We prove absolute integrability. On $(0,1]$, $|\sin (1/x)|\le 1$ and the denominator is at least one, so $|f(x)|\le 1$. On $[1,\infty )$, the standard inequality $|\sin t|\le |t|$ givesSubstitution $t=\log x$ yieldsThus ${\int }_0^{\infty }|f(x)|dx\le 1+\pi /2<\infty$, and the function is integrable despite its oscillation near zero.

For every real $x$, the numerator is at most $2e^{|x|}$, while the denominator is at least $e^{2|x|}$. ThereforeThe upper bound is integrable by part (a), so the comparison theorem shows that the function is integrable on $\mathbb{R}$.

On finite intervals the function is continuous, butConsequently its nonnegative integral on the half-line is infinite: it is not integrable. The divergence occurs at infinity, not at the origin.

The function is nonnegative, measurable, and even. Direct integration givesThus it is integrable, with integral $\boxed{2}$.

Reducing modulo $23$ gives $189\equiv 5$. Reciprocity with five has positive sign, sosince the nonzero squares modulo five are one and four. Modulo $59$, $189\equiv 12=4\cdot 3$, and $59\equiv -1(\mathrm{mod}12)$. Thus part (b) givesAs $1357=23\cdot 59$, a solution modulo $1357$ would in particular solve the modular congruence modulo $23$, contradicting the first symbol. Therefore there is no integer solution modulo $1357$.

The quadratic reciprocity states that for distinct odd primes $p,q$,With $q=3$, it gives $\left(\frac{3}{p}\right)=(-1{)}^{(p-1)/2}\left(\frac{p}{3}\right)$. The nonzero square modulo three is one, so $\left(\frac{p}{3}\right)=1$ when $p\equiv 1(\mathrm{mod}3)$, and $-1$ when $p\equiv -1(\mathrm{mod}3)$. An odd prime other than three is congruent to $1,5,7$, or $11$ modulo twelve. Evaluating the two signs givesThus

Write $|G|=p^n$, where $n\ge 1$. By part (c), each conjugacy-class size divides $p^n$ and hence is a power of $p$. Every nonsingleton class consequently has size divisible by $p$. The class equation impliesThe identity element lies in the centre, so the centre is nonempty. Its size is a positive multiple of $p$, and thereforeThis proves that a nontrivial finite p-group has a nontrivial centre.

Conjugacy is an equivalence relation: the identity element gives reflexivity, conjugation by $g^{-1}$ reverses conjugation by $g$, and composing conjugations gives transitivity. Its equivalence classes therefore partition the finite set $G$. Counting this disjoint union givesSeparating singleton classes from the others gives the class equation $|G|=|Z(G)|+{\sum }_{|C_i|>1}|C_i|$.

An element $x$ is central exactly when $gx=xg$ for every $g\in G$. By part (b), this is exactly the assertion that every $g$ stabilizes $x$. That means every conjugate of $x$ is $x$; conversely, a singleton conjugacy class has this property for every $g$.In particular, the singleton classes are in one-to-one correspondence with the elements of the centre.

Let $H=Z(x)$. The map $G/H\to C(x)$ sending $gH$ to $gx{g}^{-1}$ is well defined: multiplying $g$ on the right by an element commuting with $x$ leaves its conjugate unchanged. It is onto by the orbit definition. If $gx{g}^{-1}=hx{h}^{-1}$, then $h^{-1}g$ commutes with $x$, so $gH=hH$, proving injectivity.

Every coset has $|H|$ elements, so $|C(x)|=|G|/|H|$. ThereforeThis is the orbit-stabilizer theorem specialized to conjugation.

By definition, $g$ stabilizes $x$ exactly when $gx{g}^{-1}=x$. Multiplying on the right by $g$ gives the equivalent condition $gx=xg$. HenceThus the stabilizer in this action is the centralizer of the element.

The identity element satisfies $e\ast x=ex{e}^{-1}=x$. For $g,h\in G$,Both action axioms hold, so conjugation is a group action on $G$.

A constructible regular polygon can be translated, rotated, and rescaled by constructible operations to one inscribed in the unit circle with a vertex at $(1,0)$. Thus the real and imaginary parts of $\zeta =e^{2\pi i/p}$ are constructible. Part (c) puts both in a real field $K$ of degree $2^n$ over $\mathbb{Q}$. Consequently $\zeta \in K(i)$, and $[K(i):\mathbb{Q}]=2^{n+1}$ because $K\subset \mathbb{R}$.

For prime $p$, the supplied irreducibility of $1+x+\cdots +x^{p-1}$ shows that $[\mathbb{Q}(\zeta ):\mathbb{Q}]=p-1$. The intermediate-field tower law implies that $p-1$ divides $2^{n+1}$. Every positive divisor of a power of two is a power of two, soThis necessary condition includes $p=2$ with $r=0$. It does not assert sufficiency for arbitrary integers of this form.

Induct through the finitely many construction steps. Initially all available coordinates lie in $\mathbb{Q}$. If coordinates obtained so far lie in a real field $F$, the line through two of its points has coefficients in $F$. A circle with centre in $F^2$ and radius a distance between two such points has equationwhose coefficients lie in $F$, since the squared distance $r^2$ lies in $F$; the radius itself need not lie there.

Part (b) therefore puts every new intersection in $F$ or a real quadratic extension of $F$. If a step produces several points, adjoin their coordinates successively, omitting extensions that add nothing. A finite construction ends in a real field $K$ containing the desired coordinates, with a tower of genuine quadratic extensionsThe tower law for field extensions yieldsA construction involving no nontrivial extension has $n=0$ and degree one.

Consider an isolated intersection of the two curves. For two distinct lines, solving their linear equations by elimination gives coordinates in $F$ whenever there is an intersection.

For a line and a circle, solve the line for one coordinate with a nonzero coefficient and substitute into the circle equation. The resulting quadratic has coefficients in $F$. A real root lies in $F$ if its discriminant is a square in $F$, and otherwise in $F(\sqrt{D})$ for its positive nonsquare discriminant $D\in F$. The other coordinate is an $F$-linear expression in that root. Thus both coordinates lie in the same real quadratic extension. A tangent intersection has discriminant zero and already lies in $F$.

Write two circle equations in normalized form $x^2+y^2+u_{i}x+v_{i}y+w_i=0$. Subtracting eliminates the quadratic terms. If the circles are distinct and meet, this gives a nontrivial line, reducing the problem to the preceding case. Concentric distinct circles do not intersect. ThereforeThe isolated-intersection qualification is implicit in the construction operation. If coincident curves were allowed and every point on their common curve counted as an intersection point, the printed assertion would be false: a rational line contains points with transcendental coordinates.

Start with $O=(0,0)$ and $E=(1,0)$. A constructible point is a point obtained from these by finitely many ruler-and-compass steps. At each step one may draw the line through two previously obtained distinct points, or the circle with a previously obtained centre and radius equal to the distance between two previously obtained points, and take isolated intersection points of the drawn lines or circles. A constructible line is a line through two distinct constructible points; a constructible circle has constructible centre and a radius obtainable as such a distance. These conventions also permit copying a previously obtained segment with the compass.

Finite construction from a fixed unit segment is essential: a drawing through an arbitrary unspecified real point does not make that point constructible.