m such that 
1/m
We have been treating Lf as a commutative ring. More restrictive categories from abstract algebra are fields and integral domains. A field is essentially a commutative ring with a multiplicative identity and reciprocals for all elements. An integral domain is a commutative ring with a multiplicative identity and no divisors of 0.
We know Lf has a multiplicative identity (namely 1). We know Lq
Lf
can have elements with no reciprocals and divisors of 0, but does this apply to
all number bases?
The following are properties that Lk may or may not have for a given k.
LP1:
m such that 1/m
LP2:
m,n m
0, n0
such that mn = 0.
LP3:
m such that a,b,c
abc such that
a2 = b2 = c2 = m.
LP1 is the non-existence of all reciprocals. LP2 is the existence of divisors of 0. LP3 is the existence of more than two square roots for some number.
LP Conjecture:
LP1 
LP2 LP3 LP1 and
~LP1 ~LP2 ~LP3 ~LP1
By the Number of Square Roots Conjecture, this would suggest that Lk is a field and an integral domain if k is prime or the power of a prime and isn't a field or an integral domain otherwise.
© Scott Sherman 1999