For fixed $x\in X$, the function $a\mapsto d(x,a)$ on $A$ is continuous, indeed $1$-Lipschitz continuous by the reverse triangle inequality. A continuous real-valued function on a nonempty compact space attains its minimum by the extreme value theorem. Hence some $a_0\in A$ satisfiesThe metric is symmetric, so this is also $d(a_0,x)$. Compactness supplies attainment of the infimum.

The distance to a set is finite and nonnegative because $A$ is nonempty. For $x,x^{\prime }\in X$ and every $a\in A$, the triangle inequality givesTaking the infimum over $a$ yields $D_A(x)\le d(x,x^{\prime })+D_A(x^{\prime })$. Interchanging $x$ and $x^{\prime }$ gives the opposite inequality. Together these implyThus the distance to a set is $1$-Lipschitz continuous. No closedness, compactness or attainment of the infimum was used.

Given $ϵ>0$, choose $\delta =ϵ/K>0$. If $d_X(x,x^{\prime })<\delta$, the Lipschitz condition yieldsThis single $\delta$ works for every pair, proving the map is uniformly continuous:

The positive integers below $30$ coprime to $30$ areThere are eight, so $\phi (30)=8$. Their sum can be grouped into four pairs summing to $30$:This agrees with the formula.

Write $S={\sum }_{k\in R_n}k$. Adding the two equal sums from part (b), and using $|R_n|=\phi (n)$, givesThereforeEquivalently, the average of the positive reduced residues is $n/2$, a consequence of their symmetry under reflection about that midpoint.

Let $R_n=\{k\in \mathbb{Z}:0<k<n,\gcd (k,n)=1\}$. The map $k\mapsto n-k$ sends $R_n$ to itself becauseApplying it twice returns $k$, so it is a bijection. Reindexing the finite sum by this bijection givesThere is no need to assume that every residue belongs to a pair of distinct elements; the reindexing also handles the fixed point when $n=2$.

For a positive integer $n$, Euler's totient function counts the integers in a complete residue system that are coprime to $n$:For $n>1$ this is equally the count with $0<k<n$. The convention gives $\phi (1)=1$.

The three nonzero vectors in ${\mathbb{F}}_2^2$ are $e_1$, $e_2$ and $e_1+e_2$. An invertible linear map permutes this three-element set, so part (b) gives a group homomorphismIf a matrix acts trivially on all three vectors, it fixes both basis vectors $e_1,e_2$, so it is the identity matrix. Thus the kernel of a group homomorphism is trivial and $\rho$ is injective.

To count the domain, its first column may be any of the three nonzero vectors. The second column must be outside the two-element span of the first, leaving two choices. HenceAn injection between these finite sets of equal size is surjective, proving

Given a group action, define $\rho (g)(j)=g\cdot j$. Each $\rho (g)$ is a permutation, since $\rho (g^{-1})$ is its inverse. The action law gives, for every $j$,Thus $\boxed{\rho :G\to S_n\text{ is a group homomorphism}}$.

Conversely, given a group homomorphism $\rho :G\to S_n$, put $g\cdot j=\rho (g)(j)$. Since a homomorphism sends the identity to the identity permutation, $e\cdot j=j$. AlsoThese are exactly the two group action axioms, so the construction defines the required action.

A left group action of $G$ on a set $S$ is a map $G\times S\to S$, written $(g,s)\mapsto g\cdot s$, satisfyingfor every $s\in S$ and $g,h\in G$, where $e$ is the identity element of $G$. In particular each $g$ acts by a bijection, whose inverse is the action of $g^{-1}$.

Choose $\alpha \in K\setminus F$. Since $[K:F]=2$, the linearly independent elements $1,\alpha$ form a basis of $K$ over $F$. Thus there are $a,b\in F$ with ${\alpha }^2=a+b\alpha$. The assumption on the characteristic of a field allows division by two. SetCompleting the square givesAlso $\delta \notin F$, since otherwise $\alpha =\delta +b/2$ would lie in $F$. Because $\alpha \in F(\delta )$ and $\delta \in F(\alpha )$, these generated subfields agree. The basis representation shows $F(\alpha )=K$, henceThe square-root generator is obtained by translating an arbitrary generator to remove the linear term of its quadratic relation.

The notation $K=F(\delta )$ means that $K$ is the smallest subfield containing both $F$ and $\delta$, namely the simple field extension generated by $\delta$. Its elements can be expressed as rational functions in $\delta$ with coefficients in $F$ and nonzero denominator.

Write $d={\delta }^2\in F$. The element $\delta$ is a root of $X^2-d$, so its minimal polynomial over $F$ has degree at most two. Since $\delta \notin F$, that degree cannot be one. The degree of the simple field extension equals the degree of the minimal polynomial, givingIn particular $1,\delta$ form an $F$-basis, so every element is uniquely $a+b\delta$. This argument also works in characteristic two; separability is not needed for this part.

The degree of a field extension is the dimension of the larger field, regarded as a vector space over the smaller field:The scalar multiplication is the multiplication of $K$ restricted to scalars in $F$. A finite extension has finite dimension; otherwise the degree is infinite.

At either quadrature node, $2x_j^2=1$, so $u(x_j)=\arctan 1=\pi /4$. Thus the Gaussian quadrature approximation isThis is the requested quadrature value; the integrand is not a cubic polynomial, so the degree-three exactness established in part (b) does not make this the exact integral.

Since $T_2=2x^2-1$, take $x_1=-1/\sqrt{2}$, $x_2=1/\sqrt{2}$. Exactness for $T_0$ and $T_1$ requires $A+B=\pi$ and $(-A+B)/\sqrt{2}=0$. ThusFor $T_2$, both node values and its weighted integral vanish. For $T_3=4x^3-3x$, oddness and symmetric nodes likewise give zero on both sides. Therefore $R(T_j)=0$ for $j=0,1,2,3$. These four Chebyshev polynomials have successive degrees zero through three and form a basis of the degree-at-most-three polynomial space. By linearity, the rule is exact on that whole space: this is two-node Gauss–Chebyshev quadrature.

Set $x=\cos \theta$ with $0\le \theta \le \pi$. The cosine addition identity giveshence $T_0=1$, $T_1=x$, and $T_{n+1}=2x{T}_n-T_{n-1}$. By mathematical induction, $T_n$ is a polynomial of degree $n$: the initial cases hold, and the leading term of $2x{T}_n$ has degree $n+1$ while the subtraction has lower degree. Its leading coefficient is $2^{n-1}$ for $n\ge 1$.

The same change of variables gives the weighted integral

For $A+\gamma B$ with $\gamma \ge 0$, the three Gershgorin discs have positive real centres $a+\gamma$, $2c+2\gamma$, $6b+6\gamma$, and radii $b+c$, $a+b+\gamma$, $a+c+\gamma$. Centre minus radius is respectivelyThus none of the discs contains zero. By the Gershgorin circle theorem, zero is not an eigenvalue, so $A+\gamma B$ is nonsingular for every $\gamma \ge 0$. Consequently

The Gershgorin circle theorem states that every eigenvalue of $A$ lies in at least one discFor a nonzero complex eigenvector $v$, choose an index $i$ with $|v_i|={\max }_j|v_j|>0$. The $i$th row of $Av=\lambda v$ givesDividing by $|v_i|$ proves membership in $D_i$. This is valid for real matrices even when the eigenvalues are complex.