Left Member

The question here is to determine Θ_{2} in:
q √(10 + √20) + s √(10 − √20) Θ_{2} 0 [#2]
Observe that 10 + √20 and 10 − √20 are themselves positive, and so are their square roots.
Θ_{2} in #2
 s < 0
 s = 0
 s > 0


q < 0
 Θ_{2} is lessthan  Θ_{2} is lessthan  see case alpha


q = 0
 Θ_{2} is lessthan  Θ_{2} is equalto  Θ_{2} is greaterthan


q > 0
 see case beta  Θ_{2} is greaterthan  Θ_{2} is greaterthan




Case Alpha 
Equivalent to #2 is:
s √(10 − √20) Θ_{2} −q √(10 + √20)
Both members are positive, so square it without exchange:
s^{2}(10 − √20) Θ_{2} q^{2}(10 + √20)
Rearrange:
10(s^{2} − q^{2}) Θ_{2} √20(s^{2} + q^{2}) [#3]
Θ_{2} in #3
 s^{2} + q^{2} = 0
 s^{2} + q^{2} > 0


s^{2}
− q^{2} < 0
 Θ_{2} is lessthan  Θ_{2} is lessthan


s^{2}
− q^{2} = 0
 Θ_{2} is equalto  Θ_{2} is lessthan


s^{2}
− q^{2} > 0
 Θ_{2} is greaterthan  see below


When in #3 both members are positive, square it without exchange:
100(s^{2} − q^{2})^{2} Θ_{2} 20(s^{2} + q^{2})^{2}
which is all integers, and thus can easily be evaluated exactly.


Case Beta 
Recall:
s √(10 − √20) Θ_{2} −q √(10 + √20)
Both members are negative, so square it with exchange:
q^{2}(10 + √20) Θ_{2} s^{2}(10 − √20)
Rearrange:
√20(s^{2} + q^{2}) Θ_{2} 10(s^{2} − q^{2}) [#4]
Θ_{2} in #4
 s^{2} − q^{2} < 0
 s^{2} − q^{2} = 0
 s^{2} − q^{2} > 0


s^{2}
+ q^{2} = 0
 Θ_{2} is greaterthan  Θ_{2} is equalto  Θ_{2} is lessthan


s^{2}
+ q^{2} > 0
 Θ_{2} is greaterthan  Θ_{2} is greaterthan  see below


When in #4 both members are positive, square it without exchange:
20(s^{2} + q^{2})^{2} Θ_{2} 100(s^{2} − q^{2})^{2}
which is all integers.


Right Member 
The question here is to determine Θ_{3} in:
− r (√5 + 1) − t (√5 − 1) Θ_{3} 0
or equivalently
(t − r) Θ_{3} √5(t + r) [#5]
Θ_{3} in #5
 t + r < 0
 t + r = 0
 t + r > 0


t − r < 0
 see case gamma
 Θ_{3} is lessthan
 Θ_{3} is lessthan


t − r = 0
 Θ_{3} is greaterthan
 Θ_{3} is equalto
 Θ_{3} is lessthan


t − r > 0
 Θ_{3} is greaterthan
 Θ_{3} is greaterthan
 see case delta




Case Gamma 
In #5 both members are negative, so square it with exchange:
5(t + r)^{2} Θ_{3} (t − r)^{2}
which is all integers.


Case Delta 
In #5 both members are positive, so square it without exchange:
(t − r)^{2} Θ_{3} 5(t + r)^{2}
which is all integers.


Summary 
With Θ_{2} and Θ_{3} established, resolution of Θ_{1} can begin. This table handles many cases:
Θ_{1} in #1
 Θ_{3} is lessthan
 Θ_{3} is equalto
 Θ_{3} is greaterthan


Θ_{2} is lessthan
 see case epsilon  Θ_{1} is lessthan  Θ_{1} is lessthan


Θ_{2} is equalto
 Θ_{1} is greaterthan  Θ_{1} is equalto  Θ_{1} is lessthan


Θ_{2} is greaterthan
 Θ_{1} is greaterthan  Θ_{1} is greaterthan  see case zeta




Case Epsilon 
In #1 both members are negative, so square it with exchange:
r^{2} (6 + 2 √5) + t^{2} (6 − 2 √5) + 8 r t Θ_{1} q^{2} (10 + √20) + s^{2} (10 − √20) + 2 q s √80
Move the radicals left, and divide by two:
√5(r^{2} − t^{2} − q^{2} + s^{2} − 4 q s) Θ_{1} − 3 r^{2} − 4 r t + 5 q^{2} − 3 t^{2} + 5 s^{2}
Define
X = r^{2} − t^{2} − q^{2} + s^{2} − 4 q s
and
Y = − 3 r^{2} − 4 r t + 5 q^{2} − 3 t^{2} + 5 s^{2}
This consolidates the problem to
√5X Θ_{1} Y [#6]
Θ_{1} in #6
 Y < 0
 Y = 0
 Y > 0


X < 0
 same as Y^{2} Θ_{1} 5X^{2}  Θ_{1} is lessthan  Θ_{1} is lessthan


X = 0
 Θ_{1} is greaterthan  Θ_{1} is equalto  Θ_{1} is lessthan


X > 0
 Θ_{1} is greaterthan  Θ_{1} is greaterthan  same as 5X^{2} Θ_{1} Y^{2}




Case Zeta 
In #1 both members are positive, so square it without exchanging:
q^{2} (10 + √20) + s^{2} (10 − √20) + 2 q s √80
Θ_{1}
r^{2} (6 + 2 √5) + t^{2} (6 − 2 √5) + 8 r t
Move the radicals right, and divide by two:
− 3 r^{2} − 4 r t + 5 q^{2} − 3 t^{2} + 5 s^{2} Θ_{1} √5(r^{2} − t^{2} − q^{2} + s^{2} − 4 q s)
Again define
X = r^{2} − t^{2} − q^{2} + s^{2} − 4 q s
and
Y = − 3 r^{2} − 4 r t + 5 q^{2} − 3 t^{2} + 5 s^{2}
This consolidates the problem toY Θ_{1} √5X [#7]
Θ_{1} in #7
 X < 0
 X = 0
 X > 0


Y < 0
 same as 5X^{2} Θ_{1} Y^{2}  Θ_{1} is lessthan  Θ_{1} is lessthan


Y = 0
 Θ_{1} is greaterthan  Θ_{1} is equalto  Θ_{1} is lessthan


Y > 0
 Θ_{1} is greaterthan  Θ_{1} is greaterthan  same as Y^{2} Θ_{1} 5X^{2}



