Nothing from Something?

Discussion in 'The Cesspool' started by John J. Bannan, Jul 10, 2008.

  1. Reiku Banned Banned

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    I never said \(i\) could not be integrated, because to normal human understanding, there is always an imaginary vector that seems linear in nature. This is what Einstein attributed to his Special Law's, making his theory Observer-Dependant, when time and distance was coupled with a relative measurement.
     
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  3. Enmos Valued Senior Member

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    But then why do you keep 2aib ?
     
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  5. Sciencelovah Registered Senior Member

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    No, that's not another way of doing it. If that's just another way of doing it, it should lead to the same result, but it doesn't.

    Yours: \((a+ib)(a-ib) = a^{2}-b^{2}+2abi\)
    Enmos': \((a+ib)(a-ib) = a^{2}+b^{2}\)

    That's not true at all

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    Here is far away from bright. I just know basic algebra,
    basic trigonometry, and although I learn basic calculus, I forgot most of it already.
     
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  7. Reiku Banned Banned

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    To show that it can yield yet another congujate. It's actually a mathematical process that we can use to destinguish a negative or a positive value, so it is really a statistical field --- possibly the wave function itself... since it is a purely imaginal wave of existence as we know it, that clashes together and makes this stuff we call matter an energy, with a value that seems to always indicate a corresponding value \(\pm\), so either way, we can't have vectors that move in one direction and since this seems to be a known fact, the vector of the universe as an abstraction of imagination, leads us to calculations where it experiences an Omega Value. Hawking, if my memory serves me correctly, attributes to such a ratio of gravitational cosmic collapse to be well into trillions of years away.
     
  8. Enmos Valued Senior Member

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    But you get 2aib from aib - aib, which is zero in my book.
     
  9. Reiku Banned Banned

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    Mine: \((a+ib)(a-ib) = a^{2}-b^{2}+2abi\)

    can lead to:

    Enmos': \((a+ib)(a-ib) = a^{2}+b^{2}\)

    If we treat the \(a^{2}-b^{2}+2abi=2abi\)

    where \(2abi\) is in fact Enmoses \(a^{2}+b^{2}\) can yeild the imaginary \(i\), which is always real in a sense that it isn't actually entirely virtual, but has real effects on the universe at very small levels.
     
  10. Reiku Banned Banned

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    Exactly
     
  11. Enmos Valued Senior Member

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    Where did you get 2aib from in the first place ?
    And how can you just equate a[sup]2[/sup] - b[sup]2[/sup] + 2aib to 2aib out of the blue ?
     
  12. ashura the Old Right Registered Senior Member

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    He can't, this looks like the stuff I learned in high school but gone horribly wrong.
     
  13. Enmos Valued Senior Member

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    Then you formula reads (a+b)(a-b) = 0

    Edit: which should have been obvious from the beginning since a = b.. Hmmm

    Edit2:
    Oh wait.. it was (a+ib)(a-ib).. :bugeye:

    Since a = b, substitute a for b...

    (b+ib)(b-ib) = b[sup]2[/sup] + i[sup]2[/sup]b[sup]2[/sup] = b[sup]2[/sup] - b[sup]2[/sup] = 0

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    Last edited: Jul 11, 2008
  14. Sciencelovah Registered Senior Member

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    Reiku, that is wrong. To solve that correctly:

    \((a+ib)(a-ib) = ....\)

    here is the clue (notice the colour):

    (a+b)(c+d)
    = (axc) + (axd) + (bxc) + (bxd)
    = ac + ad + bc + bd


    Forget the imaginary number. Can you do the rest below first?

    2. (a+b)(a+b)?
    3. (a+b)(a-b)?
    4. (a+2b)(a-3b)?
     
  15. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    But you're not saying that, you're claiming that \((a-ib)(a+ib) = a^{2}-b^{2}+2abi\). If you say "But it's right when b=0" than you're just saying \((a)(a) = a^{2}\), so why don't you say that?
    Your usual attempt to throw in something complicated to get people to back off.

    Vector spaces and special relativity are irrelevent here. You haven't been able to multiply out (a+b)(c+d). If you can't do something expected of 12 year olds, you're not going to able to do vector calculus.
     
  16. Sciencelovah Registered Senior Member

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    Reiku, is my example above not clear enough? I can draw some indicating arrows on
    paper and scan it if needed.

    Anyway, I gotta logout in about half an hour. Let me know soon.
     
  17. Sciencelovah Registered Senior Member

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    You will get:

    Now please try with these ones:

    If you can solve that, you should have no problem to solve the one with imaginary
    number, because the pattern are basically the same.
     
  18. Reiku Banned Banned

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    So many people here have credited you for being a nice person. You're actually one twisted sister. Alphanumeric, i would have said that, but forgot to even put it in. So what?
     
  19. Reiku Banned Banned

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    And don't patronize me like that again, ok...
     
  20. Sciencelovah Registered Senior Member

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    Actually I didn't pay attention on the idea of the OP. The 0+0=0 got me carried away, lol.

    Anyway, did your OP mean like this:

    0 = 0 + 0
    0 = 5 - 5
    0 = 1 chicken - 1 chicken + 2 apples - 2 apples

    And then you conclude that because on the left side stays zero, whereas on
    the right side the chicken and the apples can come and go (birth and die),
    then you mean that everything (right side of the equation) is actually nothing
    (left side of the equation)?

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    And then you conclude that nothing can
    come from something, and something come from nothing. Is that what you mean?
     
  21. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Why not? I sent you a link to the FOIL rule months ago and you obviously didn't read it or if you did, you didn't understand it because you continue to make the same mistakes.

    We wouldn't patronise you if you demonstrated you didn't need to be spoken to like a brain dead 6 year old with learning disabilities.

    When you demonstrate you can learn and understand like an adult we'll treat you as such.
     
  22. Myles Registered Senior Member

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    5,553
    You are overlooking one thing. People working with higher maths. frequently take the simpler intermediate steps for granted, assumimg their readership will understand. You seem not to understand this.
     
  23. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Reiku talks about relativity and quantum mechanics. The ability to multiply out (a+b)(c+d) is assumed for anyone who is good enough at mathematical physics to be doing the afore mentioned topics. But Reiku cannot expand out (a+b)(c+d).

    If Reiku is doing what he claims, he should understand how to multiply out brackets. You cannot do relativity and quantum mechanics if you cannot multiply out brackets. But Reiku can't and yet claims he's doing those topics.

    And inzomnia is having to walk him through something so basic, it should be something he'd understood for about a decade. But he hasn't.
     

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