Construction pipeline
Textbook selection, hierarchical chapter recovery, Proof QA and Gap QA extraction, followed by five quality-control checks.
Scaling Mathematical Proof Supervisionfrom Structured Textbooks
1 University of Science and Technology of China2 Fudan University 3 Shanghai AI Laboratory4 Shanghai Jiao Tong University
* Equal contribution † Corresponding authors
(coming soon)
Dataset (coming soon)
Brief Introduction
Mathematical supervision for language models is typically flattened into isolated problem–solution pairs, discarding the curricular organization and evidential structure of the original sources. We introduce TOME-Math (Textbook-Oriented Mathematical Evidence), a large-scale corpus that converts 2,118 textbooks spanning the core areas of university-level mathematics into approximately 380K structured proof-supervision records while preserving chapter order, proposition–proof links, dependency references, and source provenance.
Leveraging these structural signals, we explore curriculum learning across multiple mathematical subjects and characterize scaling patterns with supervision volume and source breadth. At full scale, fine-tuning on TOME-Math raises the unweighted mean score across our held-out evaluations and established public benchmarks by 16.7 percentage points over the base model, while exceeding all three public instruction-data baselines evaluated in aggregate. Our findings highlight structured textbooks as both a scalable source of mathematical supervision and an empirical framework for studying the impact of data organization on mathematical reasoning.
Overview of TOME-Math. Traditional pipelines flatten heterogeneous sources into isolated QA, while TOME-Math preserves textbook structure through structured QA construction and curriculum-based supervision.
380,292QA pairs
2,118textbooks
66,537chapters
29target math subjects
Construction and characterization.
How the corpus is built, filtered, and evaluated for quality.
Composition and quality
The construction funnel shows how generated candidates become the final corpus; the radar plot compares model configurations across quality criteria.
Dataset composition
Proof QA captures complete proposition–proof units. Gap QA turns missing local proof obligations into explicit training targets.
QA composition
380Ktotal
Proof QA205,86154.1%
Gap QA174,43145.9%
Corpus hierarchy
Subjects29
Books2,118
Chapters66,537
QA records380,292
Subject distribution
Largest fields in the release candidate
Top 10 target subjects.
Representative supervision pairs
Each textbook proposition yields a complete Proof QA and focused Gap QA that make its intermediate proof obligations explicit.
Differential Geometry
Triangle inequality in the hyperbolic plane
Geometry: From Isometries to Special Relativity01 Original textbook excerpt Chapter 4.2 · \(\mathbb H^2\) shortest pathsRead the source theorem and textbook proof
Parsed chapter Markdown Chapter 4.2 · \(\mathbb H^2\) shortest paths
According to Theorem 4.9, we can formulate a triangle inequality for the hyperbolic plane.
Corollary 4.11 (Triangle inequality for the hyperbolic plane). For three points \(p_{1}, p_{2}\) , and \(p_{3}\) in the hyperbolic plane,\[d_{\mathbb{H}^{2}}(p_{1}, p_{3}) \leq d_{\mathbb{H}^{2}}(p_{1}, p_{2}) + d_{\mathbb{H}^{2}}(p_{2}, p_{3}),\]where the equality holds if and only if \(p_{2}\) lies on the \(\mathbb{H}^{2}\) - shortest path from \(p_{1}\) to \(p_{3}\) .
Proof. Let \(\gamma_{ij}\) be the \(\mathbb{H}^{2}\) - shortest path from \(p_{i}\) to \(p_{j}\) for \(1 \leq i < j \leq 3\) . Then,\[d_{\mathbb{H}^{2}}(p_{i}, p_{j}) = l_{\mathbb{H}^{2}}(\gamma_{ij}).\]Note that the union (denoted by \(\delta\) ) of \(\gamma_{12}\) and \(\gamma_{23}\) is a path from \(p_{1}\) to \(p_{3}\) (Figure 4.6). According to Theorem 4.9,\[d_{\mathbb{H}^{2}}(p_{1}, p_{2}) + d_{\mathbb{H}^{2}}(p_{2}, p_{3}) = l_{\mathbb{H}^{2}}(\delta) \geq l_{\mathbb{H}^{2}}(\gamma_{13}) = d_{\mathbb{H}^{2}}(p_{1}, p_{3}),\]and the equality holds if and only if \(\delta\) has the same path as \(\gamma_{13}\) , which means that point \(p_{2}\) lies on path \(\gamma_{13}\) . \(\square\)
02 Proof QA Recover the complete textbook proof.Open the generated question
Question
Let \(p_1,p_2,p_3\) be points in the hyperbolic plane \(\mathbb H^2\). Prove the triangle inequality\[d_{\mathbb H^2}(p_1,p_3)\le d_{\mathbb H^2}(p_1,p_2)+d_{\mathbb H^2}(p_2,p_3),\]and prove that equality holds if and only if \(p_2\) lies on the unique hyperbolic shortest path from \(p_1\) to \(p_3\).
Answer
Let \(\gamma_{ij}\) be the hyperbolic shortest path from \(p_i\) to \(p_j\), and set \(\delta=\gamma_{12}*\gamma_{23}\). Then\[l_{\mathbb H^2}(\delta)=l_{\mathbb H^2}(\gamma_{12})+l_{\mathbb H^2}(\gamma_{23})=d_{\mathbb H^2}(p_1,p_2)+d_{\mathbb H^2}(p_2,p_3).\]Since \(\gamma_{13}\) is a shortest path from \(p_1\) to \(p_3\),\[d_{\mathbb H^2}(p_1,p_3)=l_{\mathbb H^2}(\gamma_{13})\le l_{\mathbb H^2}(\delta),\]which proves the triangle inequality.
If equality holds, then \(\delta\) is also a shortest path from \(p_1\) to \(p_3\). Uniqueness implies that \(\delta\) and \(\gamma_{13}\) have the same trajectory, so \(p_2\in\gamma_{13}\). Conversely, if \(p_2\) lies on \(\gamma_{13}\), splitting \(\gamma_{13}\) at \(p_2\) gives two shortest subpaths whose lengths add, and equality follows.
03 Gap QA Make one intermediate proof obligation explicit.Open the generated question
Question
Let \(p_1,p_2,p_3\in\mathbb H^2\). For \(1\le i<j\le3\), let \(\gamma_{ij}\) denote the unique hyperbolic shortest path from \(p_i\) to \(p_j\), and let \(\delta=\gamma_{12}*\gamma_{23}\) be their concatenation through \(p_2\). Assume that\[d_{\mathbb H^2}(p_1,p_3)=d_{\mathbb H^2}(p_1,p_2)+d_{\mathbb H^2}(p_2,p_3).\]Prove that \(\delta\) has the same trajectory as \(\gamma_{13}\), and hence that \(p_2\) lies on the hyperbolic shortest path from \(p_1\) to \(p_3\).
Answer
By construction,\[l_{\mathbb H^2}(\delta)=d_{\mathbb H^2}(p_1,p_2)+d_{\mathbb H^2}(p_2,p_3).\]The assumed equality therefore gives \(l_{\mathbb H^2}(\delta)=d_{\mathbb H^2}(p_1,p_3)\). Thus \(\delta\) attains the minimum possible length among all paths from \(p_1\) to \(p_3\), so it is itself a hyperbolic shortest path.
The shortest path between two points of \(\mathbb H^2\) is unique up to reparametrization. Hence \(\delta\) and \(\gamma_{13}\) have the same trajectory. Since \(\delta\) passes through \(p_2\) by construction, \(p_2\) lies on \(\gamma_{13}\).
Discrete Mathematics
Spectrum of a strongly regular graph
Elementary Methods of Graph Ramsey Theory01 Original textbook excerpt Chapter 2.6 · Spectrum and independence numberRead the source theorem and textbook proof
Parsed chapter Markdown Chapter 2.6 · Spectrum and independence number
The spectrum of a strongly regular graph is determined completely by its four parameters as follows. As \(\lambda\) is used often to signify an eigenvalue, we change the notation \(s r g(n,d,\lambda ,\mu)\) to \(s r g(n,d,\mu_{1},\mu_{2})\) to avoid interference.
Theorem 2.15 Let \(G\) be a connected strongly regular graph \(s r g(n,d,\mu_{1},\mu_{2})\) with \(n\geq 3\) . If \(G\) is neither complete nor empty, then \(\lambda_{1} = d\) is an eigenvalue with multiplicity \(m_{1} = 1\) , and any other eigenvalue \(\lambda (\neq \lambda_{1})\) satisfies\[\lambda^{2} + (\mu_{2} - \mu_{1})\lambda +(\mu_{2} - d) = 0.\]The equation has two distinct solutions \(\lambda_{2}\) and \(\lambda_{3}\) with \(\lambda_{2} > 0 > \lambda_{3}\) , and \(\lambda_{3}\) is an eigenvalue. If \(d + (n - 1)\lambda_{3}\neq 0\) , then \(\lambda_{2}\) is also an eigenvalue. Their multiplicities \(m_{2}\) and \(m_{3}\) can be determined by\[m_{2} + m_{3} = n - 1,\mathrm{and}\ d + m_{2}\lambda_{2} + m_{3}\lambda_{3} = 0.\]Proof. Let \(A\) be the adjacency matrix of \(G\) . Let \(I\) and \(J\) be the \(n\times n\) identity matrix and all-one matrix, respectively. By the definition of \(A\) and the fact that \(A\) is symmetric with zeros on the main diagonal, the \((i,i)\)-entry of \(A^{2}\) is \(d(\nu_{i}) = d\) , which can be represented by \(dI\) . For \(i\neq j\) , the \((i,j)\)-entry of \(A^{2}\) counts common neighbors of vertices \(\nu_{i}\) and \(\nu_{j}\) , so it is \(\mu_{1}\) or \(\mu_{2}\) , which can be represented by \(\mu_{1}A\) or \(\mu_{2}(J - I - A)\) , respectively. Also the regularity of \(G\) can be represented by \(AJ = dJ\) . So \(G\) is an \(s r g(n,d,\mu_{1},\mu_{2})\) is equivalent to\[AJ = dJ,\qquad A^{2} = (d - \mu_{2})I + (\mu_{1} - \mu_{2})A + \mu_{2}J.\]From Perron-Frobenius Theorem, we have that \(\lambda_{1} = d\) is an eigenvalue of \(G\) with multiplicity 1 with eigenvector \(\mathbf{1} = (1,1,\ldots ,1)^{T}\) . Let \(\lambda \neq d\) be another eigenvalue of \(G\) and \(x\in \mathbb R^{n}\) a corresponding eigenvector. Thus \(Jx = 0\) by noting \(\mathbf{1}^{T}x = 0\) since eigenvectors corresponding to distinct eigenvalues of real symmetric matrix are orthogonal. As \(Ax = \lambda x\) and \(A^{2}x = \lambda^{2}x\) , we obtain that\[\lambda^{2}x = (d - \mu_{2})x + (\mu_{1} - \mu_{2})\lambda x,\]which implies that \(\lambda\) must satisfy\[\lambda^{2} + (\mu_{2} - \mu_{1})\lambda +(\mu_{2} - d) = 0. \quad (2.4)\]The equation has two distinct solutions \(\lambda_{2}\) and \(\lambda_{3}\) (\(\lambda_{2} > \lambda_{3}\)). Since the smallest eigenvalue of \(G\) is negative, it follows that \(\lambda_{3}\) is an eigenvalue. Thus, if \(d + (n - 1)\lambda_{3}\neq 0\) , then the multiplicity of \(\lambda_{3}\) cannot be \(n - 1\) , and \(G\) has another eigenvalue, which must be \(\lambda_{2}\) . Also, their multiplicities \(m_{2}\) and \(m_{3}\) satisfy\[1 + m_{2} + m_{3} = n,\qquad \mathrm{Tr}(A) = d + m_{2}\lambda_{2} + m_{3}\lambda_{3} = 0,\]which determine \(m_{2}\) and \(m_{3}\) completely.
02 Proof QA Recover the complete textbook proof.Open the generated question
Question
Let \(G\) be a connected strongly regular graph with parameters \((n,d,\mu_1,\mu_2)\), where \(n\ge3\), and assume that \(G\) is neither complete nor empty. Prove that \(d\) is an eigenvalue of multiplicity one and that every other eigenvalue satisfies\[\lambda^2+(\mu_2-\mu_1)\lambda+(\mu_2-d)=0.\]If the roots are \(\lambda_2>\lambda_3\), prove that \(\lambda_2>0>\lambda_3\) and that \(\lambda_3\) is an eigenvalue. Show that \(\lambda_2\) is also an eigenvalue whenever \(d+(n-1)\lambda_3\ne0\), and determine the equations satisfied by their multiplicities.
Answer
Let \(A\) be the adjacency matrix of \(G\). Strong regularity gives\[AJ=dJ,\qquad A^2=(d-\mu_2)I+(\mu_1-\mu_2)A+\mu_2J.\]Because \(G\) is connected and \(d\)-regular, the Perron–Frobenius theorem shows that \(d\) is an eigenvalue of multiplicity one with eigenvector \(\mathbf1\).
Let \(Ax=\lambda x\) with \(\lambda\ne d\). Since \(A\) is real symmetric, \(x\perp\mathbf1\), and hence \(Jx=0\). Applying the matrix identity to \(x\) yields\[\lambda^2x=(d-\mu_2)x+(\mu_1-\mu_2)\lambda x,\]so \(\lambda\) satisfies the displayed quadratic equation. The equation has two distinct real roots, with the larger root positive. Since \(\operatorname{tr}(A)=0\) while \(d>0\), a negative eigenvalue must occur; therefore \(\lambda_3<0\) and \(\lambda_3\) is an eigenvalue.
If \(d+(n-1)\lambda_3\ne0\), the multiplicity of \(\lambda_3\) cannot be \(n-1\), because otherwise the trace would equal that nonzero quantity. Thus the other root \(\lambda_2\) must also occur. If their multiplicities are \(m_2\) and \(m_3\), dimension and trace give\[m_2+m_3=n-1,\qquad d+m_2\lambda_2+m_3\lambda_3=0.\]These two equations determine \(m_2\) and \(m_3\).
03 Gap QA Make one intermediate proof obligation explicit.Open the generated question
Question
Let \(G\) be a connected strongly regular graph with parameters \((n,d,\mu_1,\mu_2)\), and assume that \(G\) is neither complete nor empty. Every nonprincipal eigenvalue of its adjacency matrix is one of the two distinct real roots \(\lambda_2>\lambda_3\) of\[\lambda^2+(\mu_2-\mu_1)\lambda+(\mu_2-d)=0.\]Prove that \(\lambda_3<0\) and that \(\lambda_3\) must occur as an eigenvalue of \(G\).
Answer
The adjacency matrix \(A\) has zero diagonal, so \(\operatorname{tr}(A)=0\) and the sum of its eigenvalues, counted with multiplicity, is zero. Its principal eigenvalue is \(d>0\). If every other eigenvalue were nonnegative, their sum together with \(d\) would be positive, a contradiction. Hence a negative nonprincipal eigenvalue exists. Since every nonprincipal eigenvalue is either \(\lambda_2\) or \(\lambda_3\) and \(\lambda_2>\lambda_3\), it follows that \(\lambda_3<0\).
Suppose that \(\lambda_3\) did not occur. Then \(A\) would act as \(\lambda_2 I\) on \(\mathbf1^\perp\) and as multiplication by \(d\) on \(\operatorname{span}\{\mathbf1\}\). Consequently,\[A=\lambda_2 I+\frac{d-\lambda_2}{n}J.\]All off-diagonal entries of \(A\) would therefore be equal. Because an adjacency matrix has off-diagonal entries only in \(\{0,1\}\), the graph would have to be empty or complete, contradicting the hypothesis. Thus \(\lambda_3\) is an eigenvalue of \(G\).
Probability
Moments of the beta distribution
Probability and Statistics, 4th ed.01 Original textbook excerpt Chapter 5.8 · The beta distributionsRead the source theorem and textbook proof
Parsed chapter Markdown Chapter 5.8 · The beta distributions
Moments of Beta Distributions
Theorem 5.8.3
Moments. Suppose that \(X\) has the beta distribution with parameters \(\alpha\) and \(\beta\) . Then for each positive integer \(k\) ,\[E(X^{k}) = \frac{\alpha(\alpha + 1)\cdot\cdot\cdot(\alpha + k - 1)}{(\alpha + \beta)(\alpha + \beta + 1)\cdot\cdot\cdot(\alpha + \beta + k - 1)}. \quad (5.8.6)\]In particular,\[E(X) = \frac{\alpha}{\alpha + \beta},\]\[\mathrm{Var}(X) = \frac{\alpha\beta}{(\alpha + \beta)^{2}(\alpha + \beta + 1)}.\]
Proof For \(k = 1,2,\ldots\)\[E(X^{k}) = \int_{0}^{1}x^{k}f(x|\alpha ,\beta)dx\]\[= \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}x^{\alpha +k - 1}(1 - x)^{\beta -1}dx.\]
Therefore, by Eq. (5.8.2),\[E(X^{k}) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\cdot \frac{\Gamma(\alpha + k)\Gamma(\beta)}{\Gamma(\alpha + k + \beta)},\]which simplifies to Eq. (5.8.6). The special case of the mean is simple, while the variance follows easily from
\[E(X^{2}) = \frac{\alpha(\alpha + 1)}{(\alpha + \beta)(\alpha + \beta + 1)}.\]
02 Proof QA Recover the complete textbook proof.Open the generated question
Question
Let \(X\sim\operatorname{Beta}(\alpha,\beta)\) with density\[f(x)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1},\qquad 0<x<1.\]Prove that for every positive integer \(k\),\[E(X^k)=\frac{\alpha(\alpha+1)\cdots(\alpha+k-1)}{(\alpha+\beta)(\alpha+\beta+1)\cdots(\alpha+\beta+k-1)},\]and use this formula to compute \(E(X)\) and \(\operatorname{Var}(X)\).
Answer
Using the density and the beta integral,\[E(X^k)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_0^1x^{\alpha+k-1}(1-x)^{\beta-1}\,dx\]\[=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\frac{\Gamma(\alpha+k)\Gamma(\beta)}{\Gamma(\alpha+\beta+k)}.\]
The identity \(\Gamma(z+k)=z(z+1)\cdots(z+k-1)\Gamma(z)\) yields the stated moment formula. Taking \(k=1\) and \(k=2\) gives\[E(X)=\frac{\alpha}{\alpha+\beta},\qquad E(X^2)=\frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)}.\]Finally,
\[\operatorname{Var}(X)=E(X^2)-[E(X)]^2=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}.\]
03 Gap QA Make one intermediate proof obligation explicit.Open the generated question
Question
Let \(X\sim\operatorname{Beta}(\alpha,\beta)\), where \(\alpha,\beta>0\). Suppose that\[E(X)=\frac{\alpha}{\alpha+\beta},\qquad E(X^2)=\frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)}.\]Use \(\operatorname{Var}(X)=E(X^2)-[E(X)]^2\) to derive the variance of \(X\).
Answer
Substitution and a common denominator give\[\operatorname{Var}(X)=\frac{\alpha(\alpha+1)(\alpha+\beta)-\alpha^2(\alpha+\beta+1)}{(\alpha+\beta)^2(\alpha+\beta+1)}.\]The numerator simplifies as\[\alpha(\alpha+1)(\alpha+\beta)-\alpha^2(\alpha+\beta+1)=\alpha\beta.\]Therefore,
\[\operatorname{Var}(X)=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}.\]
Abstract Algebra
Orbit–stabilizer correspondence
Algebra, 2nd ed. · Michael Artin01 Original textbook excerpt Chapter 6.8 · Action on cosetsRead the source theorem and textbook proof
Parsed chapter Markdown Chapter 6.8 · Action on cosets
下面的命题(有时叫轨道一稳定子定理)表明如何通过在陪集上的作用来刻画任意群作用.
【6.8.4】命题令 \(S\) 是群 \(G\) 作用的集合,且 \(s\) 是 \(S\) 的元素.设 \(H\) 与 \(O_{i}\) 分别是 \(s\) 的稳定子与轨道,则存在一个由 \([a H] \sim a s\) 定义的双射 \(\epsilon : G / H \to O_{i}\) .这个映射与群的作用相容: \(\epsilon (g[C]) = g \epsilon ([C])\) 对每一个陪集 \(C\) 与 \(G\) 的每个元素 \(g\) 成立.
例如,二面体群 \(D_{5}\) 作用于正五边形顶点.令 \(\nu\) 表示顶点集合,令 \(H\) 是一个特殊顶点的稳定子.存在双射 \(D_{5} / H \to \nu\) .在平面 \(P\) 的等距群 \(M\) 作用下,一个点的轨道是 \(P\) 的所有点的集合,原点的稳定子是正交算子群 \(O_{2}\) ,且存在一个双射 \(M / O_{2} \to P\) .类似地,如果 \(H\) 表示一条直线的稳定子,且 \(C\) 表示平面上所有点的集合,则存在双射 \(M / H \to C\) .
命题6.8.4的证明显然,命题叙述中定义的映射 \(\epsilon\) 与群的作用相容,如果它存在的话,在符号上, \(\epsilon\) 简单地用符号 \(s\) 替换了 \(H\) .究竟法则 \([g H] \sim g s\) 能否定义一个映射并不清楚,因为许多符号 \(g H\) 代表的是同一个陪集,所以我们必须证明如果 \(a\) 和 \(b\) 是群元素且如果陪集 \(a H = b H\) ,则也有 \(a s = b s\) .假设 \(a H = b H\) .则 \(a^{- 1} b\) 属于 \(H(2.8.5)\) .因为 \(H\) 是 \(s\) 的稳定子,故 \(a^{- 1} b s = s\),所以 \(a s = b s\) .我们的定义是合理的,且由反向推理知, \(\epsilon\) 是单射.由于 \(\epsilon\) 映 \([g H]\) 为 \(g s\) 且 \(g s\) 可以为 \(O_{i}\) 的任意元素,所以 \(\epsilon\) 是满射也是单射.
02 Proof QA Recover the complete textbook proof.Open the generated question
Question
设群 \(G\) 作用在集合 \(S\) 上，固定 \(s\in S\)，并令 \(H=\{g\in G:gs=s\}\) 为 \(s\) 的稳定子，\(O_s=\{gs:g\in G\}\) 为 \(s\) 的轨道。证明\[\epsilon:G/H\longrightarrow O_s,\qquad \epsilon(aH)=as\]是一个良定义的双射，并证明它与 \(G\) 的作用相容，即 \(\epsilon(g(aH))=g\epsilon(aH)\)。
Answer
首先证明良定义性。若 \(aH=bH\)，则 \(a^{-1}b\in H\)，所以 \((a^{-1}b)s=s\)，从而 \(as=bs\)。因此 \(\epsilon(aH)\) 与陪集代表元的选取无关。
若 \(\epsilon(aH)=\epsilon(bH)\)，则 \(as=bs\)，故 \((a^{-1}b)s=s\)，即 \(a^{-1}b\in H\)，于是 \(aH=bH\)。所以 \(\epsilon\) 是单射。
任取 \(x\in O_s\)。由轨道的定义，存在 \(g\in G\) 使 \(x=gs\)，而 \(x=\epsilon(gH)\)，所以 \(\epsilon\) 是满射。因此 \(\epsilon\) 是双射。
最后，对任意 \(g,a\in G\)，\[\epsilon(g(aH))=\epsilon((ga)H)=(ga)s=g(as)=g\epsilon(aH),\]故该双射与群作用相容。
03 Gap QA Make one intermediate proof obligation explicit.Open the generated question
Question
设群 \(G\) 作用在集合 \(S\) 上，固定 \(s\in S\)。令\[H=\{g\in G:gs=s\},\qquad O_s=\{gs:g\in G\}.\]定义映射 \(\epsilon:G/H\to O_s\) 为 \(\epsilon(aH)=as\)。假设该映射已经证明为良定义。证明 \(\epsilon\) 是单射；也就是说，对任意 \(a,b\in G\)，若 \(\epsilon(aH)=\epsilon(bH)\)，则 \(aH=bH\)。
Answer
设 \(\epsilon(aH)=\epsilon(bH)\)，则 \(as=bs\)。在等式两边施加 \(a^{-1}\) 的作用，得到\[s=(a^{-1}b)s.\]由稳定子 \(H\) 的定义，\(a^{-1}b\in H\)。而左陪集相等的判定条件给出\[a^{-1}b\in H\quad\Longleftrightarrow\quad aH=bH.\]因此 \(\epsilon(aH)=\epsilon(bH)\) 蕴含 \(aH=bH\)，故 \(\epsilon\) 是单射。
Mathematical Analysis
Taylor’s theorem with Peano remainder
Mathematical Analysis Lecture Notes, 6th ed.01 Original textbook excerpt Chapter 6.11 · Taylor formulaRead the source theorem and textbook proof
Parsed chapter Markdown Chapter 6.11 · Taylor formula
下面讨论用泰勒多项式 \(T_{n}(x)\) 近似代替函数 \(f(x)\) , 其泰勒余项 \(R_{n}(x)\) 的性质 (定性的和定量的). 这是本节讨论的主要问题. 关于 \(R_{n}(x)\) 的定性的性质有下面的定理:
定理1(泰勒定理) 若函数 \(f(x)\) 在 \(a\) 存在 \(n\) 阶导数 \(①\) , 则 \(\forall x \in U(a)\) , 有\[f(x) = T_{n}(x) + o\left[\left(x - a\right)^{n}\right], \quad (1)\]其中\[T_{n}(x) = f(a) + \frac{f^{\prime}(a)}{1!} (x - a) + \frac{f^{\prime \prime}(a)}{2!} (x - a)^{2} + \dots +\frac{f^{(n)}(a)}{n!} (x - a)^{n};\]\(R_{n}(x) = o\left[\left(x - a\right)^{n}\right](x \to a)\) , 即 \(R_{n}(x)\) 是 \((x - a)^{n}\) 的高阶无穷小.
(1)式称为函数 \(f(x)\) 在 \(a\) (展开)的泰勒公式.
证法 由高阶无穷小的定义, 只需证明\[\lim_{x \to a} \frac{R_{n}(x)}{(x - a)^{n}} = \lim_{x \to a} \frac{f(x) - T_{n}(x)}{(x - a)^{n}} = 0.\]这是 \(\frac{0}{0}\) 的待定型, 应用 \(n - 1\) 次洛必达法则.
证明 \(R_{n}(x) = f(x) - T_{n}(x)\)\[= f(x) - \left[f(a) + \frac{f^{\prime}(a)}{1!} (x - a) + \dots +\frac{f^{(n)}(a)}{n!} (x - a)^{n}\right],\]\[R_{n}^{\prime}(x) = f^{\prime}(x) - \left[f^{\prime}(a) + \frac{f^{\prime \prime}(a)}{1!} (x - a) + \dots +\frac{f^{(n)}(a)}{(n - 1)!} (x - a)^{n - 1}\right],\]\[R_{n}^{\prime \prime}(x) = f^{\prime \prime}(x) - \left[f^{\prime \prime}(a) + \frac{f^{\prime \prime \prime}(a)}{1!} (x - a) + \dots +\frac{f^{(n)}(a)}{(n - 2)!} (x - a)^{n - 2}\right],\]\[\dots\]\[R_{n}^{(n - 1)}(x) = f^{(n - 1)}(x) - \left[f^{(n - 1)}(a) + \frac{f^{(n)}(a)}{1!} (x - a)\right].\](对它不能再求导数.)当 \(x\rightarrow a\) 时,显然, \(R_{n}(x)\) , \(R_{n}^{\prime}(x)\) ,…, \(R_{n}^{(n - 1)}(x)\) 以及 \((x - a)^{k}\) \((k\in \mathbf{N}_{+})\) 都是无穷小.于是,由洛必达法则,有\[\lim_{x\to a}\frac{R_{n}(x)}{(x - a)^{n}} = \lim_{x\to a}\frac{R_{n}^{\prime}(x)}{n(x - a)^{n - 1}} = \dots = \lim_{x\to a}\frac{R_{n}^{(n - 1)}(x)}{(x - a)}\]\[\qquad = \frac{1}{n!}\lim_{x\to a}\left[\frac{f^{(n - 1)}(x) - f^{(n - 1)}(a)}{x - a} -f^{(n)}(a)\right] = 0.\]
特别地,当 \(a = 0\) 时,(1)式称为麦克劳林公式. 定理1给出的余项 \(R_{n}(x) = o\left[(x - a)^{n}\right]\) 称为佩亚诺余项.
02 Proof QA Recover the complete textbook proof.Open the generated question
Question
设函数 \(f\) 在点 \(a\) 的一个邻域内具有直到 \(n\) 阶的导数。令\[T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k.\]证明带佩亚诺余项的泰勒公式
\[f(x)=T_n(x)+o((x-a)^n),\qquad x\to a.\]
Answer
令 \(R_n(x)=f(x)-T_n(x)\)。只需证明\[\lim_{x\to a}\frac{R_n(x)}{(x-a)^n}=0.\]由 \(T_n\) 的构造，对 \(0\le k\le n-1\) 有 \(R_n^{(k)}(a)=0\)。连续应用 \(n-1\) 次洛必达法则，得到\[\lim_{x\to a}\frac{R_n(x)}{(x-a)^n}=\lim_{x\to a}\frac{R_n^{(n-1)}(x)}{n!(x-a)}.\]而\[R_n^{(n-1)}(x)=f^{(n-1)}(x)-f^{(n-1)}(a)-f^{(n)}(a)(x-a),\]所以\[\frac{R_n^{(n-1)}(x)}{n!(x-a)}=\frac1{n!}\left[\frac{f^{(n-1)}(x)-f^{(n-1)}(a)}{x-a}-f^{(n)}(a)\right].\]括号内第一项在 \(x\to a\) 时趋于 \(f^{(n)}(a)\)，故整个极限为零。因此 \(R_n(x)=o((x-a)^n)\)，即
\[f(x)=T_n(x)+o((x-a)^n).\]
03 Gap QA Make one intermediate proof obligation explicit.Open the generated question
Question
设 \(f\) 在 \(a\) 的邻域内具有直到 \(n\) 阶的导数，并定义 \(R_n(x)=f(x)-T_n(x)\)。已知\[R_n^{(n-1)}(x)=f^{(n-1)}(x)-f^{(n-1)}(a)-f^{(n)}(a)(x-a).\]证明\[\lim_{x\to a}\frac{R_n^{(n-1)}(x)}{n!(x-a)}=0,\]并说明这一步如何完成佩亚诺余项证明中的最后一次极限计算。
Answer
先作代数变形：\[\frac{R_n^{(n-1)}(x)}{n!(x-a)}=\frac1{n!}\left[\frac{f^{(n-1)}(x)-f^{(n-1)}(a)}{x-a}-f^{(n)}(a)\right].\]因为 \(f^{(n-1)}\) 在 \(a\) 处可微，其导数为 \(f^{(n)}(a)\)，所以\[\lim_{x\to a}\frac{f^{(n-1)}(x)-f^{(n-1)}(a)}{x-a}=f^{(n)}(a).\]由极限的线性运算，所求极限等于\[\frac1{n!}\bigl(f^{(n)}(a)-f^{(n)}(a)\bigr)=0.\]此前对 \(R_n(x)/(x-a)^n\) 连续应用 \(n-1\) 次洛必达法则后恰得到上述极限，因此这一步证明了 \(R_n(x)=o((x-a)^n)\)。
Topology
Algebra of interior operators
Lectures on General Topology01 Original textbook excerpt Chapter 2.6 · InteriorRead the source theorem and textbook proof
Parsed chapter Markdown Chapter 2.6 · Interior
推论2.31 \((X,T)\) 是拓扑空间, \(A\subset X\) . \(A\) 是开集的充要条件是 \(A = A^{\circ}\)
定理2.32 \((X,T)\) 是拓扑空间, \(A\subset X\) , \(B\subset X\) ,集合的内部具有如下性质:
(1) \(A^{\circ}\subset A\)
(2) \(X^{\circ} = X\)
(3) \((A\bigcap B)^{\circ} = A^{\circ}\bigcap B^{\circ}\)
(4) \((A^{\circ})^{\circ} = A^{\circ}\)
(5)若 \(A\subset B\) ,则 \(A^{\circ}\subset B^{\circ}\)
证明(4)可由推论2.31及定理2.30可得.(1)、(2)与(5)可由定义得到,下面只说明(3).
由于 \(A^{\circ}\subset A,B^{\circ}\subset B\) ,于是有 \(A^{\circ}\bigcap B^{\circ}\subset A\bigcap B\) ,因此 \((A^{\circ}\bigcap B^{\circ})^{\circ}\subset (A\bigcap B)^{\circ}\) 而 \((A^{\circ}\bigcap B^{\circ})^{\circ} = A^{\circ}\bigcap B^{\circ}\) ,因此 \(A^{\circ}\bigcap B^{\circ}\subset (A\bigcap B)^{\circ}\) .其次对任意 \(x\in (A\bigcap B)^{\circ}\) 存在开集 \(V_{x}\) ,使得 \(x\in V_{x}\subset A\bigcap B\) ,因此 \(x\in A^{\circ}\) , \(x\in B^{\circ}\) ,即 \(x\in A^{\circ}\bigcap B^{\circ}\) ,这样 \((A\bigcap B)^{\circ}\subset A^{\circ}\bigcap B^{\circ}\) ,因此 \((A\bigcap B)^{\circ} = A^{\circ}\bigcap B^{\circ}\) 口
但应注意, \((A\bigcup B)^{\circ}\) 不一定与 \(A^{\circ}\bigcup B^{\circ}\) 相同.例如:在 \(R\) 中, \(A = [0,1],B =\) (1,2],则 \(A^{\circ} = (0,1),B^{\circ} = (1,2)\) ,但 \(A\bigcup B = [0,2]\) ,因此 \((A\bigcup B)^{\circ} = (0,2)\) ,而 \(1\notin A^{\circ}\bigcup B^{\circ}\)
02 Proof QA Recover the complete textbook proof.Open the generated question
Question
设 \((X,T)\) 是拓扑空间，\(A,B\subset X\)。证明集合的内部满足：
1. \(A^\circ\subset A\)；
2. \(X^\circ=X\)；
3. \((A\cap B)^\circ=A^\circ\cap B^\circ\)；
4. \((A^\circ)^\circ=A^\circ\)；
5. 若 \(A\subset B\)，则 \(A^\circ\subset B^\circ\)。
并举例说明 \((A\cup B)^\circ\) 不一定等于 \(A^\circ\cup B^\circ\)。
Answer
(1) 由内部的定义，内点首先属于原集合，所以 \(A^\circ\subset A\)。
(2) \(X\) 是开集，且每个 \(x\in X\) 都有开邻域 \(X\subset X\)，故 \(X^\circ=X\)。
(3) 因 \(A^\circ\cap B^\circ\) 是开集且包含于 \(A\cap B\)，有 \(A^\circ\cap B^\circ\subset(A\cap B)^\circ\)。反之，若 \(x\in(A\cap B)^\circ\)，则存在开集 \(V\) 使 \(x\in V\subset A\cap B\)，所以 \(x\in A^\circ\cap B^\circ\)。因此两者相等。
(4) \(A^\circ\) 是开集，而开集等于其内部，故 \((A^\circ)^\circ=A^\circ\)。
(5) 若 \(x\in A^\circ\)，则存在开集 \(V\) 使 \(x\in V\subset A\subset B\)，所以 \(x\in B^\circ\)。
最后在 \(\mathbb R\) 的通常拓扑中取 \(A=[0,1]\)、\(B=(1,2]\)。则 \(A^\circ=(0,1)\)、\(B^\circ=(1,2)\)，而 \((A\cup B)^\circ=(0,2)\)。点 \(1\) 属于前者但不属于 \(A^\circ\cup B^\circ\)，故二者不一定相等。
03 Gap QA Make one intermediate proof obligation explicit.Open the generated question
Question
设 \((X,T)\) 是拓扑空间，\(A\subset X\)。已知：
1. \(A^\circ\) 是开集；
2. 集合 \(U\) 为开集当且仅当 \(U=U^\circ\)。
证明内部运算具有幂等性：
\[(A^\circ)^\circ=A^\circ.\]
Answer
由已知条件 1，\(A^\circ\) 是开集。将已知条件 2 应用于集合 \(U=A^\circ\)，立即得到\[A^\circ=(A^\circ)^\circ.\]交换等号两边即有 \((A^\circ)^\circ=A^\circ\)。
也可直接验证两边包含：对任意集合 \(B\) 都有 \(B^\circ\subset B\)，故 \((A^\circ)^\circ\subset A^\circ\)；另一方面，\(A^\circ\) 本身是开集且包含于自身，所以其每一点都是 \(A^\circ\) 的内点，故 \(A^\circ\subset(A^\circ)^\circ\)。
Textbook excerpts come directly from parsed chapter Markdown. QA wording is expert-edited for self-containment and to prevent answer leakage; each pair remains grounded in the same source record.
Results and insights
Stronger mathematical reasoning.
+16.7 ppaverage gain over the base model
| SFT corpus | TOME-LongForm | TOME-Objective | MiniF2F | ProofNet | MATH-500 | AIME-26 | Avg. |
|---|---|---|---|---|---|---|---|
| Base model No SFT | 13.25 | 32.93 | 34.45 | 17.14 | 68.80 | 8.44 | 29.17 |
| OpenMath1 1.82M | 20.16 | 33.86 | 22.87 | 19.25 | 70.40 | 8.21 | 29.13 |
| MetaMath 395K | 28.73 | 34.66 | 40.70 | 22.55 | 71.20 | 9.98 | 34.64 |
| MathInstruct 262K | 35.87 | 40.69 | 52.05 | 32.93 | 73.05 | 11.98 | 41.10 |
| TOME-Math 364.6K | 43.86 | 43.96 | 57.43 | 42.98 | 76.55 | 10.19 | 45.83 |
Same Qwen3-30B-A3B-Base backbone; one epoch per corpus. SFT = supervised fine-tuning; AIME-26 = AIME 2026. Scores (%); Avg. is the mean across six benchmarks; bold marks the best result. K / M = thousand / million training examples; pp = percentage points.
Our evaluation sets. TOME-LongForm: 300 proof-heavy exercises from held-out textbooks across six mathematical subjects, scored for proof correctness. TOME-Objective: 145 multidisciplinary objective questions, scored by final-answer equivalence.
Useful structure.
Even without fine-tuning.
Retrieving propositions and local proof obligations raises TOME-LongForm pass@1 from 13.25% to 17.38%.
Frozen base model, 300 problems. All four retrieval policies share the same context-token budget.
Single-solution success rate · Pass@1 (%)
Better order. Greater gains with scale.
Forward–Gap leads in all 18 subject–scale settings. Every schedule improves as the supervision budget grows.
Broader coverage. Transferable reasoning.
More textbook sources improve learning at a fixed budget, while subject-specific supervision transfers across mathematical fields.
More sources, same budget. Expanding from 8 books (narrow–dense) to 16 books (broad–sparse) improves all six subjects by 3.3 percentage points on average, with the same 0.75M-token budget.
Reasoning transfers across subjects. Cross-subject pass@16 averages 0.92× the target specialist’s score. Values above 1 indicate performance exceeding that specialist.
Subject key: Alg. = Abstract algebra; Ana. = Mathematical analysis; Prob. = Probability; DG = Differential geometry; Disc. = Discrete mathematics; Top. = Topology. Heatmap rows show the training subject; columns show the evaluation subject.