Math Problem, Help needed

Discussion in 'Odds & Ends' started by ibemadskillzz, Sep 11, 2004.

  1. ibemadskillzz

    ibemadskillzz Semi-Pro

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    I was looking through my Calculus book from long time ago, and I came across a problem that I could not solve. It said, not to substitute numbers to solve the problem. It wants us to show proofs like geometry. The problem is: Prove that- Square root of AB is less than or equal to 1/2(a+b).
    Sorry I do not have the square root button, but the AB is inside the square root sign. And 1/2- is One-half as in fraction. I also do not have to Less than or equal to sign to type.
    thanks for all your help.
     
    #1
  2. ibemadskillzz

    ibemadskillzz Semi-Pro

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    It's amazing how no one can solve this problem. I expected more out of you guys. Are there any Math teachers out there? help
     
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  3. aahala

    aahala New User

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    I don't know if this fits your requirements, but here's a proof involving algebra and geometry:

    "Prove that- Square root of AB is less than or equal to 1/2(a+b). "

    If we make this an equation and square both sides we get:

    AB= 1/4(A2 +2AB + B2)

    So:
    4AB=A2 +2AB + B2
    2AB=A2 + B2

    The terms of the equation can be considered areas of rectangles, such as A2 is a square of sides A etc.

    To prove the original claim, one can prove it's impossible to fit the areas represented on the right into the area on the left and have any "space" leftover.

    If A=B, then it's obvious there's nothing left, as the equation becomes some form of: 2A2= A2 +A2 or 2B2 = B2 + B2.

    We can let A or B be the larger number, it doesn't matter. Let's let B be the larger and represent it as A + X.

    So we can stack two rectangles with dimensions A and B on top of one another, resulting in a rectangle B as the base and 2A in height for the left of the equation.

    Take out the b2 rectangle from this. What's left is a rectangle(if there's anything left at all) a rectange height of A- X and base A + X.

    Can we get A2 out of such a rectangle? No.

    (A - X)(A +X)=A2

    A2 - X2=A2 (This is not possible)

    We have proven the left side can not be greater than the right side of the original equation, which was our objective.
     
    #3
  4. ibemadskillzz

    ibemadskillzz Semi-Pro

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    thanks for your help. I kind of understand it, but I was looking for a all algebraic solution. Can someone prove it without using geometry shapes and formulas ?
    thanks
     
    #4
  5. aahala

    aahala New User

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    2AB=<A2 + B2

    A+X=B
    2A(A+X)=<A2 +(A+X)2
    2A2 +2AX=<A2 +A2 +2AX +X2
    2AX=<2AX +X2
    0=<X2
     
    #5
  6. Feña14

    Feña14 Legend

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    Wow that is crazy :shock:

    Is there smoke coming out of your ears? :lol:
     
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  7. SirGus

    SirGus New User

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    What's AB?a?b?
     
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