Wednesday, September 20

All Numbers Are Equal

Theorem: All numbers are equal.

Proof: Choose arbitrary a and b, and
let t = a + b. Thena + b = t

(a + b)(a - b) = t(a - b)

a^2 - b^2 = ta - tb

a^2 - ta = b^2 - tb

a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4

(a - t/2)^2 = (b - t/2)^2

a - t/2 = b - t/2

a = bNow, we know that all numbers are not equal. So, can you figure out where this proof breaks down?
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12 comments:

  1. AnonymousSeptember 20, 2006 4:02 PM

    You can't take say if x^2=y^2 that x=y. (line 7 of your theorem)
    x could equal -y.

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  2. LukeSeptember 20, 2006 4:26 PM

    Anon:
    So, you're saying that, if the final conclusion said a = +-b, then it's true?

    That's still known to be untrue.

    So, while you have a valid point, it still leaves problems.

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  3. AnonymousSeptember 20, 2006 4:47 PM

    No, it identifies an intermediate problem which is then used to build further on a false foundation. Try it with a=3 and b=5 and you will see the invalidity of the problem.

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  4. LukeSeptember 20, 2006 4:54 PM

    when a=3 and b=5 then you get (-1)^2=(1)^2 or 1=1 which is true and it holds.

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  5. MatthewSeptember 20, 2006 6:09 PM

    Somewhere in the middle you are dividing by 0 I think, I just haven't seen it yet. I will get back to you.

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  6. LukeSeptember 20, 2006 6:32 PM

    Matthew: That was my first thought too when I looked it over. I don't think that's it this time though.

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  7. AnonymousSeptember 20, 2006 6:33 PM

    (-1)^2=(1)^2, but your theorem then says that that means -1=1, which is not true.

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  8. LukeSeptember 20, 2006 6:39 PM

    anonymous: Sorry for my thickheadedness (is that a word?). I get ya now. Good call on that.

    What's your take on line 4 not really being a quadratic since you don't really have a variable? If you solve it like a quadratic then you have to assume a and b are variables...and t is a constant.

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  9. TifaniSeptember 20, 2006 11:02 PM

    My brain hurts!

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  10. AnonymousSeptember 20, 2006 11:07 PM

    I have no idea. I thought a and b were variables.

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  11. HeidiSeptember 21, 2006 8:53 AM

    I got lost after the 3rd line.

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  12. ajpSeptember 24, 2006 2:39 PM

    What about imaginary numbers, you know, when you divide by 0.

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