A basic algorithm that can give you the quotient of two leftins is to try all posibilities for the right-most digit, then all possibilities for the next digit to the left and so on. Division of leftins isn't guaranteed to give you a unique result or in fact any result at all.
Examples of division were given in the section on rational numbers. Within Lr , all quotients exist and are unique. Next we will look at cases where this doesn't hold true.
QC1: q Conjecture 1: 0/(q+1) is not unique.
In the section on square roots, we saw that
1 base 10 has four possible values. Two of these were
1 and -1. The other two were named q and -q. Using simple algebra, we see
(q+1)*(q-1) = q2 - 1 = 0. But neither (q+1) nor (q-1) are 0.
q+1 = ...18752 and q-1 = ...18750. This means that there are divisors of 0
in L10 , so it is not an integral domain.
If we try dividing 0 by q+1, we get interesting results. With each extra digit considered, the number of possible answers doubles. The following numbers, when multiplied by q+1, all give you 0: ...00000, ...50000, ...25000, ...75000, ...12500, ...62500, ...37500, ...87500, ...31250, ...81250, ...06250, ...56250, ...43750, ...93750, ...18750, ...68750, ...03125, ...53125, ...28125, ...78125, ...40625, ...90625, ...15625, ...65625, ...34375, ...84375, ...09375, ...59375, ...21875, ...71875, ...46875 and ...96875.
When you divide 0 by q-1, the number of possibilities goes up by a factor of 5 with each additional digit. Thus there are an infinite number of results.
QC2: q Conjecture 2: 1/(q+1) does not exist in Lf .
Since not all elements of L10 have reciprocals, it is not a field.
© Scott Sherman 1999