Mathematics of MOS: Difference between revisions

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=== Definition of MOS ===

A periodic scale is ''MOS'' if it has [[maximum variety]] 2. This in particular implies that MOS scales are binary.

=== MOS scales are generated ===

See [[Recursive structure of MOS scales#Proofs]].

=== MOS scales aL bs with gcd(a, b) > 1 reduce to single-period MOS scales on a smaller period ===

Assume that gcd(m, n) > 1 and assume that (n/d)-steps came in 2 sizes, pa+qb and ra+sb. Then at least one size, say pa+qb, must differ from (a/d)L + (b/d)s. WLOG p > a/d and pa+qb occurs on degree 0. Since equave = aL + bs, on a degree k(n/d) for some integer k > 0, there must be another (n/d)-step with fewer L's than a/d. This involves more than two changes from L to s. Scooting an (n/d)-step one step at a time from degree 0 to k(n/d) changes its size one step size substitution at a time, showing that intermediate (n/d)-steps also exist. This violates the MOS property, whence (n/d)-steps have only one size, which must be the period.

=== MOS scales ''aL'' ''bs'' with gcd(a, b) > 1 reduce to single-period MOS scales on a smaller period ===

Assume that {{nowrap|gcd(''m'', ''n'') > 1}} and assume that ({{frac|''n''|''d''}})-steps came in 2 sizes, {{nowrap|''pa'' + ''qb''}} and {{nowrap|''ra'' + ''sb''}}. Then at least one size, say {{nowrap|''pa'' + ''qb''}}, must differ from {{nowrap|{{frac|''a''|''d''}} L + {{frac|''b''|''d''}} s}}. WLOG {{nowrap|''p'' > {{frac|''a''|''d''}}}} and {{nowrap|''pa'' + ''qb''}} occurs on degree 0. Since the equave is equal to {{nowrap|''aL'' + ''bs''}}, on a degree ''k''({{frac|''n''|''d''}}) for some integer {{nowrap|''k'' > 0}}, there must be another ({{frac|''n''|''d''}})-step with fewer L's than {{frac|''a''|''d''}}. This involves more than two changes from ''L'' to ''s''. Scooting an ({{frac|''n''|''d''}})-step one step at a time from degree 0 to ''k''({{frac|''n''|''d''}}) changes its size one step size substitution at a time, showing that intermediate ({{frac|''n''|''d''}})-steps also exist. This violates the MOS property, whence ({{frac|''n''|''d''}})-steps have only one size, which must be the period.

=== Maximally even scales are MOS ===

Let 1 < m < n, gcd(m, n) = d, and ME(''m'', ''n'') be as in [[Maximal evenness]]. An arbitrary k-step in ME(''m'', ''n'') has size ~~floor(~~(''i'' + ''k'')''n''/''m'') − ~~floor(~~''in''/''m''), and

Let {{nowrap|1 > ''m'' > ''n''|gcd(''m'', ''n'') {{=}} d}}, and ME(''m'', ''n'') be as in [[Maximal evenness]]. An arbitrary ''k''-step in ME(''m'', ''n'') has size {{nowrap|''u'' {{=}} &lfloor;(''i'' + ''k'') {{frac|''n''|''m''}}&rfloor; − &lfloor;{{frac|''in''|''m''}}&rfloor;}}, and

<math>\lfloor in/m \rfloor + \lfloor kn/m\rfloor - \lfloor in/m\rfloor = \lfloor kn/m\rfloor ≤ \~~lfloor(i+k)n/m~~ \rfloor ~~- \lfloor in/m~~ \~~rfloor~~

<math>\lfloor in/m \rfloor + \lfloor kn/m\rfloor - \lfloor in/m\rfloor = \lfloor kn/m\rfloor \leq u \rfloor

≤ \lfloor in/m \rfloor + \lfloor kn/m \rfloor - \lfloor in/m \rfloor + 1 = \lfloor kn/m \rfloor + 1.</math>

\leq \lfloor in/m \rfloor + \lfloor kn/m \rfloor - \lfloor in/m \rfloor + 1 = \lfloor kn/m \rfloor + 1.</math>

As ~~floor((~~''i'' ~~+ ''k'')''n''/''m'') − floor(''in''/''m'')~~ is an integer, the above proves that there are at most two possible values for it and that if there are two sizes for k-steps, the two sizes must differ by 1.

As ''u'' is an integer, the above proves that there are at most two possible values for it and that if there are two sizes for ''k''-steps, the two sizes must differ by 1.

Steps of ME(''m'', ''n'') have exactly two sizes because if it were one size, we would have m | (n/d), which is a contradiction.

Steps of ME(''m'', ''n'') have exactly two sizes because if it were one size, we would have {{nowrap|''m'' {{!}} {{frac|''n''|''d''}}}}, which is a contradiction.

This maximally even MOS has n % m large steps and m − (n % m) small steps.

This maximally even MOS has {{nowrap|''n'' % ''m''}} large steps and {{nowrap|''m'' − (''n'' % ''m'')}} small steps.

=== Binary generated scales with independent period and generator are MOS ===

This proof justifies the common description of "stack until binary" for MOS building and Erv Wilson's terminology ''moment of symmetry'' where MOS sizes are "moments" in time (when stacking) where the "symmetry" of binarity and MV2 holds.

By ''generatedness'', we mean that every interval in the scale is of the form ''jg'' + ''kp'' where ''g'' is a generator, ''p'' is the period, and ''j~~, k~~'' ∈ ''~~'Z'~~'', and that either ''g'' or ''−g'' occurs on every note. We claim that any interval class not ''p''-equivalent to 0 has ''exactly'' 2 sizes in any scale satisfying the antecedent.

By ''generatedness'', we mean that every interval in the scale is of the form {{nowrap|''jg'' + ''kp''}} where ''g'' is a generator, ''p'' is the period, and {{nowrap|''j'', ''k'' ∈ ℤ}}, and that either ''g'' or −''g'' occurs on every note. We claim that any interval class not ''p''-equivalent to 0 has ''exactly'' 2 sizes in any scale satisfying the antecedent.

Suppose that such a scale ''S'' (with ''n'' ≥ 2 notes) has ''a''~~-many~~ L steps and ''b''~~-many~~ s steps per period ''p'', and has generator ''g''. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:

Suppose that such a scale ''S'' (with {{nowrap|''n'' ≥ 2}} notes) has ''a'' L steps and ''b'' s steps per period ''p'', and has generator ''g''. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:

{(−''n'' + 1)''g'', ..., −''g'', 0, ''g'', ..., (''n'' − 1)''g''},

{{(}}{{nowrap|(−''n'' + 1)''g''}}, ..., −''g'', 0, ''g'', ..., {{nowrap|(''n'' − 1)''g''}}{{)}},

and all sizes {0, ''g'', ..., (''n'' − 1)''g''} are distinct.

and all sizes {0, ''g'', ..., {{nowrap|(''n'' − 1)''g''}}{{)}} are distinct.

We thus have:

L = ''cg'' + ''dp''

{{nowrap|L {{=}} ''cg'' + ''dp''}}

{{nowrap|s {{=}} ''eg'' + ''fp''}}

~~s =~~ ''eg'' + ''fp''

for appropriate integers ''c, d, e, f'', where {{nowrap|{{!}}''c''{{!}}, {{!}}''e''{{!}} < ''n''}}.

~~for appropriate integers~~ ''c, d, e, f'', ~~where~~ |''c''~~| <~~ ''n'' and |''e''~~| <~~ ''n''.

Now we assume that ''g'' and ''p'' are linearly independent. By assumption {{nowrap|''a''L + ''b''s {{=}} (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p''}} = ''p''. Since {{nowrap|''a''L + ''b''s}} occurs on the "brightest" mode, from generatedness we have {{nowrap|''ac'' + ''be'' ∈ {{(}}0, ..., ''n'' − 1{{)}}}}. Hence we must have {{nowrap|''ac'' + ''be'' {{=}} 0}}, and thus {{nowrap|''c'' {{=}} ±''b''}} and {{nowrap|''e'' {{=}} ∓''a''}}, from the assumption that ''a'' and ''b'' are coprime.

Now we assume that ''g'' and ''p'' are linearly independent. By assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. Since ''a''L + ''b''s occurs on the "brightest" mode, from generatedness we have ''ac'' + ''be'' ∈ {0, ..., ''n'' − 1}. Hence we must have ''ac'' + ''be'' = 0, and thus ''c'' = ±''b'' and ''e'' = ∓''a'', from the assumption that gcd(a, b) = 1.

In fact, {L, s} is another valid basis for the abelian group with basis {''p'', ''g''}, since by binarity we have {{nowrap|''p, g'' ∈ span(L, s)}}. Assume {{nowrap|''c'' {{=}} ''b''}} and {{nowrap|''e'' {{=}} −''a''}} (this corresponds to assuming that ''g'' is the "bright" generator). Let {{nowrap|χ {{=}} L − s}} > 0; then χ is ''p''-equivalent to +''ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each {{nowrap|''j'' ∈ {{(}}1, ..., ''n'' − 1{{)}}}} there exists at most one {{nowrap|''k'' {{=}} ''k''(''j'') ∈ {{(}}1, ..., ''n'' − 1{{)}}}} such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if the class of ''k''-steps has ''at least'' two sizes, the sizes must be ''j''(''k'')''g'' and {{nowrap|(''j''(''k'') − ''n'')''g''}}; any other size must leave the range {{nowrap|(1 − ''n'')''g''}}, ..., 0, ..., {{nowrap|(''n'' − 1)''g''}}. Thus the class of ''k''-steps has at most two sizes for {{nowrap|1 ≤ ''k'' ≤ ''n'' − 1}}. Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an {{nowrap|(''n'' − ''k'')}}-step equivalent to −''jg'', which by linear independence must be distinct from an {{nowrap|(''n'' − ''k'')}}-step equivalent to a positive number of ''g'' generators. (Note that the latter {{nowrap|(''n'' − ''k'')}}-step does occur in the "brightest" mode of ''S'', i.e. the mode with the most ''g'' generators stacked ''up'' rather than ''down'' from the tonic.) This completes the proof.

In fact, {L, s} is another valid basis for the abelian group with basis {''p'', ''g''}, since by binarity we have ''p, g'' ∈ span(L, s). Assume ''c'' = ''b'' and ''e'' = −''a''~~. [This~~ corresponds to assuming that ''g'' is the "bright" generator.] Let χ = L − s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' − 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' − 1}~~</sub>~~ such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if the class of ''k''-steps has ''at least'' two sizes, the sizes must be ''j''(''k'')''g'' and (''j''(''k'') − ''n'')''g''; any other size must leave the range −(''n'' ~~− 1~~)''g'', ..., 0, ..., (''n'' − 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ~~≤ (~~''n'' − 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' − ''k'')-step equivalent to ''−jg'', which by linear independence must be distinct from an (''n'' − ''k'')-step equivalent to a positive number of ''g'' generators. (Note that the latter (''n'' − ''k'')-step does occur in the "brightest" mode of ''S'', i.e. the mode with the most ''g'' generators stacked ''up'' rather than ''down'' from the tonic.) This completes the proof.

The previous argument cannot be used for the non-linearly independent case. This is because not all binary generated scales with rational step ratios are limit points of binary generated scales with irrational step ratios. A counterexample is the 13-note scale ssLsLssLsLss (s = 1\17, L = 2\17), which is obtained from LsLsLssLsLss (~~5L7s~~) by stacking one more 10\17 generator.

The previous argument cannot be used for the non-linearly independent case. This is because not all binary generated scales with rational step ratios are limit points of binary generated scales with irrational step ratios. A counterexample is the 13-note scale ssLsLssLsLss ({{nowrap|s {{=}} 1\17|L {{=}} 2\17}}), which is obtained from LsLsLssLsLss (5L 7s) by stacking one more 10\17 generator.

[[Category:Math]]

@@ Line 250: / Line 250: @@
 === Definition of MOS ===
 A periodic scale is ''MOS'' if it has [[maximum variety]] 2. This in particular implies that MOS scales are binary.
 === MOS scales are generated ===
 See [[Recursive structure of MOS scales#Proofs]].
-=== MOS scales aL bs with gcd(a, b) > 1 reduce to single-period MOS scales on a smaller period ===
-Assume that gcd(m, n) > 1 and assume that (n/d)-steps came in 2 sizes, pa+qb and ra+sb. Then at least one size, say pa+qb, must differ from (a/d)L + (b/d)s. WLOG p > a/d and pa+qb occurs on degree 0. Since equave = aL + bs, on a degree k(n/d) for some integer k > 0, there must be another (n/d)-step with fewer L's than a/d. This involves more than two changes from L to s. Scooting an (n/d)-step one step at a time from degree 0 to k(n/d) changes its size one step size substitution at a time, showing that intermediate (n/d)-steps also exist. This violates the MOS property, whence (n/d)-steps have only one size, which must be the period.
+=== MOS scales ''aL''&nbsp;''bs'' with gcd(a,&nbsp;b) > 1 reduce to single-period MOS scales on a smaller period ===
+Assume that {{nowrap|gcd(''m'', ''n'') &gt; 1}} and assume that ({{frac|''n''|''d''}})-steps came in 2 sizes, {{nowrap|''pa'' + ''qb''}} and {{nowrap|''ra'' + ''sb''}}. Then at least one size, say {{nowrap|''pa'' + ''qb''}}, must differ from {{nowrap|{{frac|''a''|''d''}}&#x200A;L + {{frac|''b''|''d''}}&#x200A;s}}. WLOG {{nowrap|''p'' &gt; {{frac|''a''|''d''}}}} and {{nowrap|''pa'' + ''qb''}} occurs on degree 0. Since the equave is equal to {{nowrap|''aL'' + ''bs''}}, on a degree ''k''({{frac|''n''|''d''}}) for some integer {{nowrap|''k'' &gt; 0}}, there must be another ({{frac|''n''|''d''}})-step with fewer L's than {{frac|''a''|''d''}}. This involves more than two changes from ''L'' to ''s''. Scooting an ({{frac|''n''|''d''}})-step one step at a time from degree 0 to ''k''({{frac|''n''|''d''}}) changes its size one step size substitution at a time, showing that intermediate ({{frac|''n''|''d''}})-steps also exist. This violates the MOS property, whence ({{frac|''n''|''d''}})-steps have only one size, which must be the period.
 === Maximally even scales are MOS ===
-Let 1 < m < n, gcd(m, n) = d, and ME(''m'', ''n'') be as in [[Maximal evenness]]. An arbitrary k-step in ME(''m'', ''n'') has size floor((''i'' + ''k'')''n''/''m'') &minus; floor(''in''/''m''), and
+Let {{nowrap|1 &gt; ''m'' &gt; ''n''|gcd(''m'', ''n'') {{=}} d}}, and ME(''m'',&nbsp;''n'') be as in [[Maximal evenness]]. An arbitrary ''k''-step in ME(''m'',&nbsp;''n'') has size {{nowrap|''u'' {{=}} &lfloor;(''i'' + ''k'')&#x200A;{{frac|''n''|''m''}}&rfloor; &minus; &lfloor;{{frac|''in''|''m''}}&rfloor;}}, and
-<math>\lfloor in/m \rfloor + \lfloor kn/m\rfloor - \lfloor in/m\rfloor = \lfloor kn/m\rfloor ≤ \lfloor(i+k)n/m \rfloor - \lfloor in/m \rfloor
+<math>\lfloor in/m \rfloor + \lfloor kn/m\rfloor - \lfloor in/m\rfloor = \lfloor kn/m\rfloor \leq u \rfloor
-≤ \lfloor in/m \rfloor + \lfloor kn/m \rfloor - \lfloor in/m \rfloor + 1 = \lfloor kn/m \rfloor + 1.</math>
+\leq \lfloor in/m \rfloor + \lfloor kn/m \rfloor - \lfloor in/m \rfloor + 1 = \lfloor kn/m \rfloor + 1.</math>
-As floor((''i'' + ''k'')''n''/''m'') &minus; floor(''in''/''m'') is an integer, the above proves that there are at most two possible values for it and that if there are two sizes for k-steps, the two sizes must differ by 1.
+As ''u'' is an integer, the above proves that there are at most two possible values for it and that if there are two sizes for ''k''-steps, the two sizes must differ by 1.
-Steps of ME(''m'', ''n'') have exactly two sizes because if it were one size, we would have m | (n/d), which is a contradiction.
+Steps of ME(''m'',&nbsp;''n'') have exactly two sizes because if it were one size, we would have {{nowrap|''m'' {{!}} {{frac|''n''|''d''}}}}, which is a contradiction.
-This maximally even MOS has n % m large steps and m &minus; (n % m) small steps.
+This maximally even MOS has {{nowrap|''n'' % ''m''}} large steps and {{nowrap|''m'' &minus; (''n'' % ''m'')}} small steps.
 === Binary generated scales with independent period and generator are MOS ===
 This proof justifies the common description of "stack until binary" for MOS building and Erv Wilson's terminology ''moment of symmetry'' where MOS sizes are "moments" in time (when stacking) where the "symmetry" of binarity and MV2 holds.
-By ''generatedness'', we mean that every interval in the scale is of the form ''jg'' + ''kp'' where ''g'' is a generator, ''p'' is the period, and ''j, k'' ∈ '''Z''', and that either ''g'' or ''&minus;g'' occurs on every note. We claim that any interval class not ''p''-equivalent to 0 has ''exactly'' 2 sizes in any scale satisfying the antecedent.
+By ''generatedness'', we mean that every interval in the scale is of the form {{nowrap|''jg'' + ''kp''}} where ''g'' is a generator, ''p'' is the period, and {{nowrap|''j'', ''k'' ∈ ℤ}}, and that either ''g'' or &minus;''g'' occurs on every note. We claim that any interval class not ''p''-equivalent to 0 has ''exactly'' 2 sizes in any scale satisfying the antecedent.
-Suppose that such a scale ''S'' (with ''n'' ≥ 2 notes) has ''a''-many L steps and ''b''-many s steps per period ''p'', and has generator ''g''. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:
+Suppose that such a scale ''S'' (with {{nowrap|''n'' &ge; 2}} notes) has ''a'' L steps and ''b'' s steps per period ''p'', and has generator ''g''. Since ''S'' is generated, the interval sizes modulo ''p'' that occur in ''S'' are:
-{(&minus;''n'' + 1)''g'', ..., &minus;''g'', 0, ''g'', ..., (''n'' &minus; 1)''g''},
+{{(}}{{nowrap|(&minus;''n'' + 1)''g''}}, ..., &minus;''g'', 0, ''g'', ..., {{nowrap|(''n'' &minus; 1)''g''}}{{)}},
-and all sizes {0, ''g'', ..., (''n'' &minus; 1)''g''} are distinct.
+and all sizes {0, ''g'', ..., {{nowrap|(''n'' &minus; 1)''g''}}{{)}} are distinct.
 We thus have:
-L = ''cg'' + ''dp''
+{{nowrap|L {{=}} ''cg'' + ''dp''}}
+{{nowrap|s {{=}} ''eg'' + ''fp''}}
-s = ''eg'' + ''fp''
+for appropriate integers ''c, d, e, f'', where {{nowrap|{{!}}''c''{{!}}, {{!}}''e''{{!}} &lt; ''n''}}.
-for appropriate integers ''c, d, e, f'', where |''c''| < ''n'' and |''e''| < ''n''.
+Now we assume that ''g'' and ''p'' are linearly independent. By assumption {{nowrap|''a''L + ''b''s {{=}} (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p''}} =&nbsp;''p''. Since {{nowrap|''a''L + ''b''s}} occurs on the "brightest" mode, from generatedness we have {{nowrap|''ac'' + ''be'' ∈ {{(}}0, ..., ''n'' &minus; 1{{)}}}}. Hence we must have {{nowrap|''ac'' + ''be'' {{=}} 0}}, and thus {{nowrap|''c'' {{=}} &#177;''b''}} and {{nowrap|''e'' {{=}} &#x2213;''a''}}, from the assumption that ''a'' and ''b'' are coprime.
-Now we assume that ''g'' and ''p'' are linearly independent. By assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. Since ''a''L + ''b''s occurs on the "brightest" mode, from generatedness we have ''ac'' + ''be'' ∈ {0, ..., ''n'' &minus; 1}. Hence we must have ''ac'' + ''be'' = 0, and thus ''c'' = ±''b'' and ''e'' = ∓''a'', from the assumption that gcd(a, b) = 1.
+In fact, {L,&nbsp;s} is another valid basis for the abelian group with basis {''p'',&nbsp;''g''}, since by binarity we have {{nowrap|''p, g'' ∈ span(L, s)}}. Assume {{nowrap|''c'' {{=}} ''b''}} and {{nowrap|''e'' {{=}} &minus;''a''}} (this corresponds to assuming that ''g'' is the "bright" generator). Let {{nowrap|&chi; {{=}} L &minus; s}} &gt;&nbsp;0; then &chi; is ''p''-equivalent to +''ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each {{nowrap|''j'' ∈ {{(}}1, ..., ''n'' &minus; 1{{)}}}} there exists at most one {{nowrap|''k'' {{=}} ''k''(''j'') ∈ {{(}}1, ..., ''n'' &minus; 1{{)}}}} such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if the class of ''k''-steps has ''at least'' two sizes, the sizes must be ''j''(''k'')''g'' and {{nowrap|(''j''(''k'') &minus; ''n'')''g''}}; any other size must leave the range {{nowrap|(1 &minus; ''n'')''g''}}, ..., 0, ..., {{nowrap|(''n'' &minus; 1)''g''}}. Thus the class of ''k''-steps has at most two sizes for {{nowrap|1 &le; ''k'' &le; ''n'' &minus; 1}}. Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an {{nowrap|(''n'' &minus; ''k'')}}-step equivalent to &minus;''jg'', which by linear independence must be distinct from an {{nowrap|(''n'' &minus; ''k'')}}-step equivalent to a positive number of ''g'' generators. (Note that the latter {{nowrap|(''n'' &minus; ''k'')}}-step does occur in the "brightest" mode of ''S'', i.e. the mode with the most ''g'' generators stacked ''up'' rather than ''down'' from the tonic.) This completes the proof.
-In fact, {L, s} is another valid basis for the abelian group with basis {''p'', ''g''}, since by binarity we have ''p, g'' ∈ span(L, s). Assume ''c'' = ''b'' and ''e'' = &minus;''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] Let χ = L &minus; s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' &minus; 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' &minus; 1}</sub> such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if the class of ''k''-steps has ''at least'' two sizes, the sizes must be ''j''(''k'')''g'' and (''j''(''k'') &minus; ''n'')''g''; any other size must leave the range &minus;(''n'' &minus; 1)''g'', ..., 0, ..., (''n'' &minus; 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ≤ (''n'' &minus; 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' &minus; ''k'')-step equivalent to ''&minus;jg'', which by linear independence must be distinct from an (''n'' &minus; ''k'')-step equivalent to a positive number of ''g'' generators. (Note that the latter (''n'' &minus; ''k'')-step does occur in the "brightest" mode of ''S'', i.e. the mode with the most ''g'' generators stacked ''up'' rather than ''down'' from the tonic.) This completes the proof.
-The previous argument cannot be used for the non-linearly independent case. This is because not all binary generated scales with rational step ratios are limit points of binary generated scales with irrational step ratios. A counterexample is the 13-note scale ssLsLssLsLss (s = 1\17, L = 2\17), which is obtained from LsLsLssLsLss (5L7s) by stacking one more 10\17 generator.
+The previous argument cannot be used for the non-linearly independent case. This is because not all binary generated scales with rational step ratios are limit points of binary generated scales with irrational step ratios. A counterexample is the 13-note scale ssLsLssLsLss ({{nowrap|s {{=}} 1\17|L {{=}} 2\17}}), which is obtained from LsLsLssLsLss (5L&nbsp;7s) by stacking one more 10\17 generator.
 [[Category:Math]]