The Ring of Hyperintegers

Subject: The real 10-adic integers
Newsgroups: sci.math
Keywords:10-adics, hyperintegers

In 1986, Anthony DeLugt--an associate of mine--was working on Fermat's Last Theorem (FLT). He wanted to find a way to prove or disprove the conjecture based on noting the effect of exponentiation on the rightmost digits of ordinary positive integers. What he discovered (or possibly rediscovered) is some theorems of the 10-adic integers. DeLugt got me interested in the project to writeup his results and publish them.

Now, DeLugt and I did not see eye-to-eye on what the 10-adics really were. I wanted to treat them as purely formal and treat them as an unordered, and without magnitude, set of formal entities. He disagreed and I found out how easy it is to generate controvery over 10-adic integers. The one thing we could agree on is that the four idempotents that he discovered in the set are fascinating.

Consider the set of all strings of digits (0..9) formed by taking a first digit (the rightmost) and a second and a third, and so on without end. Each of these strings is called a transfinite digital (td) number, and each is an ordinal number of type omega, to use Cantor's notation. We can define addition and multiplication on the set of these td's in the obvious way. Now we can add into the set the additive inverses of each td, and then we get a commutative ring, which I called the ring of Hyperintegers. Now, in spite of the fact that the ordinary integers are isomorphic to a proper subring of the hyperintegers, I do not consider this too important, and I do not attempt to form an ordering, magnitude, or valuation on the set of hyperintegers.

An obvious question is whether or not the operations of addition and multiplication we've defined on the hyperintergers are well defined. But they are, because to be well defined all we need do is to be able to establish the ith digit in the hyperinteger for any positive natural i.

Let pi(a) be a function which gives us the ith digit of the hyperinteger a. Let ri(a) be the i rightmost digits of a, maintaining their order.

THEOREM: There are four idempotents in the hyperintegers.

Definition: An idempotent is an number whose square is equal to itself. It's easy to show that. ..0000 and. ..0001 are idempotents of the hyperintegers; our task is to produce the other two. Just as in other number systems, the hyperintegers has its share of multiple numerals that represent a given hyperinteger. For instance, we can write ...000 as 0, and ...0001 as 1.

Definition: A hyperinteger numeral is said to be in proper form if each of its entries is a digit (0..9). (Of course, 0 and 1 are improper forms for hyperintegers.)

Definition: Two td's a,b are said to be equal iff when both a,b are given as proper numerals, for all i in Naturals, pi(a) = pi(b).

From this last definition we can prove that ...000 = ...999X where X is the equivalent to '10', only not precisely. Think of X as an instruction to replace itself by '0' and carry a 1 to the next left digit place.

Now we can define the complement function, comp(a), for all hyperintegers. For a a positive hyperinteger

comp(a) := ...9999X - a
and for a a negative hyperinteger, comp(a) := - (. ..9991 + a). Obviously, for all td's
x + comp(x) = 0
thus the set of td's also forms a rings without formally introducing the negative sign.

We need to list some obvious properties of the rn( ) function to help us out. Think of the rn( ) function as a mapping of hyperintegers into the integers mod 10n.

Then:

rn(x) = sign(x) pn(x) pn-1(x)..p1(x)
rn(rn(x)) = rn(x)
rn(x+y) = rn(rn(x) + rn(y)) = rn(x) + rn(y) (addition taken mod 10n)
rn(x . y) = rn(rn(x) . rn(y)) = rn(x) . rn(y) (multiplication mod 10n)

Although some liberties were taken above, it should be easy to keep from getting confused by context.

Now, if x is a positive hyperinteger, then x is idempotent if for all Naturals i, either pi(x2) = pi(x), or ri(x2) = ri(x), when both x and x2 are given in proper form. It is not difficult to show that idempotents starting with 0 or 1 are unique, and it is easy to show that no idempotents can start with {2,3,4,7,8,9}. This leaves the possibility of idempotents starting with 5 and or 6.

THEOREM: There exists a positive hyperinteger idempotent whose first (rightmost) digit is 5.

Proof:

First we generate the hyperinteger then show that it satifies the definition of idempotency.

Let e1 be given recursively by

p1(e1) = 5
pn+1(e1) = pn+1([rn(e1)]2)

Now we proceed by induction:

We wish to show that rn(e12) = rn(e1) for all positive integers n. The equation is obviously true for n = 1.

Now assume that rk(e12) = rk(e1) for arbitrary k (the inductive hypothesis), and then show that the equation holds for k -> k + 1 as well. Now,

rk+1(e12) = rk+1([rk+1(e1)]2).

Let rk+1(e1) = d . 10k + rk(e1) where d = pk+1(rk+1(e1))

then

rk+1(e12) = rk+1(d2 102k + 2d10krk(e1)+[rk(e1)]2)
     = pk+1([rk(e1)]2)10k + rk([rk(e1)]2)
     = pk+1([rk(e1)]2)10k + rk(rk(e1)) by inductive hypothesis
      = pk+1(e1)10k + rk(e1) by defn of e1
     = rk+1(e1)

The first step is justified by the fact that for any Natural number k, the first two terms are in the kernel of the mapping rk+1( ). The first term is obviously so, but the second is too because the factor of 10k gets a multiple of at least another 10 because of the product 2.rk(e1), since p1(e1) = 5. This leaves a factor of 10k+1 which makes the second term in the kernel of rk+1( ).

In 1987 I calculated the first 100 digits of e1. I present the first 50:

...57423423230896109004106619977392256259918212890625.

In a ring with identity idempotents come in pairs: Define e2 := 1 - e1. Then e22 = (1 - e1)2 = 1 - 2e1 + e1 = 1 - e1 = e2. We can get e2 in positive form by using that

e2 = 1 + comp(e1) = 1 + ...375 = ...376

The first 50 digits of e2 are:

...42576576769103890995893380022607743740081787109376

It is trivial that

e1n + e2n = 1n for all Natural numbers n.

But this is not an invalidation of number theory based on ordinary integers, and it certainly is not an invalidation of FLT, which is a theorem based on ordinary integers. But do the hyperintegers have any practical uses? I think so. One is for introducing the concept of rings to math student, particularly high school students---even at the risk of them going overboard on them as some have done. Another use is in the calculation of the rightmost digits of integers taken to large powers. Consider the problem of finding the n rightmost digits of 8573647385645 (digits chosen at random).

rn(8573647385645) = rn((8573647e1)385645) + rn((8573647e2)385645)

the advantage of which is that by "projecting" the base number onto the idempotents allows us to take the exponent mod the order of the cyclic multiplicative group formed by multiplying all integers from 0 to 10n-l times each idempotent, giving a tractable cyclic group for both e1 and e2, as long as n is not too large.

Finally, of what importance is it that the hyperintegers contain non-trivial idempotents, other than already discussed? Because these non-trivial idempotents are idempotent pairs, their product is zero, even though neither of them is zero: e1e2 = e1(l - e1) = e1 - e1 = 0. Thus the ring of hyperintegers is not a field. So what! Fields aren't the only objects worth study. Lots of rings are interesting and useful, especially rings with non-trivial zero-divisors, like the hyperintegers!

cheers

--Patrick Reany