Araara cḃ — Asoho

This is the ninth in a series of expository blog posts on Araara. The preceding post introduced the Attic alphabet. We will now return to Aglobasa and give its uppercase letters a different arithmetic.

It would be nice to have reciprocals (plu soho) as well as integers in Araara. Let us first define our alphabet:

uifddcc uifdcc uifdd uifdcca uifddcca uifdccb uifddc uifdccba uiffb uifdda uiffbb uifddb uifddca uifddba uiffca uifddbb uiffba uifddcb uiff uifddcba uifddcbb uiffa uiffcba uiffc uiffcb

Decoded:

O AE BP CJ DT FV GK HX IU LR MN SZ WY

These are the uppercase Globasa letters, arranged in the same order as Aglobasa. The alphabet has twenty-five positions and no gaps. Its signature is:

uiihhffdd

The sum of its Unicode code points is 1934.

Reciprocals

In Akurto, b represents two. What if B represented its reciprocal, one half? We will call this system Asoho, after the Globasa word soho, meaning reciprocal.

For a nonzero integer n, write r(n) = 1/n. Each uppercase letter represents the reciprocal of its lowercase counterpart:

But what about O?

We include one additional value: reciprocal zero, written ∞. Define r(0) = ∞ and r(∞) = 0. Here ∞ is a single added value, without a positive or negative sign. This extends our reciprocal operation; ordinary division by zero remains undefined.

Taking the reciprocal twice returns us to where we started: r(r(x)) = x. In particular, O represents ∞. Ordinary zero is not a value of Asoho; the value corresponding to Akurto’s zero is reciprocal zero.

Concatenation

What should concatenation mean here? In Akurto, we add the letters’ values. In Asoho, we first take their reciprocals, add those, and take the reciprocal again.

Using ⋆ to describe this operation:

x ⋆ y = r(r(x) + r(y))

The inner addition is ordinary integer addition. This definition also covers reciprocal zero.

AB = r(1 + 2) = 1/3

AA = r(1 + 1) = 1/2 = B

BB = r(2 + 2) = 1/4

Although A represents one and B represents one half, concatenating them gives one third. An Asoho term is the reciprocal of the integer represented by the corresponding lowercase Akurto term:

ba = 3    BA = 1/3

cc = 10    CC = 1/10

pj = −7    PJ = −1/7

For any number of letters, we add their lowercase values and take one reciprocal of the total.

Reciprocal zero

Our definition gives O a familiar algebraic role. Since r(∞) = 0, we have:

x ⋆ ∞ = r(r(x) + 0) = x

Thus:

AO = OA = A

BO = OB = B

OO = O

O represents the identity value under this operation, just as o represents zero in Akurto. The empty expression is still the identity for literal concatenation. Its Asoho value is ∞, whose corresponding integer is zero.

The inverse pairs also behave as before:

AE = r(1 − 1) = r(0) = O

BP = O

CJ = O

DT = O

To find a term’s inverse under concatenation, replace each letter with its partner.

The same structure

Asoho’s arithmetic may initially look unfamiliar, but the correspondence with Akurto is exact:

r(m + n) = r(m) ⋆ r(n)

Every integer has exactly one reciprocal value in Asoho, including zero through ∞. Taking reciprocals recovers the integer.

This makes Asoho an isomorphic copy of the integers under addition. We have changed the values and the operation together, preserving the structure.

Consequently, concatenation is associative and commutative. For three corresponding integers m, n, and k, either grouping gives r(m + n + k). The order of the letters therefore does not affect a term’s value:

AB = BA

ABC = CBA

Normal form

We can carry Akurto’s normal form across the same correspondence. Take the corresponding integer, write its Akurto normal form, and capitalise it.

For example:

AABB = CA

The corresponding lowercase term has value 1 + 1 + 2 + 2 = 6, whose normal form is ca. Both uppercase expressions therefore represent 1/6. For reciprocal zero, the normal form is O.

We still arrange letters by their corresponding integer magnitudes. This preserves Akurto’s ordering convention: the larger integer contributions come first, even though their reciprocals have smaller ordinary magnitudes.

Exercises

1. Find the values of AAA, BC, and DP.
2. Put AAAA and BBA into normal form.
3. Find a term which concatenates with BA to give O.
4. Why does adding O anywhere in a term leave its value unchanged?

We now have two ways to express the same additive structure: through integers in Akurto, and through their reciprocals in Asoho.

In the next post, we will combine Asoho with Akurto.