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CounterexampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09
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Distinct unit summands need not attain the sum bound

Statement refuted

The assertion “for all n>0 and all unit complex numbers z1,,zn, izi=n” is false.

Facts & Assumptions

Given: The universal equality assertion for unit complex summands.

[F1]

The upper bound is attained exactly when all unit summands agree (Equality in the unit-complex finite-sum bound).

Counterexample

technique · direct
1.1

Take n=2, z1=1 and z2=1 in C. Their moduli are 12+02=1 and (1)2+02=1, so both satisfy the unit-modulus hypothesis and n>0.

givenalgebra
2.1

Their sum is 1+(1)=0, whose modulus is zero. Consequently z1+z2=0<2=n, so this instance fails the asserted equality and refutes its universal quantifier. The summands are distinct, in agreement with F1.

step 1.1F1algebra

Sources

Etingof et al., Lemma 5.4.5 proof, p. 101, motivates the strictness interface. The pair 1,1 is the locally specified counterexample.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources