prove what you need to prove.
     
  9. Aug 31, 2009 #8
    a) Since 0 ≤ h < 1 it follows that h^2 ≤ h (I think there is an axiom but I am not sure which one)
    (x + h)^2 ≤ x^2 + h(2x + 1)
    x^2 + 2hx +h^2 ≤ x^2 + 2hx +h^2
    h^2 ≤ h

    b) From this it follows that

    (x+h)^2
    = (x+h)(x+h)
    = x(x+h) + h(x+h)
    = x^2 + xh + hx + h^2
    = x^2 + 2hx + h^2 ≤ x^2 + 2hx + h
    = x^2 + h(2x+1)

    Where the first equality is the definition of ^2, the second and third
    equalities come from "the law of compatibility with multiplication", the forth from "the associative law of addition",the fifth inequality by (a), the sixth from
    "the associative law of multiplication"

    How is this? Am I using the right axioms?
     
  10. Aug 31, 2009 #9
    I'm not sure the properties you are expected to use are truly axioms, so much as axioms and theorems. In this case (a < b, c > 0 -> ac < bc) is the Multiplicative Property of Inequality. If your instructor is expecting you to fabricate these proofs from basic principles you'll need to look up the derivations of the theorems.

    Regardless, you do not need anything after the first line here.

    I made a minor edit to your statement so that each line has a separate comparison. The steps of your proof look sound but the rules you state are slightly off. Numbering the comparisons in order (equality and inequality) 1 through 6:

    (1) Definition of second power (as you correctly stated)
    (2) Distributive Property: a(b + c) = ab + ac
    (3) Distributive again
    (4) Additive Commutativity and "Combining Like Terms" (which is related to Distribution).
    (5) From (a) (you are correct here)
    (6) Factoring (again related to Distribution).

    Associativity of Multiplication is a(bc) = (ab)c. For addition it is a + (b + c) = (a + b) + c. Neither occur here. I have not heard of the "law of compatibility with multiplication." Do you mind sharing it? I might know it by another name.

    Just so you know, I have found that it is more important to be able to use the properties to form a valid proof, than it is to know all the names of the steps, although knowing them can't hurt.


    --Elucidus
     
  11. Sep 3, 2009 #10
    Thanks a lot for your help guys. As far as the other one
    "For any x > 0 and p > 0 with x^2 < p there exists y > x with y^2 < p."
    Mark44 said to draw a number line which I did. When x=1, x^2=1, y=2, y^2=4, then y^2>x^2, p=6 then p>y^2. So I see how this makes sense. but unlike the first example I am not sure what is it I need to concentrate on. Do I need to brake down the y^2 < p and work with that (kind of like the first one).
     
  12. Sep 3, 2009 #11
    I finally figured out what your professor is doing. I think she has a very beautiful goal in mind. You are going to prove directly from first principles that the set

    {w : w^2 < p}

    has no largest member.

    You have part (1) done. To prove part (2), you will use part (1). Hint: given x^2 < p, can you somehow find an h (between 0 and 1) such that x^2 + h*(2x+1) < p?
     
  13. Sep 3, 2009 #12
    Can I ask what would make you think that my teacher is trying to do that (I just wish I could see that myself). Also am I looking for a sample number just to see if it works and then by the first equation I should prove that x^2=(x+h)^2?
     
  14. Sep 3, 2009 #13

    Mark44

    Staff: Mentor

    Don't limit yourself to integers, as you seem to be doing.

    Put x^2 on the number line to the left of a number p. Do you see that you can squeeze in a number between x^2 and p?
     
  15. Sep 3, 2009 #14
    Yes, and if we keep the number constant so they stay in the same spots on number line then we could always squeez in another number between them that would be the principle of density where x<((x+z)/2)<z<((y+z)/2)<y and so on.
    Does this have anythine to do with what I am trying to prove?
     
  16. Sep 3, 2009 #15
    If x and p are given, can't you solve x^2 + h*(2x+1) < p for h?
     
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