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==== If the steps of a CS scale are linearly independent, then the scale is ~~epimorphic~~ ====

==== If the steps of a CS scale are linearly independent, then the scale is a strong CS ====

Theorem: Suppose ''S'' is a 2/1-equivalent increasing constant structure JI scale of length ''n''. Let <math>C_1</math> be the set of 1-steps of ''S'', and suppose that <math>C_1</math> is a basis for the JI subgroup ''A'' generated by it. Then there exists an ~~epimorphism~~ <math> v: A \to \mathbb{Z}</math> which is a val of ''n''-edo (and a similar statement holds for other equaves).

Theorem: Suppose ''S'' is a 2/1-equivalent increasing constant structure JI scale of length ''n''. Let <math>C_1</math> be the set of 1-steps of ''S'', and suppose that <math>C_1</math> is a basis for the JI subgroup ''A'' generated by it. Then there exists an val <math> v: A \to \mathbb{Z}</math> which is a val of ''n''-edo (and a similar statement holds for other equaves).

(The condition of <math>C_1</math> being a basis rather than merely a generating set cannot be omitted, since the scale {5/4, 32/25, 2/1} is a CS but not ~~epimorphic~~. The converse of this conditional also fails, as {9/8, 5/4, 3/2, 25/16, 2/1} is epimorphic under [[5edo]]'s [[patent val]].)

(The condition of <math>C_1</math> being a basis rather than merely a generating set cannot be omitted, since the scale {5/4, 32/25, 2/1} is a CS but not a strong CS. The converse of this conditional also fails, as {9/8, 5/4, 3/2, 25/16, 2/1} is epimorphic under [[5edo]]'s [[patent val]].)

{{proof|contents=

Define the linear map <math>v:A \to \mathbb{Z}</math> by defining <math>v(\mathbf{s}) = 1</math> for any step <math>\mathbf{s} \in C_1</math> and extending uniquely by linearity. Then for any <math>i \in \mathbb{Z}</math> we have <math>v(S[i]) = v(S[i]/S[i-1]\cdots S[1]) = v(S[i]/S[i-1]) + \cdots + v(S[1]) = i,</math> whence ''v'' is ~~an epimorphism~~. That <math>v(2) = n</math> is also automatic.

Define the linear map <math>v:A \to \mathbb{Z}</math> by defining <math>v(\mathbf{s}) = 1</math> for any step <math>\mathbf{s} \in C_1</math> and extending uniquely by linearity. Then for any <math>i \in \mathbb{Z}</math> we have <math>v(S[i]) = v(S[i]/S[i-1]\cdots S[1]) = v(S[i]/S[i-1]) + \cdots + v(S[1]) = i,</math> whence ''v'' is a strong CS. That <math>v(2) = n</math> is also automatic.

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-==== If the steps of a CS scale are linearly independent, then the scale is epimorphic ====
+==== If the steps of a CS scale are linearly independent, then the scale is a strong CS ====
-Theorem: Suppose ''S'' is a 2/1-equivalent increasing constant structure JI scale of length ''n''. Let <math>C_1</math> be the set of 1-steps of ''S'', and suppose that <math>C_1</math> is a basis for the JI subgroup ''A'' generated by it. Then there exists an epimorphism <math> v: A \to \mathbb{Z}</math> which is a val of ''n''-edo (and a similar statement holds for other equaves).
+Theorem: Suppose ''S'' is a 2/1-equivalent increasing constant structure JI scale of length ''n''. Let <math>C_1</math> be the set of 1-steps of ''S'', and suppose that <math>C_1</math> is a basis for the JI subgroup ''A'' generated by it. Then there exists an val <math> v: A \to \mathbb{Z}</math> which is a val of ''n''-edo (and a similar statement holds for other equaves).
-(The condition of <math>C_1</math> being a basis rather than merely a generating set cannot be omitted, since the scale {5/4, 32/25, 2/1} is a CS but not epimorphic. The converse of this conditional also fails, as {9/8, 5/4, 3/2, 25/16, 2/1} is epimorphic under [[5edo]]'s [[patent val]].)
+(The condition of <math>C_1</math> being a basis rather than merely a generating set cannot be omitted, since the scale {5/4, 32/25, 2/1} is a CS but not a strong CS. The converse of this conditional also fails, as {9/8, 5/4, 3/2, 25/16, 2/1} is epimorphic under [[5edo]]'s [[patent val]].)
 {{proof|contents=
-Define the linear map <math>v:A \to \mathbb{Z}</math> by defining <math>v(\mathbf{s}) = 1</math> for any step <math>\mathbf{s} \in C_1</math> and extending uniquely by linearity. Then for any <math>i \in \mathbb{Z}</math> we have <math>v(S[i]) = v(S[i]/S[i-1]\cdots S[1]) = v(S[i]/S[i-1]) + \cdots + v(S[1]) = i,</math> whence ''v'' is an epimorphism. That <math>v(2) = n</math> is also automatic.
+Define the linear map <math>v:A \to \mathbb{Z}</math> by defining <math>v(\mathbf{s}) = 1</math> for any step <math>\mathbf{s} \in C_1</math> and extending uniquely by linearity. Then for any <math>i \in \mathbb{Z}</math> we have <math>v(S[i]) = v(S[i]/S[i-1]\cdots S[1]) = v(S[i]/S[i-1]) + \cdots + v(S[1]) = i,</math> whence ''v'' is a strong CS. That <math>v(2) = n</math> is also automatic.
 }}