Irrational Numbers in Leftins

Finding the square root of a leftin suffers from many of the same problems as division with leftins. You're not guaranteed a certain number of results or even any results at all. For a more thorough look at square roots, see the Square Roots section.

We will approach extracting roots much the way we approached division: simply try digits from right to left to see what works.

Irrational Numbers in L₀

We will start by trying to find the square root of 17 base 2. We are looking for bL₀ such that b² = 17 (or 1001 base 2). The right-most digit of b must be 1 if it's square is going to end with a 1 (by AP1).

When we try the second digit, we find b can be ...01 or ...11 since the square of these numbers is ...01. The third digit gives us the four possibilities ...001, ...101, ...011 and ...111. But we find that the fourth digit doesn't allow two of these answers, though we still end up with four possible values.

At each stage from here on out, we find that the next digit eliminates two of the previous four possibilities but ends up giving us four possibilities itself. This means that the left-most digit is ambiguous until we add another digit. Thus we have two square roots of 17.

...0110011110100110010011011101001
...1001100001011001101100100010111

Both square roots of 17 are irrational numbers, so we know we can represent at least two irrational numbers in L_f . All rational numbers can be represented in L_f ; is it possible all irrationals are present? The answer is no.

     bL² , such that b² = 10 base 2.

Proof:
By SQ1, if b exists, it must be an element of L₀ . If we try squaring all the two-digit base 2 numbers, we find the following:

...00² = ...00
...01² = ...01
...10² = ...00
...11² = ...01

By AP1 we can see that 2 doesn't exist in L².

This brings us to the conclusion that some irrationals are equivalent to members of L_f , but not all.

     I L_f .
     I L_f .

© Scott Sherman 1999