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An irreducible polynomial has positive degree, so its ideal is proper: otherwise $1=fh$ would contradict degrees. Every nonzero class $[g]$ has a representative not divisible by $f$. Irreducibility gives $\gcd (g,f)=1$, and the Euclidean algorithm supplies polynomials $u,v$ with $ug+vf=1$. Passing to the quotient gives $[u][g]=1$. Thus the quotient is a nonzero commutative ring in which every nonzero element has an inverse, hence a field. Constants embed injectively because no nonzero constant lies in the positive-degree ideal $(f)$.

The two singly marked sides of the first triangle are glued with their endpoints at $A$ matched. A cross-section parallel to the remaining side has its two endpoints identified, becoming a circle; the apex becomes its cone point. Thus this quotient triangle is a disc, whose boundary is the doubly marked edge.

In the second triangle, the two triply marked sides run consecutively along the boundary from $E$ through $D$ to $F$. Pairing these sides in the indicated directions introduces one reversal; its boundary word is $aa{b}^{-1}$, where $b$ is the remaining doubly marked edge. This is a Möbius band with boundary $b$. One can obtain the usual rectangle model by cutting from the identified triple-edge seam to the midpoint of $b$ and opening the triangle.

The double markings glue the disc boundary to the boundary of that Möbius band. Capping a Möbius band by a disc givesThis description also verifies that the quotient is a closed surface, including the vertices. As a check, the vertex classes are $\{A\}$ and $\{B,C,D,E,F\}$; there are three edge classes and two faces. The Euler characteristic is $2-3+2=1$, consistent with the projective plane. The Möbius band exhibits nonorientability directly.

A quotient map $\pi :X\to Y$ is a surjection such that $U\subseteq Y$ is open if and only if ${\pi }^{-1}(U)$ is open in $X$. Thus the topology on $Y$ is exactly the final topology induced by $\pi$.

For an open continuous surjection, continuity gives one implication. For the other, if ${\pi }^{-1}(U)$ is open then $U=\pi ({\pi }^{-1}(U))$ is open by openness and surjectivity. Hence it is a quotient map.

If $\pi :X\to Y$ and $\sigma :Y\to Z$ are quotient maps, they are surjective andThis set is open exactly when ${\sigma }^{-1}(U)$ is open, exactly when $U$ is open. Thus the composition is a quotient map.

For the factorization, continuity of $g$ implies continuity of $f=g\circ \pi$. Conversely, if $f$ is continuous and $U$ is open in $Z$, then ${\pi }^{-1}(g^{-1}(U))=f^{-1}(U)$ is open. The quotient property makes $g^{-1}(U)$ open. This holds for every open $U$, so $g$ is continuous.

Use the convention that a positive vortex has counterclockwise circulation, with complex potential of a two-dimensional potential flow $\Gamma \log (z-z_1)/(2\pi i)$. To make both coordinate axes streamlines, place images of strength $-\Gamma$ at ${\bar{z}}_1$ and $-{\bar{z}}_1$, and strength $+\Gamma$ at $-z_1$. The resulting potential isAdditive constants and consistent logarithm branches do not affect velocity. The quotient has unit modulus on either wall, so the streamfunction is constant there and there is no normal flow.

The vortex moves with the regular fluid velocity after removing its own singular field. Since $W^{\prime }=u-iv$, evaluating the three image contributions at $z_1$ gives exactlyThe overall minus sign and the two reflected-term minus signs here agree with the source scan.

Write $z_1=x+iy$, with $x,y>0$. Taking real and imaginary parts yieldsDividing gives $dy/dx=-y^3/x^3$, which integrates toPut $\ell =ab/\sqrt{a^2+b^2}$. The first-quadrant branch has vertical and horizontal asymptotes $x=\ell$ and $y=\ell$. Its motion is to the right and downward: from large $y$ near the vertical asymptote toward large $x$ near the horizontal one. Both coordinates stay strictly positive, so no wall is crossed.

We first prove that every extension of two simple modules splits. ConsiderIf $n\ne m$, the central Casimir element has distinct scalars $n(n+2)$ and $m(m+2)$. On $V$ its minimal polynomial divides the product of the corresponding two distinct linear factors. Its eigenspaces are submodules, and the eigenspace for $m(m+2)$ maps isomorphically onto the quotient. It is the required complement.

Now suppose $n=m=l$. Use generalized weight spaces for $H$; their possible weights are $l,l-2,\dots ,-l$, because the characteristic polynomial on an extension is the product of those on its two factors. Lift the quotient's highest vector to $w$ in the generalized weight space of representation weight $l$. Then $ew=0$ and $f^{l+1}w=0$, since these would lie in generalized weight spaces of weights $l+2$ and $-l-2$, which are absent. Also $(H-l)w=a{v}_0$ for a highest vector $v_0$ in the submodule $A_l$. The commutator identity givesThus $a=0$. The vector $w$ is an actual highest-weight vector. Its lowering chain maps to the full basis of the quotient, so it generates an $A_l$ complementary to the original submodule. This proves splitting also when the two simple types coincide, including $l=0$.

The same splitting holds for a semisimple kernel and simple quotient by induction on the number of kernel summands. Write the kernel as $U_1\oplus U_2$ with $U_1$ simple. Modulo $U_2$, split the extension by the simple quotient; the inverse image of that chosen quotient copy is an extension with kernel $U_2$. Split it inductively, obtaining a submodule in $V$ meeting the entire original kernel trivially.

Finally induct on $\dim V$. A maximal proper submodule has smaller dimension and hence is semisimple by induction; the quotient is simple by maximality. The just-proved splitting completes the induction. Hencewith only finitely many nonzero multiplicities. This proof does not assume the complete-reducibility theorem it is establishing.

Use the standard defining properties of Category O: finite generation over $U(\mathfrak{g})$, a direct sum of finite-dimensional weight spaces, and locally finite positive root action. In particular the weights lie in a finite union of downward root cones. Submodules and quotients retain the representation weight bounds and finite weight spaces; finite generation of submodules follows from the Poincaré-Birkhoff-Witt theorem and the noetherian property of $U(\mathfrak{g})$.

Choose finitely many weight vector generators. The sum $T$ of their full weight spaces is finite-dimensional and preserved by the centre. The kernel of its central action has finite codimension in $Z(U(\mathfrak{g}))$ and kills the generators, hence, by centrality, all of $M$. Thus only finitely many central characters ${\theta }_1,\dots ,{\theta }_r$ can occur in any subquotient: they are the characters of a fixed finite-dimensional commutative quotient algebra of the centre.

By the Harish-Chandra isomorphism each ${\theta }_i$ determines one finite Weyl orbit of shifted weights. Let $S$ be the finite union of these orbits after subtracting $\rho$. Every nonzero subquotient has a nonzero highest-weight vector, using the bounded-above weights. Within its finite-dimensional highest weight space, the commuting central operators have a common eigenvector. Part (e) and the central-character restriction show that its highest weight belongs to $S$.

Define for any subquotient $N$It is finite, additive in short exact sequences, and positive whenever $N\ne 0$. Hence each strict step of a submodule chain consumes at least one unit of $d(M)$. Refining a nonsimple factor into a proper submodule and quotient cannot continue more than $d(M)$ times. It ends with simple factors, provingThe argument proves the assertion for every object of $\mathcal{O}$, rather than just for a cyclic highest weight object.

If $p\nmid n$, the preceding part says $U=S_F^{(n-1,1)}$ is simple and has no nonzero homomorphism into $W$. Its only nonzero simple quotient is itself, so $W$ cannot contain a submodule isomorphic to such a quotient.

If $p\mid n$, use the given nonzero homomorphism $\varphi :U\to W$. The submodule chain shows that $T$ is the unique maximal proper submodule of $U$ and $U/T$ is simple. Moreover ${\mathrm{Hom}}_{S_n}(T,W)=0$: for odd characteristic this follows from part (b), and in characteristic two the divisibility condition makes $n$ even, where the corrected fixed-space calculation also gives zero. Thus $\varphi$ kills $T$ and descends to a nonzero map $U/T\to W$. Simplicity makes this map injective.

Its image is the required simple submodule, provingIn the divisible case the quotient is explicitly $U/T$, of dimension $n-2$. The argument uses the supplied existence of $\varphi$ and only the fixed-space vanishing that remains valid in that case.

Consider one iteration with $a_i>b_i>0$ and $r_i=a_i\mathrm{mod}{b}_i$. If $b_i\le a_i/2$, then $r_i<b_i\le a_i/2$. If $b_i>a_i/2$, the quotient is $1$, so $r_i=a_i-b_i<a_i/2$. In either case $r_i<a_i/2$, whenceRepeated application, or mathematical induction, provesThis also holds for the final iteration, whose left side is zero. For zero iterations the correct relation is equality $a_0{b}_0=ab$, rather than a strict inequality.

Use four working processor registers by packing the growing quotient into the vacated low bits of the shifting dividend. The following is A32 assembly language with conditional execution, an allowed instance of the question's general 32-bit Reduced instruction set computer. It is not restricted to the sixteen-bit encodings in the supplied Thumb instruction set table. On entry r0 is the dividend and r1 the positive divisor; r2 becomes the high remainder machine word and r3 counts iterations. The quotient is returned in r0 and the ordinary remainder remains in r2.

mov   r2, #0
mov   r3, #32
loop:
adds  r0, r0, r0
adcs  r2, r2, r2
cmp   r2, r1
subcs r2, r2, r1
adc   r0, r0, #0
subs  r3, r3, #1
bne   loop
bx    lr

Here subcs means conditional sub when arithmetic carry is set, without the flag-setting suffix. adc also preserves flags. Consequently the arithmetic carry from cmp survives until it is inserted as the next quotient bit. The loop has seven instructions, one of them conditional, and no data-dependent branch. Its sole branch is the backwards loop branch, taken 31 times. It uses no separate quotient processor register, zero processor register, temporary comparison processor register or saved arithmetic carry processor register. A Thumb instruction set rendering using conditional execution would need an explicit conditional-execution encoding such as it cs; that is an additional encoded instruction, not an unexplained free operation.

The packed restoring division loop invariant relates it to the displayed algorithm. Specifically, let $M=2^{32}$, let $a$ retain its original mathematical value, and let $k$ be the number of completed iterations. Define the processed prefix $p_k=\lfloor a/2^{32-k}\rfloor$, its quotient $Q_k=\lfloor p_k/b\rfloor$ and remainder $R_k=p_k-b{Q}_k$. The processor register loop invariant isThe two terms in the low machine word occupy disjoint bit positions. Its next outgoing arithmetic carry is therefore the next original dividend bit, not a quotient bit. adds shifts the low machine word; adcs doubles the high remainder and incorporates that dividend bit. cmp decides whether this trial remainder is at least the divisor; subcs performs the conditional subtraction, and adc appends its decision to the shifted quotient. This is exactly the doubling, compare/subtract and quotient-bit update, with the counter maintained separately. Before processing bit $k\le 32$, the trial remainder is at most $p_k<2^k\le M$, so the high-word addition does not require a hidden 33rd remainder bit for this 32-bit dividend problem. After 32 iterations the residual dividend portion is zero, giving the stated outputs.

There is a printed source error in the auxiliary bound. At the initial state, choosing valid inputs $a=b=1$ makes the printed strict inequality $r{2}^j<b{2}^{32}$ false. The algorithm instead preserves $0\le r<b{2}^{32}$ for its unpacked 64-bit variable, together with its printed equality loop invariant. Doubling and subtracting $b{2}^{32}$ when necessary preserves that corrected bound. At termination the equality becomes $a{2}^{32}=qb{2}^{32}+r$; since $r$ is then divisible by $2^{32}$, it gives $0\le a-qb<b$. This source error does not invalidate the algorithm or the packed implementation. The figure is mathematical pseudocode: its 2^32 denotes exponentiation, rather than C's XOR operator.

The displayed A32 routine assembled successfully with Clang and was disassembled to check the conditional and non-flag-setting forms. Its carry flag model was checked exhaustively at an eight-bit machine word size and against boundary and randomized 32-bit divisions. No native ARM architecture family execution is claimed.

In positional notation, a numeral digit contributes its value multiplied by a weight determined by its position. Fix an integer radix $b\ge 2$. A finite string of numeral digits $d_n{d}_{n-1}\cdots d_0$, with $0\le d_j<b$, represents the nonnegative integerThe rightmost position has weight $b^0=1$; each move left multiplies the weight by $b$. This weight is the place value. For a nonzero integer, the canonical string has $d_n\ne 0$. Zero is written with the single numeral digit $0$. Leading zeros are permitted in a fixed-width display but omitted in the canonical form: $00205_{10}=205_{10}$. A negative integer is written with a separate minus sign and the positional notation for its absolute value; the minus sign is not a numeral digit.^[ref]

For example,The zero is essential inside this string: $25_{10}=25$, not $205$. It records that the tens position contributes nothing while keeping the hundreds and units in their proper positions. The same integer has binary representation

Repeated Euclidean division constructs the representation exactly. Set $N_0=N$. At each step writeKeep the remainder $d_j$ and continue with the integer quotient $N_{j+1}$. A positive quotient strictly decreases, so eventually it is zero. Substituting the successive equations back into the first gives $N=d_0+b{d}_1+b^2{d}_2+\cdots$: the numeral digits are obtained from right to left. The remainder modulo $b$ uniquely fixes $d_0$, and then the quotient uniquely fixes the remaining string by the same reasoning. This proves existence and uniqueness once leading zeros are excluded. The needed unique quotient and remainder are precisely the conclusion of the Euclidean division theorem.^[ref]

This finite expansion concerns nonnegative integers. Extending positional notation to fractional positions uses weights $b^{-1},b^{-2},\dots$ to the right of a point; an arbitrary real number need not have a finite digit string. In particular, the decimal representation $0.1$ represents $1/10$, while the base-two string $0.1_2$ represents $1/2$. The radix is part of the interpretation.

The benefit is compact arithmetic: $k$ positions can encode every nonnegative integer below $b^k$. To see the upper endpoint, fill all positions with $b-1$; the finite geometric series gives $(b-1)(1+b+\cdots +b^{k-1})=b^k-1$. Carrying in addition exchanges $b$ units of one place value for one unit of the next. Thus a short string stores a large count and also supplies an algorithm for computing with it.

A ratio compares two quantities $a,b$, written $a:b$ and corresponding to division $a/b$ when $b\ne 0$. Equal ratios of corresponding lengths characterize the scaling of similar triangles.

For example, $6:4$ and $3:2$ specify the same ratio because $6/4=3/2$. Multiplying both entries by the same nonzero scale preserves the quotient: $(ca)/(cb)=a/b$. This is useful when comparing mixtures or scaling a recipe; doubling both ingredient amounts preserves their ratio, while adding the same amount to both generally does not. For quantities with the same physical units, their ratio is dimensionless, connecting this elementary calculation with dimensional analysis.

For an integer $a$ and positive integer $b$, the remainder in Euclidean division is the unique integer $r$ in $0\le r<b$ for which $a=bq+r$ with an integer $q$. It is not the full quotient $a/b$.

In a triangle whose two sides at an apex are glued with that apex fixed, cross-sections become circles and the apex becomes their cone point. This quotient is a disc. A second triangle with boundary word $aa{b}^{-1}$ becomes a Möbius band after the $a$ sides are identified. Gluing the remaining $b$ boundary to the first disc caps the band and gives the real projective plane. The resulting two-face cell structure has two vertices, three edges and Euler characteristic one.