Briefly, ARITHMETIC can be GENERATED according to THREE STRATEGIES:
SHORTCUTS Primitive humans tallied with marks or tally-sticks or knotted strings or used other COUNTABLES to ACCOUNT for possessions or observations. TEEEEEDJUS!
SHORTCUT: GENERATE NATURAL OR COUNTING NUMBERS BY RECURSION from "the seeds" of ZERO, 0, and SUCCESSOR OPERATOR,S(_). Thus, S(0), S(S(0)), S(S(S(0))), .... TEEEEDJUS! (In what follows, I'll call this "Recursion in The First Degree".)
Demanding NUMERAL SHORTCUT: S(0) ⇒ 1; S(S(0)) ⇒ 2; S(S(S(0))) ⇒ 3; ....
KIKBAK ("Backwards Counting") is cannot go "below" zero. Hence, THE NATURAL NUMBERS are CLOSED UNDER COUNTING, but not KIKBACK COUNTING.
Now, ordinary counting also led to REPEATED COUNTING -- such as COMBINING 1,2 with 1, 2, 3, for 1, 2, 3, 4, 5. TEEEDJUS!
SHORTCUT: GENERATE ADDITION (a + b) OF TWO NATURAL NUMBERS (a, b) BY RECURSION ("in The Second Degree") ON THE SUCCESSOR OPERATION: a + S(b) ≡ S(a + b). This GENERATES ANY SUM, hence, THE NATURAL NUMBER SYSTEM IS CLOSED UNDER ADDITION, that is, ADDITION IS A TOTAL OPERATION.
SHORTCUT: GENERATE MULTIPLICATION (a * b) OF TWO NATURAL NUMBERS (a, b) BY RECURSION ("in The Third Degree") on THE ADDITION OPERATION: a * 1 = a, a * S(b) ≡ a * b + a. This GENERATES ANY PRODUCT, hence, THE NATURAL NUMBERS ARE CLOSED UNDER MULTIPLICATION, that is, MULTIPLICATION IS A TOTAL OPERATION.
SHORTCUT: GENERATE EXPONENTIATION (be, where b is a BASE, e is an EXPONENT) BY RECURSION ("in The Fourth Degree") on THE MULTIPLICATION OPERATION: b0 ≡ 1, bS(p) ≡ bp * b. This GENERATES ANY POWER OF A NUMBER, hence, THE NATURAL NUMBERS ARE CLOSED UNDER EXPONENTIATION, that is, EXPONENTIATION IS A TOTAL OPERATION.
BYPASSING TO INTEGERS Is there a KIKBAK for ADDITION? Yes, TWO. Because RIGHT ADDITION is WELL DEFINED (a.k.a. CANCELLATIVE): a + b = a + c ⇒ b = c; hence, RIGHT ADDITION has a RIGHT INVERSE. Also LEFT ADDITION 1s WELL DEFINED (CANCELLATIVE): b + a = c + a ⇒ b = c; hence, LEFT ADDITION has a LEFT INVERSE. However, ADDITION IS COMMUTATIVE ( a + b = b + a), so LEFT INVERSE IS EQUIVALENT TO RIGHT INVERSE FORMING A SINGLE ADDITIVE INVERSE. (NOTE! COMMUTATIVITY fails for EXPONENTIATION, as we see below.)
Hence, we can DEFINE an ADDITION-KIKBAK ("subtraction") in terms of ADDITION: a — b = c iff ("if and only if") a = b + c, which EXISTS if, and only if, a ≥ b, that is, MINUED IS GREATER THAN OR EQUAL TO THE MINUED. Thus, 3 — 2 is an ALLOWABLE DIFFERENCE of 1 (i.e., 3 — 2 = 1), BECAUSE it fits THE DEFINITION OF SUBTRACTION IN TERMS OF ADDITION: 3 = 2 + 1. But 2 — 3 DOESN'T EXIST, BECAUSE THE DEFINITION FAILS; EQUIVALENTLY, THE ANSWER WOULD LIE "BELOW" ZERO. Hence, ONLY PARTIAL CLOSURE OF NATURALS FOR SUBTRACTION. Equivalently, ADDITION IS A TOTAL OPERATION ("ALWAYS WORKS"); but SUBTRACTION IS only A PARTIAL OPERATION ("ONLY WORKS SOMETIMES").
This is BYPASSED by FORMING AN ARITHMETIC (WITH RELATIONS AND OPERATIONS) OF DEFINED (ALLOWABLE) DIFFERENCES. This LIMITED MODEL is BYPASSED by FORMING VECTORS (2-TPLES) OF NATURALS, ITS ARITHMETIC MODELED ON THAT OF THE DIFFERENCE ARITHMETIC. Since THIS REDUCES TO 3 EQUIVALENCE CLASSES OF VECTORS, THE VECTOR NOTATION CAN BE BYPASSED WITH "+" and "-" SIGNS, ressulting in INTEGER ARITHMETC (without cheating!) See! See! BYPASS TO INTEGERS.
BYPASSING TO RATIONALS MULTIPLICATION HAS A TWO KIKBACKS, namely, DIVISION, because MULTIPLICATION IS RIGHT & LEFT WELL-DEFINED (CANCELLATIVE). Thus, for a ≠ 0, a * b = a * c IF, AND ONLY IF, , so MULTIPLICATION IS RIGHT WELL-DEFINED, WITH A RIGHT INVERSE. Also, b * a = c * a IF, AND ONLY IF, b = c, so MULTIPLICATION IS LEFT WELL-DEFINED, WITH A LEFT INVERSE. However, MULTIPLICATION IS COMMUTATIVE (a * b = b * a), so BOTH INVERSES ARE EQUIVALENT, AND MULTIPLICATION IS WELL-DEFINED WITH A SINGLE INVERSE, DIVISION: d ≠ 0, m / d = q IF, AND ONLY IF, m = d * D. But QUOTIENT q EXISTS IF, AND ONLY IF, m = k * d, for some INTEGER k, that is, q = m / d = (k * d) / d. Such QUOTIENTS ARE "ALLOWABLE". But DIVISION IS ONLY A PARTIAL OPERATION. This is BYPASSED by FORMING AN ARITHMETIC (RELATIONS & OPERATIONS) OF ALLOWABLE QUOTIENTS. This RESTRICTION is BYPASSED by FORMING AN ARITHMETIC OF VECTORS (2-TPLES) OF INTEGERS, WITH RELATIONS & OPERATIONS MODELED ON QUOTIENT ARITHMETIC. Since THE VECTORS REDUCE TO 3 EQUIVALENCE CLASSES, VECTORS ARE BYPASSED BY RATIONAL NOTATION (without cheating!). See! See! BYPASSING TO RATIONALS.
BYPASSING TO REAL NUMBERS EXPONENTIATION HAS 2 KIKBAKS. fg = fh IF, AND ONLY IF, g = h: "UPPER" EXPONENTIATION, ALLOWING AN "UPPER" INVERSE. Also, uw = vw IF, AND ONLY IF, u = v: "LOWER" EXPONENTIATION, ALLOWING A "LOWER" INVERSE.
WARNING! EXPONENTIATION IS NOT COMMUTATIVE! as a single example shows: 23 = 8 but 32 = 9. Hence, unlike the cases for ADDITION and MULTIPLICATION, EXPONENTIATION HAS TWO INVERSES: LOGARITHM ("UPPER") and ROOT EXTRACTION ("LOWER"). EACH LEADS TO A DIFFERENT NUMBER SYSTEM, RESPECTIVELY, REAL NUMBERS AND COMPLEX NUMBERS.
Operation log b m = n IF, AND ONLY IF, bn = m. But this is DEFINED ONLY FOR n > 0: ALLOWABLE LOGARITHMS. AN ARITHMETIC OF ALLOWABLE LOGARITHMS PROVIDES A MODEL, which can be BYPASSED IN A VECTOR OF INFINITE COMPONENTS. And this is BYPASSES BY AN ARITHMETIC OF DECIMAL NUMBERS. See! See! BYPASSING WITH DECIMALS.
BYPASSING BY COMPLEX NUMBERS (s)1/r = t IF, AND ONLY IF, tr = s. This is RATIONAL or REAL IF, AND ONLY IF, s > 0: "ALLOWABLE ROOTS". This BYPASSED BY AN ARITHMETIC OF ALLOWABLE ROOTS, providing MODEL for a BYPASS in VECTORS (2-TPLES) OF REALS.
However, in all previous VECTORS, the VECTOR COMPONENTS ARE NOT SEPARATED IN THE INVERSES, WHICH STARTS WITH SUBTRACTION. But, in the present case, THE FIRST AND SECOND COMPONENTS REMAIN SEPARATE IN SUBTRACTION, forming a BIMODUL. THIS MEANS THAT COMPLEX NUMBERS AS VECTORS CANNOT BE BYPASSES. And this leads to GENERATION OF THE ARITHMETIC OF CLIFFORD NUMBERS, Ci, for INTEGRAL i ≥ 2. END OF BYPASSING IN GENERATING ARITHMETIC.
Then, how we GENERATE INTEGERS