Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Would you have fallen for these?

Each is a real option from a previous-year paper. Read it cold and decide before opening it.

• Option A

  $x^{4} + 4$

  Why this is wrong, and why it looks right

  By the Sophie Germain identity $a^{4}+4b^{4} = (a^{2}+2b^{2}+2$ab$)(a^{2}+2b^{2}-2$ab), with $b=1: x^{4}+4 = (x^{2}+2x+2)(x^{2}-2x+2), a$ genuine factorisation into two integer-coefficient quadratics — so $x^{4}+4$ is reducible.

  From this question

  Which of the following polynomials are irreducible over $\mathbb{Q}$?

  Part C · free

• Option B

  cl(A ∩ B) = cl(A) ∩ cl(B) for any two subsets

  Why this is wrong, and why it looks right

  False — only the inclusion cl(A ∩ B) ⊆ cl(A) ∩ cl(B) holds in general (immediate, since A ∩ B ⊆ A and A ∩ B ⊆ B). For the reverse to fail, take A = (0, 1) and B = (1, 2) in $\mathbb{R}: A \cap B = \emptyset$ so cl(A ∩ B) = ∅, but cl(A) = [0, 1] and cl(B) = [1, 2] share the point 1, so cl(A) ∩ cl(B) = {1} $\ne \emptyset. (\mathbb{Q}$ and $\mathbb{R}$ ∖ $\mathbb{Q}$ give an even more extreme version: cl$(\mathbb{Q} \cap (\mathbb{R}$ ∖ $\mathbb{Q})) =$ cl(∅) = ∅, while cl$(\mathbb{Q}) \cap$ cl$(\mathbb{R}$ ∖ $\mathbb{Q}) = \mathbb{R} \cap \mathbb{R} = \mathbb{R}.)$ The mechanism is that closure can manufacture new shared boundary points that were never anywhere near the actual overlap of A and B — a point can be a limit of A and a limit of B without being a limit of A ∩ B at all.

  From this question

  Which of the following are true?

  Part C · free

• Option A

  f(x) = sin(1/x)

  Why this is wrong, and why it looks right

  False. sin(1/x) has no limit as x → 0⁺ — SymPy samples it on (0.001, 0.05] and finds values ranging down to −0.99998 and up near +1, oscillating without settling. A continuous function on a bounded interval is uniformly continuous exactly when it extends continuously to the closure, and this one cannot.

  From this question

  Which one of the following functions is uniformly continuous on the interval (0, 1)?

  See the full question

  Part B · june 2023 · free

• Option C

  f(x) = eˣ cos(1/x)

  Why this is wrong, and why it looks right

  False. The factor eˣ tends to 1 at x = 0 and is harmless, but cos(1/x) oscillates exactly as sin(1/x) does. Multiplying an oscillating factor by a well-behaved one leaves the oscillation intact.

  From this question

  Which one of the following functions is uniformly continuous on the interval (0, 1)?

  See the full question

  Part B · june 2023 · free

• Option D

  $f(x) = \cos x \cdot \cos(\pi/x)$

  Why this is wrong, and why it looks right

  False, for the same reason. cos x → 1 as x → 0, while $\cos(\pi/x)$ oscillates without limit — the $\pi$ in the numerator changes the frequency but not the behaviour. Three of the four options oscillate at 0; the exponential in (B) is the only one that damps.

  From this question

  Which one of the following functions is uniformly continuous on the interval (0, 1)?

  See the full question

  Part B · june 2023 · free

• Option D

  U is necessarily symmetric.

  Why this is wrong, and why it looks right

  Orthogonality only requires UᵀU = I, not U = Uᵀ. A rotation $R_\theta = [[\cos\theta, -\sin\theta], [\sin\theta, \cos\theta]]$ is orthogonal for every $\theta$ but symmetric only when $\sin\theta = 0$, so generic rotations are counterexamples.

  From this question

  Let U be a real orthogonal n × n matrix (UᵀU = I). Which of the following are true?

  Part C · free

• Option B

  If |$a_{n}$₊$_{1} - a_{n}$| → 0, then {$a_{n}$} is Cauchy.

  Why this is wrong, and why it looks right

  Consecutive terms shrinking to 0 is much weaker than Cauchy, which needs ALL sufficiently late pairs close, not just adjacent ones. $a_{n} = \sqrt{n}$ has $\sqrt{n+1} - \sqrt{n} = 1/(\sqrt{n+1}+\sqrt{n}) \to 0$, yet $a_{n} \to \infty$, so the sequence is unbounded and certainly not Cauchy.

  From this question

  Let {$a_{n}$} be a sequence of real numbers. Which of the following statements are true?

  Part C · free

• Option D

  Every bounded sequence is Cauchy.

  Why this is wrong, and why it looks right

  $a_{n} = (-1)^{n}$ is bounded between −1 and 1, but |$a_{n}$₊$_{1}-a_{n}$| = 2 for every n, so it never settles down and is not Cauchy. Boundedness only guarantees a convergent subsequence by Bolzano–Weierstrass, not convergence of the whole sequence.

  From this question

  Let {$a_{n}$} be a sequence of real numbers. Which of the following statements are true?

  Part C · free

• Option B

  $f(x) = \sin(x^{2})$ on $\mathbb{R}$

  Why this is wrong, and why it looks right

  False: at $x_{n}=\sqrt{2\pi{}n}, y_{n}=\sqrt{2\pi{}n+\pi/2}, y_{n}-x_{n}=(\pi/2)/(\sqrt{2\pi{}n+\pi/2}+\sqrt{2\pi{}n})\to0$, yet $\sin(x_{n}^{2})=0$ and $\sin(y_{n}^{2})=1$ always. Points that close should have close outputs under uniform continuity, but here they never do.

  From this question

  Which of the following functions are uniformly continuous on the given domain?

  Part C · free

• Option C

  sup(A ∩ B) = min{sup A, sup B} whenever A ∩ B ≠ ∅

  Why this is wrong, and why it looks right

  Only sup(A∩B) ≤ min{sup A, sup B} holds in general, since A∩B is a subset of both A and B. Take A = [0,1]∪[10,11] and B = [0,1]∪[20,21]: sup A = 11, sup B = 21, but A∩B = [0,1] has sup 1, nowhere near min{11,21} = 11.

  From this question

  Let $A, B \subset \mathbb{R}$ be non-empty and bounded. Which of the following are true?

  Part C · free

• Option A

  G is a union of proper subgroups.

  Why this is wrong, and why it looks right

  False. Any cyclic group refutes it: a generator lies in no proper subgroup, so the proper subgroups cannot cover G. For $\mathbb{Z}/6$, SymPy gives the divisors 1, 2, 3, 6 — so the proper subgroups have orders 1, 2, 3, and their union misses the totient(6) = 2 generators, namely 1 and 5.

  From this question

  Let G be any finite group. Which one of the following is necessarily true?

  See the full question

  Part B · december 2023 · free

• Option B

  G is a union of proper subgroups if |G| has at least two distinct prime divisors.

  Why this is wrong, and why it looks right

  False, and $\mathbb{Z}/6$ is the counterexample the option seems designed to survive. |$\mathbb{Z}/6$| = 6 has two distinct prime divisors, yet $\mathbb{Z}/6$ is cyclic and so is not a union of proper subgroups. Having several prime factors says nothing about being cyclic.

  From this question

  Let G be any finite group. Which one of the following is necessarily true?

  See the full question

  Part B · december 2023 · free

• Option C

  If G is abelian, then G is a union of proper subgroups.

  Why this is wrong, and why it looks right

  False. $\mathbb{Z}/p$ is abelian and cyclic, so it is not a union of proper subgroups — in fact it has none besides the trivial one. Abelian is much weaker than non-cyclic.

  From this question

  Let G be any finite group. Which one of the following is necessarily true?

  See the full question

  Part B · december 2023 · free

• Option A

  $\mathbb{Q}$ with the usual metric

  Why this is wrong, and why it looks right

  $\mathbb{Q}$ is not complete: the decimal truncations of $\sqrt{2} (1, 1.4, 1.41, 1.414, ...)$ form a Cauchy sequence in $\mathbb{Q}$ with no limit in $\mathbb{Q}$, since $\sqrt{2} \notin \mathbb{Q}. A$ Cauchy sequence need not converge to a point of the space it lives in.

  From this question

  Which of the following metric spaces are complete?

  Part C · free

• Option D

  C[0,1] with the metric $d(f,g) = \int_{0}^{1}$ |f − g|

  Why this is wrong, and why it looks right

  C[0,1] is not complete under $d(f,g) = \int_{0}^{1}$|f−g|: let $f_{n}$ ramp linearly from 0 to 1 across [1/2, 1/2+1/n] and be constant outside that interval. This is Cauchy in the $L^{1}$ metric (the shrinking ramp width bounds $\int$|$f_{n}-f_{m}$|) but converges only to a discontinuous step function, which is not in C[0,1].

  From this question

  Which of the following metric spaces are complete?

  Part C · free

• Option B

  int(A ∪ B) = int(A) ∪ int(B)

  Why this is wrong, and why it looks right

  Only ⊇ holds in general. Take $A = \mathbb{Q}$ and $B = \mathbb{R}$∖$\mathbb{Q}$: int(A) = int(B) = ∅ so int(A)∪int(B) = ∅, but $A\cup{}B = \mathbb{R}$ has int$(\mathbb{R}) = \mathbb{R} \ne \emptyset$.

  From this question

  Let $A, B \subseteq \mathbb{R}$. Which of the following always hold?

  Part C · free

• Option D

  $\mathbb{Q} \cap [0, 1]$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option A

  B(v, v) = 0 if and only if v = 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  f is an injective function

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option D

  f is a bijective function

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  Open connected subsets of $\mathbb{R}^{3}$ need not be path-connected.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  $(P_{n})_{n}\ge_{1}$ converges to $P_{0}$ in the topological space $(X, \tau)$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  The topological space $(X, \tau)$ is compact.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  f has a removable singularity at 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option A

  f(X) is irreducible in $\mathbb{Q}[X]$, but g(X) is not.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option C

  Both f(X) and g(X) are irreducible in $\mathbb{Q}[X]$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  Neither f(X) nor g(X) is irreducible in $\mathbb{Q}[X]$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option B

  The closure of a path-connected set is path-connected

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option B

  A is finite.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  Every connected subset of $\mathbb{R}^{2}$ is path-connected.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  If X is compact and $C_{1},\dots,C_{k}$ are non-empty closed sets of X such that $\cap{}C_{i}=\emptyset$, then the minimum value of $x\mapsto\sum{}d(x,C_{i})$ on X is 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option A

  $\mathbb{Q}^{2}$ is connected.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option C

  If f is Riemann integrable, then f is continuous.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option D

  f is onto

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option C

  f is not differentiable at 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  The subsequences {$x_{kn}$}$_{n}$ are convergent for every k ≥ 2.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2023

• Option D

  lim |$x_{n}$₊$_{1} - x_{n}$| = 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2023

• Option A

  connected and path-connected

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  neither connected nor path-connected

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option A

  H contains an orthonormal basis of $\ell^{2}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option B

  H is a linear subspace of $\ell^{2}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option C

  f is differentiable on $\mathbb{R}^{2}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  A is symmetric with negative diagonal entries

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  $\int$₋$_{1}^{1} f(x)x^{n}$ dx = 0 for all n ≥ 0 odd.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option D

  $\int$₋$_{1}^{1} f(x)x^{n}$ dx = 0 for all n ≥ 0 even.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option A

  f = 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option B

  f is a constant function.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option C

  There exists a compact subset K of $\mathbb{C}$ such that $f^{-1}(K)$ is not compact.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option B

  X + I is invertible.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option D

  Y is invertible.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option A

  For every bounded subset $A \subseteq \mathbb{R}, f$⁻$^{1}(A)$ is a bounded subset of $\mathbb{R}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  There exists $x \in \mathbb{R}$ such that f(x) = x.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  Every p-group is cyclic.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option A

  f is a Lipschitz continuous function on D

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  X is the line segment joining −1 and −i.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option A

  S is in bijection with the set of rational numbers.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2023

• Option B

  S is in bijection with the set of real numbers.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2023

• Option A

  $A_{n}$ is finite for every integer n ≥ 1.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option B

  $A_{n}$ is finite for some integer n ≥ 1.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option C

  $A_{n}$ is infinite for some integer n ≥ 1.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option C

  f is differentiable at (0, 0).

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  A ⊊ C and B ⊊ C

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option C

  A ⊊ B ⊊ C

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option A

  The series $\sum(n\ge1) f_{n}$ converges only at z = 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  The series $\sum(n\ge1) f_{n}$ converges pointwise only on a countable set E ⊆ 𝔻 but not on 𝔻 \ E.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  The ring of 2 × 2 real matrices.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  The ring of continuous functions from [0, 1] to $\mathbb{R}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  For every non-zero vector $v \in \mathbb{R}^{4}$, the set {2v, 3Tv} is linearly independent

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  V is the union of 3 proper subspaces.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option D

  I_(m,n) = 1 if m = n + 2

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option A

  R is never an integral domain.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  $f_{4} \in V$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option B

  There is a bijection between S and $\mathbb{N}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option D

  There is a bijection between S and a non-empty finite set

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option D

  If g is identically 0 and f is continuous on the set S, then f is identically 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option D

  There do not exist non-zero $\mathbb{Q}-$linear transformations $T : \mathbb{R} \to \mathbb{Q}[x]$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option C

  For all x ∈ X, S_G(x) is a normal subgroup of G.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option D

  Increasing the number of equally spaced interpolation nodes always reduces the maximum interpolation error.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  The design is orthogonal

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option C

  If the sum Re(f) + Im(f) and the product Re(f)Im(f) are bounded, then f is constant

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option B

  $\mathbb{Q}[x]/(x^{5} + x^{4} + x^{3} + x^{2} + x + 1)$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  f is increasing.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option A

  {(x, y) : xy = 1}

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option B

  f is differentiable on (0, 1).

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  f is monotone.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  f is Lipschitz.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  {$x \in \mathbb{R}$ | $x \ge 0, \sqrt{x} \in \mathbb{Q}$}

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  {$x \in \mathbb{R}$ | $x^{2} \in \mathbb{Q}$}

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  |X| divides n.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  The model with intercept passes through at least one of the observed data points

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option B

  f is Lipschitz

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  f is differentiable almost everywhere

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  The group $\mathbb{Z}$ has property (P).

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  A has at least two distinct eigenvalues.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  The minimal polynomial of A cannot have degree 2.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  p ≡ 3 (mod 4), G = GL$_{2}($𝔽_p)/SL$_{2}($𝔽_p) and H is a subgroup of order 2.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  p ≡ 1 (mod 4), G = GL$_{2}($𝔽_p)/SL$_{2}($𝔽_p) and H is a subgroup of order 2.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  f(z̄) is an entire function.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2023

• Option B

  conj(f(z)) is an entire function.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2023

• Option A

  C is countable with measure zero.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  C is uncountable with positive measure.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option B

  G has a normal Sylow 2-subgroup.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option A

  The Dirichlet function: f(x) = 1 for x rational, f(x) = 0 for x irrational.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  f is differentiable only at 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option D

  f is not differentiable at 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option A

  E is the real axis.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  E contains an open subset of $\mathbb{C}$, but $E \ne \mathbb{C}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option B

  W ⊊ ker(T)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  W ∩ ker(T) ⊊ ker(T)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option A

  There exists an infinite set X such that f(z) = 0 for all z ∈ X

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  There exists a non-empty closed set X such that f(z) = 0 for all z ∈ X

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  If A is finite and B = [0, 1], then there is no bijection between S and T.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  There is no bijection between S and U for any choice of A.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  If A ≠ B, then there is no bijection between T and U.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option B

  $\mathbb{R}$ is compact in the topology $\tau$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option C

  Topology $\tau$ is the same as the Euclidean topology

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  If there are exactly two elements of order 4 in G, then G is abelian.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option C

  If there are exactly six elements of order 4 in G, then G is abelian.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option D

  There is an element of order 4 in G.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option B

  Let $\tau$ be a topology on $\mathbb{R}$. If the identity function of $\mathbb{R}$ is a continuous map from $(\mathbb{R},\tau_E)$ to $(\mathbb{R},\tau)$, then $(\mathbb{R},\tau)$ is a regular space.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option D

  $\mathbb{R}$ is a regular space in the finite-complement topology.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option C

  $lim_{n}\to\infty P(X_{n} = 0) = 1/2024$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option A

  If f(z) = f(z + 1) for all $z \in \mathbb{C}$ then f is a constant function.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  If f is a non-constant function then f(1/z) has a pole at 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option B

  There exist $\lambda,\mu\in\mathbb{C}$ and linearly independent vectors v,w∈V such that Av$=\lambda{}v$ and Aw$=\mu{}w$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option C

  f(x) = x for all $x \in \mathbb{R}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option D

  If f is one-to-one, then f is onto.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option C

  The cyclic group of order 5

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  The symmetric group $S_{5}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  If $B \in M_{4}(\mathbb{C})$ is such that AB+BA=0, then B is diagonalisable.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option C

  (Q) is true.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option D

  Systematic sampling is always at least as precise as SRS.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  For each non-zero v∈V, {v,Tv,$T^{2}v$} is linearly dependent.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option A

  If d is a positive integer that divides |G|, then G has a subgroup of order d.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option D

  |G| ≠ 16

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option B

  If $B = \mathbb{N}$, then there exists a one-to-one map from $S_{2}$ to B.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option D

  If A is finite, then $S_{2}$ is finite for any B.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option C

  rank(AB) = rank(BA).

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option A

  p ∤ 1 + (p − 1)! for some odd prime p.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option D

  State 1 is transient.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  Both $\sum_{n}$₌$_{1}^\infty A_{n}$ and $\sum_{n}$₌$_{1}^\infty B_{n}$ are convergent

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  K is an algebraic extension of $\mathbb{Q}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2023

• Option C

  $\mathbb{C}$ is an algebraic extension of K.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2023

• Option B

  There exists an open neighbourhood U of (0, 0) in $\mathbb{R}^{2}$ such that $f_{3}$ is a function of $f_{1}$ and $f_{2}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option C

  The symmetric group $S_{3}$ has property (J).

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  f is necessarily a constant function.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  The symmetric group $S_{5}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  |f(z)| ≤ $|z|^{3/2}$ for all |z| ≥ 1.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  S \ $\mathbb{Q}$ is discrete with subspace topology on S.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  S is connected.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  If A and B are upper triangular, then M is diagonalizable over $\mathbb{R}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  If A and B are diagonalizable over $\mathbb{R}$, then M is diagonalizable over $\mathbb{R}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  If A and B are diagonalizable over $\mathbb{R}$, then there exists $\lambda \in \mathbb{R}$ such that $M = \lambda{}I_{2}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  The element 4 mod 101 in $(\mathbb{Z}/101\mathbb{Z})$ˣ is a primitive root.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  S is an uncountable set.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option A

  Let G and G′ be finite groups such that S(G) = S(G′). Then G is isomorphic to G′.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option B

  If S(G) = 4, then |G| = pᵐ for some prime number p and positive integer m.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option C

  If S(G) = 5, then G is a cyclic group.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option A

  $lim_{z\to0}$ f(z) does not exist.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option C

  R is an integral domain.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option D

  R is a euclidean domain.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option A

  If G(n) = 1, then n is prime.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  Every trajectory tends to an equilibrium point as $t \to \infty$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  The statistic $(U_{n}, V_{n})$ is complete

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  state 2 is aperiodic and null-recurrent

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option C

  Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function. Then there exists an uncountable subset $T \subseteq \mathbb{R}$ such that $f(T) \subseteq \mathbb{R}$ is uncountable.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option D

  Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous function. If R ⊆ image(f) is a countable set, then f⁻$^{1}(R)$ is countable.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option B

  $\mathbb{Z}[\surd-5]$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  S has to be a finite set

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option A

  f never vanishes in [0, 1].

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2023

• Option A

  The set of all irrational numbers is dense in $(\mathbb{R}, \tau)$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  [0, 1] is compact in $(\mathbb{R}, \tau)$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  $(\mathbb{R}, \tau)$ is Hausdorff.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  W^⊥ = {0}

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option D

  du/dx $- \lambda{}u = 0, u(0) = 0 = u(1)$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option A

  The product of the numbers is zero

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · september 2022

• Option B

  At most two of these numbers are positive

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · september 2022

• Option C

  There cannot be exactly one zero

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · september 2022

• Option A

  $M^{2} = N^{2}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option B

  $M^{3} = N^{3}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option D

  The minimal polynomials of M and N are the same.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option C

  If ST − TS has only one eigenvalue, then ST − TS $= \lambda{}I$ for some $\lambda \in \mathbb{C}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  {$(x, y) \in \mathbb{R}^{2}$ : xy = 1}

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option B

  If $g(\alpha)=0$, then |$\alpha$|≠1.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option C

  The function g has only finitely many zeros.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option A

  There is a bijective map from $\mathbb{N}$ to S.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option C

  There is a bijective map from $\mathbb{N}$ to $(0,1)\S$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option D

  S is a countable set and $(0,1)\S$ is uncountable.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option A

  There exists $(a,b,c) \in \mathbb{R}^{3}${(0,0,0)} such that f(a,b,c) = 0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option B

  If there exists $(a,b,c) \in \mathbb{R}^{3}${(0,0,0)} such that f(a,b,c) = 0, then $(a,b,c) \in \mathbb{Q}^{3}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option C

  {$(a,b,c) \in \mathbb{Z}^{3}$ ∣ f(a,b,c) = 0} is a finite set.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option C

  For every finite group G, there exists a non-negative integer n such that |H_G| = 4n+1.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option D

  For every finite group G, there exists a non-negative integer n such that |H_G| = 4n+3.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2025

• Option C

  There exists a nonzero constant c such that f(z) = ce^(−1/z) for all z ∈ 𝔻 \ {0}.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  There exists a nonzero constant c and a positive integer n such that f(z) = cz$^{n}e^(-1/z)$ for all z ∈ 𝔻 \ {0}.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  The set {a ∈ A : 1/a ∉ A} ∪ {0} is an additive subgroup of $\mathbb{Q}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option B

  There exists a countably infinite set in C

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option C

  There exists an uncountable set in C

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option C

  If 0 is the only eigenvalue of A and of B, then 0 is the only eigenvalue of AB.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  If AB and BA have the same minimal polynomial, then either A or B is invertible.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  H need not be normal.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  If A is connected, then B and C are connected.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  $(f_{n})_{n}\ge_{1}$ converges pointwise on [0, 1] to a discontinuous function f.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option A

  The composition f ∘ g is Riemann integrable.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  There exists a vector $v \in \mathbb{R}^{10}$ such that Av ≠ 0 and $A^{2}v = 0$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option C

  A must have a non-zero eigenvalue

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option B

  (‖$f_{n}$‖$_{2})$ is a convergent sequence.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  Both (‖$f_{n}$‖$_{1})$ and (‖$f_{n}$‖$_{2})$ are convergent sequences.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  $x^{2} + 2y^{2} + 3z^{2} - 2$xy − 2yz

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2024

• Option D

  For every $\mathbb{R}-$linear transformation $S : \mathbb{R}^{4} \to \mathbb{R}^{4}$, we have ST = TS.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  If $f(X)\in\mathbb{Q}[X]$ is such that $f(\lambda_{i})=0$ for all 1≤i≤3, then f(A)=0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option D

  If $f(X)\in\mathbb{Q}[X]$ is a cubic polynomial such that $f(\lambda_{i})=0$ for all 1≤i≤3, then f(A)=0.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2025

• Option A

  The ideal $2\mathbb{Z}[i]$ is maximal in $\mathbb{Z}[i]$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

The counterexamples that settle these

One example is enough to kill a false statement. These are the standard ones for the claims above.

• Completeness is a topological property
• Connected ⇒ path-connected
• Continuous on a bounded interval ⇒ bounded
• Uniformly continuous ⇒ Lipschitz
• Differentiable ⇒ continuously differentiable
• Pointwise limit of continuous functions is continuous

Other mistakes that cost marks