Directly,Thus take $\alpha =aC+A$, $\beta =aD+B$, $\gamma =C$, and $\delta =D$. They are nonnegative and

Each finite block of a continued fraction acts by a fractional linear transformation with an integral matrix of determinant $\pm 1$. If the block $a_1,\dots ,a_N$ repeats, removing one period leaves the same tail $x$, sofor integers $A,B,C,D$. Hence $C{x}^2+(D-A)x-B=0$, with the linear case allowed when $C=0$.

For period $(a,b)$,soThe continued fraction is positive, so it is the positive rootwhile the other root is negative.

For an irreducible algebraic variety $X$ and $p\in X$, the local ring ${\mathcal{O}}_{X,p}$ consists of germs of rational functions regular on a neighbourhood of $p$. If $X\subseteq {\mathbb{A}}^n$ has ideal $I(X)$, its Zariski tangent space is

The given variety is the blowup of the affine plane at the origin. On the chart $W\ne 0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne 0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.

If $(X,Y)\ne (0,0)$, the equation forcesConsequently $\pi$ restricts to an isomorphism away from the origin, and is therefore birational. Above the origin, however,so $\pi$ is not injective and cannot be an isomorphism of algebraic varieties. Birationality gives

For any morphism $\phi :V\to V^{\prime }$ with $V^{\prime }$ affine, its restriction to the exceptional curve $E\cong {\mathbb{P}}^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V^{\prime }$ therefore have constant pullbacks. Since $E$ has more than one point, $\phi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.

Over $\mathbb{C}$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex projective variety is Euclidean compact. Since compactness is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.

The Lebesgue differentiation theorem states that if $f\in L_{\mathrm{loc}}^1({\mathbb{R}}^n)$, thenfor Lebesgue almost every $x$. The Radon-Nikodym theorem states that if two sigma-finite measures satisfy $\nu \ll \mu$, then there is a nonnegative measurable function $h=d\nu /d\mu$, unique $\mu$-almost everywhere, such that $\nu (E)={\int }_{E}hd\mu$.

For any $t\in [0,1]$, take $z=0$ and let $B$ be a planar sector of angle $2\pi t$. Every disc centred at the origin meets this sector in the proportionso its measure density at $0$ is $t$.

Apply the differentiation theorem to the indicator function $1_A$. Its averages over balls are exactly ${\rho }_{\lambda ,A}(x)$, so the Lebesgue density theorem givesalmost everywhere. If $\lambda (A)=0$, every numerator vanishes, so the density is zero wherever its denominator is nonzero. Conversely, if the density vanishes almost everywhere, the displayed identity gives $1_A=0$ almost everywhere, hence $\lambda (A)=0$.

Finally suppose $\nu$ and $\lambda$ are mutually absolutely continuous measures. Write $d\nu =hd\lambda$. Then $0<h(x)<\infty$ almost everywhere by the positive Radon-Nikodym derivative result. At almost every $x$, the differentiation theorem applied to both $h$ and $h{1}_A$ givesThus the $\nu$-density also exists and belongs to $\{0,1\}$ at $\lambda$-almost every point.