Ternary scale theorems: Difference between revisions
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Suppose Q = (α, β, γ) ≠ R = (α, β′, γ′) are the two ''k''-steps in ''S'' that project to P. Then T = (α′, β′′, γ′′) projects to I. Here the values in each component differ by at most 1, and α ≠ α′. Then the circular word Λ<sub>1</sub> formed by the a-components of the ''k''-steps in P is α...αα′. Since Σ<sub>2</sub> is a primitive mos pattern of βb + (''n'' − β)(a~c) and β′a + (''n'' − β′)(a~c), the circular word Λ<sub>2</sub> = the pattern of β and β′ must be a primitive mos. Similarly, Λ<sub>3</sub> = the pattern of γ and γ′ is a primitive mos. | Suppose Q = (α, β, γ) ≠ R = (α, β′, γ′) are the two ''k''-steps in ''S'' that project to P. Then T = (α′, β′′, γ′′) projects to I. Here the values in each component differ by at most 1, and α ≠ α′. Then the circular word Λ<sub>1</sub> formed by the a-components of the ''k''-steps in P is α...αα′. Since Σ<sub>2</sub> is a primitive mos pattern of βb + (''n'' − β)(a~c) and β′a + (''n'' − β′)(a~c), the circular word Λ<sub>2</sub> = the pattern of β and β′ must be a primitive mos. Similarly, Λ<sub>3</sub> = the pattern of γ and γ′ is a primitive mos. | ||
Suppose Λ<sub>2</sub> is the mos λβ μβ′. Then Λ<sub>3</sub> is the mos (λ ± 1)γ (μ ∓ 1)γ′. Since | Suppose Λ<sub>2</sub> is the mos λβ μβ′. Then Λ<sub>3</sub> is the mos (λ ± 1)γ (μ ∓ 1)γ′. Since both Λ<sub>2</sub> and Λ<sub>3</sub> are primitive, and at least one of μ and (μ ∓ 1) are even, it is now immediate that ''n'' is odd. | ||
Either β′′ = β or β′′ = β′. Assume β′′ = β′. Then γ′′ = γ, and Λ<sub>3</sub> is (λ + 1)γ (μ − 1)γ′. Also assume that the first ''k''-step in Σ is Q. Then we have: | Either β′′ = β or β′′ = β′. Assume β′′ = β′. Then γ′′ = γ, and Λ<sub>3</sub> is (λ + 1)γ (μ − 1)γ′. Also assume that the first ''k''-step in Σ is Q. Then we have: | ||