Collapse the equator $E=\{(u,v,t)\in S^2:t=0\}$ to one point and identify no other points:Each closed hemisphere is a disk; collapsing its boundary gives a sphere. The two boundaries are the same equator, so the resulting spheres meet at the single collapsed point. This is the sphere with its equator collapsed.

An explicit verification avoids relying only on the picture. Away from the poles, use $p=(\sqrt{1-t^2}\cos \theta ,\sqrt{1-t^2}\sin \theta ,t)$ and defineAt the north and south poles use $(2,0,0)$ and $(-2,0,0)$, respectively. The angular terms tend uniformly to zero at both poles, making these extensions continuous. Both hemisphere formulas agree at the equator, with value $(0,0,0)$. Their images lie on the two specified unit spheres and cover them; away from the common origin the height parameter and angle are recovered uniquely. Thus the fibres are exactly the stated equivalence classes. The preceding compact-to-Hausdorff argument gives $\boxed{S^2/\mathcal{R}\cong Y}$.

A basis of a topology $\mathcal{T}$ is a collection $\mathcal{B}\subseteq \mathcal{T}$ such that every open set is a union of members of $\mathcal{B}$. Equivalently, for each open $U$ and every $x\in U$, there exists $B\in \mathcal{B}$ with $x\in B\subseteq U$. The empty union represents the empty set. In particular, the basis covers $X$, and any intersection of two basis elements is itself a union of basis elements.

First use the preceding isometry $t$ to align the first two points, and put $R_3=t(P_3)$. It remains to align $R_3$ with $Q_3$ while fixing $Q_1,Q_2$. Use orthonormal coordinates in which $Q_1=(0,0)$ and $Q_2=(d,0)$, $d>0$, and write $R_3=(r,s)$, $Q_3=(u,v)$. The prescribed distances giveSubtracting yields $r=u$, and then $s^2=v^2$. Therefore $Q_3$ is either $R_3$ or its reflection across the line through $Q_1,Q_2$. In the first case take $s=t$; in the second compose $t$ with that reflection. Both constructions fix the first two images and send the third point to its required image. The argument also covers three collinear points, when the two possible third images coincide. Thus all three points can be matched by one plane isometry.

Put $v=P_2-P_1$ and $w=Q_2-Q_1$. They are nonzero vectors with the same Euclidean norm. Choose the rotation matrix $R$ taking the direction $v/\Vert v\Vert$ to $w/\Vert w\Vert$; equal lengths then give $Rv=w$. DefineThis is a translation by $Q_1-P_1$ followed by a rotation about $Q_1$, so it is an isometry. Direct substitution gives $t(P_1)=Q_1$ and $t(P_2)=Q_1+w=Q_2$.

An isometry of the Euclidean plane is a map $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ preserving all Euclidean distances, $\Vert T(x)-T(y)\Vert =\Vert x-y\Vert$. Such a map of the whole plane is automatically bijective, and has the form $T(x)=Rx+b$ with $R$ an orthogonal matrix.

The classification is identity, translations, rotations about a point, reflections in a line, and glide reflections. Here a glide reflection is a reflection followed by a nonzero translation parallel to its mirror. The cases preserving orientation have $\det R=1$ and are translations or rotations; the orientation-reversing cases have $\det R=-1$ and are reflections or glide reflections. The identity is included by allowing a zero translation or rotation angle.

The tower law for field extensions states that, for $F\subseteq K\subseteq L$ with both successive degrees finite,Indeed, if $u_i$ form an $F$-basis of $K$ and $v_j$ form a $K$-basis of $L$, their products $u_i{v}_j$ form an $F$-basis of $L$: expansion gives spanning, and independence follows first over $K$ and then over $F$.

Starting from a unit segment, every straightedge and compass construction introduces coordinates by solving linear or quadratic equations over the coordinates already available. Thus a constructible number belongs to a tower of field extensions whose successive degrees are one or two; the total degree is a power of two. Doubling the volume of a unit cube would require the edge length $\sqrt[3]{2}$. Its minimal polynomial over $\mathbb{Q}$ is $x^3-2$, irreducible by Eisenstein's criterion at the prime number two. Thus $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. If that number lay in a construction field of degree $2^r$, the tower law would make three divide $2^r$, a contradiction. Therefore doubling the cube by straightedge and compass is impossible.

Let $[L:F]=n<\infty$. The intermediate field $K$ is an $F$-linear linear subspace of $L$, so the finite-dimensional linear subspace theorem gives $[K:F]\le n$. To handle the other field extension, take an $F$-basis $v_1,\dots ,v_n$ of $L$. Every $\ell \in L$ is a sum $\sum f_i{v}_i$ with $f_i\in F\subseteq K$, so the same list spans $L$ as a $K$-vector space. It need not remain linearly independent, but spanning suffices to give $[L:K]\le n$. Hence both intermediate extensions are finite.

The degree of a field extension is the dimension of the larger field as a vector space over the smaller one:The field extension is a finite field extension precisely when this dimension is finite, equivalently when finitely many elements of $K$ span $K$ over $F$. Finiteness here concerns the dimension, not the number of elements of the fields; for example, $\mathbb{C}$ has degree two over $\mathbb{R}$ although both fields are infinite.

Take the invertible matrix $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, whose determinant is $-1$. Suppose it had an LU factorization with any lower triangular $L$ and upper triangular $U$:The $(1,2)$ entry gives ${\ell }_{11}{u}_{12}=1$, so ${\ell }_{11}\ne 0$. The $(1,1)$ entry gives ${\ell }_{11}{u}_{11}=0$, so $u_{11}=0$. But then the $(2,1)$ entry would be ${\ell }_{21}{u}_{11}=0$, contradicting $a_{21}=1$. Thus this nonsingular matrix has no unpermuted LU factorization, even without imposing a unit diagonal on $L$. A row permutation resolves the zero pivot, explaining the role of partial pivoting.

The elimination multipliers are $m_{21}=2$, $m_{31}=4$, and $m_{32}=3$. The unit lower triangular matrix in the LU factorization is thereforeDirect multiplication checksThe positive signs in $L$ undo the row subtractions used during Gaussian elimination.

The first pivot is two. Gaussian elimination without partial pivoting uses $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3-4R_1$, givingThe second pivot is one; $R_3\leftarrow R_3-3R_2$ gives the upper triangular matrixBoth pivots used for division are nonzero, so no row exchange is needed.

Integrate the Lagrange interpolation polynomial and exchange the integral with the finite sum:Therefore the interpolatory quadrature weights areThe resulting quadrature rule is exact for every polynomial of degree at most $n$, since its interpolant is itself. For a general continuous $f$, replacing $f$ by its interpolant supplies the approximation.

If $p$ and $q$ both interpolate the data and have degree at most $n$, the polynomial $r=p-q$ has the $n+1$ distinct roots $x_0,\dots ,x_n$. A nonzero polynomial of degree at most $n$ has at most $n$ roots: repeated use of the factor theorem would otherwise make a product of $n+1$ linear factors divide it. Hence $r=0$, proving the interpolating polynomial is unique.

The Lagrange interpolation polynomial isThe distinct nodes make all denominators nonzero. Each Lagrange basis polynomial has degree $n$ and satisfies ${\ell }_j(x_i)={\delta }_{ij}$, so the sum has degree at most $n$ and interpolates every specified value.