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=== temperament states ===
=== cases temperaments can be regarding which of their mapping forms equal each other ===


[[File:Temperament states for matrix forms.png|300px|thumb|right|The three possible temperament states w/r/t equivalence of established matrix forms, plus in the bottom right, a state which is only possible for enfactored temperaments AKA temperoids.]]
[[File:Cases for temperament mapping forms.png|300px|right]]


All considered, three different states are possible for a given temperament:
Considering only full-rank, integer mappings, we find three cases for a given temperament which is not enfactored. In all three cases, HNF is the same as DCF:
# The RREF, IRREF, HNF, and DCF are all the same. Example: [[Meantone_family#Meantone_.2812.2619.2C_2.3.5.29|meantone]] with all four equal to {{vector|{{map|1 0 -4}} {{map|0 1 4}}}}.
# The RREF, IRREF, and HNF are all ''different''. Example: [[Porcupine_family#Porcupine|porcupine]] with RREF of {{vector|{{map|1 0 <span><math>-\frac13</math></span>}} {{map|0 1 <span><math>\frac53</math></span>}}}}, IRREF of {{vector|{{map|3 0 -1}} {{map|0 3 5}}}}, and HNF of {{vector|{{map|1 2 3}} {{map|0 3 5}}}}.  
# The RREF, IRREF, and HNF are all different, but HNF is the same as DCF. Example: [[Porcupine_family#Porcupine|porcupine]] with RREF of {{vector|{{map|1 0 <span><math>-\frac13</math></span>}} {{map|0 1 <span><math>\frac53</math></span>}}}}, IRREF of {{vector|{{map|3 0 -1}} {{map|0 3 5}}}}, and HNF and DCF of {{vector|{{map|1 2 3}} {{map|0 3 5}}}}, respectively.  
# The RREF, IRREF, HNF are all ''the same''. Example: [[Meantone_family#Meantone_.2812.2619.2C_2.3.5.29|meantone]] with all equal to {{vector|{{map|1 0 -4}} {{map|0 1 4}}}}. This case is quite rare.
# The IRREF, HNF, and DCF are the same, but the RREF is different. Example: [[Kleismic_family#Hanson|hanson]] with IRREF, HNF, and DCF of {{vector|{{map|1 0 1}} {{map|0 6 5}}}} but RREF of {{vector|{{map|1 0 1}} {{map|0 1 <span><math>\frac56</math></span>}}}}.
# The IRREF and HNF are the same, but the ''RREF is different''. Example: [[Kleismic_family#Hanson|hanson]] with IRREF and HNF of {{vector|{{map|1 0 1}} {{map|0 6 5}}}} but RREF of {{vector|{{map|1 0 1}} {{map|0 1 <span><math>\frac56</math></span>}}}}.


A fourth state is possible, but only for enfactored mappings, i.e. for temperoids. This is where the RREF and IRREF are the same, but the HNF is different, and the DCF is different still. This is only possible for enfactored mappings because:
And there are three corresponding cases when a temperament is enfactored. In all three cases, the key difference is that HNF is no longer the same as DCF, with the only difference being that the common factor is not removed. In all cases below, the examples are shown with a common factor of 2 introduced in their second row, which stays behind in the second row of the HNF:
* Whenever the RREF and IRREF match, the IRREF has all 1's for pivots.  
# Now ''all four'' are different. Example: enfactored porcupine, e.g. {{vector|{{map|15 24 35}} {{map|14 22 32}}}} causes the HNF to be {{vector|{{map|1 2 3}} {{map|0 6 10}}}}.
* But in that case, the IRREF would also be the HNF, because there's only one possible nonnegative integer value less than 1, which is zero, and so this form would also satisfy the normal constraint.
# Everything is still the same now, ''except HNF''. Example: enfactored meantone, e.g. {{vector|{{map|5 8 12}} {{map|14 22 32}}}} causes the HNF to be {{vector|{{map|1 0 -4}} {{map|0 2 8}}}}. This case, like the corresponding unenfactored state, is also quite rare.
* The only way to avoid this would be to ensure that at least one pivot of the HNF was not 1, so that it could have some positive value somewhere in the entries above that pivot in its pivot column, to distinguish it from IRREF. For example, we could use {{vector|{{map|1 1 0}} {{map|0 2 1}}}}.
# The ''only match'' now is between IRREF and DCF. In other words, the HNF and DCF diverged, and it was the DCF which remained the same as IRREF. Example: enfactored hanson, e.g. {{vector|{{map|15 24 35}} {{map|38 60 88}}}} causes the HNF to be {{vector|{{map|1 0 1}} {{map|0 12 10}}}}.
* But now if we look at the RREF of this mapping with this HNF, we have to divide rows until all pivots are 1. So that second row would be changed to {{map|0 1 <span><math>\frac12</math></span>}}. So now RREF won't match IRREF, because it contains a noninteger. The only way to prevent this would be if a pivot of the HNF was not 1, but still every entry in that row was evenly divisible by that value, such as {{vector|{{map|1 1 0}} {{map|0 2 4}}}}. But now that row is enfactored by its pivot's value.
 
* If the mapping is enfactored, that is the case when the HNF ≠ DCF.
There is also a final case which is incredibly rare. It can be compared to the #3 cases above, the ones using hanson as their example. The idea here is that when the HNF and DCF diverge, instead of DCF remaining the same as IRREF, it's the HNF that remains the same as IRREF. I haven't found any practical temperoids with this case, but {{vector|{{map|165 264 393}} {{map|231 363 524}}}} will do it<ref>AKA 165b⁴c¹⁹&231b⁶c²⁴, which tempers out the 7.753¢ comma {{vector|-131 131 -33}}!</ref>, with IRREF and HNF of {{vector|{{map|33 0 -131}} {{map|0 33 131}}}}, DCF of {{vector|{{map|1 1 0}} {{map|0 33 131}}}}, and RREF of {{vector|{{map|1 0 <span><math>\frac{-131}{33}</math></span>}} {{map|0 1 <span><math>\frac{131}{33}</math></span>}}}}.


=== sum-and-difference defactoring ===
=== sum-and-difference defactoring ===
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