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For every $v\in V$, $(\alpha +\beta )v=\alpha v+\beta v$, so $\mathrm{im}(\alpha +\beta )\subseteq \mathrm{im}\alpha +\mathrm{im}\beta$. Taking dimensions and applying the preceding dimension formula for a sum of subspaces givesThis proves the upper rank bounds for a sum of linear maps, without assuming the two images have zero intersection.

Choose a basis $e_1,\dots ,e_k$ for $U\cap U^{\prime }$. Extend it to a basis $e_1,\dots ,e_k,u_1,\dots ,u_r$ of $U$, and separately to a basis $e_1,\dots ,e_k,v_1,\dots ,v_s$ of $U^{\prime }$. The combined list spans $U+U^{\prime }$.

To prove linear independence, suppose a linear combination of the combined list is zero. Then the combination of the $v_j$ lies in $U\cap U^{\prime }$, so it can be expressed using the $e_i$. Independence of the chosen basis of $U^{\prime }$ forces every $v_j$ coefficient to vanish. Independence of the chosen basis of $U$ then forces the remaining coefficients to vanish. Thus the combined list is a basis, with $k+r+s$ elements. This proves the dimension formula for a sum of subspaces:

First use a homogeneous solution to meet the nonzero endpoint values:It satisfies $y_h^{\prime \prime }+k^2{y}_h=0$, $y_h(0)=a$ and $y_h(\pi )=b$. The Green-function representation then givesApplying the differential operator to the integral gives $f(x)$ by the defining Dirac delta property of $G$, and that integral vanishes at the endpoints. For continuous $f$ this is a classical solution; integrable $f$ also gives the corresponding weak solution. The same nonresonant homogeneous argument as in part (a) proves uniqueness.

Away from the source point, the Green function obeys the homogeneous equation and must satisfy the appropriate endpoint condition. Thus it has the form $A\sin (kx)$ to the left of $\xi$ and $B\sin (k(\pi -x))$ to the right. It must be continuous at $\xi$: otherwise its second distributional derivative would contain an unwanted derivative of a Dirac delta. Integrating across $\xi$ gives the jump condition $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$.

The left and right homogeneous solutions $u=\sin (kx)$, $v=\sin (k(\pi -x))$ have constant WronskianThis is nonzero for real noninteger $k$. Dividing the product of the left and right solutions by that Wronskian therefore gives the Dirichlet Helmholtz Green function on an intervalThe formula vanishes at both endpoints, is continuous at the source and has derivative jump one, verifying all distributional and boundary conditions. A difference of two such Green functions would be a homogeneous Dirichlet solution; it vanishes because $\sin (k\pi )\ne 0$, proving uniqueness. Although zero is excluded by the stated noninteger condition, the removable $k\to 0$ limit is $-\min (x,\xi )(\pi -\max (x,\xi ))/\pi$.

The same recurrence matrix hasTherefore ${\mathbf{v}}_{n+5}\equiv 3{\mathbf{v}}_n(\mathrm{mod}5)$, and in particular $F_{n+5}\equiv 3F_n(\mathrm{mod}5)$. Starting from $F_0=0$ and iterating,Consequently every nonnegative index divisible by five has a Fibonacci number divisible by five, including index zero. This directly proves the requested divisibility without assuming a general Fibonacci divisibility theorem.

Put ${\mathbf{v}}_n=(F_{n+1},F_n{)}^T$. The Fibonacci recurrence matrix modulo an integer is $A=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$, so ${\mathbf{v}}_{n+r}=A^r{\mathbf{v}}_n$. Direct matrix multiplication givesTaking the second coordinate provesFor explicit arithmetic behind the second power, squaring $A^2=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ gives $A^4=\left(\begin{array}{cc}5& 3\\ 3& 2\end{array}\right)$, whose square is the displayed $A^8$. Thus the modular period assertions follow from calculations and the recurrence, not from quoting a period theorem.

Write $F_n=x_n$. For $n\ge 1$, the recurrence and the previously proved greatest common divisor invariance giveRepeating reaches $\gcd (F_1,F_0)=\gcd (1,0)=1$. This also gives the $n=0$ case directly. Applying the same invariance once more,Thus both requested greatest common divisors equal one for every $n\ge 0$. This is the adjacent-index instance of Fibonacci gcd reduction, derived directly rather than assuming the stronger general formula.

For $p\equiv 1(\mathrm{mod}4)$, part (i) constructs a solution $a=q!$ to $a^2\equiv -1$. It is nonzero modulo $p$, so $a$ and $-a$ are distinct for odd $p$. Any other solution $x$ obeysThe Euclid lemma forces $x\equiv a$ or $-a$. Hence there are exactly two solutions. For $p\equiv 3(\mathrm{mod}4)$, a solution to $x^2\equiv -1$ would have $x^4\equiv 1$, but part (ii) would imply $x^2\equiv 1$, impossible because $p$ cannot divide two. ThereforeThese two cases exhaust all odd primes.

The modular congruence $x^4\equiv 1$ first implies $x\not\equiv 0$. Multiplication by such an $x$ permutes the nonzero residue classes modulo the prime $p$: if $xa\equiv xb$, cancellation gives $a\equiv b$. Multiply all these residues and cancel their nonzero product to obtain $x^{p-1}\equiv 1$, the usual short proof of Fermat little theorem.

Now $(x^2-1)(x^2+1)\equiv 0$. By the Euclid lemma, either $x^2\equiv 1$ or $x^2\equiv -1$. In the latter case, because $p-1=4k+2$,contradicting the preceding product argument, since $p$ is odd. ThusNo unproved assertion about the number of fourth roots is needed.

For a prime number $p$, every nonzero residue class has a unique multiplicative inverse modulo $p$: Bezout identity supplies one because its representative is coprime to $p$, and cancellation modulo a prime gives uniqueness. Pair each residue with its inverse. The only unpaired residues satisfy $a^2\equiv 1$, so $p\mid (a-1)(a+1)$; the Euclid lemma gives $a\equiv 1$ or $-1$. For odd $p$, these are distinct. Every other pair contributes one to the product of all nonzero residues, leavingFor $p=2$, $1!\equiv -1(\mathrm{mod}2)$ as well. This proves Wilson theorem in all cases.

For an odd prime, set $q=(p-1)/2$. Pair the upper half of the factorial with the negatives of the lower half:If $p\equiv 1(\mathrm{mod}4)$, then $q$ is even. ThereforeThis constructs a square root of minus one modulo a prime explicitly.

This is not an equivalence relation. With the usual convention that zero is a perfect square, $x^2-x^2=0$ gives reflexivity, and changing the sign of $x^2-y^2$ shows that the relation is symmetric. However $3\sim 5$ because $3^2-5^2=-16=-4^2$, and $5\sim 13$ because $5^2-13^2=-144=-12^2$, whereas $3^2-13^2=-160$ is neither a positive nor a negative perfect square: $12^2<160<13^2$. Thus transitivity fails. If “perfect square” were restricted to positive squares, reflexivity would fail as well, leaving the same negative classification.

This is not an equivalence relation. For $x=1/2$, $x\overline{x}=1/4\notin \mathbb{Z}$, so reflexivity fails. It is symmetric, because $y\overline{x}$ is the conjugate of $x\overline{y}$, and an integer is real. It also fails transitivity: take $x=z=1/2$ and $y=2$. Then $x\overline{y}=y\overline{z}=1$ are integers but $x\overline{z}=1/4$ is not. A single failure would suffice, but these examples identify exactly which properties break.

For nonzero complex numbers, $x\overline{y}=(x/y)|y{|}^2$. Since $|y{|}^2$ is positive real, the condition is equivalent to $x/y\in \mathbb{R}\setminus \{0\}$. This is an equivalence relation. The ratio $x/x=1$ gives reflexivity; the reciprocal of a nonzero real ratio shows that the relation is symmetric; and $(x/y)(y/z)=x/z$ gives transitivity. An equivalence class is a real line through zero with zero removed, or equivalently all nonzero complex numbers with the same argument modulo $\pi$.

This is an equivalence relation. The difference $x-x=0$ is an even integer, proving reflexivity. If $x-y=2k$ then $y-x=-2k$, proving that the relation is symmetric. If $x-y=2k$ and $y-z=2\ell$ then $x-z=2(k+\ell )$, proving transitivity. Its equivalence classes are the translates $x+2\mathbb{Z}$; two real numbers need not be integers themselves to belong to the same class.

An integer square is divisible by four exactly when its base is even: an even base has square $4j^2$, while an odd base has square $(2j+1{)}^2\equiv 1(\mathrm{mod}4)$. Apply this observation to the triangular number $T_n$ obtained above. Since $n(n+1)$ is even,Checking the four residue classes of $n$ modulo four, the products $n(n+1)$ have residues $0,2,2,0$, respectively. ThereforeThe original PDF has two alternatives modulo four; the TeX transcription incorrectly turns the second alternative into a modulus-three expression.

Let $T_n=n(n+1)/2$, the $n$th triangular number. Pairing the terms of the arithmetic progression proves ${\sum }_{r=1}^{n}r=T_n$. We prove the cube identity by mathematical induction. For $n=1$ both sides equal one. If ${\sum }_{r=1}^n{r}^3=T_n^2$, thenAdding $(n+1{)}^3$ establishes the next case. ThusEquivalently, the same displayed difference telescopes from $T_0=0$, providing a direct proof without induction.

The characteristic polynomial has only the eigenvalue $\mu$. If $A$ is diagonalizable, there is an invertible matrix $P$ and a diagonal matrix $D$ with $A=PD{P}^{-1}$. Every diagonal entry of $D$ is an eigenvalue, hence equals $\mu$. Thus $D=\mu I$ and $A=P(\mu I)P^{-1}=\mu I$. Conversely, $\mu I$ is already diagonal in every basis. HenceThis argument uses the existence of an eigenvector basis rather than merely the repeated root of the polynomial; repeated eigenvalues alone do not imply diagonalizability.

Using the monic convention for the characteristic polynomial, expand the determinant along the last column:For the eigenvalue $1$, the equations $(A-I)v=0$ give $v_2=v_1$ and $v_3=0$. Thus every corresponding nonzero eigenvector is a nonzero multiple of $(1,1,0{)}^T$. For the eigenvalue $2$, the first two equations give $v_2=2v_1$ and $v_1=0$, so every corresponding nonzero eigenvector is a nonzero multiple of $(0,0,1{)}^T$. ThereforeThese eigenspaces together have dimension $2$, so they cannot provide an eigenvector basis of ${\mathbb{R}}^3$. The matrix is not diagonalizable; the eigenvalue $1$ has algebraic multiplicity $2$ but geometric multiplicity $1$.

Each element of $H$ permutes the three-point set, giving a group homomorphism $\rho :H\to S_3$. To prove injectivity, let an element fix all three points. Conjugate it by the normalizing Möbius transformation from part (i). The conjugate fixes $0,1,\infty$. A Möbius transformation $z\mapsto (az+b)/(cz+d)$ fixing $0$ has $b=0$, and fixing infinity has $c=0$. It is then $z\mapsto kz$; fixing $1$ forces $k=1$. Thus the conjugate, and hence the original transformation, is the identity. The kernel of a group homomorphism is trivial.

For surjectivity, let $(x_1,x_2,x_3)=(\alpha ,\beta ,\gamma )$ and choose any permutation $\sigma$. Let $g$ normalize the ordered triple $(x_1,x_2,x_3)$ and let $g_{\sigma }$ normalize $(x_{\sigma (1)},x_{\sigma (2)},x_{\sigma (3)})$. Then $g_{\sigma }^{-1}g$ sends $x_j$ to $x_{\sigma (j)}$, lies in $H$, and induces the chosen permutation. Thus $\rho$ is onto andFor the normalized set, the six Möbius transformations permuting three points are $z$, $1-z$, $1/z$, $1/(1-z)$, $z/(z-1)$, and $(z-1)/z$. Conjugating by $g$ gives the corresponding six maps for the original set.

If all three points are finite, define the Möbius transformationIts numerator vanishes at $\alpha$, its pole is at $\gamma$, and substitution of $\beta$ gives $1$. The three points are distinct, so the numerator and denominator are not proportional and the associated $2\times 2$ matrix has nonzero determinant.

If one point is infinite, the limiting formulas areIn each case the finite values and the limit at infinity give $g(\alpha )=0$, $g(\beta )=1$, $g(\gamma )=\infty$, and the corresponding determinant is nonzero. Distinctness permits at most one of the points to be infinite. Hence every ordered triple of distinct points on the Riemann sphere can be normalized to $(0,1,\infty )$.