Semester 1 Answer Key and Calibration

Open after preserving the first attempt. Accept equivalent correct proofs and constructions. A final number without the requested reasoning is partial evidence.

E01: Logic and proof audit

There exists x in X such that no y in Y is valid for x. For X={a,b}, Y={0}, relation {(a,0)} violates the original statement at b. Consistency of P and Q establishes neither all P cases nor exclusion of a P-and-not-Q counterexample.

E02: Repair and prove

At n=4 both sides are 16. Repair to n>=5: 32>25. Assume 2n>n squared for n>=5. Then 2(n+1)>2n squared>(n+1) squared because n squared-2n-1>0 for n>=5. State base, hypothesis and step; accepting the false original statement fails this gate.

E03: Count with roles

Choose chair in 12 ways, vice-chair in 8 ways outside that group, then the remaining unordered pair in C(10,2)=45 ways. Total 12845=4320. Every outcome determines its two ordered officers and remaining unordered members uniquely.

E04: Graph certificate

Label vertices by three-bit strings and join strings differing in exactly one bit (the cube). Each vertex has three distinct neighbours and there are no loops. Degree sum is 24, so edges=12. The construction proves existence; an even degree sum is merely a necessary condition in general.

E05: Probability model and bounds

A fixed bucket is empty with probability (99/100)**20, approximately 0.817907. Linearity gives about 81.7907 empty buckets without assuming independent indicators. At most 20 buckets can be occupied, so at least 80 are empty: fewer than 15 has probability exactly zero. Two fixed buckets are simultaneously empty with probability (98/100)**20, which differs from (99/100)**40.

E06: Inference assumptions

Standard error 0.4/5=0.08; margin 2.064*0.08=0.16512. Interval [2.13488,2.46512]. Under the model, the procedure covers the fixed population mean in 95% of repeated samples; this is not a posterior probability for this realised interval. Serial dependence invalidates the stated independent-sample standard error; redesign independent replication or model dependence rather than reuse the interval uncritically.

E07: Exact versus unique solutions

Subtract the third equation from the first: z=6-c. Then x+2y=3c-12. The second equation requires b=6c-24+a(6-c). If this fails there is no solution. If it holds, all solutions are (3c-12-2t, t, 6-c) for real t. Columns one and two are dependent, so no choice has a unique solution.

E08: Numerical judgment

Original solution (1,1); perturbed solution (1,2). Infinity norms of A and inverse A are 1 and 108, so the condition number is 108. Residual of (1,2) against original b is (0,10**-8); relative residual divided by norm(b) is 10**-8. Relative infinity-norm forward error is 1. Conditioning explains amplification; do not confuse residual, backward error and forward error.

E09: Synthesis: triangle estimator

Common-neighbour counts on edges 01,12,02,23 are 1,1,1,0. Sum is 3, so T=1; trace(A cubed)=6 counts six oriented closed walks per triangle. A sampled c is Bernoulli(3/4), with variance 3/16. Multiplying its k-sample mean by 4/3 gives expectation 1 and variance 1/(3k). Chebyshev bounds P(abs(estimate-1)>=epsilon) by 1/(3k epsilon squared), so integer k>=ceil(1/(3 delta epsilon squared)) is sufficient. This is not a necessary lower bound or a relative-error guarantee.

E10: Changed assumptions and handoff

Canonicalise undirected edges or reject duplicates before defining the simple graph and uniform sample space. Without replacement, independent-draw variance no longer applies; for a uniform fixed-size k<=4 sample the E09 variance gains factor (4-k)/3. Adaptive stopping requires separate analysis. A stopped calculation is unknown unless completeness is established. Reproduce a tiny duplicate-edge fixture, report completed checks and pending assumptions, and avoid claiming unbiasedness for an unanalysed sampling/stopping policy.

Project calibration

For the triangle-with-tail fixture in E09, the exact target is 1 and variance is 1/(3k) under the specified with-replacement model. A square without diagonals has zero triangles; adding one diagonal gives two. Exact enumeration, common-neighbour sum divided by 3, and trace(A cubed) divided by 6 should agree for simple undirected graphs. Self-loops, multiedges and directed edges require different formulas.

The project uncertainty report must distinguish a proved expectation from a particular random run. Across independent seeds, report signed error, absolute error and squared error; compare empirical mean squared error with the theoretical variance only under the stated unbiased model. Record the tested graph distribution. Choosing graphs because they look favourable is a selection effect, not evidence of general performance or a causal improvement.

For geometry, the rotation matrix [[cos(theta),-sin(theta)], [sin(theta),cos(theta)]] preserves lengths and dot products in exact arithmetic. Check approximate orthogonality and distance preservation numerically with a declared scale-aware tolerance. Repeated rotations can accumulate rounding error; report observations, not assumed failure magnitudes. Reuse Module 4's conditioning experiment to explain a second numerical limitation without building another numerical library.

Checkpoint calibration

The checkpoint consolidates existing module gates rather than introducing hidden new prompts. The reviewer checks evidence paths, reproduces one result and uses the defense prompts. A missing model assumption, unsupported general claim, unreproducible experiment or fabricated observation requires repair regardless of other scores. Apply the rubric and record review status honestly.

E11 calibration: complete reachability

gcd(14,22)=2 with 8*14 - 5*22 = 2. Divide 14t ≡ 10 (mod 22) by 2 to get 7t ≡ 5 (mod 11). The inverse of 7 modulo 11 is 8, so t ≡ 7 (mod 11) and canonical times are [7,18]. They both verify; every solution has this reduced residue, proving completeness. The cycle length is 11. Slot 9 is unreachable because 2 does not divide 9. Checking only listed witnesses would fail to detect an omitted time or an empty list.

E12 — Calculus calibration

Normalization gives (9/2)c=1, so c=2/9. CDF is zero below 0, t²/9 on [0,3], one above 3. Mean is ∫₀³ 2t²/9 dt=2; P(T>1)=8/9. Point probability is zero under this density model. With |f″|=2 for the integration target t², the trapezoid bound is 2·27/(12n²)=4.5/n². Thus n≥22 suffices for error ≤0.01. Distinguish this polynomial integration target from the preceding density, and distinguish exact integration from numerical approximation.