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Math Help - Neutral Operations

  1. #1
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    Neutral Operations

    If we neutralize the operations of ADDITION and MULTIPLICATION, we arrive at a set -- let's call it it a subset of the "Neutral Set."

    FOR EXAMPLE, LET A+B = A*B = C .... THEN THE OPERATIONS OF ADDITION AND MULTIPLICATION ARE HELD NEUTRAL TO EACH OTHER. WE CAN SOLVE THIS EQUATION:
    A+B-B = AB-B
    A = B(A-1)
    A/(A-1) = B AND SINCE BOTH OPERATIONS ARE COMMUTATIVE,
    B/(B-1) = A .... (WE MIGHT JUST AS EASILY HAVE STARTED BY SUBTRACTING A FROM BOTH SIDES). AT ANY RATE,
    A HAS NOW BEEN ISOLATED AND DEFINED IN TERMS OF B AND 1 AND B MAY NOT EQUAL 1 OR THE DENOMINATOR GOES TO ZERO, AND
    B HAS NOW BEEN ISOLATED AND DEFINED IN TERMS OF A AND 1 AND A MAY NOT EQUAL 1 OR THE DENOMINATOR GOES TO ZERO.

    FURTHER, SINCE AB = C AND B = A/(A-1), then C = A^2/(A-1) = B^2/(B-1).

    It is a simple matter to then see that for the number A=5, if B=5/4, THEN A+B = A*B = C.

    It is also simple to do A-B = A*B, A+B = A/B, and A-B = A/B.

    WE CAN SOLVE FOR AB = A^B, THUS NEUTRALIZING MULTIPLICATION VS. EXPONENTIATION.

    FURTHER, BY A SEQUENCE OF STEPS, WE CAN CALCULATE A+B+C+D+... = A*B*C*D*...

    BY THE NEWTON-RAPHE METHOD, WE CAN DO A+B = A^B, BUT I HAVE YET TO FIND A SIMPLER METHOD, AND THAT'S WHERE I'M ASKING FOR HELP.
    THE SAME APPLIES TO A+B = A SQRT(B) AND A-B = A SQRT(B) AND A-B = A^B (FOR FRACTIONS).
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  2. #2
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    Re: Neutral Operations

    Oh yes, the set {a,b} such that a<>1, b<>1 becomes {a,a/(a-1} and {b/(b-1), b} for a+b = a*b exists and is real; however, if the tangent of 90 degrees is perpendicular to the tangent of 0 degrees and they equal infinity and 0 respectively, such that y/x = infinity when x approaches zero as close as we like and y/x approaches 0 when y=0, then for two lines to be perpendicular, if their slopes be multiplied together and equal one, they are said to be perpendicular, then infinity * 0 = 1 defines the cartesian plane by defining the perpendicularity of the x and y axes, and division by 0 is allowed in the denominator after all.
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  3. #3
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    Re: Neutral Operations

    oh yes, -1, that's right.
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