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Take points $p=(3,0,4)$ and $q=(-2,3,3)$ with directions $u=(1,3,-1)$ and $v=(0,1,-1)$. A common normal is $u\times v=(-2,1,1)$, of length $\sqrt{6}$. The planes through the two lines with this common normal have separationThis is a lower bound for the distance between skew lines. It is attained at $s=-1$ and $t=-4$: the two points are $(2,-3,5)$ and $(-2,-1,7)$, whose difference $(4,-2,-2)$ is perpendicular to both directions and has length $2\sqrt{6}$. Thus the shortest distance is $\boxed{2\sqrt{6}}$.

The plane through $a_i$ has the constant normal coordinate $r\cdot \hat{n}=a_i\cdot \hat{n}$. Since $\hat{n}$ is a unit vector, the separation of these normal coordinates is a length. Every displacement joining the planes has this fixed normal component, so its length is at least its absolute value; a displacement parallel to $\hat{n}$ attains that bound. Hence the distance between parallel planes isThe absolute value, present in the PDF, is necessary because distance is nonnegative.

The first coordinate of the vector triple product isThe other two coordinates follow by cyclically permuting the indices, proving the identity.

A point on the line has the form $r=b+sm$. Substituting in the plane equation gives $s(m\cdot n)=(a-b)\cdot n$. When $m\cdot n\ne 0$ the parameter, and therefore the intersection point, is unique. The vector triple product gives $n\times (b\times m)=(n\cdot m)b-(n\cdot b)m$, so the unique intersection isFor the parallel case $m\cdot n=0$, the line is entirely in the plane if $(b-a)\cdot n=0$, and otherwise has no intersection. As usual, a genuine line and plane require nonzero direction and normal vectors. This is the line-plane intersection criterion.

Let $a_n=n!/n^n$. Its consecutive coefficient ratio isThe ratio test for the full terms therefore has limit $|z|/e$: the series has absolute convergence when $|z|<e$ and its terms eventually grow in magnitude when $|z|>e$. Thus the radius of convergence is $\boxed{R=e}$. No boundary decision is needed to determine the radius of convergence.

Expanding the characteristic polynomial givesSolving the three nullspace systems gives the eigenvalues and eigenspacesFor example, at $\lambda =1$ the first two equations force $z=0$ and $x=y$; at $\lambda =2$ they force $x=0$ and $y=-z$; at $\lambda =-1$ they force $x=y=z$.

Putting these eigenvectors in columns proves their linear independence:Hence $A$ is a diagonalizable matrix over the reals.

An eigenvector of a matrix $A$ is a nonzero vector $v$ for which $Av=\lambda v$. The scalar $\lambda$ is its eigenvalue. For a fixed eigenvalue, the zero vector together with all corresponding eigenvectors forms the eigenspace $\ker (A-\lambda I)$. The requirement $v\ne 0$ prevents every scalar from being an eigenvalue merely because $A0=\lambda 0$.

For a real line $ux+vy+w=0$, with $(u,v)\ne (0,0)$, choose $B=(u-iv)/2$ and $C=w$. Direct multiplication gives $Bz+\overline{B}\overline{z}=ux+vy$, so every line has the required form with $B\ne 0$.

For a circle of centre $c$ and radius $\rho >0$, expand its modulus equation:Thus the required circle parameters are $\boxed{B=-c,C=|c{|}^2-{\rho }^2}$. Conversely, completing the square in the circle form gives $|z+B{|}^2=|B{|}^2-C$, which describes a genuine circle when $|B{|}^2-C>0$. The complex conjugate terms make both equations real-valued.

The exponent in the PDF is $2z^2$. A complex exponential equals one precisely when its exponent is an integer multiple of $2\pi i$: its real part must be zero, and its imaginary part must be a full rotation. Therefore $z^2=\pi ik$ for $k\in \mathbb{Z}$. Taking both square roots yields all solutionsThe first nonzero family corresponds to positive exponent index, and the second to negative index. Both signs are required.

For a nondecreasing function, the endpoint bounds show it is bounded. On the uniform partition into $N$ intervals, its Darboux sums satisfyFor a nonincreasing function apply the same calculation to $-f$, or use $|f(1)-f(0)|/N$. Thus every monotone function on $[0,1]$ is Riemann integrable.

For the unheaded rational-weight example, each summand is nonnegative, and increasing $x$ only adds indices. Hence its sum is nondecreasing and lies between zero and one, including the separately specified value at zero. It is a left-continuous cumulative function of an atomic measure.

If $q_j\in [0,1)$ and $q_j<x\le 1$, the index $j$ appears in the sum at $x$ but not at $q_j$. All other changes are nonnegative, so $f(x)-f(q_j)\ge 2^{-j}$. This remains true for arbitrarily small positive $x-q_j$, proving discontinuity at $q_j$, including a right discontinuity at zero when that rational is encountered. Every interval of positive length contains such a rational in its interior. The discontinuities are therefore dense, although the function is Riemann integrable by the monotone-function result just proved.

More concretely, truncate after the first $N$ weights. The resulting finite sum of step functions is integrable, and the uniform tail is at most $2^{-N}$. This also proves integrability and gives, if its value is desired,The integral formula follows by integrating the finite sums and using the uniform tail bound. Here “every interval” has the usual nondegenerate-interval meaning; a singleton is not being asserted to contain a discontinuity.

For a bounded real function on $[0,1]$, take a partition $P$ with nodes $0=x_0<\cdots <x_N=1$. Its lower Darboux sum and upper Darboux sum areThe function is Riemann integrable when ${\sup }_{P}L(f,P)={\inf }_{P}U(f,P)$; their common value is its Riemann integral. Equivalently, for every $\eta >0$ some partition has $U-L<\eta$. Indeed a common refinement of nearly extremizing upper and lower partitions proves necessity, while arbitrarily small gaps force equality of the two integrals. This is the Riemann integrability criterion.

A continuous function on a compact interval is bounded and uniformly continuous. Choose a mesh small enough that its oscillation on each subinterval is less than $\eta$. Then $U-L\le \eta {\sum }_j(x_j-x_{j-1})=\eta$; using $\eta /2$ first makes the desired inequality strict. Thus every continuous function on $[0,1]$ is Riemann integrable.

From $f^{\prime \prime }=f$, differentiating successively gives $f^{(2r)}=f$ and $f^{(2r+1)}=f^{\prime }$ whenever those derivatives have been constructed. Starting with twice differentiability, this relation supplies the next two derivatives at each stage. Thus $f$ is infinitely differentiable, and its origin series isFor a fixed $x$, both $f$ and $f^{\prime }$ are bounded on the compact interval between zero and $x$, say by $M_x$. Every higher derivative is one of these two functions. Taylor's theorem, with the remainder just proved, therefore bounds the order-$n$ remainder by $M_x|x{|}^n/n!$, which tends to zero: its consecutive positive-term ratio is $|x|/(n+1)$. Hence the series converges to $f(x)$ for every real $x$. It also gives $f(x)=A\cosh x+B\sinh x$ by the usual hyperbolic-function series.

At any other centre $a$, apply the proved origin result to $g(t)=f(a+t)$, which also satisfies $g^{\prime \prime }=g$. Its even derivatives at zero are $f(a)$, and its odd ones are $f^{\prime }(a)$. ConsequentlyThis proves convergence at every centre without assuming that smoothness alone implies equality to a Taylor series.

For $x\ne 0$, put $P(t)={\sum }_{k=0}^{n-1}{f}^{(k)}(0)t^k/k!$ and $K=(f(x)-P(x))/x^n$. The auxiliary function $F(t)=f(t)-P(t)-K{t}^n$ satisfiesRolle's theorem first gives a zero of $F^{\prime }$ strictly between zero and $x$. Apply Rolle's theorem to $F^{\prime }$ between that zero and zero, where $F^{\prime }(0)=0$, to obtain a zero of $F^{\prime \prime }$. Continue in this way. The derivatives through order $n-1$ are continuous, since they are differentiable; hence every application is valid even if $f^{(n)}$ is not continuous. After $n$ applications there is $c$ strictly between zero and $x$ with $F^{(n)}(c)=0$.

But $F^{(n)}(c)=f^{(n)}(c)-n!K$, so $K=f^{(n)}(c)/n!$. Writing $c=\theta x$ yields the Taylor formula with Lagrange remainderThe same construction using repeated applications of Rolle's theorem applies when $x<0$, with intervals ordered from $x$ to zero. For $x=0$ the identity is immediate and any $\theta \in (0,1)$ works. This proves the Taylor theorem with Lagrange remainder using only the allowed theorem and elementary differentiation.

Let $D=f_{+}^{\prime }(a)$ and write $g(t)=(f(t)-f(a))/(t-a)$ for $t>a$. By the assumed right derivative, choose $\delta >0$ with $|g(t)-D|<ϵ/2$ whenever $a<t<a+\delta$. Choose $b_0\in (a,\min (b,a+\delta ))$.

Apply the mean value theorem to $f$ on $[a,b_0]$. It gives $x\in (a,b_0)$ with $f^{\prime }(x)=g(b_0)$. Both $g(x)$ and $g(b_0)$ are within $ϵ/2$ of $D$, soThis proves the endpoint secant-tangent approximation without any continuity assumption on $f^{\prime }$. The continuity required by the theorem is only that of $f$ itself.

For continuity at any $x$, choose $\delta ={\eta }^{1/\alpha }$ for a desired error $\eta >0$. If $|y-x|<\delta$, the stated bound gives $|f(y)-f(x)|<\eta$. Thus $f$ is continuous, in fact uniformly continuous.

When $\alpha >1$, the difference quotient satisfiesTherefore the derivative exists at every point and is zero. Part (i) proves $f$ is constant. This is the lemma that Hölder exponent greater than one forces constancy; no differentiability was assumed in advance.

The mean value theorem states that a function continuous on $[p,q]$ and differentiable on $(p,q)$ has some $c\in (p,q)$ with $f(q)-f(p)=f^{\prime }(c)(q-p)$.

Take any $p<q$ inside the given open interval. Differentiability implies the continuity needed on $[p,q]$, and the derivative is zero throughout it. The mean value theorem therefore gives $f(q)=f(p)$. Since the two points were arbitrary, the function is constant on the interval.

Let $C$ and $D$ be the connected components of $x$ and $y$. First, $C\times D$ is connected in the product topology. For each $c\in C$, the connected slice $\{c\}\times D$ meets the connected slice $C\times \{y\}$. Their union over $c$ is connected by the common-intersection argument, and is exactly $C\times D$.

Now let $K$ be the connected component of $(x,y)$ in $X\times Y$. The coordinate projections are continuous, so their images of $K$ are connected and contain $x$ and $y$. Maximality therefore gives $K\subseteq C\times D$. Conversely, the connected set $C\times D$ contains $(x,y)$, so it lies in $K$. $K=C\times D$. This is connected components of a product space; it requires neither local connectedness nor path connectedness.

Fix $x$, and let $C_x$ be the union of all connected subsets containing $x$. Their union is connected: a separation of the union would place $x$ on one side, and every connected member meeting that side would have to lie entirely there. The other side would be empty. By the defining equivalence relation this union is exactly the connected component of $x$.

The closure of a connected set is connected as well. Indeed, if its closure had a separation into relatively open nonempty pieces, density would make both pieces meet the original set and give a separation of that connected set. Therefore $\overline{C_x}$ is a connected subset containing $x$, so maximality gives $\overline{C_x}\subseteq C_x$. The reverse inclusion is automatic. Every connected component is connected and closed.

The map $\phi :M\to N$, $x\mapsto rx$, is a module homomorphism, and it is injective because $r\ne 0$ and $M$ is a torsion-free module. The linear independence in a module of $T$ makes the map $R^{|T|}\to N$ sending the standard basis to $T$ an isomorphism. Thus $N$ is a finite free module, and $M\cong \phi (M)\subseteq N\cong R^{|T|}$. This proves embedding a finitely generated torsion-free module in a finite free module. No claim that the submodule itself is free over an arbitrary integral domain is needed.