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Let $B=\{b_1,b_2,b_3\}$ be a blue triangle and $Y=\{y_1,y_2,y_3\}$ a vertex-disjoint yellow triangle. If some $b_i$ has two yellow neighbours in $Y$, those three vertices form a yellow triangle meeting $B$. Otherwise every $b_i$ has at least two blue neighbours in $Y$, giving at least six blue cross-edges. Some $y_j$ then has two blue neighbours in $B$, and these form a blue triangle meeting $Y$. This proves the overlap assertion.

The triangle-packing Ramsey number isFor completeness, its induction starts from the stated $K_{10}$ result. At each step, remove two disjoint monochromatic triangles supplied on ten currently unused vertices. If their colour disagrees with the packing already constructed, apply the overlap assertion to an oppositely coloured pair; replacing that pair by the overlapping pair is the augmenting move that shifts the boundary between the two packings. Repeating the move either adds a triangle to the existing colour or converts all pairs needed for the opposite-colour packing. Thus $K_{5t}$ contains $t$ pairwise vertex-disjoint monochromatic triangles of one colour.

The bound is sharp. For $K_{5t-1}$, split the vertices into $A,B_1,B_2$ withColour edges inside $A$ and between $B_1$ and $B_2$ yellow; colour all edges from $A$ to $B_1\cup B_2$ and all edges within each $B_i$ blue. Every yellow triangle lies wholly in $A$, so there are at most $t-1$ disjoint yellow triangles. Every blue triangle uses at least two vertices of $B_1\cup B_2$, and $t$ such triangles would require a blue perfect matching on those $2t$ vertices, impossible because the isolated one-vertex blue component $B_1$ and the odd component $B_2$ have no such matching. Hence there is no $t$-triangle packing of either colour for any $t\ge 2$.

Use the two-colour matching Ramsey numberIts upper bound follows by induction on $s$. If there is a blue edge, remove its endpoints and apply the induction hypothesis to the remaining $K_{2r+s-3}$: it contains either a red $r$-edge matching or a blue $(s-1)$-edge matching, and in the latter case the removed edge completes a blue $s$-edge matching. If there is no blue edge, the graph itself contains a red $r$-edge matching. For $r=s=t$, this gives the required monochromatic matching in $K_{3t-1}$.

The result never remains true for $K_{3t-2}$, including $t=1$. Partition its vertices into $A$ and $B$ with $|A|=2t-1$ and $|B|=t-1$. Colour edges within $A$ blue and every other edge yellow. A blue matching has at most $t-1$ edges, while every yellow edge meets $B$, so a yellow matching also has at most $t-1$ edges.

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 $2.2$.