{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-vector"},"kind":"guides","total":3,"pages":1,"page":1,"entries":[{"id":"derivation-mko-vector-01","title":{"zh-Hant":"固定基底下的平面向量：逐分量運算、相等判定與減法方向","en":"Plane vectors in a fixed basis: componentwise operations, equality and the direction of subtraction"},"summary":{"zh-Hant":"在固定標準基底下，設 u=(3,−1)、v=(−2,4)、w=(1,3)、k=−2。逐分量計算 u+v、u−v、v−u、ku、0u、2u+v；求滿足 u+x=v 的向量 x；核對 k(u+v)=ku+kv；並判定 u=w 與 u=(−1,3) 是否成立。","en":"In the fixed standard basis let u=(3,−1), v=(−2,4), w=(1,3) and k=−2. Compute u+v, u−v, v−u, ku, 0u and 2u+v componentwise; find the vector x with u+x=v; check k(u+v)=ku+kv; and decide whether u=w and whether u=(−1,3)."},"mko_ids":["mko-vector"]},{"id":"derivation-mko-vector-subspace-01","title":{"zh-Hant":"以全稱封閉認證過原點直線","en":"Universal closure of a line through zero"},"summary":{"zh-Hant":"U={(x,2x):x∈R}是R²子空間，仿射平移U+(0,1)不是。","en":"U={(x,2x):x in R} is a subspace of R2, whereas U+(0,1) is not."},"mko_ids":["mko-vector-subspace"]},{"id":"derivation-mko-vector-space-dimension-01","title":{"zh-Hant":"先證基底再讀取維度","en":"Prove a basis before reading dimension"},"summary":{"zh-Hant":"B=((1,1),(1,−1))是R²有序基底，故dim R²=2。","en":"B=((1,1),(1,-1)) is an ordered basis of R2, so dim R2=2."},"mko_ids":["mko-vector-space-dimension"]}],"title":"Concept derivation guides"}