The quintright integral domain.
Version of Saturday 19 July 2014.
Dave Barber's other pages.
Section 1. A quintright is an ordered quadruple of integers, the components of which are written, separated by commas, between shallow angle brackets. Examples are ⟨ 3, −1, −5, 6 ⟩ and ⟨ 0, 0, 0, 0 ⟩. If A is a quintright, its components are denoted by the subscripts q, r, s, t. Hence this notational identity:

A ≡ ⟨ Aq, Ar, As, At

Another identity defines the meaning of a quintright in terms of real numbers; the factor of 2 greatly simplifies the multiplication table to be encountered later:

A ≡ 2 Aq cos (18°) + 2 Ar cos (36°) + 2 As cos (54°) + 2 At cos (72°)

It is convenient to give names to certain quintrights:

Q ≡ ⟨ 1, 0, 0, 0 ⟩
R ≡ ⟨ 0, 1, 0, 0 ⟩
S ≡ ⟨ 0, 0, 1, 0 ⟩
T ≡ ⟨ 0, 0, 0, 1 ⟩
Z ≡ ⟨ 0, 0, 0, 0 ⟩

With these, one can write:

AAqQ + ArR + AsS + AtT

The quintrights are patently a subset of the real numbers. Note that the integers are denumerable, and ordered quadruplets of them are also denumerable. Since the real numbers are not denumerable, there must be some reals that cannot be written as quintrights, therefore the quintrights are a proper subset of the real numbers.

Quintright comes from quint, meaning one-fifth, and right, meaning right angle. The angles in this formula are all multiples of eighteen degrees, which is one-fifth of a right angle. The first subscript, q, happens to be the first letter of quintright, and the last subscript, t, is also the last letter of quintright.

Available is a demonstrator program in the C++11 language.


Section 2. The rules for arithmetic derive from the cosine identity above. Addition and subtraction are obvious:

A + B = ⟨ Aq + Bq, Ar + Br, As + Bs, At + Bt
AB = ⟨ AqBq, ArBr, AsBs, AtBt

Multiplication by an integer scalar n is not a problem:

nA = An = ⟨ Aqn, Arn, Asn, Atn

Here is the rather complicated rule for general multiplication (derivation):

(AB)q = (AqBr + ArBq) + (ArBs + AsBr) + (AsBt + AtBs)
(AB)r = (AqBs + AsBq) + (ArBt + AtBr) + (3AqBq + 2ArBr + 2AsBs + AtBt)
(AB)s = (AqBr + ArBq) + (AqBt + AtBq) − (AsBt + AtBs)
(AB)t = (AqBs + AsBq) − (ArBt + AtBr) − (2AqBq + ArBr + 3AsBs + 2AtBt)

This table may help to clarify:

Multipli-
cation
Q
1.902113…
R
1.618034…
S
1.175571…
T
0.618034…
Q
1.902113…
3R − 2T
3.618034…
Q + S
3.077684…
R + T
2.236068…
S
1.175571…
R
1.618034…
Q + S
3.077684…
2RT
2.618034…
Q
1.902113…
RT
exactly 1
S
1.175571…
R + T
2.236068…
Q
1.902113…
2R − 3T
1.381966…
QS
0.726543…
T
0.618034…
S
1.175571…
RT
exactly 1
QS
0.726543…
R − 2T
0.381966…
Similar tables can be produced when the right angle is divided into other than five parts.

Addition is commutative and associative, and Z is the additive identity:

A + B = B + A
(A + B) + C = A + (B + C)
A + Z = A

Similarly, multiplication is commutative and associative, and RT is the multiplicative identity:

AB = BA
(AB)C = A(BC)
A⟨ 0, +1, 0, −1 ⟩ = A

Further, multiplication distributes over addition, and the identity for addition is the annihilator for multiplication:

A(B + C) = AB + AC
AZ = Z

Suppose that a, b, c and d are ordinary integers, and n is a positive integer. Then a basic result from modulo arithmetic is that if both of these are true:

then all three of these will be true:

Now define modulo equality for quintrights: A = B [mod n] if and only if all four of these are true:

It is easy to prove that if both of these are true:

then all three of these will be true:
Section 3. In general, division does not work, because it often leads to non-integers as elements of the quotient; and division by zero always fails. However, there is a scheme that will find the quotient of A and B whenever it exists. Let C = A ÷ B; then CB = A, and the multiplication rule can be written this way:

Aq = Cq(Br) + Cr(Bq + Bs) + Cs(Br + Bt) + Ct(Bs)
Ar = Cq(3Bq + Bs) + Cr(2Br + Bt) + Cs(Bq + 2Bs) + Ct(Br + Bt)
As = Cq(Br + Bt) + Cr(Bq) + Cs(− Bt) + Ct(BqBs)
At = Cq(Bs − 2Bq) + Cr(− BrBt) + Cs(Bq − 3Bs) + Ct(− Br − 2Bt)

Regard this as a system of four linear equations in the four unknowns Cq, Cr, Cs, Ct and solve. If the four results are integers, division was successful. Much can be said about the reciprocals of quintrights.

Although quintrights do form an integral domain, they do not form a division ring because many of them fail to have multiplicative inverses. For instance, Q−1 would have been (2QS) ÷ 5, and S−1 would have been (Q + 2S) ÷ 5. Were the quintrights defined as a rational domain (in other words, a field) rather than an integer domain, this problem would evaporate. Quintrights of the form ⟨ 0, n, 0, m ⟩ constitute an integral subdomain.


Section 4. Some quintrights, but not very many, have square roots. For instance, ⟨ +8, +125, −54, −152 ⟩ ≈ +60.049179 is irrational when viewed as a real number, but as a quintright it has two square roots, namely ±⟨ +1, −3, +7, +4 ⟩ ≈ ±7.749141.

Let E = D2; then the components of E, after application of the general multiplication rule (above), appear thus:

Eq = 2DqDr + 2DrDs + 2DsDt
Er = 2DqDs + 2DrDt + (3Dq2 + 2Dr2 + 2Ds2 + Dt2)
Es = 2DqDr + 2DqDt − 2DsDt
Et = 2DqDs − 2DrDt − (2Dq2 + Dr2 + 3Ds2 + 2Dt2)

Rearrange:

Eq = 2(DqDr + DrDs + DsDt)
Er = 2Dq2 + (Dq + Ds)2 + Ds2 + Dr2 + (Dr + Dt)2
Es = 2(DqDr + DqDtDsDt)
Et = Dq2 + (DqDs)2 + 2Ds2 + (Dr + Dt)2 + Dt2

Note that each of Er and −Et can be written as the sum of squares. If E is going to have a square root, four conditions will have to be satisfied:

No algorithm for calculating square roots is obvious, but the formulas for Er and Et do place bounds on the values that might be searched. For instance, 2Dq2, Ds2 and Dr2 can never exceed Er. Similarly, Dq2, 2Ds2 and Dt2 can never exceed −Et.


Section 5. The cosines can be written in radical form:

Q= 2 cos(18°)= ½ √(10 + √20) = 1.902113…
R= 2 cos(36°)= ½ (√5 + 1) = 1.618034…
S= 2 cos(54°)= ½ √(10 − √20)= 1.175571…
T= 2 cos(72°)= ½ (√5 − 1) = 0.618034…

Observe from these that ⟨ 0, +n, 0, −n ⟩ = n, thus the integers are a subset of the quintrights. Because cos(18°) is irrational, so is ⟨ 1, 0, 0, 0 ⟩, hence the integers are a proper subset (more specifically a subdomain) of the quintrights.

The golden ratio is R = 1.618034…, while T = 0.618034… is the golden mean. Powers of the golden mean exemplify a sequence of quintrights where the terms approach zero, but where the components of the terms are unbounded:

T 1= ⟨ 0, 0,0, +1 ⟩= 0.618034…
T 2= ⟨ 0, +1,0, −2 ⟩= 0.381966…
T 3= ⟨ 0, −1,0, +3 ⟩= 0.236068…
T 4= ⟨ 0, +2,0, −5 ⟩= 0.145898…
T 5= ⟨ 0, −3,0, +8 ⟩= 0.090170…
T 6= ⟨ 0, +5,0, −13 ⟩= 0.055728…
T 7= ⟨ 0, −8,0, +21 ⟩= 0.034442…
T 8= ⟨ 0, +13,0, −34 ⟩= 0.021286…
T 9= ⟨ 0, −21,0, +55 ⟩= 0.013156…
T10= ⟨ 0, +34,0, −89 ⟩= 0.008131…
T11= ⟨ 0, −55,0,+144 ⟩= 0.005025…
T12= ⟨ 0, +89,0,−233 ⟩= 0.003106…
T13= ⟨ 0, −144,0,+377 ⟩= 0.001919…
T14= ⟨ 0, +233,0,−610 ⟩= 0.001186…
T15= ⟨ 0, −377,0,+987 ⟩= 0.000733…
et cetera

This shows that is not generally possible to approximate a quintright by approximating its components.

If the quintrights are viewed as being embedded among the real numbers, this sequence does provide a means to show that within any neighborhood of a quintright A there exists another quintright: simply use A ± Tn for sufficiently large n.


Section 6. Given any two quintrights B and C, if these four conditions apply:then B clearly must equal C.

To prove the converse, however, is more complicated. First, it can be established that if A = 0, then Aq = Ar = As = At = 0. Second, to investigate B = C, write BC = 0. Then

(BC)q = (BC)r = (BC)s = (BC)t = 0
BqCq = BrCr = BsCs = BtCt = 0
Bq = Cq, Br = Cr, Bs = Cs, Bt = Ct

Component unicity is the name of this property that, when two quintrights are equal, their respective components must also be equal. It does not necessarily exist when the right angle is divided into other than five parts. For instance, it fails when the right angle is divided into nine parts because one of the angles created is sixty degrees, the cosine of which is rational. Here is an example of two "nonarights" that have different components but which are nonetheless equal:

2 (0) cos (10°) + 2 (+1) cos (20°) + 2 (0) cos (30°) + 2 (−1) cos (40°) +
2 (0) cos (50°) + 2 (+1) cos (60°) + 2 (0) cos (70°) + 2 (−1) cos (80°) = 1

2 (0) cos (10°) + 2 (0) cos (20°) + 2 (0) cos (30°) + 2 (0) cos (40°) +
2 (0) cos (50°) + 2 (+1) cos (60°) + 2 (0) cos (70°) + 2 (0) cos (80°) = 1

More concisely,

⟨ 0, +1, 0, −1, 0, +1, 0, −1 ⟩ = 1
⟨ 0, 0, 0, 0, 0, +1, 0, 0 ⟩ = 1

Because quintrights are real numbers, the notions of greater-than and less-than do apply; the comparison procedure, although elementary, is rather intricate.