In the file, "ALLOWABLE DIFFERENCE MODEL", we have a limited ARITHMETIC OF ALLOWABLE DIFFERENCES (MINUENDS NOT LESS THAN SUBTRAHEND), provided with EQUALITY-INEQUALITY RELATIONS and OPERATIONAL RULES, such that THE PRIMARY OPERATIONS (ADDITION, MULTIPLICATION, EXPONENTIATION) are TOTAL (ALWAYS ALLOWABLE), whereas THE INVERSE OPERATION OF SUBTRACTION EXISTS ONLY for ALLOWABLE DIFFERENCES. So, we wish to GENERATE AN ENTIRE NUMBER SYSTEM with TOTAL SUBTRACTION.

Noting that SUBTRACTION IS A BINARY ORDERED OPERATION, we consider a BINARY ORDERED STRUCTURE: THE 2-VECTOR or 2-TUPLE or ORDERED PAIR, with NATURAL NUMBERS as its COMPONENTS. Thus, [p, q], where p, q are NATURALS.

Please note that this is a BYPASS. 2 — 3 is NOT MEANINGFUL for NATURAL NUMBERS 2, 3. But nothing prevents us from writing [2, 3]. And we'll see that this BYPASSES our SUBTRACTION PROBLEM in another way. Also, we MODEL VECTOR RULES on RULES for ALLOWABLE DIFFERENCES.

1. EQUALITY:
   • The ALLOWABLE DIFFERENCES fitted an ANTITONIC (MIXING) EQUALITY RULE: a₁ — b₁ = a₂ — b₂ → a₁ + b₂ = b₁ + a₂.

   • Similarly, VECTORS OF NATURALS, we have an ANTITONIC (MIXING) RULE: [a₁, b₁] = [a₂, b₂] → a₁ + b₂ = b₁ + a₂.     Thus, [8, 5] = [12, 9] because 8 + 9 = 5 + 12 = 17.
   • We now can INVOKE EQUIVALENCE TO SIMPLIFY COMPONENTS: [8, 5] = [8 — 5, 5 — 5] = [3, 0], that is, [8, 5] = [3, 0] because 8 + 0 = 5 + 3. Similarly, [12, 9] = [12 — 9, 9 — 9] = [3, 0].
   • And this has the CONSEQUENCE OF A NEW ALLOWABLE! For NATURALS, 2 - 3 is MEANINGLESS. But, for VECTORS OF NATURALS, we can write [2, 3]. And EQUIVALENCE finds that [2, 3] = [2 - 2, 3 - 2] = [0, 1]. That is, [2, 3] = [0, 1] BECAUSE 2 + 1 = 3 + 0. Charlotte?

2. Assignment: Find INEQUALITY RULES for VECTORS from THE EQUALITY RULE. We need only the latter rule below.

3. ADDITION:

   • For ALLOWABLE DIFFERENCES, we have an ISOTONIC (MATCHING) RULE: (a₁ — b₁) + (a₂ — b₂) = (a₁ + b₁) — (a₂ + b₂).

   • Similarly, for VECTORS OF NATURALS, we have an ISOTONIC (MATCHING) RULE: [a₁, b₁] + [a₂, b₂] = [a₁ + a₂, b₁ + b₂]. (CLOSURE!).     Thus (using cases from above), [8, 5] + [12, 9] = [8 + 12, 5 + 9] = [20, 14].     However, using EQUIVALENCE to simplify, [20, 14] = [20 — 14, 14 — 14] = [6, 0].    (Do you see how this agrees with what we found above?)

4. SUBTRACTION:
   • For ALLOWABLE DIFFERENCES, we have an ANTITONIC (MIXING) RULE: (a₁ — b₁) — (a₂ — b₂ ) = ((a₁ + b₂) — (a₂ + b₁)).

   • Similarly, for VECTORS OF NATURALS, we have an ANTITONIC (MIXING RULE): [a₁, b₁] — [a₂, b₂] = [a₁ + b₂, a₂ + b₁].     Thus (using the vector addition case), [20, 14] — [8, 5] = [20 + 5, 14 + 8] = [25, 22] = [25 — 22, 22 — 22] = [3, 0]. Charlotte?

     But HEY! We can also turn that around: [8, 5] — [20, 14] = [8 + 14, 20 + 25] = [22, 25] = [22 — 22, 25 — 22] = [0, 3].     We get results both ways with VECTORS, which we can't do with SINGLE NATURAL NUMBERS. And we did it without CHEATING!

     Why? Because the RULE TURNED SUBTRACTION INTO ADDITION FOR NATURALS, WHICH IS ALWAYS ALLOWED OR TOTAL!

5. MULTIPLICATION:

   • For ALLOWABLE DIFFERENCES, we have a MATCH-&=MIX ISOTONIC-&-ANTITONIC) RULE: (a₁ — b₁) · (a₂ — b₂) = ((a₁ · a₂ + b₁ · b₂) — (a₁ · b₂ + a₂ · b₁)).

   • Similarly, for VECTORS OF NATURALS, we have a MATCH-&-MIX (ISOTONIC-&-ANTITONIC) RULE: [a₁, b₁] · [a₂, b₂] = [a₁ · a₂ + b₁ · b₂, a₁ · b₂ + a₂ · b₁].     Thus, [9,5] · [11,8] = [9 · 11 + 5 · 8, 9 · 8 + 5 · 11] = [99 + 40, 72 + 55] = [139, 127].

ASSIGNMENT: Show this is equivalent to [4,0] · [3,0] = [12, 0].

I'll now show you some GENERAL RESULTS, which lead to "the hiding".

Students of abstract algebra will realize that our VECTOR EQUIVALENCE RULE (MODELED ON OUR ALLOWABLE DIFFERENCE EQUIVALENCE RULE) can do "what an Equivalence Rule does best": REDUCE ITS SYSTEM INTO A NUMBER OF EQUIVALENCE CLASSES (often a considerable reduction). Here, we'll see that THE VECTOR EQUIVALENCE RULE YIELDS JUST THREE EQUIVALENCE CLASSES.

PROOF: Given NONZERO NATURAL NUMBERS a, b for VECTORS of the form, [a, b]:

1. If a > b, then, for some NONZERO NATURAL, a — b, [a, b] = [a — b, b — b] = [a — b, 0], -- which I label "A CANONICAL UPPER PAIR".

2. If a < b, then, for some NONZERO NATURAL, b — a, [a, b] = [a — a, b — a] = [0, b — a], -- which I label "A CANONICAL LOWER PAIR".

3. If a = b, then [a, b] = [a — b, a — b] = [0, 0] -- which I label "A CANONICAL NULL PAIR".

The VECTOR EQUIVALENCE RULE (MODELED ON OUR ALLOWABLE DIFFERENCE EQUIVALENCE RULE) REDUCES the infinity of TYPES OF NATURAL NUMBER VECTORS TO JUST THREE EQUIVALENCE CLASSES, which I choose to label CANONICAL TYPES: CANONICAL UPPER and LOWER PAIRS and NULL PAIR!

Before THE VECTOR-INTEGER HIDING, let's see VECTOR-VECTOR HIDING. Thus, a particular CANNONICAL UPPER PAIR, such as [2, 0] HIDES AN INFINITY OF EQUIVALENCE VECTORS: [2, 0] = [3, 1] = [4, 2] = [5, 3] = ...., AD INFINITUM, for any TWO NATURAL DIFFERING BY 2. Dig?

This shows us how to HIDE VECTORS BY SIGNS:

1. For any NONZERO NATURAL NUMBER, n, [n, 0] => ⁺n. Example: [2, 0] → ⁺2, or simply 2.

2. [0, n] → ¯n. Example: [0, 2] = ¯2.

3. [0, 0] → 0.

Behold! THE INTEGERS! (Please note: I've written the +,- signs as SUPERSCRIPTS to AVOID THEIR CONFUSION with THE OPERATIONAL SIGNS FOR ADDITION AND SUBTRACTION. So please stop blathering about "minus 2", OR I'LL HAUNT YOU! Label it "negative 2"!) Also, please note that, while there is a 1-1 correspondence, such as [2, 0] → 2, THE INFINITY OF EQUIVALENCE VECTORS CARRIES OVER: 2 → [2, 0] = [3, 1] = [4, 2] = [5, 3] = ...., AD INFINITUM, for any TWO NATURAL DIFFERING BY 2. Etsettery.

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SUMMARY:

• We didn't CHEAT. For NATURALS 2, 3, the operation, 2 — 3, is still MEANINGLESS. But, for INTEGERS 2, 3, we have 2 — 3 = ¯1.

• Notice how we did that SUBTRACTION! Behind it was [2,0] — [3,0] = [2 + 0, 0 + 3] = [2, 3]. But, by THE EQUIVALENCE RULE, [2, 3] = [2 — 2, 3 — 2] = [0, 1] → ¯1. However, look inside the brackets. WE NEVER NEVER NEVER PERFORMED A FORBIDDEN SUBTRACTION! BYPASS ADDITION DID THE TRICK.

• So, now, we have an INTEGER SYSTEM in which ADDITION, MULTIPLICATION, EXPONENTIATION, AND SUBTRACTION ARE TOTAL. (DIVISION, LOGARITHM, ROOT EXTRACTION ARE STILL PARTIAL!)

• And "The Law of Signs" is shown not to be a WEIRD RULE. We need a MULTIPLICATION RULE FOR ALLOWABLE DIFFERENCES SATISFYING TWO CONDITIONS: CLOSURE (PRODUCT OF DIFFERENCES EQUALS A DIFFERENCE ALREADY IN THE SYSTEM), and DISTRIBUTIVITY (SHORT AND LONG WAYS OF CALCULATING WHICH SHOULD BE EQUIVALEN ARE INDEED EQUIVALENT). Satisfying these two conditions FORCE ON US A "MIX-&-MATCH" PRODUCT RULE -- WHICH BECOMES OUR "Law of Signs" in INTEGERS. But IT IS NATURAL NUMBER ARITHMETIC WHICH REQUIRES IT! NOT SOME KIND OF "NEW MATH"!

• But INTEGERS ARE SIMPLY ORDERED PAIRS OF NATURAL NUMBERS. That is, EVERY INTEGER IS EQUIVALENT TO A VECTOR [n₁, n₂], where n₁, n₂ ARE NATURAL NUMBERS. We didn't leave THE NATURAL NUMBER SYSTEM for good. We just went onto a "higher level" with it -- sort of "NATURALS in 2-D".

  Well, that explains INTEGERS in RENDERING TOTAL THE OPERATION OF SUBTRACTION AS INVERSE OF ADDITION. But another problem about ADDITION arises.

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  Back in NATURAL NUMBERS, just as REPEATED COUNTING BECAME TEEDJUS LEADING TO GENERATING ADDITION, so also REPEATED ADDING -- such as 5 + 5 + 5 -- becomes TEEEDJUS.

  SHORTCUT: GENERATE MULTIPLICATION (·) OF TWO NATURAL NUMBERS (a, b) BY RECURSION ("in The Third Degree") ON ADDITION OPERATION: a · 1 = a, S(a · b) = a · b + a. This GENERATES ANY PRODUCT, hence, THE NATURAL NUMBER SYSTEM IS CLOSED UNDER MULTIPLICATION.

  Is there a KIKBAK? Yes, because MULTIPLICATION is WELL DEFINED. For NONZERO a: a · b = a · c Þ b = c; and, COMMUTATIVITY OF MULTIPLICATION -- a · b = b · a -- MEANS THAT THE INVERSE IS UNIQUE (again, see failure of this, below).

  Hence, can DEFINE an MULTIPLICATION-KIKBAK ("in The Third Degree": "division") in terms of MULTIPLICATION: for NONZERO b, a ÷ b = c iff ("if and only if") a = b · c.

  Good News: DIVISION MAKES SENSE WHEN IT EXISTS. Bad News: THE QUOTIENT a ÷ b EXISTS iff a IS MULTIPLE of b. (Such quotients are ALLOWABLE QUOTIENTS.) And the SAME FOR INTEGERS! Hence, ONLY PARTIAL CLOSURE OF NATURALS AND INTEGERS FOR DIVISION. Equivalently, MULTIPLICATION IS A TOTAL OPERATION ("ALWAYS WORKS"); DIVISION IS A PARTIAL OPERATION ("ONLY WORKS SOMETIMES").

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  Again, the VECTORINGBYPASS&HIDING-SHORTCUT STRATEGY "saves the day" -- at this point, MAKING (NONZERO) DIVISION TOTAL. SHORTCVT by a VECTOR or ORDERED PAIR OF NATURALS and INTEGERS BECAUSE DIVISON IS ORDERED (NONCOMMUTATIVE), that is, although 4 · 2 = 2 · 4, but 4 ÷ 2 ¹ 2 ÷ 3.

  VECTOR&HIDING-SHORTCUT

  1. FORM ALLOWABLE QUOTIENT MODEL (FOR VECTORS) OF EQUALITY, INEQUALITY, SUM, DIFFERENCE, PRODUCT, QUOTIENT, EXPONENTIAL, ROOT OF ALLOWABLE QUOTIENTS.
  2. WRITE ARITHMETIC FOR VECTOR OF INTEGERS -- [a, b] -- IN TERMS OF AN ALLOWABLE QUOTIENT MODEL.
  3. VECTORS OF INTEGERS FALL INTO 3 BASIC TYPES:

     1. [a, b], WITH a MULTIPLE of b ("WHOLE NUMBERS");
     2. [a, b], WITH a NONMULTIPLE of b ("FRACTIONS");

     3. [a, b], WITH a = b (UNIT).
  4. The DIVISION PROBLEM of INTEGERS is BYPASSED: [a,b] ÷ [c,d] = [a · d, b · c] -- DIVISION IS BYPASSED FOR MULTIPLICATION, WHICH ALWAYS WORKS!

  5. Hence, VECTOR FORM CAN BE HIDDEN BY SHORTCUT OF SLASH SIGNS or NO SIGN -- SHORTCUT TO RATIONAL NUMBERS (VECTORS OF INTEGERS WITH ARITHMETIC OF ). Thus, [4,2] Þ 2; [2,4] Þ 2/4 = 1/2; [4,4] = [2,2] Þ 1.

  Let's see THE GENERATION OF RATIONAL NUMBERS AS VECTORS OF INTEGERS, WITH RULES MODELED ON ALLOWABLE QUOTIENTS OF INTEGERS.