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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 $X$.

The product topology on $X\times Y$ consists of arbitrary unions of rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. Equivalently, these rectangles form a basis.

The empty set is the empty union, and $X\times Y$ is itself an allowed rectangle. An arbitrary union of sets of this form is still a union of allowed rectangles. Finally,which is an allowed rectangle because the factor topologies are closed under finite intersections. Distributing intersection over unions shows that the intersection of any two product-open sets is product-open; induction handles all finite intersections. These verify every topology axiom. Rectangles alone are generally not a topology, which is why their arbitrary unions are included.

A collections library can specify a List<E> Java interface, with different implementations such as ArrayList<E> and LinkedList<E>. A method accepting List<E> can iterate or access elements according to the documented list contract without requiring one particular storage representation. Each implementation may also extend its own appropriate class superclass.

The benefit is separating a usable contract from its implementation. A parameter restricted to a concrete ArrayList<E> would exclude a linked list implementation and unnecessarily expose the storage choice. Requiring all providers to extend one concrete class would couple them to its implementation and consume Java's single class-inheritance relationship. Interfaces permit multiple implemented contracts, though clients must rely on the stated contract rather than assuming every optional operation is supported or has the same time complexity.

A growable-array library should store its backing Java array and logical size in private fields in Java. Public methods such as add, get and remove can enforce bounds and keep the size consistent with the stored elements. The representation can later change without requiring clients to modify direct field accesses.

The benefit is preserving invariants through encapsulation. If clients could assign the size or replace the backing Java array directly, they could create negative sizes, report elements that do not exist, or modify entries without the library's checks. Accessor methods should not expose a mutable backing object indiscriminately, since a private field alone does not prevent that indirect route. Private fields also do not provide a general security boundary against privileged reflection or native code; the point here is ordinary library access discipline.

A collections library can supply a genuinely generic method in Java whose own type parameter connects the input element and the result element type:For example, List<String> xs = oneElement("sample"); is type-checked without a cast. The method can also construct a List<Integer> from an integer object. The standard library's Collections.singletonList is another example of this type relationship, though its returned list is unmodifiable.

The benefit is one reusable implementation with compile-time argument/result consistency. A raw List result would lose element-type information, invite unchecked calls and require casts that may fail later. Separate methods for every element type would duplicate code. Merely enclosing a concrete type in angle brackets, such as returning List<String>, does not make a method generic; the method here declares <T>. Type erasure means the guarantee is primarily a source-level type check, not a distinct runtime class for each type argument.

An immutable library value such as java.lang.String can be a final class in Java. No class can extend it, so a client receiving a String cannot secretly receive a subclass that changes the behavior of its methods. This protects assumptions about value semantics in code using strings as keys or trusted identifiers.

The benefit is a closed implementation contract. Without final, subclasses might override methods inconsistently or introduce behavior incompatible with the library's intended guarantees. Making a class final does not by itself make its fields immutable; a correct immutable implementation also controls its state. The cost is that clients cannot customize it by inheritance and must instead use composition or a different abstraction.