Axiomitizations of algebras in terms of - and /.

[Mark Hopkins]

The properties for the operators (+) and (x) in groups, rings and
fields are well-known and well-studied and it is common to define the
inverse operations (/) and (-) in terms of these. Thus, for instance,
the additive group of integers can be thought of as being generated from
the set { +1, -1 } using the addition operation (or: as the free Abelian
group generated by one element), and the rationals from the set { +1, -1 }
using the addition and the reciprocal operation.

What is not as well-known is that the integers can actually be generated
using nothing more than the number 1 and the single operation: -, or the
rationals from just the number 1 with subtraction and division.

More generally, what is not well-known, or maybe not even known at all,
is that these algebras can be more efficiently defined in terms of their
inverse operators.

Contents:
(1) Groups in 3 Axioms
(2) Abelian Groups in 2 Axioms
(3) Axiomitizations for Rings and Fields, in terms of the inverse
operators

(1) Groups in 3 Axioms
AXIOM A: (a/c)/(b/c) = a/b
AXIOM B: (a/a)/(b/b) = b/b
AXIOM C: a/(a/a) = a

THEOREM 1: The following basic properties hold:
(D) a/a = b/b
(E) (a/(b/c))/d = a/(d/(c/b))
(F) (a/a)/(b/c) = c/b
(G) a/(b/b) = a
Proof:
From A and B we can prove D:
a/a = (b/b) / (a/a) by B
= ((b/b)/(b/b))/((a/a)/(b/b)) by A
= (a/a)/(b/b) by B
= b/b by B
From A alone, we can prove E:
(a/(b/c))/d = ((a/(b/c))/(c/b))/(d/(c/b))
= ((a/(b/c))/((c/c)/(b/c)))/(d/(c/b))
= (a/(c/c))/(d/(c/b))
= (a/(c/c))/((d/(c/b))/((c/b)/(c/b)))
= (a/(c/c))/((d/(c/b))/(c/c))
= a/(d/(c/b))
From A and D, we can prove F:
(a/a)/(b/c) = (c/c)/(b/c) by D
= c/b by A
From C and D, we can prove G:
a/(b/b) = a/(a/a) by D
= a by C

THEOREM 2: Let V be a set equipped with the operation A/B. Then the following
interrelations exist:
(1) F -> B,
(2) G -> C,
(3) F -> D,
(4) E & G -> A
Proof:
(1), (2) are trivial
(3) a/a = (b/b)/(a/a)
= (b/b)/((b/b)/(a/a))
= (a/a)/(b/b)
= b/b
where the next to last step makes essential use of F, as opposed to
just B.
(4) Let a, b and c be given and let d = c/c. Then by G (or C), it follows
that c/d = c. Therefore,
(a/c)/(b/c) = (a/(c/d))/(b/c) by G (or C)
= a/((b/c)/(d/c)) by E
= a/((b/(c/d))/(d/c)) by G (or C)
= a/(b/((d/c)/(d/c))) by E
= a/b by G
The last step makes essential use of G, instead of just C.

A review of all the possible tables of sizes 1-8 will reveal that there
are no other interrelations between (A)-(G) than those described ni the
previous two theorems.

DEFINITIONS: a' = (a/a)/a, ab = a/b', 1 = a/a

THEOREM 3: Basic properties
a' = 1/a, a/1 = a, a'' = a, (a/b)' = b/a, a/b = a b', 1' = a
Proof:
a' = (a/a)/a = 1/a
a/1 = a/(a/a) = a
a'' = 1/(1/a) = (a/a)/(1/a) = a/1 = a
(a/b)' = 1/(a/b) = (b/b)/(a/b) = b/a
a b' = a/b'' = a/b
1' = 1/1 = 1

THEOREM 4: Let V be any set satisfying axioms A-C. Then V is a group
with inverse A' and product AB defined as above.
Proof:
a 1 = a/1' = a/1 = a
1 a = 1/a' = a'' = a
a a' = a/a'' = a/a = 1
a' a = a'/a' = 1
a(bc) = a/(b/c')'
= a/(c'/b)
= a/((1/c)/(b/1))
= (a/(1/b))/(1/c)
= (a/b')/c'
= (ab)c

THEOERM 5: Let G be any group. Then G satisfies axioms A-C with the
operation a/b defined as a b'.
Proof:
(a/c)/(b/c) = (a c') (b c')' = a c' c'' b' = a b' = a/b
(a/a)/(b/b) = (a a') (b b')' = 1 1' = 1 1 = 1 = b b' = b/b
a/(a/a) = a (a a')' = a 1' = a 1 = a
where 1 is the group's identity.

(2) Abelian Groups in 2 Axioms
The following system of axioms will completely characterise an Abelian
group, defining it in terms of the subtraction operator.

AXIOM A: a - (a - b) = b
AXIOM B: a - (b - c) = c - (b - a)

THEOREM 1: a - a = b - b
Proof:
a - a = b - (b - (a - a))
= b - (a - (a - b))
= b - b

DEFINITIONS: -a = (a - a) - a, a + b = a - (-b), 0 = a - a

THEOREM 2: Basic properties
a - 0 = a, --a = a, -(a - b) = b - a, -0 = 0
Proof:
a - 0 = a - (a - a) = a
--a = 0 - (0 - a) = a
-(a - b) = 0 - (a - b) = b - (a - 0) = b - a
-0 = 0 - 0 = 0

THEOREM 3: Let V be a set with the operation A - B, satisfying axioms (A)
and (B). Then V is an Abelian group with the sum and negative
defined as above, and with a - b = a + (-b).
Proof:
a + 0 = a - -0 = a - 0 = a
0 + a = 0 - -a = --a = a
a + -a = a - --a = a - a = 0
-a + a = -a - -a = 0
a + b = a - (0 - b) = b - (0 - a) = b + a
(a + b) + c = (a - (0 - b)) - (0 - c)
= c - (0 - (a - (0 - b)))
= c - ((0 - b) - (a - 0))
= c - ((0 - b) - a)
= a - ((0 - b) - c)
= a - ((0 - b) - (c - 0))
= a - (0 - (c - (0 - b)))
= a - (0 - (b - (0 - c)))
= a + (b + c)
a + (-b) = a - (--b) = a - b

THEOREM 4: Let G be any Alebian group. Then G satisfies the axioms (A)-(B)
above, with subtraction defined by a - b = a + -b.
Proof:
a - (a - b) = a + -(a + -b)
= a + -a + b = 0 + b = b
a - (b - c) = a + -(b + -c)
= a + -b + c
= c + -b + a
= c + -(b + -a)
= c - (b - a)

(3) Axiomitizations for Rings and Fields, in terms of the inverse operators
Using the results above, we can define rings and field in terms of
the inverse operators. A ring can be characterised as a set with a
subtraction and multiplication operator satisfying the properties:

RING AXIOMS: a - (a - b) = b
a - (b - c) = c - (b - a)
(ab)c = a(bc)
a(b - c) = ab - ac
(b - c)a = ba - ca

The last two identities are clearly equivalent to distributivity. Therefore
we have an abelian group with respect to subtraction (and therefore addition)
with an associative product that distributes over it -- a ring.

A field can be represented using only subtraction and division, with
the following axioms

FIELD AXIOMS: a - (a - b) = b
a - (b - c) = c - (b - a)
a/(a/b) = b, if a and b are non-singular
a/(b/c) = c/(b/a), if a, b and c are non-singular
(a - b)/c = a/c - b/c, if c is non-singular
there exists a non-singular element

where an element, a, is called singular if a = a - a (i.e., if a = 0).

This provides us with an algebra which is an abelian group with respect
to subtraction (thus: addition), whose non-zero elements form an abelian
group with respect to division (thus: multiplication), which distributes
over the additivie operators + and - (i.e., a field).