A proper finite-index subgroup of a torsion-free group cannot be a direct summand when the complement is a subgroup of that same free abelian group. Indeed, if $L=M\oplus K$, every $k\in K$ has $18k\in M$ because the quotient has order eighteen. But also $18k\in K$, so $18k=0$ by the trivial intersection. The ambient group ${\mathbb{Z}}^3$ is torsion-free, forcing $k=0$ and hence $K=0$. This would give $M=L$, contrary to index eighteen. Thus $M$ is not a direct summand.

Put the generators into the rows of the integer matrixIts determinant is $5-2-21=-18$. A full-rank integer lattice generated by its rows has index equal to the absolute determinant, henceFor a direct verification, $u,v,e_3$ form a unimodular basis because their row determinant is $-1$, and $w=-7u+5v+18e_3$. In these integer coordinates $M=\mathbb{Z}u\oplus \mathbb{Z}v\oplus 18\mathbb{Z}{e}_3$, visibly leaving eighteen cosets.

An example in dimension four isBoth are invertible, so each has rank four. Both have characteristic polynomial $(x-1{)}^4$, and each has minimal polynomial $(x-1{)}^2$, since its nonzero nilpotent part squares to zero. However $\mathrm{rank}(A-I)=2$ and $\mathrm{rank}(B-I)=1$. If $B=PA{P}^{-1}$, then $B-I=P(A-I)P^{-1}$, which would preserve these ranks. The contradiction proves that the matrices are not similar, despite all three specified invariants agreeing.

Take a size-three Jordan block at one and a scalar block at three:Triangularity gives ${\chi }_A(x)=(x-1{)}^3(x-3)$. On the first block, $(A-I{)}^3$ is zero but $(A-I{)}^2$ is not; on the last component the eigenvalue is three. Thus both factors with exactly those powers are needed, and their product annihilates $A$. Its minimal polynomial is therefore also $(x-1{)}^3(x-3)$.

Let $m_{\alpha }(x)$ be the monic minimal polynomial. The Cayley-Hamilton theorem says ${\chi }_{\alpha }(\alpha )=0$. Divide ${\chi }_{\alpha }$ by $m_{\alpha }$; its remainder also annihilates $\alpha$, and a nonzero remainder of smaller degree would contradict minimality. Thus $m_{\alpha }$ divides ${\chi }_{\alpha }$.

Every characteristic root ${\lambda }_i$ is an eigenvalue, with a nonzero eigenvector $v_i$. Since $m_{\alpha }(\alpha )v_i=m_{\alpha }({\lambda }_i)v_i=0$, it must be a root of the minimal polynomial as well. This proves there are no possibilities beyondEvery such choice really occurs. For each $i$, take one Jordan block of size $r_i$ at ${\lambda }_i$, together with $n_i-r_i$ scalar blocks at that eigenvalue, and form their direct sum. The characteristic polynomial has the prescribed multiplicities. The nilpotent part of a block of size $r_i$ has its $r_i$th power zero and its $(r_i-1)$st power nonzero, so this component's minimal exponent is exactly $r_i$. A polynomial annihilates the direct sum precisely when it annihilates every component, establishing the displayed minimal polynomial without any extra restrictions.

There is a small defect in the printed definition: the listed family omits $\mathbb{R}$ and is not a topology, since ${\bigcup }_{n\ge 1}(-n,\infty )=\mathbb{R}$. We use the intended upper-ray topology on the real line, obtained by adjoining $\mathbb{R}$ to that family.

Take $f(n)=n$. For every real $a$,has finite complement and is therefore open in the cofinite topology. The inverse images of the empty set and the whole line are also open. Hence this nonconstant function is a continuous map for the corrected upper-ray topology on the real line. The same inverse-image calculation verifies the condition for every ray literally listed in the question.

Any two nonempty open sets in the cofinite topology on $\mathbb{N}$ intersect: their complements have finite union, which cannot exhaust $\mathbb{N}$. Thus this is a hyperconnected space. Suppose a continuous map $f$ took two distinct values $a$ and $b$. The usual real line is a Hausdorff space, so there are disjoint open intervals $U$ and $V$ containing $a$ and $b$. Their inverse images under the continuous map are disjoint, nonempty open sets, contradicting hyperconnectedness. Every such $f$ is constant.

The weighted binary-sequence metric is finite because $0\le d(ϵ,\eta )\le {\sum }_{n\ge 1}{2}^{-n}=1$. It is symmetric and nonnegative. If $d(ϵ,\eta )=0$, every nonnegative summand is zero, so every coordinate agrees; conversely agreement makes the sum zero. For three sequences, apply $|{ϵ}_n-{\zeta }_n|\le |{ϵ}_n-{\eta }_n|+|{\eta }_n-{\zeta }_n|$ in each coordinate and sum. This proves the triangle inequality, so $d$ is a metric.

If two sequences differ in one of their first $n$ coordinates, their distance is at least $2^{-n}$. Thus the open ball $B_d(\eta ,2^{-n})$ is contained in $C_n(\eta )$. For every $\eta \in C_n(ϵ)$, the latter cylinder equals $C_n(\eta )$, so each cylinder set is open in the metric topology. Every set in the specified topology, being a union of cylinder sets, is therefore metric-open.

Conversely, agreement in the first $n$ coordinates givesGiven a metric-open set containing $ϵ$, choose a ball $B_d(ϵ,r)$ inside it, then choose $n$ with $2^{-n}<r$. The bound gives $C_n(ϵ)\subseteq B_d(ϵ,r)$, establishing the required prefix-neighbourhood condition. The two topologies coincide. The strict inequality $2^{-n}<r$ avoids incorrectly treating the closed tail bound as a strict ball bound.

For a sequence $ϵ$, let $C_n(ϵ)$ be its prefix cylinder set, consisting of sequences agreeing with $ϵ$ in the first $n$ coordinates. The specified collection consists exactly of unions of such cylinder sets.

The empty set meets the defining condition vacuously, and $X$ meets it because every cylinder set is contained in $X$. For a union ${\bigcup }_{\lambda }{Y}_{\lambda }$, a point lies in one $Y_{\lambda }$ and has a prefix cylinder set contained in that set, hence contained in the union. For a finite intersection, a point has a cylinder neighbourhood in every constituent. Taking the largest of the finitely many prefix lengths gives one cylinder set contained in all of them. Therefore the collection contains the empty and whole sets, is closed under arbitrary unions and finite intersections, and is a topology on $X$, the prefix topology of Cantor space.

Factor the denominator as $z^2(1+z^2)$. Near zero,so zero is a double pole with residue zero. At $z=\pm i$ the poles are simple and the denominator derivative is $2z+4z^3=-2z$. ThereforeThese are all the poles.

The finite geometric series gives $F_n(x)=(1-x^{n+1})/(1-x)$, so the pointwise limit is $F(x)=1/(1-x)$. The error is $x^{n+1}/(1-x)$, unbounded as $x\uparrow 1$ for every fixed $n$. Thus convergence is not uniform on $(0,1)$. It is uniform on any smaller interval bounded away from one, since the error is at most $b^{n+1}/(1-b)$ for $0<x\le b<1$.

For every fixed $x>0$, $n{x}^2\to \infty$, so the pointwise limit is zero. But ${\sup }_{0<x<1}{e}^{-n{x}^2}=1$ for every $n$, since $x$ can approach zero. Hence convergence is not uniform. For an explicit moving-point test take $x_n=n^{-1/2}$ for $n>1$, which gives $F_n(x_n)=e^{-1}$.

At every fixed $x$, $F_n(x)=e^x/n\to 0$. Moreover ${\sup }_{0<x<1}{e}^x/n=e/n\to 0$. Thus the pointwise limit is zero and convergence is uniform.

An irreducible Markov chain has the property that for every pair of states $i,j$ there is an integer $n\ge 0$ with $(P^n{)}_{ij}>0$. Equivalently every state can reach every other state along a finite path of positive-probability transitions.

Write $x=11+100t$. Reduction of the other integer congruence modulo $79$ gives $21t\equiv -3(\mathrm{mod}79)$. The Bézout's identity $19\cdot 79-15\cdot 100=1$ gives $-15\cdot 21\equiv 1(\mathrm{mod}79)$, so $t\equiv 45(\mathrm{mod}79)$. Consequently,Indeed, $4511=8+79\cdot 57=11+100\cdot 45$. The Chinese remainder theorem guarantees one class modulo $79\cdot 100$, since the two moduli are coprime.

Divide the linear congruence by the common factor $3$, also dividing its modulus:The Bézout's identity from part (a) makes $19$ the inverse of $79$ modulo $100$. Multiplication by $19$ therefore gives $x\equiv 133\equiv 33(\mathrm{mod}100)$. Thus all integer solutions are $\boxed{x=33+100k,k\in \mathbb{Z}}$. Equivalently, the three solution classes modulo $300$ are $33,133,233$; using modulus $300$ must not discard two of these classes.

The Euclidean algorithm givesBack substitution produces the Bézout's identityHence a positive integer solution is $\boxed{m=19,n=15}$. More generally, subtracting this identity from any other solution shows $79(m-19)=100(n-15)$. Since $79$ and $100$ are coprime, the integer solutions are $m=19+100k$, $n=15+79k$; both are positive precisely when $k\ge 0$.

False. Unequal mixed partial derivatives can exist when the continuity hypotheses of Clairaut's theorem fail.

False. Differentiability does not imply continuous derivatives.

False. Directional derivatives do not imply differentiability; the resulting dependence on direction need not be a linear derivative.