Bring ideas together without being told which method to use. Show the reasoning that makes your result valid.
Suggested session length: 120 minutes, as an unvalidated planning estimate. This collection contains 4 tasks; choose a manageable set and return for further sessions. Completing the whole collection may take substantially longer. This is a unit assessment, not a full course or AP examination.
State domains, allowed moves and stopping conditions before proving a claim.
Separate a necessary obstruction from a sufficient construction, and vertex occurrences from distinct labels.
Attempt each changed case before opening its public solution. Written arguments require mathematical review; submission alone is not mastery.
Loading saved assessment responses…
0 of 4 tasks have recorded responses. No overall mastery score is assigned.
Let A={0,1,2,3}, B={u,v,w}, C={red,blue}. Define f:A→B by f(0)=f(1)=u, f(2)=v, f(3)=w. A partial g:B⇀C has domain {v,w}, with g(v)=red and g(w)=blue. Determine the exact domain of g∘f and prove it is injective there although f is not. Formulate and prove a sufficient domain condition under which injectivity of g∘f would imply injectivity of f on dom(f), for arbitrary partial f and g. Does that condition force g to be defined on all B? Give a counterexample or proof.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
A process starts with positive integer pile sizes 2,3,4. A merge of a,b earns 3ab credits. Prove a formula for accumulated credits at any stopping configuration using the sum of squared pile sizes. Calculate credits if the final sizes are 5,4 and if they are 9. Prove that every full merge earns the same amount, and explain why this does not say all early stopping configurations earn the same amount. Now restrict legal merges to pairs with a+b≤6. Give two terminal configurations reachable from the same start with different earned credits, and prove both are terminal.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
A token starts at (1,−2). Allowed forward moves are u=(3,1) and v=(1,3); a second version also allows −u and −v. Let (a,b) denote displacement from the start. Audit this claim: the reverse-enabled version reaches a target exactly when a+b is divisible by4 and a−b is even. Prove it or provide a counterexample and a corrected criterion. Find the unique signed numbers of u and v moves. For the forward-only version derive the additional conditions. Apply both criteria to targets (5,2), (6,−3) and (3,0), checking the actual displacements.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
A finite rooted full ternary tree has one label per vertex and exactly three children at every internal vertex. Let V count vertex occurrences and S be the set of labels. Derive |S(T)| at a root r with child label sets A,B,C, including all intersections and whether r occurs below. Prove that pairwise disjoint child label sets and a fresh root at every constructor imply |S(T)|=V(T)=3i(T)+1, where i counts internal vertices. Find a smallest counterexample if root freshness alone is kept. A proposed shortcut checks only A∩B∩C=∅ at each node: does this suffice? Finally prove or refute that |S(T)|=V(T) forces both constructor requirements everywhere.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Review and revisit
Use each rubric to identify a valid argument, a missing condition or a step to revise. For independent evaluation, ask a teacher or reviewer to examine the complete reasoning. Self-review and recorded responses are not verified assessment results.
Coverage and targeted lesson returns
Let A={0,1,2,3}, B={u,v,w}, C={red,blue}. Define f:A→B by f(0)=f(1)=u, f(2)=v, f(3)=w. A partial g:B⇀C has domain {v,w}, with g(v)=red and g(w)=blue. Determine the exact domain of g∘f and prove it is injective there although f is not. Formulate and prove a sufficient domain condition under which injectivity of g∘f would imply injectivity of f on dom(f), for arbitrary partial f and g. Does that condition force g to be defined on all B? Give a counterexample or proof.
A process starts with positive integer pile sizes 2,3,4. A merge of a,b earns 3ab credits. Prove a formula for accumulated credits at any stopping configuration using the sum of squared pile sizes. Calculate credits if the final sizes are 5,4 and if they are 9. Prove that every full merge earns the same amount, and explain why this does not say all early stopping configurations earn the same amount. Now restrict legal merges to pairs with a+b≤6. Give two terminal configurations reachable from the same start with different earned credits, and prove both are terminal.
A token starts at (1,−2). Allowed forward moves are u=(3,1) and v=(1,3); a second version also allows −u and −v. Let (a,b) denote displacement from the start. Audit this claim: the reverse-enabled version reaches a target exactly when a+b is divisible by4 and a−b is even. Prove it or provide a counterexample and a corrected criterion. Find the unique signed numbers of u and v moves. For the forward-only version derive the additional conditions. Apply both criteria to targets (5,2), (6,−3) and (3,0), checking the actual displacements.
A finite rooted full ternary tree has one label per vertex and exactly three children at every internal vertex. Let V count vertex occurrences and S be the set of labels. Derive |S(T)| at a root r with child label sets A,B,C, including all intersections and whether r occurs below. Prove that pairwise disjoint child label sets and a fresh root at every constructor imply |S(T)|=V(T)=3i(T)+1, where i counts internal vertices. Find a smallest counterexample if root freshness alone is kept. A proposed shortcut checks only A∩B∩C=∅ at each node: does this suffice? Finally prove or refute that |S(T)|=V(T) forces both constructor requirements everywhere.
