Thread: 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).

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.

oh yes, -1, that's right.