Next we will try division. Unlike addition, subtraction and multiplication, division isn't a nice simple right-to-left algorithm that ensures an obviously unique result or even any result at all when leftins are involved. For more on the implications of leftin division, see the Divison section.

To find the quotient of two leftins, a/b, we will simply try all possible digits for the right-most digit to see what multiplied by b will give us a, making use of the Multiplication Accuracy Principle. If there are no possibilities, there are no answers. Once we have narrowed down the possibilities, we will try all possible digits for the next digit to the left and continue the process.

In the following example, we will try to find 1/3.

__________ 3 | ...00001

If we assume 1/3 is an element of ** L_{0}** (i.e., it's of
the form infinite.0), then clearly the only possibility for the right-most digit
is 7, since 3*7=21 and no other product would give us a right-most digit of 1.

7 __________ 3 | ...00001 21 _____ ...99980

The next digit must produce a right-most digit of 8 when multiplied against 3. Again there's a single possibility - in this case, 6.

67 __________ 3 | ...00001 21 _____ ...99980 18 _____ ...9980

At this point, you can clearly see that the remaining digits must all be 6's,
so 1/3 = ...(6)7. This is the only possible result in ** L_{0}**.
If there was another possibility in

Next we will try 1/2. If we first work within ** L_{0}**,
the right-most digit of the answer should give us a right-most of 1 when multiplied
against 2. Since there is no such digit, we instead need to try digits to the right
of the decimal point. Now we must find a digit which gives us 0 for the right-most
digit of the product - the only possibility is 5.

0.5 _____ 2 | 1 1.0 ___ 0.0

So we can conclude 1/2 = 0.5. We can see that we won't find other answers by looking at extra digits since we would need to generate an odd number when multiplying by 2 which obviously can't happen.

0.05 _____ 2 | 1.00 .10 ___ 0.90

We found 1/2 has the same representation in ** L_{f}** as it
does in normal decimal notation. However 1/3 was different since as a real number
it takes an infinite number of digits to the right of the decimal point to represent.
It's important to note that both were unique results however.

The Table of Rationals can give you a feel for what rational numbers look like expressed as leftins in various number bases.

Negative rational numbers can also be expressed in ** L_{f}** .
The simplest case in base 10 is to divide -1, which is ...(9), by 3 or 9.
We can easily see that -1/3 = ...(3) and -1/9 = ...(1). In general, -1 divided by
any power of 10 minus 1 gives us an obvious result.

-1/99 = ...(9)/99 = ...(01) = ...010101

-1/999 = ...(9)/999 = ...(001) = ...001001

This is closely related to expressing rational numbers in normal decimal notation.

1/9 = 0.(1)... = 0.111111...1/99 = 0.(01)... = 0.010101...

1/999 = 0.(001)... = 0.001001...

From these results, you can see there's a simple method for converting a rational leftin into a rational number.

...(53)76.1 = ...(53) + 23.1 = -53/99 + 23.1 = 22 559/990 = 22.5(64)... **Rational Leftins:**
** L_{r}** = {a

*L _{r}*

Stated simply, all rational numbers can be uniquely represented as a leftin, and all rational leftins are equivalent to a unique rational number.

**Proof Outline:**

- Every rational number can uniquely be represented as finite.infinite (setting aside the case where 1 = 0.999...) which can be uniquely decomposed into 0.({digits})... + finite.finite. The first term only consists of the portion that repeats infinitely.
- 0.({digits})... is equal to {digits} divided by 10^(number of digits) - 1. For example 0.(123)... = 123 / 999.
- It is simple to uniquely translate 0.({digits})... to an infinite.0 leftin. In the previous example, 123 / 999 = 1 + (-876 / 999) = 1 + ...(876) = ...(876)877.
- A finite.finite number can only be represented as finite.finite in
.*L*_{f} - Thus the rational can be uniquely represented by an infinite.0 + finite.finite which produces an infinite.finite leftin.

- Every rational leftin can be uniquely decomposed into ...({digits}).0 + finite.finite.
- ...({digits}).0 is equal to -{digits} divided by 10^(number of digits) - 1. For example, ...(123) = -123 / 999.

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© Scott Sherman 1999