Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 20 (Diophantine Analysis and Transcendence Theory)

We first give an auxiliary-integral proof, rather than deducing both requested results from a theorem only stated later. The following symmetric Hermite integral obstruction will do both jobs. Supposewhere $k\ne 0$ and $b_j$ are integers, the distinct nonzero algebraic exponents form a Galois-stable set, and their weights are invariant under conjugation. Choose $P\in \mathbb{Z}[X]$ vanishing at those exponents, with $P(0)\ne 0$, and a positive integer $c$ for which every $c{\beta }_j$ is an algebraic integer. For a large prime number $p$, letAt zero, derivatives below $p-1$ vanish; the derivative of order $p-1$ equals $c^{D}P(0{)}^p$. All higher derivatives at zero are integers divisible by $p$. At each ${\beta }_j$, derivatives below $p$ vanish and all higher derivatives belong to $p{\mathcal{O}}_L$ in a Galois splitting field $L$. Indeed their coefficients contain $r!/(p-1)!$, divisible by $p$ for $r\ge p$, and every $c^D{\beta }_j^u$ is integral for $u\le D$. ConsequentlyGalois invariance makes the sum rational, and integrality makes it an integer. For prime numbers not dividing $kcP(0)$ it is nonzero.

But $F_p-F_p^{\prime }=f_p$, so integration along the straight segment to $\beta$ givesMultiply by the weights and use the assumed exponential relation. This expresses $-I_p$ as the sum of these integrals. On the finitely many fixed segments, their absolute values are at most $C^p/(p-1)!$ for a fixed $C$: all polynomial and exponential factors have fixed bounds, and $D$ is linear in $p$. Thus $|I_p|\to 0$, contradicting its being a nonzero integer. The obstruction is proved.

If $e$ were algebraic, an integral polynomial relation would give $a_0+{\sum }_{j=1}^s{a}_j{e}^j=0$ with $a_0\ne 0$. Integer exponents and weights satisfy the obstruction, so $e$ is transcendental.

If $\pi$ were algebraic, put $\alpha =i\pi$ and let ${\alpha }_1,\dots ,{\alpha }_d$ be its algebraic conjugates. Since $1+e^{\alpha }=0$, the product ${\prod }_j(1+e^{{\alpha }_j})$ vanishes. Expand it and collect equal subset sums. The zero sums contribute a positive integer $k\ge 1$; the nonzero sums give a Galois-stable list ${\beta }_j$ with positive integral multiplicities. This is precisely the prohibited relation. Hence $\pi$ is transcendental.

The general Lindemann–Weierstrass theorem states that exponentials of distinct algebraic numbers are linearly independent over $\overline{\mathbb{Q}}$. Suppose the sine quotient were algebraic, say $\gamma$. The denominator is nonzero: $\sin \beta =0$ would give a forbidden relation between the exponentials of the distinct algebraic numbers $i\beta ,-i\beta$. The quotient identity would giveThe four exponents are distinct under the stated nonzero and non-opposite assumptions, so this also contradicts linear independence. Therefore

We prove the Mahler alternative in full. Let $f(z)={\sum }_{n\ge 0}{z}^{l^n}$. Its series converges locally uniformly in the unit disc and satisfiesIt is transcendental as a function. At a root of unity $\zeta$ with ${\zeta }^{l^r}=1$, the radial tail ${\sum }_{n\ge r}{t}^{l^n}$ in $f(t\zeta )$ is real and tends to infinity as $t\uparrow 1$, while the initial terms remain bounded. These roots are dense on the unit circle. An algebraic function can have singularities only at the finitely many zeros of its leading coefficient and polynomial discriminant: elsewhere its defining polynomial and the implicit-function theorem give local analytic branches. Hence $f$ cannot be algebraic over $\mathbb{C}(z)$.

Assume now that $\alpha$ and $f(\alpha )$ are algebraic, with $0<|\alpha |<1$, and put both in a number field $K$ of degree $d$. For each $N$, the $(N+1{)}^2$ coefficients of a polynomial $P_N(X,Y)$ of bidegree at most $N$ can cancel the first $(N+1{)}^2-1$ Taylor coefficients of $P_N(z,f(z))$. The equations have integer coefficients, so a nonzero integer solution exists after clearing denominators. Functional transcendence makes $R_N(z)=P_N(z,f(z))$ nonzero. Its order $T_N$ at zero satisfies $T_N\ge (N+1{)}^2-1$.

At ${\alpha }_r={\alpha }^{l^r}$ the functional equation gives $f({\alpha }_r)=f(\alpha )-{\sum }_{j<r}{\alpha }^{l^j}\in K$. Thus ${\eta }_r=R_N({\alpha }_r)\in K$ is nonzero for all sufficiently large $r$ andUse the absolute logarithmic height $h$. The elementary inequalities for sums and polynomial evaluation giveHere $h(f({\alpha }_r))\le h(f(\alpha ))+(l^r-1)h(\alpha )/(l-1)+r\log 2$. The Liouville height inequality gives $\log |{\eta }_r|\ge -dh({\eta }_r)$. Divide by $l^r$ and let $r\to \infty$ to obtainThe left grows quadratically in $N$, the right only linearly. Choose $N$ large to contradict this. Hence every such algebraic $\alpha$ has transcendental $f(\alpha )$. This is the auxiliary-value height argument for the Fredholm series.

For the rational continuation, let $a_n={\sum }_{j=0}^n(p/q{)}^{l^j}$ and $Q_n=q^{l^n}$. This is a rational number with reduced denominator $B_n\le Q_n$, and its positive tail satisfies, for large $n$,A rational value $a/b$ would instead have distance at least $1/(b{Q}_n)$ from each distinct $a_n$, already impossible. If the value were algebraic irrational, choose $ϵ>0$ with $2+ϵ<\kappa$. The distinct truncations have unbounded reduced denominators, and the displayed errors are eventually smaller than $B_n^{-2-ϵ}$. This contradicts Roth theorem. Thus the requested rational values are transcendental by the Roth criterion for rational Fredholm values.

For completeness, the elliptic alternative's conclusion also has a short auxiliary-divisor certificate. If $\Phi$ is Frobenius on $E/{\mathbb{F}}_q$, then $\mathrm{deg}\Phi =q$ and $\mathrm{deg}(1-\Phi )=N=\#E({\mathbb{F}}_q)$: the latter map is separable and its kernel is exactly the rational points. The divisor identity for $x(P)-x(Q)$ on $E\times E$ gives the parallelogram rule $\mathrm{deg}(u+v)+\mathrm{deg}(u-v)=2\mathrm{deg}u+2\mathrm{deg}v$. Polarizing it yieldsAll these degrees are nonnegative. In particular $\mathrm{deg}(-t+2\Phi )=4q-t^2\ge 0$, so $|t|\le 2\sqrt{q}$. The Frobenius characteristic polynomial is $X^2-tX+q$; its complex roots have modulus $\sqrt{q}$, including the double-root case. The zeta function has numerator $1-tT+q{T}^2$, whose zeros consequently have modulus $q^{-1/2}$, the elliptic Riemann hypothesis. This supplements the fully proved Mahler branch with the degree-form proof of the Hasse bound; it is an auxiliary-divisor proof, not a claim that an unspecified transcendence theorem proves the bound.

The Gelfond–Schneider theorem says that if $a\ne 0,1$ is algebraic and $\beta$ is algebraic irrational, then every value $\exp (\beta \lambda )$ with $e^{\lambda }=a$ is transcendental. Fix such a nonzero logarithm $\lambda$, and suppose $\gamma =e^{\beta \lambda }$ is algebraic. The field $K=\mathbb{Q}(a,\beta ,\gamma )$ then contains all values needed in Schneider lattice proof of the Gelfond–Schneider theorem.

Here is an outline including the parameter balance. For a large $L$, choose polynomial degrees $M\asymp L\log L$, $N\asymp L/\log L$, with constants large enough that $(M+1)(N+1)>[K:\mathbb{Q}]L^2$. ImposeExpand these $K$-linear equations in a rational basis and clear denominators. Siegel lemma gives a nonzero integer polynomial $P$ with $\log H(P)=O(L^2/\log L)$. This follows because each coefficient in the grid equations has arithmetic height $O(M\log L+NL)$ and the ratio of unknowns to constraints stays bounded away from one. The exponential polynomial $F(z)=P(z,e^{\lambda z})$ is not identically zero: distinct exponentials are independent over the polynomial ring, as one sees by successively applying powers of $D-j\lambda$ to kill all but one term.

The irrationality of $\beta$ makes the $L^2$ grid points distinct. Their zeros make subsequent grid values extremely small. More precisely, if a whole $S\times S$ grid has been obtained, take an outer radius $R\asymp S\log S$. Growth gives $\log {\max }_{|z|=R}|F(z)|\le \log H(P)+O(M\log R+NR)$. The product form of Schwarz's lemma contributes $(C/\log S{)}^{S^2}$ at the next grid points. Thereforefor large initial $L$, unless the value is zero. But every such value is algebraic in the same $K$, since $e^{\lambda (s+t\beta )}=a^s{\gamma }^t$. Its arithmetic height is $O(\log H(P)+M\log S+NS)$, so the Liouville lower bound is incompatible with the displayed upper bound for a nonzero value. Induction extends the zero grid indefinitely.

A fixed nonzero exponential polynomial has only $O(R)$ zeros in a disc of radius $R$, by its exponential-type growth and Jensen's formula centered at a nonzero value. The grids give $S^2$ distinct zeros in a disc of radius $O(S)$, a contradiction. Thus $a^{\beta }$ is transcendental for every choice of logarithm, completing the Schneider lattice proof of the Gelfond–Schneider theorem.

A standard form of the Schneider-Lang theorem is this. Let $f_1,\dots ,f_s$ be meromorphic functions with $f_j^{\prime }\in K[f_1,\dots ,f_s]$ for a fixed number field $K$. Suppose $f_1,f_2$ have algebraic independence over $\mathbb{C}$ and have finite growth orders ${\rho }_1,{\rho }_2$. There are at most $[K:\mathbb{Q}]({\rho }_1+{\rho }_2)$ points at which all the functions are finite and all their values lie in $K$.

Take $f_1(z)=z$ and $f_2(z)=e^z$, of orders zero and one. Their derivatives belong to $K[z,e^z]$. They have algebraic independence: a polynomial relation ${\sum }_j{p}_j(z)e^{jz}=0$ is impossible by exponential-polynomial independence, or by letting real $z$ tend to infinity and considering the largest $j$.

If nonzero algebraic $\alpha$ had algebraic $e^{\alpha }$, put $K=\mathbb{Q}(\alpha ,e^{\alpha })$. At each of the distinct points $z=n\alpha$, both $z$ and $e^z=(e^{\alpha }{)}^n$ belong to $K$. Infinitely many points contradict the theorem's finite bound. Therefore

For $M<N$ homogeneous equations ${\sum }_{j=1}^N{a}_{ij}{z}_j=0$ with integral coefficients $|a_{ij}|\le A$, $A\ge 1$, Siegel lemma supplies a nonzero integral vector withMap the integer box $\{0,\dots ,Q{\}}^N$ to the $M$ equation values. There are $(Q+1{)}^N$ inputs, while each output coordinate lies in $[-NAQ,NAQ]$, so there are at most $(2NAQ+1{)}^M$ outputs. Since $Q+1>(2NA{)}^{M/(N-M)}$ and $2NAQ+1\le 2NA(Q+1)$, the input count is strictly larger. Two inputs have the same image; their nonzero difference solves every equation and has the stated norm bound. This proves the lemma by the pigeonhole proof of integer Siegel lemma.

Rational coefficients are handled by clearing denominators. For coefficients in a number field of degree $d$, expansion in a rational basis gives at most $dM$ rational equations; if $N>dM$ the same argument gives an integer solution with the corresponding effective bound after clearing the basis denominators. The lemma concerns homogeneous equations; arbitrary inhomogeneous integer equations need not be soluble.

In Siegel's original convention, an E-function is an entire series $F(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$ with coefficients in a fixed number field, satisfying a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$, together with these arithmetic bounds: for every $ϵ>0$, all conjugates of $a_0,\dots ,a_n$ have size $O_{ϵ}(n^{ϵn})$, and some positive integer $d_n=O_{ϵ}(n^{ϵn})$ makes every $d_n{a}_j$, $j\le n$, an algebraic integer. A frequently used stricter convention replaces both bounds by $C^{n+1}$. The original and strict growth definitions should not simply be declared equivalent; the exponential examples below satisfy the strict one.

For sums, work in the compositum of the coefficient fields. Coefficient conjugate bounds add. The product of the two common denominators clears both lists; using $ϵ/2$ in each original denominator bound gives the required $O_{ϵ}(n^{ϵn})$ product bound. Finally all derivatives of $F+G$ lie in the sum of the two finite-dimensional derivative spaces over that rational function field, so they satisfy a nontrivial linear dependence, giving a differential equation for the sum. Thus sums are again E-functions, the addition closure of Siegel E-functions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $F^{\prime }=A(z)F$ with rational-function matrix over $\overline{\mathbb{Q}}$, then at every nonzero algebraic $\xi$ which is not a pole of $A$,In particular algebraically independent functions give algebraically independent values.

Take $F_j(z)=e^{{\gamma }_{j}z}$ for algebraic numbers ${\gamma }_j$ linearly independent over $\mathbb{Q}$. Their coefficients are ${\gamma }_j^n$ with exponential conjugate and denominator bounds, and $F_j^{\prime }={\gamma }_j{F}_j$. A polynomial relation among the functions expands into a relation among exponentials with distinct exponents $\sum m_j{\gamma }_j$ and polynomial coefficients after clearing rational-function denominators; exponential-polynomial independence excludes it. At $\xi =1$ the theorem gives algebraic independence of the $e^{{\gamma }_j}$. For arbitrary distinct algebraic ${\alpha }_i$, choose a rational basis of their span and divide that basis by a common denominator. The $e^{{\alpha }_i}$ are distinct Laurent monomials in the corresponding algebraically independent exponential values, so are linearly independent over $\overline{\mathbb{Q}}$. This recovers the general Lindemann–Weierstrass theorem. Its simplest instance uses just $F(z)=e^z$ at nonzero algebraic $\xi$.

Fix fundamental units ${\epsilon }_1,\dots ,{\epsilon }_r$ and write $x=\zeta \prod {\epsilon }_i^{b_i}$. Rank zero already gives a finite unit group. Otherwise put $B=\max (1,|b_i|)$. The logarithmic embedding in Dirichlet's theorem has a weighted sum-zero image and is injective on the free exponent lattice. Norm equivalence and that sum-zero condition give an archimedean embedding $\sigma$ withThis is the small archimedean value of a large unit. From the equation, $\sigma (\beta y)=1-\sigma (\alpha x)$ is exponentially close to one. Also $h(y)\le h(x)+O(1)$ by the arithmetic height inequalities for $y=(1-\alpha x)/\beta$, so the fundamental-unit exponents of $y$ have size $O(B)$.

For large $B$, take the small-branch logarithm of $\sigma (\beta y)$. With $y=\eta \prod {\epsilon }_i^{c_i}$ it is the nonzero formusing fixed logarithms and $\log (-1)=\pi i$. The integer $k$ corrects the branch, and nonzero follows since $\beta y=1$ would force $\alpha x=0$. Its upper bound is $|\Lambda |\le C^{\prime }{e}^{-cB}$. The assumed effective logarithmic-form estimate for the fixed algebraic numbers gives $\log |\Lambda |\ge -C^{\prime \prime }\log (2B)$. Thus $cB\le C^{\prime \prime }\log (2B)+O(1)$, effectively bounding $B$. The finitely many embeddings and torsion choices give uniform constants. Therefore there are only finitely many unit solutions, and their exponents are effectively bounded.

For the Thue application, take an irreducible integral binary form $F$ of degree $d\ge 3$ and $m\ne 0$. In a splitting field, write $F(X,Y)=a{\prod }_i(X-{\rho }_{i}Y)$ and put ${\theta }_i=a{\rho }_i$, which are algebraic integers. The integral factors $L_i=aX-{\theta }_{i}Y$ satisfy ${\prod }_i{L}_i=a^{d-1}m$. Each ideal $(L_i)$ therefore divides one fixed ideal, giving finitely many possibilities. Choose a generator ${\delta }_i$ for each principal possibility, so $L_i={\delta }_i{u}_i$ with units $u_i$.

For three distinct roots,Division by the third term gives a fixed-coefficient unit equation in $u_1/u_3,u_2/u_3$. Its solutions are finite, hence there are finitely many values of $L_1/L_2$. This ratio determines the rational direction $X:Y$, since the roots are distinct. In a fixed direction, homogeneity and $F(X,Y)=m\ne 0$ leave only finitely many integer scales. This proves Thue theorem, and the argument also works for a form with at least three distinct projective roots even if reducible.

The hyperelliptic alternative uses a related factorization for the usual nondegenerate case with squarefree $f$ of degree at least three. For $y^2=f(x)$, ideals of two different root factors can share primes only above the fixed root differences and the leading coefficient. At other primes their valuations are even. Finite exceptional-prime and ideal-class choices, together with units modulo squares, reduce the factors to finitely many expressions ${\delta }_i{z}_i^2$. Relations between distinct root factors then give simultaneous norm or Pell-type equations. Writing their solutions in fundamental units and applying the same small-value/logarithmic-form estimates bounds the unit exponents, hence $x,y$. The common-factor and square-class reductions are essential; a single quadratic norm equation alone can have infinitely many Pell solutions. This outlines the effective hyperelliptic treatment as well as the explicit Thue reduction.

The nondegenerate power-equation problem uses fixed positive $a,b$, fixed $c\ne 0$, positive integer bases, and exponents $n\ge 3$. Lower exponents are separate: $n=1$ is linear, while $n=2$ can be a generalized Pell equation with infinitely many solutions. For example $x^2-2y^2=1$ has the infinite sequence generated by $(3+2\sqrt{2}{)}^j$. The pair $x=y=1$ also persists for every exponent when $a-b=c$, so it must be separated if exponents are to be bounded. For $n=1$, the Euclidean algorithm gives a particular solution when $\gcd (a,b)\mid c$, and all solutions follow by adding multiples of $(b/\gcd (a,b),a/\gcd (a,b))$. For $n=2$, put $X=ax$ to obtain $X^2-ab{y}^2=ac$ with the congruence $a\mid X$. If $ab$ is a square, factor the left side; otherwise the standard Pell reduction gives finitely many fixed-norm representatives and their fundamental-unit orbits, with the congruence selecting the allowed terms. This is an effective description even when the set is infinite.

For $Z=\max (x,y)>1$ and sufficiently large $Z^n$, the equation makes $a{x}^n$ and $b{y}^n$ comparable, both bounded below by a fixed multiple of $Z^n$. Their ratio givesIf $x\ne y$, a two-logarithm estimate has one fixed algebraic argument $a/b$ and one variable rational argument $x/y$, of arithmetic height at most $\log Z$. It gives $\log |\Lambda |\ge -C(a,b)\log Z\log (2n)$. A zero fixed logarithm is omitted, leaving the valid one-logarithm case. Therefore $n\le C\log (2n)+O(1)$, effectively bounding $n$. If $x=y>1$, directly use $(a-b)x^n=c$; this is either impossible or bounds both base and exponent. Cases where $Z^n$ stays small are finite and directly enumerable.

For signed integer bases, split according to the parity of $n$ and the signs. Cases reducing to a sum of positive powers are elementary because their sum is fixed; zero or absolute-value-one bases give separately detectable constant-power families.

For each bounded $n\ge 3$, the form $a{X}^n-b{Y}^n$ has $n$ distinct projective roots. The three-factor unit-equation argument of Question 5 gives an effective finite list of pairs, even if the binomial is reducible. Consequently all nondegenerate solutions can be effectively found, with the linear/Pell families and constant-base exception treated separately. This is the exponent bound for a binomial power equation.

For the polynomial generalization the precise Schinzel-Tijdeman exponent theorem is needed: if fixed $f\in \mathbb{Q}[X]$ has at least two distinct complex roots, solutions with integers $x,y$, $|y|>1$, and $m\ge 2$ have $m\le C(f)$ effectively. The hypotheses cannot be dropped. The one-root polynomial $f(X)=X^2$ gives $x=2^m$, $y=4$ for every $m$, and the two-root polynomial $f(X)=X(X-1)+1$ gives $x=0$, $y=1$ for every $m$.

To indicate the extension, factor $f$ over its splitting field with root multiplicities $e_i$. Distinct factors have common ideal divisors only over a fixed finite collection of primes. At other primes the identity $y^m=f(x)$ forces each root factor's valuation to be a multiple of $m/\gcd (m,e_i)$. Ideal classes and fundamental units supply the remaining finitely specified factors. Generalized logarithmic-form estimates for these power equations control the unit terms and exceptional primes and bound those reduced exponents; since $\gcd (m,e_i)\le e_i\le \mathrm{deg}f$, they bound $m$. This is why distinct roots and nontrivial perfect powers are indispensable to the asserted finiteness of exponents.

The abc conjecture says that for every $ϵ>0$ there is $C_{ϵ}$ such that coprime positive integers $A+B=C$ satisfy $C\le C_{ϵ}\mathrm{rad}(ABC{)}^{1+ϵ}$, where the radical is the product of distinct prime divisors.

The Catalan alternative yields finiteness of actual nontrivial solutions, with $x,y\ge 2$ and $p,q\ge 3$. Put $M=x^p$, so $y^q=M-1$ and $x,y$ are coprime. The triple $(y^q,1,x^p)$ has radical at most $xy<M^{2/3}$. Take $ϵ=1/6$ to obtainBoth bases are bounded, and $2^p\le M$, $2^q\le M-1$ also bound both exponents. Thus only finitely many nontrivial Catalan solutions exist under abc, the Catalan power bound from abc.

For the Fermat alternative, the usual arithmetic height statement concerns primitive solutions, or equivalently solutions modulo common scaling. A primitive positive triple is pairwise coprime, and $x,y<z$. Applying abc to $(x^n,y^n,z^n)$ givesThis is uniform even in $n$. If $h=\max (x,y)<z$, then $z^n=x^n+y^n\le 2h^n$, so $n\le \log 2/\log (z/h)$ is bounded after $z$ is bounded. There are therefore finitely many primitive triples and exponents, the primitive Fermat bound from abc. Without the primitive convention, any one nonzero homogeneous solution would produce infinitely many common multiples. The Catalan proof above supplies the requested literal finiteness alternative without that normalization issue.

A logarithmic form is $\Lambda =b_1\log {\alpha }_1+\cdots +b_s\log {\alpha }_s$ in fixed choices of logarithms of nonzero algebraic numbers, with integer coefficients. The indispensable condition is $\Lambda \ne 0$. Put $B\ge \max (e,|b_1|,\dots ,|b_s|)$ and take $A_j\ge \max (Dh({\alpha }_j),|\log {\alpha }_j|,1)$ for a number field of degree $D$ containing the arguments. A typical effective estimate iswith an explicitly computable constant. For fixed arguments this is a power-type lower bound $|\Lambda |>B^{-C}$. It is fundamentally stronger for applications than a crude norm estimate exponential in $B$.

The basic estimates follow the auxiliary-function pattern already used here, in a multivariable form. Choose a polynomial or an interpolation determinant in powers of the algebraic arguments; impose many derivative or interpolation conditions by Siegel lemma. Analytic extrapolation near the chosen logarithms makes the determinant tiny. Arithmetic denominator and norm estimates prevent a nonzero algebraic determinant from being so small, while a multiplicity or zero estimate ensures the necessary determinant does not vanish identically. Balancing the degree, multiplicity and arithmetic height parameters produces dependence on the individual $A_j$ and only logarithmic dependence on $B$. The nonvanishing argument and the arithmetic arithmetic height control are both necessary parts of this outline.

For practical equations the output is a finite, certified computation. Factor the equation in a number field, record the finitely many ideal and torsion possibilities, and express its factors in fundamental units. At a place where the original equation gives an exponentially small remainder, choose the principal logarithm and include a multiple of $2\pi i$ for the branch. Compare that upper bound with the effective lower bound to obtain an initial exponent bound. Question 5 shows this for a unit equation, and Question 6 shows why a rational argument of variable arithmetic height can still bound a common power exponent. Zero forms and degenerate families are separated before any estimate is applied.

A concrete two-logarithm example is $2^u-3^v=1$, with positive exponents. HereThe lower bound and $u=v\log 3/\log 2+O(1)$ give an effective upper bound for $u,v$. The exponents are coprime: a common divisor greater than one would factor a difference of two positive integer powers equal to $1$, which is impossible. For large $v$ the same upper estimate makes $u/v$ a continued-fraction convergent of $\log 3/\log 2$, since its error is smaller than $1/(2v^2)$. One checks only the finitely many convergents up to the proven bound, with small cases checked directly. Higher-dimensional problems use the LLL algorithm to obtain analogous improvements from the first large bound.

Every numerical logarithm must be evaluated with a certified error interval, so that a proposed reduction really is an inequality and an exact integer check can certify each surviving candidate. Continued fractions or lattice reduction accelerate the search; the prior effective upper bound is what makes the search complete. This explains how linear forms in logarithms of algebraic numbers turn qualitative finiteness into practical Diophantine algorithms, while also outlining the auxiliary machinery behind the estimates.