Choose a vertex $v$ of $K_6$. At least three of its five incident edges have one colour, say blue, by the pigeonhole principle. Let their other endpoints be $x,y,z$. If one of $xy,yz,zx$ is blue, it completes a blue triangle with $v$; if none is blue, $xyz$ is a yellow triangle. Thus every colouring of $K_6$ has a monochromatic triangle.

The statement fails for $K_5$. Colour the edges of a five-cycle blue and its complementary five-cycle yellow. Neither colour contains a triangle.

False. Take $\alpha =\omega$ and $\beta =\omega \cdot 2$. Thenbut associativity and the limit clause for ordinal multiplication give

Always true. Under the synthetic definition of ordinal multiplication, $\alpha \beta$ is the order type of $\beta$ consecutive copies of $\alpha$. Grouping the copies first into $\gamma$ blocks of $\beta$ copies, or first into single blocks of type $\alpha \beta$, gives the same lexicographic well-order. Hence

False. With $\alpha ={\omega }_1$ and $\beta ={\omega }_1+1$, both ordinals are uncountable, butOrdinal addition is therefore not commutative even for uncountable ordinals.

Always true. The synthetic definition identifies both sides with the order type obtained by concatenating well-orders of types $\alpha$, $\beta$, and $\gamma$ in that order. Thus ordinal addition is associative:

Let $|{\alpha }_1^{\perp }⟩$ be a unit vector in $S$ orthogonal to $|{\alpha }_1⟩$, and chooseOutcome zero is then impossible when Alice sent $|{\alpha }_1⟩$, so it identifies $|{\alpha }_0⟩$ without error. This measurement succeeds with probabilityThe optimum under the extra requirement is at least the performance of this admissible measurement, so

By the Rayleigh-Ritz variational principle, $\mathrm{tr}({\Pi }_{0}D)$ is maximized by taking ${\Pi }_0$ to project onto the positive-eigenvalue eigenspace of $D$. Put $|⟨{\alpha }_0|{\alpha }_1⟩|=\cos \theta$. On $S$, the characteristic polynomial of $D$ isso its positive eigenvalue isPart (b) then gives the Helstrom-Holevo bound

Since ${\Pi }_1=I-{\Pi }_0$, the success probability is

The operator $D$ vanishes on the orthogonal complement of $S=\mathrm{span}\{|{\alpha }_0⟩,|{\alpha }_1⟩\}$, so it has rank at most two. The states are distinct rays, hence they are linearly independent and $S$ has dimension two. In an orthonormal basis of $S$, its determinant isThus its two eigenvalues on $S$ are nonzero and have opposite signs. All remaining eigenvalues are zero.

Under the gauge transformation,because the curl of a gradient vanishes, andThe transformed Lagrangian isA total time derivative changes the action only by endpoint terms, whose variation vanishes for fixed endpoints, so the Euler-Lagrange equations are unchanged.

The new momentum isUse the type-two generating function for a canonical transformationIt gives $\mathbf{p}=\partial F_2/\partial \mathbf{r}=\tilde{\mathbf{p}}-q\nabla f$, $\tilde{\mathbf{r}}=\partial F_2/\partial \tilde{\mathbf{p}}=\mathbf{r}$, andThis equals $|\tilde{\mathbf{p}}-q\tilde{\mathbf{A}}{|}^2/(2m)+q\tilde{\varphi }$, so the gauge change is a time-dependent canonical transformation.

The canonical momentum is $\mathbf{p}=m\dot{\mathbf{r}}+q\mathbf{A}$, and the Legendre transform givesIts Hamilton's equations areand, componentwise,Combining the second equation with $\mathbf{p}=m\dot{\mathbf{r}}+q\mathbf{A}$ recovers the Lorentz force law.

The canonical momentum isIn components, the Euler-Lagrange equation isThusThe last term is the $i$th component of $\mathbf{v}\times (\nabla \times \mathbf{A})$, sothe Lorentz force law.

Apply the Hankel representation with $z$ replaced by $1-z$:Enlarge $H$ to a modified Hankel contour that encloses the simple poles $t=2\pi in$, $n\in \mathbb{Z}\setminus \{0\}$. In a strip where the deformation and residue sum converge, the outer pieces vanish and the orientation gives the Hankel integral as minus $2\pi i$ times the sum of the enclosed residues. Part (b) givesSubstitution therefore yieldsBoth sides are meromorphic functions of $z$, so the identity theorem extends the equality from the initial strip to every point where either side is defined.

For $n\ne 0$, the denominator has derivativeHence the simple-pole residue is

Take the Hankel contour about the negative real axis: it starts at $-\infty$ just below the cut, circles the origin counterclockwise, and returns to $-\infty$ just above the cut. The notation $(0+)$ records this return side of the branch point. Choose the branch $-\pi <\arg t<\pi$ for $t^{z-1}$.

For $\mathrm{Re}z>1$, collapse the contour onto the two banks of the cut. Their phase difference is determined byand the resulting sine factor, together with the Gamma reflection formula $\Gamma (z)\Gamma (1-z)=\pi /\sin (\pi z)$, recovers the integral representation (‡). The Hankel integral converges exponentially at its unbounded end and, after the standard local subtraction of the singular part at zero, depends holomorphically on $z$. The apparent poles of $\Gamma (1-z)$ are cancelled by zeros of the contour integral except at $z=1$. It therefore gives the analytic continuation of the Riemann zeta function to $\mathbb{C}\setminus \{1\}$.

True. The Huffman tree is a full binary tree, so its leaves form a complete prefix code. The Kraft inequality is therefore an equality:

False. For probabilities $(0.30,0.25,0.25,0.20)$, every length-two code is optimal. A code in which the $0.30$ and $0.20$ symbols are sibling leaves is optimal, but no Huffman construction produces that labelled tree, because its first merge must pair $0.20$ with a $0.25$ symbol.

False. For probabilities $(0.4,0.2,0.2,0.1,0.1)$, first merge the two $0.1$ symbols. Subsequent ties allow Huffman trees with length multisets $(2,2,2,3,3)$ and $(1,3,3,3,3)$, both of expected length $2.2$.

The conclusions of (b) and (c) remain true. Indeed, the unit circle is a path-connected space, and the upper semicircle supplies a path from $-1$ to $1$; restricting the proposed closed cover or two-valued continuous function to that path reduces either claim to the corresponding result on $I$.

The fixed-point conclusion (d) fails. For example, any nontrivial spatial rotation $f(z)=e^{i\theta }z$ of the unit circle has no fixed point when $e^{i\theta }\ne 1$; the antipodal map $f(z)=-z$ is the simplest example.

Suppose $I=A\cup B$, where $A$ and $B$ are disjoint closed sets with $-1\in A$ and $1\in B$. Since each is the complement of the other, both are also open in $I$. Define $f=-1$ on $A$ and $f=1$ on $B$. This locally constant function is continuous and contradicts part (c). Therefore $A\cap B\ne \varnothing$.