(a + b √c) / d

Given a, b, c and d such that n = (a + b √c)/d, constuct a QI corresponding to n.

>>> qi 3 4 5 6
qi 3 4 5 6

The fractions are reduced:

>>> qi 30 40 5 60
qi 3 4 5 6

If b = 0 then c is zeroed and vice versa:

>>> qi 3 0 42 1
qi 3 0 0 1

>>> qi 3 42 0 1
qi 3 0 0 1

The b √c term is simplified:

>>> qi 0 1 (5*5*6) 1
qi 0 5 6 1

If c = 1 (after simplification) then b is moved to a:

>>> qi 1 5 (2*2) 1
qi 11 0 0 1

Given a, b and c such that n = a + b √c, constuct a QI corresponding to n.

>>> qi' 0.5 0.7 2
qi 5 7 2 10

Given n and f such that n = (a + b √c)/d, run f a b c d.

>>> runQI (qi 3 4 5 6) (\a b c d -> (a,b,c,d))
(3,4,5,6)

Given n and f such that n = a + b √c, run f a b c.

>>> runQI' (qi' 0.5 0.7 2) (\a b c -> (a, b, c))
(1 % 2,7 % 10,2)

Given n such that n = (a + b √c)/d, return (a, b, c, d).

>>> unQI (qi 3 4 5 6)
(3,4,5,6)

Given n such that n = a + b √c, return (a, b, c).

>>> unQI' (qi' 0.5 0.7 2)
(1 % 2,7 % 10,2)

Given a QI corresponding to n = (a + b √c)/d, access (a, b, c, d).

>>> view _qi (qi 3 4 5 6)
(3,4,5,6)

>>> over _qi (\(a,b,c,d) -> (a+10, b+10, c+10, d+10)) (qi 3 4 5 6)
qi 13 14 15 16

Given a QI corresponding to n = a + b √c, access (a, b, c).

>>> view _qi' (qi' 0.5 0.7 2)
(1 % 2,7 % 10,2)

>>> over _qi' (\(a,b,c) -> (a/5, b/6, c*3)) (qi 3 4 5 6)
qi 9 10 15 90

Given a QI corresponding to n = (a + b √c)/d, access (a, b, d). Avoids having to simplify b √c upon reconstruction.

>>> view _qiABD (qi 3 4 5 6)
(3,4,6)

>>> over _qiABD (\(a,b,d) -> (a+10, b+10, d+10)) (qi 3 4 5 6)
qi 13 14 5 16

Given a QI corresponding to n = (a + b √c)/d, access a. It is more efficient to use _qi or _qiABD when modifying multiple terms at once.

>>> view _qiA (qi 3 4 5 6)
3

>>> over _qiA (+ 10) (qi 3 4 5 6)
qi 13 4 5 6

Given a QI corresponding to n = (a + b √c)/d, access b. It is more efficient to use _qi or _qiABD when modifying multiple terms at once.

>>> view _qiB (qi 3 4 5 6)
4

>>> over _qiB (+ 10) (qi 3 4 5 6)
qi 3 14 5 6

Given a QI corresponding to n = (a + b √c)/d, access c. It is more efficient to use _qi or _qiABD when modifying multiple terms at once.

>>> view _qiC (qi 3 4 5 6)
5

>>> over _qiC (+ 10) (qi 3 4 5 6)
qi 3 4 15 6

Given a QI corresponding to n = (a + b √c)/d, access d. It is more efficient to use _qi or _qiABD when modifying multiple terms at once.

>>> view _qiD (qi 3 4 5 6)
6

>>> over _qiD (+ 10) (qi 3 4 5 6)
qi 3 4 5 16

The constant zero.

>>> qiZero
qi 0 0 0 1

The constant one.

>>> qiOne
qi 1 0 0 1

Check if the value is zero.

>>> map qiIsZero [qiZero, qiOne, qiSubR (qi 7 0 0 2) 3.5]
[True,False,True]

Convert a QI number into a Floating one.

>>> qiToFloat (qi 3 4 5 6) == ((3 + 4 * sqrt 5)/6 :: Double)
True

Add an Integer to a QI.

>>> qi 3 4 5 6 `qiAddI` 1
qi 9 4 5 6

Subtract an Integer from a QI.

>>> qi 3 4 5 6 `qiSubI` 1
qi (-3) 4 5 6

Multiply a QI by an Integer.

>>> qi 3 4 5 6 `qiMulI` 2
qi 3 4 5 3

Divice a QI by an Integer.

>>> qi 3 4 5 6 `qiDivI` 2
qi 3 4 5 12

Add a Rational to a QI.

>>> qi 3 4 5 6 `qiAddR` 1.2
qi 51 20 5 30

Subtract a Rational from a QI.

>>> qi 3 4 5 6 `qiSubR` 1.2
qi (-21) 20 5 30

Multiply a QI by a Rational.

>>> qi 3 4 5 6 `qiMulR` 0.5
qi 3 4 5 12

Divice a QI by a Rational.

>>> qi 3 4 5 6 `qiDivR` 0.5
qi 3 4 5 3

Negate a QI.

>>> qiNegate (qi 3 4 5 6)
qi (-3) (-4) 5 6

Compute the reciprocal of a QI.

>>> qiRecip (qi 5 0 0 2)
Just (qi 2 0 0 5)

>>> qiRecip (qi 0 1 5 2)
Just (qi 0 2 5 5)

>>> qiRecip qiZero
Nothing

Add two QIs if the square root terms are the same or zeros.

>>> qi 3 4 5 6 `qiAdd` qiOne
Just (qi 9 4 5 6)

>>> qi 3 4 5 6 `qiAdd` qi 3 4 5 6
Just (qi 3 4 5 3)

>>> qi 0 1 5 1 `qiAdd` qi 0 1 6 1
Nothing

Subtract two QIs if the square root terms are the same or zeros.

>>> qi 3 4 5 6 `qiSub` qiOne
Just (qi (-3) 4 5 6)

>>> qi 3 4 5 6 `qiSub` qi 3 4 5 6
Just (qi 0 0 0 1)

>>> qi 0 1 5 1 `qiSub` qi 0 1 6 1
Nothing

Multiply two QIs if the square root terms are the same or zeros.

>>> qi 3 4 5 6 `qiMul` qiZero
Just (qi 0 0 0 1)

>>> qi 3 4 5 6 `qiMul` qiOne
Just (qi 3 4 5 6)

>>> qi 3 4 5 6 `qiMul` qi 3 4 5 6
Just (qi 89 24 5 36)

>>> qi 0 1 5 1 `qiMul` qi 0 1 6 1
Nothing

Divide two QIs if the square root terms are the same or zeros.

>>> qi 3 4 5 6 `qiDiv` qiZero
Nothing

>>> qi 3 4 5 6 `qiDiv` qiOne
Just (qi 3 4 5 6)

>>> qi 3 4 5 6 `qiDiv` qi 3 4 5 6
Just (qi 1 0 0 1)

>>> qi 3 4 5 6 `qiDiv` qi 0 1 5 1
Just (qi 20 3 5 30)

>>> qi 0 1 5 1 `qiDiv` qi 0 1 6 1
Nothing

Exponentiate a QI to an Integer power.

>>> qi 3 4 5 6 `qiPow` 0
qi 1 0 0 1

>>> qi 3 4 5 6 `qiPow` 1
qi 3 4 5 6

>>> qi 3 4 5 6 `qiPow` 2
qi 89 24 5 36

Compute the floor of a QI.

>>> qiFloor (qi 10 0 0 2)
5

>>> qiFloor (qi 10 2 2 2)
6

>>> qiFloor (qi 10 2 5 2)
7

Convert a (possibly periodic) continued fraction to a QI.

[2; 2] = 2 + 1/2 = 5/2.

>>> continuedFractionToQI (2,NonCyc [2])
qi 5 0 0 2

[2; 1, 1, 1, 4, 1, 1, 1, 4, …] = √7.

>>> continuedFractionToQI (2,Cyc [] 1 [1,1,4])
qi 0 1 7 1

>>> continuedFractionToQI (0,Cyc [83,78,65,75,69] 32 [66,65,68,71,69,82])
qi 987601513930253257378987883 1 14116473325908285531353005 81983584717737887813195873886

Convert a QI into a (possibly periodic) continued fraction.

5/2 = 2 + 1/2 = [2; 2].

>>> qiToContinuedFraction (qi 5 0 0 2)
(2,NonCyc [2])

√7 = [2; 1, 1, 1, 4, 1, 1, 1, 4, …].

>>> qiToContinuedFraction (qi 0 1 7 1)
(2,Cyc [] 1 [1,1,4])

>>> qiToContinuedFraction (qi 987601513930253257378987883 1 14116473325908285531353005 81983584717737887813195873886)
(0,Cyc [83,78,65,75,69] 32 [66,65,68,71,69,82])