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Using the Levi-Civita symbol and Einstein summation convention, the scalar triple product isInterchange the dummy indices $i,j$. Antisymmetry gives $S=-a_j{\epsilon }_{ijk}{a}_i{b}_k=-S$, since $a_i{a}_j=a_j{a}_i$. Over the real numbers this proves

The normal vector to the first plane and the specified positive direction give $a=(1,1,1)$. Then $a\cdot m=3$ and $a\times m=(1,1,-2)$, soNext $a\times b=(-9,9,0)$. A vector perpendicular to both $a$ and $b$ is a scalar multiple of their cross product; the required length gives the two possibilities $c=(-2,2,0)$ and $c=(2,-2,0)$. The original PDF specifies approach to the x-axis. From $(4,4,-5)$ the original squared distance to that axis is $4^2+(-5{)}^2=41$. The two candidate endpoint squared distances are $6^2+(-5{)}^2=61$ and $2^2+(-5{)}^2=29$. Hence $c=(2,-2,0)$.

Adding the given $d$ yields $(7,0,-3)$. By the proved scalar triple product identity, $a\cdot b=3a\cdot (a\times m)=0$, so the only vector of the specified length for $e$ is the zero vector. Finally set $f=sm=(2s,0,s)$. The last point is $(7+2s,0,-3+s)$, and its plane condition becomes $2+8s=10$. Thus $s=1$ and $f=(2,0,1)$.

All waypoints, in the given distance units, areTherefore the final location is $(9,0,-2)$. The converted TeX omits two vector instructions and changes the axis; the computation above follows the original PDF.

A countable set is a set that is finite or can be put in bijection with the natural numbers; equivalently, it admits an injective function into the natural numbers. To enumerate the rational numbers, list pairs $(p,q)$ with $p\in \mathbb{Z}$, $q\ge 1$ in increasing order of $|p|+q$, and then list $p/q$, discarding repetitions. Each level is finite and every rational eventually appears. An enumeration $q_1,q_2,\dots$ of the rationals then enumerates $\mathbb{Q}\times \mathbb{Q}$ by listing $(q_i,q_j)$ in diagonals of constant $i+j$. ThusTo prove that the real numbers are uncountable, suppose $(0,1)$ had an enumeration $x_1,x_2,\dots$. Write each $x_n$ in its decimal expansion that does not end in an infinite string of nines. Define a new decimal $y=0.d_1{d}_2\cdots$ by taking $d_n=2$ if the $n$th digit of $x_n$ is $1$, and $d_n=1$ otherwise. Then $y\in (0,1)$, has no ambiguous terminating/nines representation, and differs from $x_n$ at digit $n$. The Cantor diagonal argument contradicts the supposed enumeration. Consequently the real numbers are uncountable.

For the discs, use the enumeration of ${\mathbb{Q}}^2$ just constructed. Every positive-radius disc contains a rational-coordinate point: its interior contains a small open rectangle, and each coordinate interval contains a rational number. For each disc $D$, let $j(D)$ be the least index of an enumerated rational-coordinate point in its interior. This minimum exists and requires no arbitrary simultaneous choice. Disjoint discs cannot receive the same point, so $D\mapsto j(D)$ is an injective function into the natural numbers. This proves the countability of pairwise disjoint open disks and also the same conclusion for closed discs, by using their disjoint interiors.

For circles, take all concentric circles centred at the origin with radii $r\in (1,2)$. Two different radii give disjoint circles, and every radius is positive. The radius map is a bijection from the uncountable interval $(1,2)$ to this family. An uncountable disjoint family of nontrivial circles is therefore possible: the rational-point argument for discs relies on their nonempty interiors, which circumferences do not have.

Begin with two points at distance one. A ruler draws the line through two known points; a compass draws a circle with a known centre and a constructible radius. New points are intersections of such lines and circles. A real number is constructible when it occurs as a coordinate of a point obtained by finitely many such operations; a complex point is constructible when both coordinates are.

If current coordinates lie in a field $K\subseteq \mathbb{R}$, a line-line intersection solves linear equations over $K$, while line-circle or circle-circle intersections reduce to a quadratic equation over $K$. Thus each step requires at most adjoining a square root. Consequently every constructible number lies in a towerConversely field sums and differences are obtained by laying off segments, products and quotients by similar triangles, and a positive square root by the geometric-mean construction in a semicircle. Hence every real number in such a tower is constructible. This proves the quadratic-tower criterion, not merely its necessity.

In particular a constructible algebraic number has degree a power of two. Degree alone is not sufficient in general; the exact criterion is containment in a quadratic tower. Equivalently the Galois group of its normal closure is a finite two-group: a tower's normal closure remains a two-extension, and a two-group admits a chain of subgroups with successive index two, yielding the converse by Galois correspondence.

The standard impossibility results follow. Doubling a unit-volume cube requires $\sqrt[3]{2}$, whose polynomial $X^3-2$ is irreducible by Eisenstein and has degree three. Trisecting an arbitrary angle would in particular trisect $60$ degrees, requiring $2\cos 20^{\circ }$, a root of $X^3-3X-1$. This cubic has no rational root and is irreducible, so this construction is impossible, although some particular angles can be trisected. Squaring the unit circle requires side length $\sqrt{\pi }$, impossible because $\pi$ is transcendental whereas constructible numbers are algebraic.

For a regular $n$-gon, constructibility is equivalent to that of $e^{2\pi i/n}$. The cyclotomic extension is Galois of degree $\phi (n)$, so it is constructible exactly when this degree is a power of two. Writing the prime factorization of $n$ shows the equivalent conditionIndeed odd prime exponents must be one, and each $p_i-1$ must be a power of two. If $2^m+1$ is prime, $m$ must itself be a power of two by factoring when $m$ has an odd divisor. Conversely this factorization makes the cyclotomic Galois group a two-group and provides a quadratic tower. This gives both positive constructions, such as the regular seventeen-gon, and obstructions, such as the regular seven-gon.

The key telescoping series identity isFor the requested mathematical induction, when $n=1$ the proposed sum is $q/(q+1)=1-q!/(q+1)!$. If the identity holds for $n$, adding $c_{n+1}=q!/(q+n)!-q!/(q+n+1)!$ cancels its last term and yields $1-q!/(q+n+1)!$. Thus it holds for every $n\ge 1$. Since $q!/(q+n)!\to 0$ for fixed positive integer $q$, the infinite series has sum $\boxed{1}$.

For the second series, the integral digit condition is $0\le a_n\le n-1$, and in particular $a_1=0$. Its partial sums are increasing and satisfyThus the bounded monotone sequence theorem proves convergence to a real number $S\in [0,1]$. The same telescoping calculation gives, for every $N\ge 1$,If positive digits occur infinitely often, then every tail contains a positive term, so $R_N>0$. If $a_n\le n-2$ occurs infinitely often, then every tail contains a digit smaller than its maximum. Indeed,Consequently the two conditions together give $0<R_N<1$ for every $N$. Were $S=A/B$ a rational number, with positive integer $B$, choose $N\ge B$. Because $B\mid N!$, the number $N!S$ is an integer, as is ${\sum }_{n=1}^N{a}_{n}N!/n!$. Their difference $R_N$ would be an integer strictly between zero and one, a contradiction.

Conversely, if positive digits occur only finitely often, then $S$ is a finite sum of rational numbers. If digits below their maximum occur only finitely often, then $a_n=n-1$ for every $n>N$ for some $N$, givingThese are exactly the two ways the conjunction of infinitude conditions can fail. Hence the irrationality criterion for factorial series is proved in both directions: $S$ is irrational exactly when both kinds of digit occur infinitely often.

For a real number $\alpha$, set ${\alpha }_0=\alpha$ and $a_j=\lfloor {\alpha }_j\rfloor$. If ${\alpha }_j-a_j\ne 0$, define ${\alpha }_{j+1}=({\alpha }_j-a_j{)}^{-1}$. The resulting simple continued fraction is $[a_0;a_1,a_2,\dots ]$, where $a_0$ is an integer and subsequent $a_j$ are positive integers. It terminates when a complete quotient is integral; otherwise its convergents tend to $\alpha$.

For the continued fraction of the square root of nineteen, the complete quotients after $a_0=4$ areTheir integer parts are $2,1,3,1,2,8$. Taking the reciprocal after subtracting $8$ from the last quotient returns the first one, so

The integers are countable: $j(z)=2z+1$ for $z\ge 0$ and $j(z)=-2z$ for $z<0$ is an injection $\mathbb{Z}\to \mathbb{N}$. Hence $\mathbb{Z}\times \mathbb{N}$ is a countable set. The map $(a,b)\mapsto a/b$ is a surjection onto the rational numbers, so the image result already proved gives $\mathbb{Q}$ is countable. Unique reduced fractions are not required for this surjection argument.

For the final unheaded geometry request, suppose for contradiction that no two distinct circles in the collection intersect. Restrict to the subcollection ${\mathcal{C}}_0$ of circles tangent to the $x$-axis. Each such nondegenerate circle has a unique tangency point, and the tangency-point map ${\mathcal{C}}_0\to \mathbb{R}$ is a surjection by the hypothesis.

The closed disk of a circle tangent at $(a,0)$ meets the $x$-axis only at $(a,0)$. Two distinct members of ${\mathcal{C}}_0$ cannot have nested disks: if one disk were inside the other, its tangency point would belong to the larger disk, so both tangency points would be equal and both circle boundaries would pass through it. That already contradicts nonintersection. Circles with disjoint boundaries and no nesting have disjoint disks. Thus their open disks are pairwise disjoint.

By the density of the rational numbers, every nonempty open disk contains a point of ${\mathbb{Q}}^2$, which is a countable set. Fix an enumeration of ${\mathbb{Q}}^2$ and assign to each disk the first enumerated point inside it. Disjointness makes this assignment an injection. This is the countability of pairwise disjoint open disks, and proves that ${\mathcal{C}}_0$ is a countable set. Its surjection onto $\mathbb{R}$ would make the real numbers countable, contrary to the Cantor diagonal argument. ThereforeThe proof allows circles on either side of the axis and counts tangential contact as intersection; it does not assume that all circles in the original collection are tangent to the axis.

A set $S$ of prime numbers has Dirichlet density $\delta$ if the following limit exists:Equivalently, the denominator can be ${\sum }_p{p}^{-s}$, since that sum is $\log (1/(s-1))+O(1)$. Adding or removing finitely many prime numbers leaves the Dirichlet density unchanged.

Put $\alpha =\sqrt[4]{2}>0$ and $L=\mathbb{Q}(\alpha ,i)$. This is the splitting field of $X^4-2$, whose four roots of a polynomial are $\alpha ,i\alpha ,-\alpha ,-i\alpha$. The polynomial is an Eisenstein polynomial at $2$, so $[\mathbb{Q}(\alpha ):\mathbb{Q}]=4$. Since $\mathbb{Q}(\alpha )$ is contained in the real numbers, it does not contain $i$, giving $[L:\mathbb{Q}]=8$. Thus $L/\mathbb{Q}$ is a Galois extension of degree $8$. Its Galois group is the dihedral group of order $8$: the field automorphisms $r(\alpha )=i\alpha$, $r(i)=i$ and $s(\alpha )=\alpha$, $s(i)=-i$ satisfy $r^4=s^2=1$ and $srs=r^{-1}$.

For an odd prime number $p$, the condition $p\equiv 1(\mathrm{mod}4)$ means that the finite field ${\mathbb{F}}_p$ contains all fourth roots of unity. Under this condition, $2$ is a quartic residue exactly when $X^4-2$ has a root in ${\mathbb{F}}_p$; multiplying that root by $1,i,-1,-i$ gives all four distinct roots. Conversely, complete splitting of $X^4-2$ over ${\mathbb{F}}_p$ gives both a fourth root of $2$ and a primitive fourth root of unity, so it also forces $p\equiv 1(\mathrm{mod}4)$.

The polynomial discriminant of $X^4-2$ is $-2^{11}$. Hence every odd prime number is unramified in $L$, and the root-splitting criterion is equivalent to its Frobenius automorphism acting trivially on all the roots. Since the roots generate $L$, this is equivalent to the Frobenius automorphism being the identity, or to $p$ being a completely split prime of $L/\mathbb{Q}$. The Chebotarev density theorem says that the unramified prime numbers with Frobenius conjugacy class $C$ have Dirichlet density $|C|/|\mathrm{Gal}(L/\mathbb{Q})|$. Here $C=\{1\}$, so

For a nonempty set $S\subseteq \mathbb{R}$, an upper bound in a partially ordered set is a real number $u$ such that $s\le u$ for every $s\in S$. Its supremum, or least upper bound, is an upper bound $L$ with $L\le u$ for every upper bound $u$. The least-upper-bound property asserts that every nonempty set of real numbers bounded above has a real supremum.

Let $(x_n)$ be a bounded, increasing sequence; here increasing allows $x_{n+1}=x_n$. Apply the least-upper-bound property to its set of terms and put $L=\sup \{x_n:n\ge 1\}$. Given $\epsilon >0$, the number $L-\epsilon$ cannot be an upper bound, so some $x_N$ satisfies $x_N>L-\epsilon$. The monotone sequence property then givesHence $|x_n-L|<\epsilon$ for all $n\ge N$. This proves the monotone bounded sequence result, with limit $\boxed{x_n⟶L}$.

For the two series, define their partial sums $A_N={\sum }_{n=1}^N{a}_n$ and $B_N={\sum }_{n=1}^N{b}_n$. Positivity makes both partial-sum sequences increasing, and termwise comparison gives $0<B_N\le A_N$. Since $(A_N)$ is a convergent sequence, it is bounded above, and therefore so is $(B_N)$. Applying the monotone bounded sequence result already proved shows that $(B_N)$ converges. By the definition of a convergent series, ${\sum }_{n=1}^{\infty }{b}_n$ converges. This is the comparison test for series for positive terms.

The Freiling axiom of symmetry isThe values of $f$ are countable subsets of the real numbers, including finite ones. One may require $x\ne y$ without changing the axiom: apply the displayed version to $x\mapsto f(x)\cup \{x\}$.

The Generalized continuum hypothesis statesHere ${\kappa }^{+}$ is the successor cardinal; equivalently $2^{{\aleph }_{\alpha }}={\aleph }_{\alpha +1}$ for every ordinal $\alpha$.

The Freiling theorem givesFirst assume the Continuum hypothesis and fix a well-order $(r_{\alpha }{)}_{\alpha <{\omega }_1}$ of the real numbers. Define $f(r_{\alpha })=\{r_{\beta }:\beta \le \alpha \}$. Each value is countable. For any pair, one index is at most the other, so at least one of $x\in f(y)$ or $y\in f(x)$ holds. The same holds for $x=y$, since $x\in f(x)$. This violates the Freiling axiom of symmetry.

Conversely, assume $|\mathbb{R}|>{\aleph }_1$ and let $f$ assign a countable set to each real. Choose $X\subseteq \mathbb{R}$ of cardinality ${\aleph }_1$. By infinite cardinal arithmetic,Choose $y$ outside this union. Since $f(y)$ is countable and $X$ is uncountable, choose $x\in X\setminus f(y)$. Then $y\notin f(x)$ and $x\notin f(y)$, with $x\ne y$. This proves the Freiling axiom of symmetry and completes both implications.

A subgroup $H$ of a group $G$ is a normal subgroup when $gH{g}^{-1}=H$ for every $g\in G$, equivalently $gH=Hg$. The quotient group $G/H$ has the left cosets as elements, with $(gH)(kH)=gkH$, identity $H$ and inverse $(gH{)}^{-1}=g^{-1}H$. Normality makes this multiplication independent of representatives.

The first isomorphism theorem for groups says that for a group homomorphism $\phi :G\to K$, its kernel of a group homomorphism is normal andThe image is taken with the induced group operation; surjectivity onto all of $K$ is not required.

For the given upper triangular matrices, take diagonal entries:The target is a direct product of groups, with multiplication of nonzero real numbers in each coordinate. Indeed the product of two such matrices has diagonal $(a{a}^{\prime },d{d}^{\prime })$, proving the group homomorphism property. Any pair of nonzero diagonal entries is attained by a diagonal matrix, so $\phi$ is surjective. Its kernel of a group homomorphism consists precisely of the matrices with $a=d=1$, namely $H$. Hencewith coordinatewise multiplication, not addition on ${\mathbb{R}}^2$.

For each positive integer $n$, consider the threshold set $A^{(n)}=\{a\in A:a\ge 1/n\}$. If it were infinite, one could choose recursively a sequence of distinct elements from it. Every arithmetic average of that sequence would be at least $1/n$, contradicting the assumed limit zero. Hence every $A^{(n)}$ is a finite set.

Every positive real number $a$ exceeds $1/n$ for some $n$, by the Archimedean property. ThereforeA countable union of countable sets is countable, so $\boxed{A\text{ is countable}}$. In fact, under the stipulated infinitude, $A$ is a countably infinite set. Positivity matters: it ensures that these thresholds cover $A$. This is countability from vanishing averages of distinct sequences; ordinary Cesaro convergence of a sequence of one chosen enumeration would not suffice to control the whole set.

Pullback of a differential form commutes with the exterior derivative, so $f^{\ast }$ gives a linear map on de Rham cohomology. Functoriality and $f\circ f=\mathrm{id}$ imply $(f^{\ast }{)}^2=I$. Over the real numbers, define the complementary projectionsThey obey $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. Every cohomology class therefore decomposes asThe first term is fixed and the second negated by $f^{\ast }$. Their intersection is zero, since a class in both equals its own negative. ConsequentlyThis is the eigenspace decomposition of a linear involution. The factor $1/2$ uses the real coefficient field; a corresponding argument over characteristic two would fail. At the level of differential forms, the same averaging projections give representatives of either parity whenever the class has that parity.

Take a transitive model $M\models \mathsf{ZFC}+2^{{\aleph }_0}={\aleph }_2$. In $M$, choose a bijection $e:{\omega }_2\to \mathcal{P}(\omega )$ and encode its graph by a set $A\subseteq {\omega }_2$, using a fixed bijection between ${\omega }_2\times \omega$ and ${\omega }_2$. The relative constructible universe $L(A)$ can decode $e$, and therefore contains every real number of $M$; being an inner model of $M$, it has no additional reals.

Models with the same reals have the same first uncountable ordinal, because their reals code exactly the same countable well-orders. If $L(A)$ satisfied the Continuum hypothesis, its bijection between ${\omega }_1$ and the reals would also belong to $M$, contradicting $M\models 2^{{\aleph }_0}={\aleph }_2$. This is the construction in relative constructible universe can violate the continuum hypothesis, and it gives

Apply Arzelà-Ascoli theorem on $[-m,m]$. Pointwise boundedness and equicontinuity give a subsequence converging uniformly there. Successively extract subsequences for $m=1,2,\dots$, and take the diagonal subsequence. It converges uniformly on every closed bounded interval. Its limits on nested intervals agree, so they define $f$ on the real numbers; the locally uniform limit of continuous functions is continuous.

Let $f_1,f_2\in V^{\ast }$ satisfy $f_1(v)f_2(v)=0$ for every vector $v$. If $f_1=0$ there is nothing to prove. Otherwise choose $u$ with $f_1(u)\ne 0$. For every $v\in \ker f_1$ and every real number $t$,so the hypothesis gives $f_2(u+tv)=0$. Taking two values of $t$ shows that $f_2(u)=f_2(v)=0$. Sincewe obtain $f_2=0$. Thus one of the two linear functionals is zero.

Sylvester's law of inertia says that, in a suitable basis, every real quadratic form isThe rank is $r=p+m$, and the signature is $\sigma =p-m$; both are independent of the chosen basis.

If $q=f_1{f}_2$, its polar form has image contained in $\mathrm{span}\{f_1,f_2\}$, so $r\le 2$. If $r=2$, the two functionals must be independent and, after taking their sum and difference, $q$ is a difference of two squares. Hence $\sigma =0$. If $r=1$, then $|\sigma |=1$. In every case,Conversely, this inequality leaves onlyup to a linear change of coordinates. These factor respectively as $0$, $(\pm x_1)x_1$, and $(x_1-x_2)(x_1+x_2)$. Therefore

Finally suppose that $q$ takes both positive and negative values. Sylvester's law supplies normalized basis vectorswith $q(e_i)=1$, $q(f_j)=-1$, and the $z_l$ spanning the radical. The $p+m+k$ vectorsare linearly independent and all satisfy $q(v)=0$. They form the required basis of isotropic vectors.

The deterministic boundholds for every real number $x$. Thereforefor every $T$, so $f_{ϵ}(B)$ converges to $|B|$ uniformly on every compact time interval, and hence ucp.

For the linear map $T:V\to W$, its image of a linear map and kernel of a linear map areBoth contain the zero vector, since $T(0)=0$. If $y_1=T(v_1)$ and $y_2=T(v_2)$ belong to the image of a linear map, then for any real numbers $a,b$,Thus $\mathrm{im}T$ is a vector subspace of $W$. If $u_1,u_2\in \ker T$, thenso $\ker T$ is a vector subspace of $V$.

Writing the unknown vector as $(x,y,z{)}^T$, the first two linear equations give $x=z-1$ and $y=z-\beta$. Substitution into the third givesThus, when $\alpha \ne 2$, put $t=-\beta /(\alpha -2)$. There is exactly one solution:When $\alpha =2$, the remaining equation is $0=-\beta$. HenceThe second case is an affine line in a vector space, so it contains infinitely many solutions. These cases exhaust all real numbers $\alpha ,\beta$.

Using the convention ${\chi }_T(x)=\det (xI-A)$, the characteristic polynomial isSince $x^2+1>0$ for every $x\in \mathbb{R}$, the only eigenvalue over the real numbers is $1$. Solving $(A-I)v=0$ givesThus the only eigenspace over the real numbers is the axis of the rotation.