How to Do the Sub Steps of Neutral Operations

Three Parts:The TutorialExponential OperationsHelpful Guidance

There is some simple algebra involved in Neutral Operations. Learn to do the algebraic sub-steps in this article. Neutral Operations are important for several reasons: they can be symmetrical (which is a hot topic in math these days!), they comprise a "tipping point" where one may choose between one operation between two values versus a different operation and they work well in graphics situations because of their unique property of holding operators neutral to one another. Their theory and sets were invented by this editor, as far as he knows.

Part 1
The Tutorial

  1. 1
    Enter the following into a new worksheet in a New Workbook or just use pencil and paper.
  2. 2
    To be discussed are the 5 basic Neutral Operations:
    • 1) a+b = a*b = c;
    • 2) a-b = a*b = c;
    • 3) a+b = a/b = c;
    • 4) a-b = a/b = c; and
    • 5) a*b = a^b = c.
    • In previous wikiHows, more complicated relations have been solved, e.g. a+b+c+...+z = abc...z and abcdefghi = a^(bcdefghi). Those will be referred to at the bottom in Related wikiHows.
  3. 3
    The first equation neutralizes Addition versus Multiplication, i.e. for the same two numbers a and b, a+b = a*b. The algebraic sub-steps to write down and learn are:
    • 1) a+b = a*b = c;
    • 2) a+b-b = ab - b;
    • 3) a = b*(a-1) as b is factored out on the right side while the left side is simplified;
    • 4) a/(a-1) = b as both sides are divided by (a-1) and the right side's (a-1)/(a-1) reduces to simply 1.
    • Now b has been isolated and defined in terms of a and 1. Since it was also possible to begin by subtracting a instead of b, and because both addition and multiplication are commutative, it's also true that b/(b-1) = a. And since ab = c, it is true also that a^2 / (a-1) = c and b^2 / (b-1) = c, which is very symmetrical.
    • It is well worth noting that the formula works wherever in Math or Physics one finds the relationship of c = ab, such as F = ma, or E = mc^2 or for a+b = c as in the Pythagorean Theorem, as a^2 + b^2 = c^2 can be set equal to a^2 * b^2 (or even a^2/b^2).
    • As a = b/(b-1), a is set to 5 and b then = 5/4. a+b = ab becomes 5+ 5/4 = 5 * 5/4 which both = 25/4. Check: the hypothesis holds. Neither a nor b may equal 1 lest 0 result in the denominator, for division by 0 is undefined.
  4. 4
    The second equation neutralizes Subtraction versus Multiplication and is handled in much the same way as the first. The algebraic steps to write down and learn are:
    • 1) a-b = a*b;
    • 2) a-b+b = ab + b;
    • 3) a = b*(a+1) as b is again factored out on the right while the left is simplified;
    • 4) a/(a+1) = b after both sides are divided by (a+1 and simplified. Now b has again been isolated and defined in terms of a and 1 but subtraction is not commutative so there is not a symmetrical relationship. If a is subtracted to start with instead of adding b, one obtains
    • 2) a-a-b =ab-a;
    • 3) -b = a*(b-1);
    • 4) -b/(b-1) = a and a has been isolated and defined in terms of b and 1.
    • 5) -b/(b-1) * -1/-1 = b/(1-b) = a.
    • Letting b equal 5 again then with b/(1-b) = a, a = -5/4 and -5/4 - 5 = -5/4 *5 because -5/4 - 20/4 = -25/4, check.
  5. 5
    The third equation is handled somewhat differently when neutralizing Addition versus Division. We need to be able to factor out a from the numerator, so we lose discretion. The steps to write down and learn are as follows:
    • 1) a+b = a/b and since we must factor out a on the right, we must subtract it on the left;
    • 2) a-a+b = a/b -a;
    • 3) b = a*(1/b -1) factoring out a on the right and simplifying the left;
    • 4) b/(1/b -1) = a as both sides have been divided by (1/b -1) and a has been isolated and defined in terms of b and 1. There is no symmetry. Letting b = 5, for b/(1/b -1) = a then, 5/(1/5 -1) = a = 5/(-4/5) = -25/4 = a. However, since a/b = c, it's possible to set c equal to [b/(1/b -1)]/b so that c = 1/(1/b - 1). Letting b = 5 and for c = 1/(1/b - 1), we have c = 1/(1/5 - 1) or 1/((-4/5) = -5/4 = c.
    • Testing the hypothesis of a+b = a/b = c, -25/4 + 5 = (-25/4)/5 becomes -25/4 + 20/4 = -25/20 which is -5/4 = c, check.. And c = 1/(1/b - 1) = 1/(1/5 -1) = 1/(-4/5) = -5/4, check also.
  6. 6
    The fourth principle equation, neutralizing Subtraction versus Division, operates much like the third where we lose some discretion. The steps to write down and learn are as follows:
    • 1) a-b = a/b and since a must be factored out on the right, it must be subtracted on the left;
    • 2) s-s-b = a/b - a;
    • 3) -b - a*(1/b - 1) factoring out a on the right and simplifying the left;
    • 4) -b//(1/b - 1) = a can become, by multiplying the left by -1/-1, b/(1 - 1/b) = a and a has been isolated and defined in terms of b and 1 by dividing through by (1/b -1). There is no symmetry.
    • However, since a/b = c, it's possible to set c equal to [b/(1 - 1/b)]/b = c so that c = 1/(1 - 1/b) ... sort of the opposite of the above. Letting b = 5 and for a = b/(1 - 1/b), a = 5/(1 -1/5) = 5/(4/5) = 25/4 (whereas above a equaled -25/4).
      • Testing the hypothesis of a-b = a/b = c, 25/4 - 5 = (25/4)/5 = 25/4 - 20/4 = 25/20 = 5/4, check. And c = 1/(1 - 1/b) or 1/(1 - 1/5) = 1/(4/5) = 5/4, check again.

Part 2
Exponential Operations

  1. 1
    Lastly, we have the principle neutral operation of a*b = a^b, which may be written down and learned from as follows:
    • 1) a*b = a^b;
    • 2) ab/a = (a^b)/a;
    • 3) b = a^(b-1);
    • 4) The root of (b-1) is taken of both sides so that b^(1/(b-1)) = a and a has been isolated and defined in terms of b and 1. Letting b=9, and for b^(1/(b-1)) = a then, 9^(1/(9-1)) = a and the left side evaluates in Excel to the value 1.31607401295249 = a.
    • Testing the hypothesis of a*b = a^b gives 1.31607401295249 * 9 = 1.31607401295249 ^ 9 and the first answer = 11.8446661165724 and the second value = 11.8446661165722, which is off just a smidgen, but acceptably so because of how Excel cuts off digits when calculating.
    • Because a*b = c and a = b^(1/(b-1)), then it is also true that b^(1/(b-1))*b or b^(1+(1/(b-1))) = c. 9^(1+(1/(9-1))) = 9^(9/8) and this answer = 11.8446661165724 = c, check.
  2. 2
    Many other side-excursions are possible, e,g, a+b = -a/b, etc, etc. Perhaps another time. Or perhaps you will want to work those out. The thrust of this article has been to simply cover the basics and now you have them. Good going!
  3. 3
    What does f = m+a or E = m+c^2 mean? It means the point of first tangence when acceleration first touches mass, or light speed squared does. It is the instantaneous addition and multiplication state and does nothing to invalidate the basic law involved.

Part 3
Helpful Guidance

  1. 1
    Make use of helper articles when proceeding through this tutorial:
    • See the Related wikiHows below and the article How to Do the Sub Steps of Neutral Operations for a list of articles related to Excel, Geometric and/or Trigonometric Art, Charting/Diagramming and Algebraic Formulation relating to Neutral Operations.
    • For more art charts and graphs, you might also want to click on Category:Algebra, Category:Mathematics, Category:Spreadsheets or Category:Graphics to view many Excel worksheets and charts where Trigonometry, Geometry and Calculus have been turned into Art, or simply click on the category as appears in the upper right white portion of this page, or at the bottom left of the page.


  • the introductory graph is of Neutral Operations asymptotes when a and b are in the form of x and y; more specifically, when x and y are defined as spheroids. By letting x and y's spheroidal values vary between say 2 and -2, while solving for y and while at no point letting either equal 1, the graphs are obtained. Learn more about this from the article, How to Create Spheroidal Asymptotes and Skewed Sphere Ring, included below under Related wikiHows.
  • n.b. A new Part, The Quadratic Relation of Neutral Operations / Symmetry by Commutation, has been added to the article How to Convert a Quadratic Formula to Roots Form by Completing the Square
  • Here is a list of articles relating in some way to neutral operations:

How to Do Neutral Operations with Exponents, How to Do Neutral Operations with Square Roots, How to Solve a Difficult Neu Ops Problem by Goal Seeking in Excel, How to Solve a Medium Difficulty Neutral Operations Problem with Excel, How to Solve aB=a^B in Neutral Operations Using Algebra, How to Solve a+b+c+d=abcd=e in Neutral Operations, How to Create a Dark Matter Box Pattern in Microsoft Excel, How to Do a Simple Rate Distance Time Problem with Neutral Operations, How to Create Floral and Other Images with Trig and Neutral Operations, How to Do a Short Goalseeking Neutral Operations Problem in Excel, How to Arrive at ‐1 and 1 from Spaces or Zeroes

Sources and Citations

  • Try googling NeuOps-Graphs01 for information on Neutral Operations.

Article Info

Categories: Mathematics | Algebra