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Claim. For all positive integers $p,q$ with $q>p^2$ the rational $\alpha=1+\frac pq$ is a Turán exponent (Theorem 1.3), a corollary of Theorem 1.2, which gives $1+\frac{p}{kp+b}$ for all positive integers $p,k,b$ with $k\ge b$. The engine is Theorem 1.10, an upper bound $\mathrm{ex}(n,t\cdot S^s_{b,k})=O(n^{1+(s-1)/((s-1)k+b)})$ for subdivisions of $K_{s,t}$ whose edges are subdivided unequally, matched by the Bukh--Conlon lower bound; Corollaries 1.4 and 1.5 transfer the exponents to the form $2-\frac pq$ for all positive integers $q>p$ with $q\bmod p\le\sqrt p$. The statements are recorded on the library's source card.

Covers. The instances $\alpha=1+\frac pq$ with $q>p^2$ (and the finer family $1+\frac{p}{kp+b}$, $k\ge b$), each realized by a single bipartite graph. The statement for every rational $\alpha\in[1,2)$ is settled by the accepted claim page Adamczewski 2026.

Depends on. Nothing in this wiki; the result is the paper's own theorem.

Acceptance. Refereed: T. Jiang and Y. Qiu, Many Turán exponents via subdivisions, Combin. Probab. Comput. 32 (2023), no. 1, 134--150, doi:10.1017/S0963548322000177, a refereed journal. First posting: arXiv:1908.02385, v1 6 August 2019 (the date this page is named by). No reviewed evidence is listed: the site's commentary lists these exponents among the Turán exponents known before 2026 and credits the paper, but its label credits GPT-6 Astra with the full proof and is not an acceptance of this result.

Read depth. The statements are taken from the paper's abstract and the result list on the library card; no proof was read, and nothing is independently reviewed in this corpus.