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Take the parallel at height one,It has constant speed $\sqrt{2}$. Its distance from the axis is the constant $\rho =\sqrt{2}$, and its tangent is parallel to the parallel, so $\theta =0$ and it satisfies the Clairaut first integral for a surface of revolution.

However,whereas a normal vector to $x^2+y^2-z^2=1$ at $\gamma (t)$ isThese vectors are not parallel. By the ambient acceleration criterion for a surface geodesic, $\gamma$ is not a geodesic. Thus constant speed together with Clairaut's relation is not sufficient for the geodesic equations.

Choose a coprime presentationFor a polynomial $r$, write $r^{\ast }(X,Y)=r(Y,X)$. Symmetry of $f$ givesSince $\mathbb{C}[X,Y]$ is a unique factorization domain and $\gcd (p,q)=1$, one has $p\mid p^{\ast }$ and $q\mid q^{\ast }$. Exchanging the variables preserves total degree, sofor nonzero constants $\lambda ,\mu$. The displayed identity gives $\lambda =\mu$, and applying the exchange twice gives ${\lambda }^2=1$.

If $\lambda =-1$, then $p(X,X)=q(X,X)=0$, so $X-Y$ divides both polynomials, contradicting coprimality. Therefore $\lambda =1$, and both $p$ and $q$ are symmetric. Taking $g=p$ and $h=q$ proves the claim, which is the symmetric rational function in two variables result.

To deduce the final assertion, use the characterization that an integral domain is a unique factorization domain when it is atomic and every irreducible element is prime. Factoring the contents in $R$ and then the primitive parts by degree shows that $R[X]$ is atomic.

An irreducible constant of $R$ is prime in $R[X]$. Every irreducible polynomial of positive degree is, up to a constant unit, primitive. Part (i) makes it irreducible in $F[X]$; because $F[X]$ is a principal ideal domain, it is prime there, and part (ii) makes it prime in $R[X]$. Thus every irreducible of $R[X]$ is prime, andThis is the polynomial ring over a unique factorization domain theorem.

Let $N⊴G$ be proper. If $7$ divides $|N|$, then $N$ contains a Sylow $7$-subgroup of $G$, and normality makes it contain all eight of them. Thus $n_7(N)=8$, so the Sylow divisibility theorem givesSince $|N|$ divides 168 and $N$ is proper, this forces $|N|=56$.

Every nontrivial normal subgroup is transitive by normal subgroup orbits in a faithful prime-degree action. HenceBut $N\cap G_x$ is normal in $G_x\cong S_4$, whereas $S_4$ has three, rather than one, Sylow $2$-subgroups. This contradiction proves that no proper normal subgroup has order divisible by seven.

If instead $3$ divides $|N|$, normality makes $N$ contain all 28 Sylow $3$-subgroups of $G$. Hence $n_3(N)=28$, so $28$ divides $|N|/3$ and therefore $84$ divides $|N|$. A proper such subgroup would have order 84, which is divisible by seven and has just been ruled out.

Finally, if $N$ is any nontrivial normal subgroup, the prime-degree orbit argument makes $N$ transitive, so seven divides $|N|$. This is impossible for a proper normal subgroup. Therefore $G$ is simple.

Let $Q$ act by left multiplication on the set $G/P$ of left cosets. Every orbit has size a power of $p$, whileis not divisible by $p$. Therefore at least one orbit has size one. If $gP$ is fixed, then $qgP=gP$ for every $q\in Q$, so $g^{-1}Qg\le P$, equivalentlyThis is Sylow containment from a coset fixed point.

The remaining Sylow theorems state that Sylow $p$-subgroups exist, that any two are conjugate, and that their number $n_p$ satisfies

As $n\to \infty$,whileThus the credible interval and confidence interval have asymptotically equal centres and half-widths, and both contract around the true parameter. Their interpretations remain different: the confidence statement concerns repeated samples, whereas the credible statement concerns posterior probability. This is normal-normal conjugacy with known observation variance.

Write the posterior mean and variance asThe central 95 percent posterior credible interval is

Sincean exact 95 percent confidence interval is

Apart from constants, the log-likelihood isDifferentiating gives ${\ell }^{\prime }(\theta )=n(\overline{X}-\theta )/{\sigma }^2$, so

A streamline $x(s)=(x(s),y(s),z(s))$ satisfieswith $(x(0),y(0),z(0))=(1,0,0)$. ThereforeFor $\beta \ne 0$ this is a helix, with $x=\cos (\alpha z/\beta )$ and $y=-\sin (\alpha z/\beta )$; for $\beta =0$ it is the unit circle in the plane $z=0$.

Conversely, if $R\cong R_1\times R_2$ with both factors nontrivial, the inverse image of $(1_{R_1},0)$ is an idempotent. It is neither zero nor one, so (i) holds. This completes the ring product decomposition by an idempotent equivalence.

Let $e^2=e$ with $e\ne 0,1$. Then $e(1-e)=0$, and the ideals $eR$ and $(1-e)R$ are nonzero rings with identities $e$ and $1-e$. DefineThis is a ring homomorphism. Its inverse is $(x,y)\mapsto x+y$, because $ex=x$, $(1-e)y=y$, and the cross terms vanish. Hence $R\cong eR\times (1-e)R$, proving (ii).

False. Again use two states, but now takeThe support condition holds. Under $P$, the first return time to state one is identically one, so ${\mu }_1=1$. Under $Q$, the chain is irreducible with stationary distribution $(1/2,1/2)$. The mean recurrence time formula givesThus enlarging transition support does not imply a smaller mean return time.

False. On the two-state space, takeEvery positive entry of $P$ remains positive in $Q$. Under $P$, both singleton states are closed communicating classes, so neither state is a transient state. Under $Q$, however, state one eventually moves to the absorbing state two and can never return after doing so. State one is therefore transient.