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Let $g(n,k)=r(n)$, where $r$ is the total primitive-recursive parity function from part (ii), and apply unbounded minimization in $k$:If $n$ is even, then $g(n,0)=0$ and $p(n)=0$. If $n$ is odd, then $g(n,k)=1$ for every $k$, so the search never terminates and $p(n)$ is undefined. Hence $p$ is partial recursive. Because every function built without minimization is total, $p$ cannot be defined without minimization; the functions $s$ and $r$ can.

It remains to characterize the functions built using only the initial functions and composition. They are exactly the functionsfor some $c\in \mathbb{N}$ and some input coordinate $i$.

Indeed, the zero function is the first form, the successor function and each projection function have the second form, and composing functions of these forms preserves the classification: a constant outer function remains constant, while an outer function $y_j+c$ selects one inner function and adds $c$. This proves necessity by structural induction. Conversely, applying the successor function $c$ times to zero produces the constant $c$, and applying it $c$ times to ${\pi }_i^k$ produces $x_i+c$.

Every function in this class is directly computable: an algorithm either writes the fixed constant $c$, or copies the $i$th input and performs $c$ successor steps.

Using the zero test $s$ from part (i), define by primitive recursionSince $s(0)=1$ and $s(1)=0$, the values alternate:Thus $r$ is the required parity function and, like $s$, is primitive recursive without minimization.

A partial recursive function is obtained from the three kinds of initial function of recursion theory—
the zero function, successor function, and projection function—by finitely many applications of function composition in recursion theory, primitive recursion, and unbounded minimization. Explicitly, primitive recursion has the formwhile minimization takes the least $y$ for which $g(\mathbf{x},y)=0$ and is undefined if no such $y$ is found.

For the function in part (i), defineThe initial value $1=S(Z(0))$ and the identically zero recursion step are primitive recursive, so this is a primitive-recursive definition of the required zero test. It uses no minimization.

The no-cloning theorem for two pure states states that, for distinct nonorthogonal states $|c_0⟩$ and $|c_1⟩$, there is no unitary operator $U$ and fixed blank state $|0⟩$ such that

To derive this from state discrimination, letand suppose such a cloner existed. Applying it repeatedly would produce $N$ copies. The inner product of the two possible $N$-copy states has magnitude $s^N$, so their optimal success probability under the Helstrom-Holevo bound would beAlready for $N=2$,because $0<s<1$ implies $s^4<s^2$. But cloning followed by the two-copy Helstrom measurement for two pure states would itself be a quantum procedure applied to the original single state, contradicting the optimal one-copy bound $P_1$. Hence the assumed unitary cloner cannot exist.

Equivalently, clone-assisted asymptotic state discrimination would give $P_N\to 1$, although two nonorthogonal states are not perfectly distinguishable. Thus the Helstrom--Holevo theorem implies the no-cloning theorem.

With equal priors, the probability of a correct result from this quantum measurement isThe two required amplitudes areThereforeIts maximum occurs at $\beta =-\pi /8$ and is

Here $|⟨a_0|a_1⟩|=1/\sqrt{2}$, so $\theta =\pi /4$ and $\sin \theta =1/\sqrt{2}$. The stated Helstrom-Holevo bound is consequently exactly the same value, and the basis above is the Helstrom measurement for the zero and plus states.

By the real spectral theorem, the real symmetric matrix $A$ has an orthonormal basis of eigenvectors. In that basis the associated quadratic form isRescaling the coordinates belonging to nonzero eigenvalues changes every positive coefficient to $+1$ and every negative coefficient to $-1$. Hence the diagonal normal form has one positive square for each positive eigenvalue and one negative square for each negative eigenvalue. By Sylvester's law of inertia, these counts do not depend on the diagonalizing basis, so

The numerical eigenvalues are not invariant under a general change of basis, because the matrix changes by congruence rather than similarity. For example, on a one-dimensional space let $q(e)=1$. Its matrix in the basis $(e)$ is $(1)$, whereas in the basis $(2e)$ it is $(4)$. The eigenvalue changes from $1$ to $4$, although its sign, and therefore the signature, is unchanged.

The symmetric matrix of the quadratic form in the standard basis isbecause the off-diagonal entries contribute twice to $x^{T}Ax$.

Apply Gram-Schmidt orthogonalization for a symmetric bilinear form toThese vectors are pairwise orthogonal for $B(u,v)=u^{T}Av$, andThus the form is diagonal in the basis $(u_1,u_2,u_3)$, with matrixIt follows that its rank is $3$ and its signature is

A Hermitian form is a sesquilinear form $H:V\times V\to \mathbb{C}$ such thatUse the convention that $H$ is linear in its first argument and conjugate-linear in its second. Its matrix in the basis $(v_1,\dots ,v_n)$ isIf $v={\sum }_i{x}_i{v}_i$ and $w={\sum }_j{y}_j{v}_j$, then

Since $w_i={\sum }_j{p}_{ij}{v}_j$, the coordinate row of each new basis vector is a row of $P$. Direct substitution givesso the new matrix is

The subspace $N=V^{\perp }$ is the radical of $H$. In coordinates it is the kernel of $A$, and the rank-nullity theorem gives

Appending the node $-2$ to the existing Newton interpolation polynomial givesAt the new node,The condition $p_4(-2)=10$ therefore givesHence

Use the nodes in the order $0,1,2,3$. Their first divided differences are $4,-2,6$, their second divided differences are $-3,4$, and their third divided difference is $7/3$. The Newton interpolation polynomial is thereforeThis is the unique interpolating polynomial of degree at most three through the four data points.

No. Take the Hausdorff space $X=\mathbb{R}$ and declare all nonzero points equivalent, leaving $0$ in its own class. The quotient has two points,The nonzero class is open because its inverse image $\mathbb{R}\setminus \{0\}$ is open. However, the only saturated set containing $0$ but not the other class is $\{0\}$, which is not open. Thus every neighbourhood of $[0]$ is the whole quotient, so the two quotient points cannot have disjoint neighbourhoods. Therefore $\tilde{X}$ need not be Hausdorff.

If $f:\tilde{X}\to Y$ is continuous, then $f\circ \pi$ is continuous because the quotient map $\pi$ is continuous by definition.

Conversely, suppose $f\circ \pi$ is continuous. For every open set $V\subseteq Y$,is open in $X$. The definition of the quotient topology then says that $f^{-1}(V)$ is open in $\tilde{X}$. HenceThis is the universal property of the quotient topology.

The quotient topology on $\tilde{X}$ isSince inverse images preserve arbitrary unions and finite intersections,Also ${\pi }^{-1}(\varnothing )=\varnothing$ and ${\pi }^{-1}(\tilde{X})=X$. The displayed collection therefore satisfies all the topology axioms.

Fordirect multiplication givesWith respect to the ordered standard basis $(E_{11},E_{12},E_{21},E_{22})$, its matrix is therefore

This nilpotent operator satisfies ${\varphi }_A^3=0$, while ${\varphi }_A^2\ne 0$. Moreover,These ranks determine one nilpotent block of size three and one of size one. Hence its Jordan normal form is

Let $E_{rs}$ be the matrix unit with its only nonzero entry in row $r$, column $s$. Forone hasThereforeThe $n^2$ matrix units form an eigenbasis, with $E_{rs}$ having eigenvalue $r-s$. The zero eigenspace consists exactly of the diagonal matrices and has dimension $n$. The rank-nullity theorem now gives