A ≡ ⟨ A_{q}, A_{r}, A_{s}, A_{t} ⟩
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 A_{q} cos (18°) + 2 A_{r} cos (36°) + 2 A_{s} cos (54°) + 2 A_{t} 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:
A ≡ A_{q}Q + A_{r}R + A_{s}S + A_{t}T
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.
A + B =
⟨ A_{q} + B_{q}, A_{r} + B_{r},
A_{s} + B_{s}, A_{t} + B_{t} ⟩
A − B =
⟨ A_{q} − B_{q}, A_{r} − B_{r},
A_{s} − B_{s}, A_{t} − B_{t} ⟩
Multiplication by an integer scalar n is not a problem:
nA = An = ⟨ A_{q}n, A_{r}n, A_{s}n, A_{t}n ⟩
Here is the rather complicated rule for general multiplication (derivation):
(AB)_{q} | = (A_{q}B_{r} + A_{r}B_{q}) | + (A_{r}B_{s} + A_{s}B_{r}) | + (A_{s}B_{t} + A_{t}B_{s}) |
(AB)_{r} | = (A_{q}B_{s} + A_{s}B_{q}) | + (A_{r}B_{t} + A_{t}B_{r}) | + (3A_{q}B_{q} + 2A_{r}B_{r} + 2A_{s}B_{s} + A_{t}B_{t}) |
(AB)_{s} | = (A_{q}B_{r} + A_{r}B_{q}) | + (A_{q}B_{t} + A_{t}B_{q}) | − (A_{s}B_{t} + A_{t}B_{s}) |
(AB)_{t} | = (A_{q}B_{s} + A_{s}B_{q}) | − (A_{r}B_{t} + A_{t}B_{r}) | − (2A_{q}B_{q} + A_{r}B_{r} + 3A_{s}B_{s} + 2A_{t}B_{t}) |
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… | 2R − T 2.618034… | Q 1.902113… | R − T exactly 1 |
S 1.175571… | R + T 2.236068… | Q 1.902113… | 2R − 3T 1.381966… | Q − S 0.726543… |
T 0.618034… | S 1.175571… | R − T exactly 1 | Q − S 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 R − T 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:
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:
A_{q} | = C_{q}(B_{r}) | + C_{r}(B_{q} + B_{s}) | + C_{s}(B_{r} + B_{t}) | + C_{t}(B_{s}) |
A_{r} | = C_{q}(3B_{q} + B_{s}) | + C_{r}(2B_{r} + B_{t}) | + C_{s}(B_{q} + 2B_{s}) | + C_{t}(B_{r} + B_{t}) |
A_{s} | = C_{q}(B_{r} + B_{t}) | + C_{r}(B_{q}) | + C_{s}(− B_{t}) | + C_{t}(B_{q} − B_{s}) |
A_{t} | = C_{q}(B_{s} − 2B_{q}) | + C_{r}(− B_{r} − B_{t}) | + C_{s}(B_{q} − 3B_{s}) | + C_{t}(− B_{r} − 2B_{t}) |
Regard this as a system of four linear equations in the four unknowns C_{q}, C_{r}, C_{s}, C_{t} 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 (2Q − S) ÷ 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.
Let E = D^{2}; then the components of E, after application of the general multiplication rule (above), appear thus:
E_{q} | = 2D_{q}D_{r} | + 2D_{r}D_{s} | + 2D_{s}D_{t} |
E_{r} | = 2D_{q}D_{s} | + 2D_{r}D_{t} | + (3D_{q}^{2} + 2D_{r}^{2} + 2D_{s}^{2} + D_{t}^{2}) |
E_{s} | = 2D_{q}D_{r} | + 2D_{q}D_{t} | − 2D_{s}D_{t} |
E_{t} | = 2D_{q}D_{s} | − 2D_{r}D_{t} | − (2D_{q}^{2} + D_{r}^{2} + 3D_{s}^{2} + 2D_{t}^{2}) |
Rearrange:
E_{q} | = 2(D_{q}D_{r} + D_{r}D_{s} + D_{s}D_{t}) |
E_{r} | = 2D_{q}^{2} + (D_{q} + D_{s})^{2} + D_{s}^{2} + D_{r}^{2} + (D_{r} + D_{t})^{2} |
E_{s} | = 2(D_{q}D_{r} + D_{q}D_{t} − D_{s}D_{t}) |
− E_{t} | = D_{q}^{2} + (D_{q} − D_{s})^{2} + 2D_{s}^{2} + (D_{r} + D_{t})^{2} + D_{t}^{2} |
Note that each of E_{r} and −E_{t} 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 E_{r} and E_{t} do place bounds on the values that might be searched. For instance, 2D_{q}^{2}, D_{s}^{2} and D_{r}^{2} can never exceed E_{r}. Similarly, D_{q}^{2}, 2D_{s}^{2} and D_{t}^{2} can never exceed −E_{t}.
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… |
T^{10} | = ⟨ 0, | +34, | 0, | −89 ⟩ | = 0.008131… |
T^{11} | = ⟨ 0, | −55, | 0, | +144 ⟩ | = 0.005025… |
T^{12} | = ⟨ 0, | +89, | 0, | −233 ⟩ | = 0.003106… |
T^{13} | = ⟨ 0, | −144, | 0, | +377 ⟩ | = 0.001919… |
T^{14} | = ⟨ 0, | +233, | 0, | −610 ⟩ | = 0.001186… |
T^{15} | = ⟨ 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 ± T^{n} for sufficiently large n.
To prove the converse, however, is more complicated. First, it can be established that if A = 0, then A_{q} = A_{r} = A_{s} = A_{t} = 0. Second, to investigate B = C, write B − C = 0. Then
(B − C)_{q} =
(B − C)_{r} =
(B − C)_{s} =
(B − C)_{t} = 0
B_{q} − C_{q} =
B_{r} − C_{r} =
B_{s} − C_{s} =
B_{t} − C_{t} = 0
B_{q} = C_{q},
B_{r} = C_{r},
B_{s} = C_{s},
B_{t} = C_{t}
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.