DIVISION for NATURAL NUMBERS and INTEGERS is DEFINED in terms of MULTIPLICATION, as its INVERSE: For NATURALS or INTEGERS a, b, c, a ÷ b = c iff a = b · c -- that is, DIVIDEND (a) DIVIDED BY DIVISOR (b) EQUALS QUOTIENT (c) iff DIVIDEND (a) EQUALS DIVISOR (b) TIMES QUOTIENT (c). Thus, 12 ÷ 3 = 4, because 3 · 4 = 12.
This RESTRICTS NATURAL NUMBER and INTEGER DIVISION TO THOSE QUOTIENTS that are MULTIPLES OF THE DIVISOR! Restricting DIVISION AS A PARTIAL INVERSE FOR MULTIPLICATION IN THE NATURALS or INTEGERS, WHEREAS IT SHOULD BE TOTAL (AS ARE ADDITION AND MULTIPLICATION IN NATURALS and INTEGERS, and SUBTRACTION in INTEGERS).
Typical teaching of DIVISION looks like CHEATING! "You can't divide 11 by 3 -- except you can by putting a funny slash mark between them -- 11/3." To say that this is a new and different NUMBER SYSTEM -- THE RATIONAL NUMBERS ("FRACTIONS") -- does little for CREDIBILITY, since it then invokes what seems to be a WEIRD RULE: "to divide a fraction by a fraction, invert the divisor fraction and multiply it times the dividend fraction". Whahhhhh?
As with the problem of SUBTRACTION, STUDENTS AREN'T SHOWN HOW ALL DIVISION ARITHMETIC CAN BE GENERATED WITHIN THE INTEGER SYSTEM -- WITHOUT CHEATING or WEIRDNESS!
We seek CLOSURE FOR (NONZERO) DIVISION -- which FAILS in the NATURAL NUMBER SYSTEM and in THE INTEGER SYSTEM.
NOW HEAR THIS!!! IN ALL THE FOLLOWING DIVISION OPERATIONS, THE DIVISOR IS NONZERO! (If you think "Don't divide by zero!" is mathematical prudery, try it on a calculator. But be ready to turn it off quickly!)
Let o denote ANY OPERATION (ADDITION, MULTIPLICATION, SUBTRACTION, etc.) Then, let p, q belong to the SAME SYSTEM ("family") OF ELEMENTS (say, both are NATURAL NUMBERS or INTEGERS); then, for p o q = r, CLOSURE requires r to be of the same type ("family") as p, q.
As in discovering amid NATURAL NUMBERS the RULES FOR ALLOWABLE DIFFERENCES (leading to INTEGER ARITHMETIC), we use the 2-WAY STRATEGY in DISCOVERING within INTEGERS THE RULES FOR ALLOWABLE QUOTIENTS -- leading to RATIONAL NUMBER ARITHMETIC, wherein SUBTRACTION "ALWAYS WORKS", and without CHEATING and encountering WEIRD RULES.
We start with the RESTRICTED PART of THE INTEGER SYSTEM (DIVIDEND EQUALS MULTIPLE OF DIVISOR), treating it as a SPECIAL NUMBER SYSTEM: THE ALLOWABLE (INTEGRAL) QUOTIENT SYSTEM. And proceed by FULFILLING TWO CONDITIONS:
- CLOSURE ("keep it in the Family");
- 2-WAY STRATEGY (the "longer CLOSURE way" of solving a problem yields the same answer as the familiar way).
QUOTIENT: <<dividend>> ÷ <<divisor>> = <<quotient>>. (Please note that we don't state the Standard Division Definition, with a Condition on The Remainder, because we need to consider only CASES WITHOUT REMAINDER.)
ALLOWABLE QUOTIENT: (aD ÷ bd) = rq, where a, b, r are ALL NATURAL NUMBERS or INTEGERS. (Note subscripts, "D", "d", "q", respectively for "dividend", "divisor", "quotient".)
Let o DENOTE ANY OPERATION (ADDITION, SUBTRACTION, ETC.). Then we consider the CLOSURE form: (a ÷ b) o (c ÷ d) = (e ÷ f) -- "a quotient, (a ÷ b) combined o with a quotient, (c ÷ d) yields a quotient, (e ÷ f)".
- You start by DERIVING EQUIVALENCE FOR ALLOWABLE QUOTIENTS ("When are two allowable quotients equal?") FROM THE DIVISION DEFINITION. Write, gD ÷ hd =jD ÷ 1d => gD · 1d = jD · hd. ("Dividend of one quotient multiplies divisor of other".) Hence, we have an EQUIVALENCE RULE: (gD ÷ hd) = (jD ÷ kd) iff gD · kd = hD · jd. (Thus, 12 ÷ 4 = 21 ÷ 7, since 12 · 7 = 4 · 21 = 84.)
As with THE ALLOWABLE EQUIVALENCE RULE, give a cookie to this ALLOWABLE QUOTIENT RULE, since IT WILL DO THE MAJOR PART OF OUR WORK FOR US!
- Assignment: change = to < or > to find the two INEQUALITY RULES.
Now, we need ADDITION and MULTIPLICATION OPERATIONS for ALLOWABLE QUOTIENTS. CLOSURE: (eD ÷ fd) o (gD ÷ hd) = (jD ÷ kd).
- ADDITION and SUBTRACTION RULES: Given (gD ÷hd) ± (jD ÷ kd) = (? ÷ ?). Let's arrange for the relata to have a common denominator. Multiply first relatum by (kD ÷ kd), and second relatum by (hD ÷ hd), then we have (gD÷ hd) · (kD ÷ kd) ± (jD ÷ kd) · (hD ÷ hd) = (gD · kd ± jD · hd) ÷ (hd · kd).
Our ADDITION-SUBTRACTION RULES FOR ALLOWABLE QUOTIENTS! AN ALLOWABLE QUOTIENT PLUS/MINUS AN ALLOWABLE QUOTIENT EQUALS AN ALLOWABLE QUOTIENT.
- The MULTIPLICATION RULE is easy: (gD ÷ hd) · (jD ÷ kd) = ((gD · jD) ÷ (hd · kd)). AN ALLOWABLE QUOTIENT TIMES AN ALLOWABLE QUOTIENT EQUALS AN ALLOWABLE QUOTIENT.
- For the DIVISION RULE, we'll appeal to CLOSURE and 2-WAY STRATEGY ("short" and "closure" calculations agree): (gD ÷ hd) ÷ (jD ÷ kd) = (gD · kd ÷ hd · jD).
AN ALLOWABLE QUOTIENT DIVIDED BY AN ALLOWABLE QUOTIENT EQUALS AN ALLOWABLE QUOTIENT -- which is what CLOSURE demands. Let's check it by 2-WAY. Say, (24 ÷ 3) ÷ (12 ÷ 6) = (24 · 6 ÷ 3 · 12) = (144 ÷ 36) = 4. (24 ÷ 3) = 8; (12 ÷ 6) = 2; and 8 ÷ 2 = 4. CHECKS.
Now we form ordered pairs or 2-tuples or vectors of integers with these rules to see how to create a new NUMBER SYSTEM in which NONZERO DIVISION IS TOTAL.