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For the real quadratic field $K=\mathbb{Q}(\sqrt{d})$ with squarefree $d>1$, Dirichlet unit theorem givesfor a fundamental unit $ϵ>1$, since the roots of unity in a real field are $\pm 1$.

If $p\mid d$ and $p\equiv 3(\mathrm{mod}4)$, a unit of norm $-1$ would have the form $(a+b\sqrt{d})/2$ with integers $a,b$ and satisfy $a^2-d{b}^2=-4$. Reducing modulo the odd prime $p$ gives $(a/2{)}^2\equiv -1(\mathrm{mod}p)$, impossible because $-1$ is not a quadratic residue for such a prime. Hence every unit has norm $+1$ in this case. In contrast, $\boxed{d=2,u=1+\sqrt{2}}$ has norm $-1$.

For a number field with $r$ real embeddings ${\sigma }_i$ and $s$ conjugate pairs of complex embeddings ${\tau }_j$, use the logarithmic embedding of number field unitsIt is a group homomorphism from multiplication to addition. A unit has field norm $\pm 1$, so the coordinates sum to zero. Deleting one coordinate identifies this hyperplane with ${\mathbb{R}}^{r+s-1}$ and gives the desired map.

If its image coordinates remain in a bounded set, the omitted coordinate is bounded too, and every conjugate of $u$ has bounded absolute value. The coefficients of its monic minimal polynomial are integers and elementary symmetric expressions in at most $[K:\mathbb{Q}]$ bounded conjugates. They are therefore bounded integers. Only finitely many such polynomials, and finitely many roots of them, exist. Thus the inverse image of every bounded set is finite. In particular the kernel is finite and the image is discrete. This argument includes the zero-dimensional target case; it does not yet assert the full-rank part of Dirichlet unit theorem.

Suppose $\chi$ is an irreducible character of a group of odd order and $\chi =\overline{\chi }$. Write $d=\chi (1)$. Every nonidentity element pairs with its distinct inverse, since there is no element of order $2$. Also $\chi (g^{-1})=\overline{\chi (g)}=\chi (g)$. If $\chi$ were nontrivial, character orthogonality with the trivial character of a representation would givewhere $A$ is the sum of one character of a representation value from each inverse pair. This $A$ is an algebraic integer, so $A=-d/2\in \mathbb{Q}$ is an integer. Hence $d$ is even.

But an irreducible character degree divides the group order, so $d$ must be odd. One quick justification of that divisibility is that the central class sum of a conjugacy class $C$ acts by the algebraic integer ${\lambda }_C=|C|\chi (g_C)/d$. Its integrality follows from its integer matrix on the regular group representation. Character orthogonality then givesa rational algebraic integer, hence an integer. The parity contradiction proves $\boxed{\chi =1_G}$: the trivial character is the only self-conjugate irreducible character of an odd-order group.

Let $g$ have order $n$ in a finite group. In a complex group representation, its matrix is diagonalizable with eigenvalues ${\zeta }_j$ satisfying ${\zeta }_j^n=1$. If each $g^i$ with $\gcd (i,n)=1$ is conjugate to $g$, the character of a representation is constant on these powers. Henceusing the stated integrality of each Ramanujan sum. Therefore $\chi (g)$ is rational. It is also an algebraic integer, being a sum of roots of unity; a rational algebraic integer is an integer. Thus $\boxed{\chi (g)\in \mathbb{Z}}$.

In a symmetric group, raising a permutation to a power coprime to its order preserves each cycle length and therefore its entire cycle type. Since cycle type determines the conjugacy class, the hypothesis holds, and all character values of symmetric groups are integers.

In characteristic $p$, let $K={\mathbb{F}}_p(s,t)$ and $L={\mathbb{F}}_p(s^{1/p},t^{1/p})$, with $s,t$ having algebraic independence. The monomials $s^{i/p}{t}^{j/p}$, $0\le i,j<p$, form a $K$-basis, so $[L:K]=p^2$. Their independence follows by grouping polynomial monomials according to their exponents modulo $p$, after clearing rational-function denominators.

Every $\alpha \in L$ has ${\alpha }^p\in K$, so its minimal polynomial divides $T^p-{\alpha }^p$ and $[K(\alpha ):K]\le p$. Thus this finite purely inseparable extension has no primitive element. It demonstrates why the separable field extension hypothesis cannot simply be removed.

Put $a=\sqrt{p}$, $b=\sqrt{-q}$ and $\alpha =a+b$. Since $a^2-b^2=p+q$, the identity $(a+b)(a-b)=p+q$ givesThus $\mathbb{Q}(\alpha )=\mathbb{Q}(\sqrt{p},\sqrt{-q})$. The latter has degree $4$: $p$ is not a square and $\mathbb{Q}(\sqrt{p})$ is real, whereas $\sqrt{-q}$ is not real. Squaring ${\alpha }^2=p-q+2\sqrt{-pq}$ gives its monic degree-$4$ minimal polynomial

A finite field extension $L/K$ is a separable field extension if every element of $L$ has a minimal polynomial over $K$ with distinct roots in an algebraic closure. The primitive element theorem says that a finite separable field extension $L/K$ is $K(\alpha )$ for some $\alpha \in L$.

If $K$ is finite, then so is $L$. The multiplicative group of a finite field is cyclic, so a generator of $L^{\times }$ generates $L$ as a field. To justify the group fact, a finite subgroup of a field's multiplicative group is cyclic: in a finite abelian group its exponent is attained as an element order, while the polynomial root bound forces its size to be at most that exponent.

If $K$ is infinite, first consider $L=K(a,b)$. The minimal polynomials of $a$ and $b$ have distinct roots $a_i,b_j$ in an algebraic closure. Choose $c\in K$, $c\ne 0$, avoiding the finitely many values for which $a_i+c{b}_j=a+cb$ for a pair other than $(a,b)$. Set $\alpha =a+cb$. Over $K(\alpha )$, the polynomials $f(T)$ and $g((\alpha -T)/c)$ have precisely one common root, $a$. Their monic greatest common divisor is therefore $T-a$, so $a\in K(\alpha )$ and $b=(\alpha -a)/c\in K(\alpha )$. Thus $K(a,b)=K(\alpha )$. Induct over finitely many generators of the finite separable field extension. This proves every finite separable extension is simple.

True. A useful fact is that squaring is strictly increasing on ordinals. If $A<B$, monotonicity in both arguments gives $A^2\le BA$, and strict increase in the right argument for the positive left factor $B$ gives $BA<B^2$. Hence $A^2<B^2$.

Suppose, for a contradiction, that $\alpha \beta <\beta \alpha$. Again using monotonicity and associativity of ordinal multiplication,This contradicts the hypothesized equality of the outer terms. Reversing the roles excludes $\beta \alpha <\alpha \beta$. If either factor is zero, they already commute. Therefore the commuting squares of ordinals property isThe proof uses monotonicity, which follows directly from the ordered-copy definition; it does not assume a cancellation law for ordinals.

True. If $\alpha \beta =\beta \alpha$, repeated use of associativity givesNo unlicensed commutativity of ordinal multiplication is being used: only the assumed commuting pair is swapped.

True. The valid distributive law for ordinal multiplication isThe synthetic definition of ordinal multiplication takes a sequence of copies of the left factor, indexed by the right factor. A sequence indexed first by $\beta$ and then by $\gamma$ is exactly the concatenation of the two corresponding sequences. Taking order types gives the identity. Equivalently, use transfinite induction on $\gamma$, with the successor rule and the supremum rule at limit ordinals.

False. Put $\alpha =\beta =1$ and $\gamma =\omega$. ThenThe two ordinals differ. The order of the factors matters in the distributive law for ordinal multiplication.

The first assertion is false: $\boxed{1+\omega =\omega <\omega +1}$. Thus ordinal addition is not commutative.

The dashed curves are contours of constant Cook's distance, labelled 0.5 and 1, in the plot of standardized residual against regression leverage. Writing $h_{ii}$ for the diagonal of the hat matrix, $s^2$ for the residual variance estimate, and $p=8$ for the number of fitted coefficients,Thus a contour $D_i=d$ has $r_i=\pm \sqrt{dp(1-h_{ii})/h_{ii}}$. On the actual PDF plot, observation 579 has leverage about 0.084 and standardized residual about $-8$, lying beyond the $D=0.5$ contour and below $D=1$ in influence magnitude, approximately $D\simeq 0.7$ to $0.8$ from the plotted coordinates. It is unusually light relative to its fitted log-weight and potentially influential because deleting it can materially change fitted coefficients. It warrants checking, not automatic deletion. The solid smooth red curve is distinct from the dashed Cook's distance contours.

A residual-versus-fitted plot for the untransformed fit places $e_i=y_i-{\hat{y}}_i$ on the vertical axis and ${\hat{y}}_i$ on the horizontal axis. A fan-shaped cloud, with larger residual spread at larger fitted weight, suggests heteroscedasticity and approximately multiplicative errors. A logarithmic transformation may stabilize this spread when the conditional standard deviation is roughly proportional to the mean.

One can also use the scale-location plot of $\sqrt{|r_i|}$ against fitted values, where $r_i$ is a standardized residual. A systematic rise suggests nonconstant error variance. Neither plot proves that taking logarithms is sufficient: diagnostics of the transformed fit should be checked as well, and dependence between repeated chicken measurements is a separate issue.

The comparison is a nested-model F-test of the three time-by-diet interaction coefficients, with null hypothesis that all are zero. The reduced linear regression model has a common time slope and four diet-specific intercepts; the full model permits four slopes. It has eight fitted coefficients and $579-8=571$ residual degrees of freedom.The rounded sums of squares give the displayed approximation; the software reports $F=16.824$ using unrounded values. The quoted p-value $1.744\times 10^{-10}$ gives overwhelming evidence against equal log-weight growth slopes across diets under the normal linear model assumptions. Thus reject the common-slope hypothesis. Because repeated measurements on the same chicken may be correlated, the exact reference distribution requires independent, homoscedastic Gaussian errors; the displayed ordinary regression output by itself does not establish that assumption.

Treat the fixed letters $a_1,\dots ,a_n$, the empty-language and empty-word symbols, and the punctuation of expressions as a finite token alphabet. In the usual syntax allowing parentheses, union, concatenation, and star, the following context-free grammar generates exactly the syntactically valid regular expressions:Structural induction proves that every generated string is an expression and that every expression has a derivation. Grammar ambiguity does not affect the context-free language assertion.

If the language of these expressions were a regular language, its intersection with the regular token language consisting of some opening parentheses, then $a_1$, then some closing parentheses would be regular. That intersection is exactlyThe pumping lemma for regular languages contradicts regularity: a sufficiently long opening-parenthesis prefix forces a pumpable segment wholly within that prefix, and pumping it changes the two parenthesis counts unequally. Hence the expression syntax is context-free but not regular. This conclusion concerns the strings describing expressions, not the languages those expressions denote; it uses the usual concrete syntax with arbitrary nested parentheses.

Let a deterministic finite automaton recognizing the regular language have states $Q$, start state $q_0$, accepting states $F$, and transition function $\delta$. Construct a context-free grammar with one nonterminal $A_q$ for every state, start symbol $A_{q_0}$, and productionsInduction on the length of a word shows that $A_q$ derives $w{A}_r$ exactly when the automaton reaches $r$ from $q$ after reading $w$. The terminal production can then end the derivation precisely at an accepting state. Thus the grammar generates exactly the recognized language, proving every regular language is a context-free language. This is actually a right-linear grammar, a special case of a context-free grammar.

Away from a transmitted grazing or critical angle, ${\rho }_1{c}_1\gg {\rho }_2{c}_2$ gives a small normal acoustic impedance ratio $\delta$. ThusThe reflected energy is nearly all the incident energy, while the pressure phase depends on the plate response. Large $|\beta |$ gives $R\simeq +1$, resembling a rigid boundary; $\beta \simeq 0$ gives $R\simeq -1$, resembling a pressure-release boundary. Bulk impedance contrast alone does not ensure $\delta \ll 1$ arbitrarily close to a transmitted critical angle, since the factor $1/\cos {\theta }_2$ can be large.

For fixed frequency and finite plate parameters, $\beta$ contains the incident normal wavevector and therefore tends to zero as $\theta \to \pi /2$. If $c_1\ne c_2$, the outgoing $l_2$ tends to a nonzero real or imaginary value, whereas $l_1\to 0$, so $\delta \to 0$ too. ConsequentlyThe lower-side boundary pressure $1+R$ tends to zero, regardless of how large the fixed $m,B$ are. The angle must become closer to grazing as those parameters grow; the assertion is not uniform in an unbounded plate mass or rigidity.

If $c_1=c_2$, however, $l_2=l_1$ and $\delta ={\rho }_2/{\rho }_1$ remains constant. The exact grazing limit is $(\delta -1)/(\delta +1)$, rather than exactly $-1$ for finite contrast. Under the stated large-contrast approximation it is still arbitrarily close to the pressure-release value, with boundary pressure $2\delta /(1+\delta )$. Thus the printed pressure-release description is an approximation in that equal-speed case, not an exact zero-pressure identity. If a transmitted wave becomes evanescent, the outgoing square-root convention above continues to give the unequal-speed grazing limit.

Linearized mass conservation, momentum balance and the isentropic pressure-density relation are ${\rho }_t^{\prime }+{\rho }_0\nabla \cdot \mathbf{u}=0$, ${\rho }_0{\mathbf{u}}_t=-\nabla \tilde{p}$, and $\tilde{p}=c_0^2{\rho }^{\prime }$. Differentiate mass conservation in time and take the divergence of momentum balance to eliminate velocity. This gives the acoustic pressure wave equationFor the convention $\tilde{p}=\mathrm{Re}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, momentum balance gives the complex velocity amplitude $\mathbf{k}/({\rho }_0\omega )$, with $\omega =c_0|\mathbf{k}|$. Thus the incident velocity field is $\mathrm{Re}[{\mathbf{k}}_I{e}^{i({\mathbf{k}}_I\cdot \mathbf{r}-\omega t)}/({\rho }_1\omega )]$ in the lower gas.

Let $a=k_I\sin \theta$, $l_1=k_I\cos \theta >0$, $\omega =c_1{k}_I$, and $l_2=\sqrt{(\omega /c_2{)}^2-a^2}$ with nonnegative real or imaginary part. This chooses an outgoing propagating wave or an evanescent wave decaying as $y\to +\infty$. Conservation of tangential wavevector is the acoustic Snell law for elastic and acoustic waves. Write the complex pressures asInviscid kinematic matching requires equal normal velocities at the plate, not equal tangential velocities:Define the normal acoustic impedance ratio $\delta ={\rho }_2{l}_1/({\rho }_1{l}_2)$, so $T=\delta (1-R)$. For a propagating transmitted wave this is ${\rho }_2{c}_2\cos \theta /({\rho }_1{c}_1\cos {\theta }_2)$. The dynamic condition giveswhich is exactly the printed dimensionless parameter. Solving yields the pressure reflection coefficient and transmission coefficientAt $l_2=0$ use the preceding velocity equations, or take their limit, rather than substituting an infinite $\delta$. For real $l_2>0$, the reflected energy fraction is $|R{|}^2$, the transmitted fraction is $|T{|}^2/\delta$, and these sum to one. An evanescent transmitted wave carries no mean normal energy flux.

When the gases match, $\delta =1$, so $R=-i\beta /(2-i\beta )$ and $T=2/(2-i\beta )$. For a nongrazing incident wave of nonzero frequency, perfect transmission means $R=0$, equivalent toThe plate's inertial and bending terms cancel: the incident frequency and tangential wavevector coincide with its free flexural wave dispersion relation. Then $T=1$, including its phase. With positive $m,B$, this condition is possible only at angles and frequencies satisfying that relation; for normal incidence at nonzero frequency it cannot occur unless the mass is zero.

For $k\ne 0$, put $t=k^2>0$ and use the positive-definite Lyapunov function $V=x^2+t{y}^2$. Its derivative isFor $V>0$ set $u=x^2/V\in [0,1]$. The brace divided by $V^2$ is $A{u}^2+2Bu(1-u)+C(1-u{)}^2$, where $A=1$, $B=\beta (1+t)/(2t)$ and $C=1/t$. Since $\beta >2$, $B>A,C$. Completing the square gives its maximumHence $\dot{V}\le -2V+2M{V}^2$. If $V(0)<1/M$, this sublevel region is positively invariant: $V$ decreases, and $\dot{V}\le -2(1-MV(0))V$. The exponential bound makes $V\to 0$. The trajectory is bounded in a compact ellipse, so smoothness of the vector field also guarantees existence for every positive time. ThereforeFor $k=0$ the displayed strict inequality has no solutions; negative $k$ gives the same ellipse as $-k$. Taking the union of the nonempty ellipses is valid because each point needs only one Lyapunov certificate.

At $k=1$ the certified disk has squared radius $2/(\beta +1)$. Along $y=0$, the limiting threshold as $k^2\to \infty$ is $4(\beta -1)/{\beta }^2$, which is strictly larger, since their difference is $2({\beta }^2-2)/({\beta }^2(\beta +1))>0$ for $\beta >2$. Choose $x^2$ strictly between these thresholds and then a sufficiently large finite $k$. This point is certified by that ellipse but lies outside the $k=1$ disk. Thus the union gives a strictly larger certified region. It need not equal the whole basin of attraction.