The matrix trace is the sum of diagonal entries, $\mathrm{tr}A={\sum }_{i=1}^n{A}_{ii}$. For any two square matrices,Apply this cyclic property of the trace with $B=T$ and $C=A{T}^{-1}$:Thus matrix similarity preserves the matrix trace; no assumption that either matrix is diagonalizable is needed.

The map $(x,y)\mapsto (x,y,x^2+y^2)$ is a continuous surjection from the path-connected plane onto this surface. It is a homeomorphism because coordinate projection is its continuous inverse. Thus the paraboloid is path-connected and connected. Explicitly, lifting any straight planar segment by this map produces a path on the surface.

Factorization gives $y=x$ or $y=-x$. Each point can be joined to the origin by a straight segment lying on its own line. Two such segments join any pair of points, so the union is path-connected and hence connected. Alternatively it is the union of two connected lines with a common point.

The equation implies $|z|\ge 1$. Its intersections with $\{z>0\}$ and $\{z<0\}$ are disjoint, nonempty subsets open in the subspace topology, and together cover the entire surface. They therefore form a separation, proving the surface is disconnected. More precisely its two sheets are $z=\pm \sqrt{1+x^2+y^2}$; each is homeomorphic to the connected plane by projection, so these are exactly its two connected components.

The pointwise limit is the constant function $1$. For $0<x<1$, the inequality $1-e^{-u}\le u$ for $u\ge 0$ givesHence the sequence converges uniformly to one on $(0,1)$. Its exact supremum error is $1-e^{-1/n}$, although the supremum is not attained on the open interval. Compactness of the interval is not needed for this uniform convergence bound.

For each fixed $x\in (0,1)$, $1/(nx)\to 0$, so the only possible limit is zero. But taking $x_n=1/(2n)$ gives $f_n(x_n)=2$ for every $n$. In fact the supremum of $1/(nx)$ over the interval is infinite for each $n$. Thus the convergence is pointwise but not uniform; uniform convergence cannot be inferred from convergence at each fixed point.

Change the generators to $a=m_1$, $b=m_3-m_2$ and $c=m_2$. This is an invertible integer change of generators, since $m_1=a$, $m_2=c$, $m_3=b+c$. The relations become $2b=0$ and $4c=0$, with no relation involving $a$. ConsequentlyThis is already the invariant-factor order required by the structure theorem for finitely generated modules over a principal ideal domain, because $2\mid 4$. The inverse change of generators shows that there are no hidden extra relations.

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. In invariant-factor form, every finitely generated module $M$ over such a domain $R$ has a decompositionwhere the $d_i$ are nonzero nonunits, $r\ge 0$, and either list of summands may be empty. The free rank $r$ and the invariant factors $d_i$ are unique up to multiplication of each $d_i$ by a unit. Equivalently the torsion part is a direct sum of cyclic modules with prime-power annihilators. The nonunit convention omits zero summands $R/(1)$.

A Euclidean domain is a commutative integral domain $R$ with a function $\delta :R\setminus \{0\}\to {\mathbb{Z}}_{\ge 0}$ such that, for $a,b\in R$ with $b\ne 0$, there are $q,r\in R$ satisfying $a=qb+r$, where either $r=0$ or $\delta (r)<\delta (b)$. One may also impose the usual monotonicity $\delta (a)\le \delta (ab)$ for nonzero $a,b$; the example below has it.

The integers form a Euclidean domain that is not a field. Use $\delta (a)=|a|$. Integer division by $|b|$ supplies a remainder with $0\le r<|b|$, and changing the sign of the quotient handles negative $b$. The integers are an integral domain, while $2$ has no multiplicative inverse in them. This verifies both the Euclidean property and the failure to be a field.

One excursion from zero escapes forever on the positive branch with probability $a_{+}=\frac{1}{2}(1-\frac{1}{2})=1/4$, and on the negative branch with probability $a_{-}=\frac{1}{2}(1-\frac{1}{3})=1/3$. The remaining probability is the return probability $r=5/12$. By the last-excursion direction law at a transient junction, the eventual positive escape probability isEquivalently, if $p$ denotes the desired probability, the Strong Markov property gives $p=1/4+(5/12)p$. The negative escape probability is $4/7$, and their sum is one, as independently established in part (b).

On the negative branch set $Y=-X$. Its outward step probability is $3/4$ and inward probability is $1/4$. The same finite gambler's ruin recurrence as in part (a), with root $(1/4)/(3/4)=1/3$, gives ${\mathrm{Pr}}_{-1}(T_0<\infty )=1/3$. Hence, following one departure from zero, the probability of returning isAt each return the Strong Markov property restarts the same excursion experiment. The probability of at least $m$ returns is $r^m$, tending to zero. Thus almost surely there is a last visit to zero.

After that visit the walk remains strictly positive or strictly negative, because nearest-neighbour steps cannot change sign without crossing zero. On the positive branch it can be coupled, up to any return, to independent steps of mean $2/3-1/3=1/3$. The strong law of large numbers gives a walk tending to $+\infty$ almost surely. This conclusion remains true on the event that the excursion never returns: conditioning on a positive-probability event cannot make a probability-zero failure of the law occur. On the negative branch the mean step is $1/4-3/4=-1/2$, giving $-\infty$ in the same way. Apply this at each possible departure time, a countable family, or use the Strong Markov property at the successive returns. ThereforeFinite visits to zero alone would not prove divergence on an arbitrary chain; the outward-drift argument is the additional step needed here.

Until the first visit to zero, the positive-side chain is a nearest-neighbour random walk with upward probability $2/3$ and downward probability $1/3$. For an integer $m>1$, let $h_i^{(m)}={\mathrm{Pr}}_i(T_0<T_m)$, $0\le i\le m$. The Markov property givesThe characteristic roots of this recurrence are $1$ and $1/2$. Solving its two boundary equations gives the biased gambler's ruin probabilityA finite path that reaches zero has a finite maximum before doing so, and is counted by $\{T_0<T_m\}$ for every sufficiently large $m$. Conversely each such event implies a visit to zero. Therefore these events increase to $\{T_0<\infty \}$, and continuity of probability givesMore generally the hitting probability from $i>0$ is $2^{-i}$. The finite-boundary derivation avoids choosing an unjustified solution of the infinite recurrence.

Write $W={\int }_{-1}^{1}w(x)dx$, finite and positive. Exactness on the constant polynomial gives ${\sum }_i{a}_i^{(n)}=W$. Positivity therefore implies the uniform operator bound $|I_n(h)|\le W\Vert h{\Vert }_{\infty }$, and the integral has the same bound.

For a continuous $f$ and $ϵ>0$, the Weierstrass approximation theorem gives a fixed polynomial $Q$ of degree $m$ with $\Vert f-Q{\Vert }_{\infty }<ϵ$. For every $n\ge m$, polynomial exactness gives $I_n(Q)=I(Q)$. ThusSince $ϵ$ is arbitrary, $\boxed{I_n(f)\to I(f)}$ for every continuous $f$. This is convergence of positive quadrature on continuous functions. The argument requires nodes in $[-1,1]$, implicit in a rule defined on all continuous functions there; otherwise their function values would not even be specified by the stated domain. Exactness alone would not provide the norm bound if arbitrarily signed weights were allowed.

For node $x_i$, let ${\ell }_i$ be the Lagrange interpolation polynomial with ${\ell }_i(x_j)={\delta }_{ij}$. It has degree $n$, so ${\ell }_i^2$ has degree $2n$. The Gaussian quadrature with $n+1$ nodes is exact through degree $2n+1$, henceThe integrand is nonnegative. Since ${\ell }_i$ is a nonzero polynomial, it is nonzero on some open subinterval; the positive weight has positive integral there. Consequently $\boxed{a_i>0}$ for every node. This proves the positive weights of Gaussian quadrature property without assuming any coefficient sign beforehand.

Relabel the two primes so that $p<q$. By the Sylow theorems, the number $n_q$ of Sylow $q$-subgroups divides $p$, so it is either $1$ or $p$. But $n_q\equiv 1(\mathrm{mod}q)$ and $1<p<q$, ruling out $n_q=p$. Thus there is a unique Sylow subgroup of order $q$, which is a normal subgroup. It is nontrivial and proper because $1<q<pq$. Consequently no group of order $pq$ is simple. No additional assumption on divisibility of $q-1$ by $p$ is required.

The remaining Sylow theorems state that every $p$-subgroup of a finite group is contained in a Sylow subgroup, and that all Sylow $p$-subgroups are conjugate. If $|G|=p^{n}r$ with $p\nmid r$ and $n_p$ is the number of Sylow $p$-subgroups, thenIn particular $n_p=1$ if and only if that Sylow subgroup is a normal subgroup.

Let $\mathcal{S}$ be the set of all subsets of $G$ with $p^n$ elements. First show that $|\mathcal{S}|=\left(\genfrac{}{}{0ex}{}{p^{n}r}{p^n}\right)$ is not divisible by $p$. In the polynomial ring over the field of $p$ elements, $(1+X{)}^p=1+X^p$, since the intermediate binomial coefficients are divisible by $p$. Iterating givesComparison of coefficients of $X^{p^n}$ yields $\left(\genfrac{}{}{0ex}{}{p^{n}r}{p^n}\right)\equiv r\not\equiv 0(\mathrm{mod}p)$.

Let $G$ act on $\mathcal{S}$ by left translation. Because its orbit sizes sum to a number not divisible by $p$, some orbit has size prime to $p$. Choose a subset $S$ in that orbit, with stabilizer $H$. The orbit-stabilizer theorem gives the orbit size $[G:H]$, so $p\nmid [G:H]$. On the other hand $H$ acts freely on the elements of $S$: if $hs=s$, cancellation in the group gives $h=1$. Every $H$-orbit in $S$ therefore has $|H|$ elements, and $|H|$ divides $|S|=p^n$. Sinceand the second factor has no factor of $p$, the first must contain all $p^n$. Hence $\boxed{|H|=p^n}$, proving the first of the Sylow theorems. This is Sylow existence by subset action and includes the trivial case $n=0$.

At the first step of Gaussian elimination, the first column of a nonsingular matrix cannot be zero. Choose a nonzero pivot there using a row interchange and eliminate below it. The resulting matrix has block formElementary elimination and row permutations preserve nonsingularity, so $p\det B\ne 0$ and the trailing matrix $B$ is nonsingular. Its first column again has a nonzero entry. Repeating this argument inductively provides a nonzero pivot at every step and produces an upper-triangular factor. The recorded elimination multipliers, with previous entries permuted when necessary, form the unit lower-triangular factor. Every nonsingular square matrix admits $PA=LU$ under columnwise partial pivoting.

For the column-interchange convention from part (a), use the same induction with a nonzero entry in the first row of each nonsingular trailing matrix. Column interchanges put it in the pivot position, and elimination gives $AQ=LU$. These existence statements are in exact arithmetic; they do not assert that all choices have equally good numerical stability.

Use columnwise partial pivoting: select the largest absolute entry in the current column and interchange rows. In the first column the pivot is $4$, so interchange rows 1 and 2. The multipliers are $1/2$ and $-1/2$, and elimination givesThe second pivot is $5/2$, so interchange the two remaining rows. Their previously stored first-column multipliers must also be interchanged. The final multiplier is $(1/2)/(5/2)=1/5$, giving the LU decompositionThe permutation matrix orders the original rows as $2,3,1$. Multiplication verifies the displayed identity.

If “column pivoting” instead denotes interchanging columns rather than searching within a column, that convention gives $AQ=\tilde{L}\tilde{U}$ withThis second identity uses a different pivot convention; it does not change the row-pivoted answer above.

Parametrize the logarithmic spiral byThe map is continuous into the ambient plane, hence into $X$ with its subspace topology, and it is onto by the definition of the spiral. It is injective because its radius $|h(t)|=e^t$ determines $t$ uniquely, even when the angular coordinate winds through repeated turns.

Every point of $X$ is nonzero. The inverse is thereforethe restriction of a continuous function on the punctured plane. Hence $h$ is a homeomorphism and the spiral is homeomorphic to the real line. Its accumulation at the origin causes no failure of this argument: the origin is not included in $X$.