### Division of Leftins

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 *L*_{r} , all quotients exist
and are unique. Next we will look at cases where this doesn't hold true.

#### Non-uniqueness

**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) = q^{2} - 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 *L*^{10} , 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.

#### Non-existence

**QC2: q Conjecture 2:** 1/(q+1) does not exist in
*L*_{f} .

Since not all elements of *L*^{10} have reciprocals,
it is not a *field*.

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© Scott Sherman 1999