Alphabeta Math
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The ARS(2,2,2) coefficients satisfy the coupled second-order identities

Statement

For γ=1−1/2 and δ=−1/2 in The ARS(2,2,2) additive tableau and its three stages, one has γ>0 and δ=1−12γ,γ(1−δ)=12,γ(2−γ)=12,γ(1−γ−δ)=12−γ2.

Facts & Assumptions

Given: The coefficient definitions in the statement.

[F1]

The ARS(2,2,2) additive tableau and its three stages: γ=1−1/2 and δ=−1/2.

[F3]

Squaring is monotone on the nonnegatives: squaring is strictly increasing on nonnegative reals.

[F4]

Inverses of positives are positive, and reciprocation reverses order: if 0<a<b, then 0<b−1<a−1.

Proof

1.1F2F3F4algebra

Let r=2. If r≤1, then 0≤r≤1 and [F3] gives r2≤1, contradicting r2=2 from [F2]; hence r>1. Applying [F4] to 0<1<r gives 0<1/r<1, so γ=1−1/r>0.

2.1F1F2step 1.1algebra

By [F1] and γ=(r−1)/r from step 1.1, 1−1/(2γ)=1−r/(2(r−1))=(r−2)/(2(r−1)). Since r2=2, r(r−2)=2−2r=−2(r−1); dividing by the nonzero 2r(r−1) gives (r−2)/(2(r−1))=−1/r=δ.

2.2F1F2step 1.1algebra

Using δ=−1/r and γ=(r−1)/r, γ(1−δ)=((r−1)/r)(1+1/r)=(r2−1)/r2=1/2, and γ(2−γ)=1−(1−γ)2=1−1/r2=1/2.

3.1step 2.1step 2.2algebra∎

The identity in step 2.1 gives γ(1−γ−δ)=γ(1/(2γ)−γ)=1/2−γ2. This proves all coefficient identities.

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