Catalan Constant: source and evidence

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1 \chapter{A constant with many exact faces}
2 \label{ch:constant}
3 
4 Catalan's constant begins with an unusually accessible definition. Its terms are rational, their signs alternate, and its value can be enclosed without assuming anything about irrationality. The difficulty appears when one asks what kind of number the limit is. A definition, an integral, a recurrence and a long decimal expansion each give access to the same number. None of these descriptions alone supplies the arithmetic conclusion.
5 
6 This chapter fixes the analytic object before we introduce rational approximants. All integral identities and interval bounds below are proved directly. The finite computations in the accompanying bundle use these bounds to decide signs. They do not use a floating point approximation to make an exact decision.
7 
8 \section{Definition and the question being asked}
9 
10 \begin{definition}[Catalan's constant]
11 \label{def:c01-G}
12 The real number studied in this book is
13 \begin{equation}
14 G=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}.
15 \label{eq:c01-definition}
16 \end{equation}
17 The series converges absolutely because
18 \[
19 \sum_{k=0}^{\infty}\frac1{(2k+1)^2}
20 \leq\sum_{j=1}^{\infty}\frac1{j^2}<\infty.
21 \]
22 \end{definition}
23 
24 The familiar decimal display begins
25 \[
26 G\approx0.91596559417721901505460351493238411077.
27 \]
28 The approximation sign matters. A decimal is a coordinate for seeing the number, while a rational enclosure is evidence for a particular numerical assertion. A proof of irrationality requires a different assertion again.
29 
30 \begin{conjecture}[The irrationality question]
31 \label{conj:c01-irrationality}
32 Catalan's constant does not belong to $\Q$.
33 \end{conjecture}
34 
35 The checked material for this edition does not establish this conjecture. The literature record includes a September 2026 preprint claiming a proof and a subsequent primary note identifying an incomplete asymptotic derivation \cite{Sun2026,Wachs2026}. The later chapters state that record with its date and scope. The constructions developed here do not depend on accepting the claimed proof.
36 
37 Irrationality would mean that no integers $a,b$ with $b\ne0$ satisfy $G=a/b$. Transcendence would mean that no nonzero polynomial with integer coefficients vanishes at $G$. These are different questions. The second implies the first, because a rational number $a/b$ is a zero of the polynomial $bx-a$. Our target throughout is the first assertion.
38 
39 \begin{figure}[tbp]
40 \centering
41 \includegraphics[width=0.94\linewidth]{../visual/figures/01_constant.pdf}
42 \caption{The same constant appears through a convergent series and through exact integral representations. The displayed decimal supplies orientation. The formulas, rather than the printed digits, define the object.}
43 \label{fig:c01-constant}
44 \end{figure}
45 
46 \subsection{The Dirichlet beta family}
47 
48 For real $s>1$, the Dirichlet beta value is
49 \begin{equation}
50 \beta(s)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^s}.
51 \label{eq:c01-beta}
52 \end{equation}
53 Absolute convergence follows from comparison with $\sum j^{-s}$. Thus $G=\beta(2)$. The alternating series also defines $\beta(s)$ for $s>0$, though absolute convergence is no longer available throughout that wider range. At $s=1$, integration of the finite geometric expansion of $(1+x^2)^{-1}$ gives $\beta(1)=\pi/4$. The integrated remainder is bounded by $\int_0^1x^{2N}\dd x=(2N+1)^{-1}$ and tends to zero, so this passage does not require absolute convergence of the beta series at $1$.
54 
55 The family gives useful context without collapsing its separate arithmetic questions. Zudilin proved that at least one of
56 \[
57 \beta(2),\ \beta(4),\ \beta(6),\ \beta(8),\ \beta(10),\ \beta(12)
58 \]
59 is irrational \cite{Zudilin2019}. This is a disjunction. It permits the irrational member to be a value other than $\beta(2)$. A proof about the collection therefore cannot be inserted as a proof about $G$. The precise source contract is LC09 in the evidence bundle.
60 
61 \section{Alternating partial sums give rational intervals}
62 
63 Write
64 \begin{equation}
65 s_N=\sum_{k=0}^{N-1}\frac{(-1)^k}{(2k+1)^2},\qquad N\geq0,
66 \label{eq:c01-partial-sum}
67 \end{equation}
68 where $s_0=0$ is the empty sum. Thus $N$ counts retained terms, as it does in the visual guide. Each $s_N$ is an exact rational number. The terms decrease strictly in magnitude to zero. Grouping successive terms shows that the even partial sums increase and the odd partial sums decrease. These two sequences approach the same limit.
69 
70 \begin{proposition}[Alternating enclosure]
71 \label{prop:c01-alternating}
72 For every integer $r\geq0$,
73 \begin{equation}
74 s_{2r}<G<s_{2r+1}.
75 \label{eq:c01-parity-enclosure}
76 \end{equation}
77 For every integer $N\geq0$,
78 \begin{equation}
79 0<(-1)^N(G-s_N)<\frac1{(2N+1)^2}.
80 \label{eq:c01-alternating-error}
81 \end{equation}
82 \end{proposition}
83 
84 \begin{proof}
85 The difference between two consecutive even sums is
86 \[
87 s_{2r+2}-s_{2r}
88 =\frac1{(4r+1)^2}-\frac1{(4r+3)^2}>0.
89 \]
90 Similarly,
91 \[
92 s_{2r+3}-s_{2r+1}
93 =-\frac1{(4r+3)^2}+\frac1{(4r+5)^2}<0.
94 \]
95 Consecutive sums differ by a term tending to zero, so the increasing even sums and decreasing odd sums have the same limit $G$. Strictness follows because the next sum of the same parity lies strictly between the current endpoint and the limit on the other side.
96 
97 To obtain the error bound, apply this enclosure to $s_N$ and $s_{N+1}$. The limit lies strictly between them. Their distance is $(2N+1)^{-2}$, and the direction from $s_N$ to $s_{N+1}$ has sign $(-1)^N$.
98 \end{proof}
99 
100 For example,
101 \[
102 s_0=0,\qquad s_1=1,\qquad s_2=\frac89,\qquad
103 s_3=\frac{209}{225},\qquad
104 s_4=\frac{10016}{11025}.
105 \]
106 Hence
107 \begin{equation}
108 \frac{10016}{11025}<G<\frac{209}{225}.
109 \label{eq:c01-small-bracket}
110 \end{equation}
111 These statements use only rational arithmetic. The strict inequality also proves $0<G<1$, which will simplify later tail estimates.
112 
113 \begin{figure}[tbp]
114 \centering
115 \includegraphics[width=0.94\linewidth]{../visual/figures/02_alternating_series.pdf}
116 \caption{Even and odd partial sums approach the common limit from opposite sides. Their errors are controlled by the next omitted term. Every partial sum represented in the figure is rational.}
117 \label{fig:c01-alternating}
118 \end{figure}
119 
120 The error bound is transparent, but its decay is slow. To make the next omitted term smaller than $10^{-d}$, one needs $2N+1>10^{d/2}$. Increasing numerical accuracy this way rapidly becomes expensive. The Euler transform below produces a much faster certified enclosure. Before using it, we establish the positive integral that controls its terms.
121 
122 \section{Three exact integral descriptions}
123 
124 \subsection{A logarithmic integral}
125 
126 \begin{lemma}[A basic moment]
127 \label{lem:c01-moment}
128 For every real $a>-1$,
129 \begin{equation}
130 \int_0^1x^a(-\log x)\,\dd x=\frac1{(a+1)^2}.
131 \label{eq:c01-moment}
132 \end{equation}
133 The integral is understood as an improper integral at $0$.
134 \end{lemma}
135 
136 \begin{proof}
137 Set $x=e^{-t}$. Then $-\log x=t$, $\dd x=-e^{-t}\dd t$, and the integral becomes
138 \[
139 \int_0^{\infty}t e^{-(a+1)t}\,\dd t.
140 \]
141 Integration by parts gives $(a+1)^{-2}$. The boundary term at infinity is zero because an exponential with positive decay rate dominates $t$. The boundary term at zero is also zero.
142 \end{proof}
143 
144 \begin{proposition}[Single integral]
145 \label{prop:c01-log-integral}
146 Catalan's constant satisfies
147 \begin{equation}
148 G=\int_0^1\frac{-\log x}{1+x^2}\,\dd x.
149 \label{eq:c01-log-integral}
150 \end{equation}
151 \end{proposition}
152 
153 \begin{proof}
154 For $0\leq x<1$, the geometric expansion is
155 \[
156 \frac1{1+x^2}=\sum_{k=0}^{\infty}(-1)^kx^{2k}.
157 \]
158 The integrated absolute values of the proposed terms have finite sum:
159 \[
160 \sum_{k=0}^{\infty}\int_0^1x^{2k}(-\log x)\,\dd x
161 =\sum_{k=0}^{\infty}\frac1{(2k+1)^2}<\infty.
162 \]
163 Therefore the series of integrable functions can be summed under the integral. The moment identity gives the defining series for $G$. The point $x=1$ has measure zero and does not affect the integral.
164 \end{proof}
165 
166 The singularity in $-\log x$ is integrable. This is a useful example of why an unbounded integrand is not the same as an infinite integral. In fact, $\int_0^1-\log x\,\dd x=1$. The factor $(1+x^2)^{-1}$ is bounded between $1/2$ and $1$, so the integral representation independently places $G$ between $1/2$ and $1$.
167 
168 \subsection{An arctangent and a logarithmic tangent}
169 
170 \begin{proposition}[Arctangent and tangent forms]
171 \label{prop:c01-tangent}
172 The following equalities hold:
173 \begin{equation}
174 G=\int_0^1\frac{\arctan x}{x}\,\dd x
175 =-\int_0^{\pi/4}\log(\tan\theta)\,\dd\theta.
176 \label{eq:c01-tangent-forms}
177 \end{equation}
178 The integrand $\arctan x/x$ is assigned its continuous value $1$ at $x=0$.
179 \end{proposition}
180 
181 \begin{proof}
182 For $0<\delta<1$, integration by parts gives
183 \[
184 \int_\delta^1\frac{\arctan x}{x}\,\dd x
185 =-\arctan(\delta)\log\delta
186 -\int_\delta^1\frac{\log x}{1+x^2}\,\dd x.
187 \]
188 The boundary expression tends to zero as $\delta\downarrow0$, since $0\leq\arctan\delta\leq\delta$ and $\delta\abs{\log\delta}\to0$. The first equality follows from Proposition~\ref{prop:c01-log-integral}.
189 
190 For the second equality, substitute $x=\tan\theta$ into the logarithmic integral. On $0<\theta\leq\pi/4$,
191 \[
192 \frac{\dd x}{1+x^2}=\dd\theta.
193 \]
194 The substitution is valid first on a truncated interval. Its limit exists by the integrability already established for $-\log x$.
195 \end{proof}
196 
197 The logarithmic tangent form will later identify an integral normalization at index zero. That check is especially valuable because a constant factor in a linear form changes its normalization even when its qualitative decay is unaffected. We will use the equality just proved, rather than assume the normalization from a printed formula.
198 
199 \subsection{A double integral over a square}
200 
201 \begin{proposition}[Elementary double integral]
202 \label{prop:c01-double}
203 Catalan's constant also satisfies
204 \begin{equation}
205 G=\int_0^1\int_0^1\frac{\dd x\,\dd y}{1+x^2y^2}.
206 \label{eq:c01-double}
207 \end{equation}
208 \end{proposition}
209 
210 \begin{proof}
211 One proof integrates with respect to $y$. For $x>0$,
212 \[
213 \int_0^1\frac{\dd y}{1+x^2y^2}=\frac{\arctan x}{x}.
214 \]
215 The value at $x=0$ is $1$. Proposition~\ref{prop:c01-tangent} now gives the result.
216 
217 There is also a direct series justification. Expanding $(1+x^2y^2)^{-1}$ on the square except at the corner $(1,1)$ gives geometric terms $(-1)^kx^{2k}y^{2k}$. Their integrated absolute values sum to
218 \[
219 \sum_{k=0}^{\infty}\left(\int_0^1x^{2k}\,\dd x\right)^2
220 =\sum_{k=0}^{\infty}\frac1{(2k+1)^2}<\infty.
221 \]
222 Thus summation and integration may be interchanged, giving the defining series again.
223 \end{proof}
224 
225 The double integral is bounded and positive on a compact square. The logarithmic integral has an integrable endpoint singularity. Both represent exactly the same number. Their different analytic forms permit different transformations, but their arithmetic classification remains a separate obligation.
226 
227 \section{An Euler transform with an exact tail bound}
228 
229 Let
230 \[
231 a_k=\frac1{(2k+1)^2},\qquad k\geq0,
232 \]
233 and define the positive difference convention
234 \begin{equation}
235 \Delta a_k=a_k-a_{k+1}.
236 \label{eq:c01-difference-convention}
237 \end{equation}
238 The sign convention is fixed here because the other common convention would change every odd difference. Iteration gives
239 \begin{equation}
240 \Delta^j a_0=\sum_{k=0}^{j}(-1)^k\binom jk\frac1{(2k+1)^2}.
241 \label{eq:c01-finite-difference}
242 \end{equation}
243 This is an exact rational number for every $j$.
244 
245 \begin{lemma}[Positive differences]
246 \label{lem:c01-positive-differences}
247 For every integer $j\geq0$,
248 \begin{equation}
249 \Delta^j a_0
250 =\int_0^1(1-x^2)^j(-\log x)\,\dd x,
251 \qquad 0<\Delta^j a_0\leq1.
252 \label{eq:c01-positive-differences}
253 \end{equation}
254 \end{lemma}
255 
256 \begin{proof}
257 Insert the moment formula into the finite sum in \eqref{eq:c01-finite-difference}. A finite sum can be interchanged with an integral without any convergence issue. The binomial theorem gives
258 \[
259 \sum_{k=0}^{j}(-1)^k\binom jkx^{2k}=(1-x^2)^j.
260 \]
261 The integrand is positive for $0<x<1$. It is at most $-\log x$, whose integral is $1$.
262 \end{proof}
263 
264 \begin{theorem}[Certified Euler enclosure]
265 \label{thm:c01-euler}
266 Define
267 \begin{equation}
268 E_N=\sum_{j=0}^{N}\frac{\Delta^j a_0}{2^{j+1}},\qquad N\geq0.
269 \label{eq:c01-E}
270 \end{equation}
271 Then
272 \begin{equation}
273 G=\sum_{j=0}^{\infty}\frac{\Delta^j a_0}{2^{j+1}},
274 \qquad
275 E_N<G<E_N+2^{-N-1}.
276 \label{eq:c01-euler-enclosure}
277 \end{equation}
278 More precisely,
279 \begin{equation}
280 G-E_N
281 =\int_0^1\frac{-\log x}{1+x^2}
282 \left(\frac{1-x^2}{2}\right)^{N+1}\,\dd x.
283 \label{eq:c01-euler-remainder}
284 \end{equation}
285 \end{theorem}
286 
287 \begin{proof}
288 By the preceding lemma all terms of the series are nonnegative. Monotone convergence permits the sum to pass under the integral. For $0\leq x\leq1$,
289 \[
290 \sum_{j=0}^{\infty}\frac{(1-x^2)^j}{2^{j+1}}
291 =\frac1{1+x^2}.
292 \]
293 The resulting integral is $G$ by Proposition~\ref{prop:c01-log-integral}.
294 
295 The finite geometric identity is
296 \[
297 \sum_{j=0}^{N}\frac{(1-x^2)^j}{2^{j+1}}
298 =\frac{1-((1-x^2)/2)^{N+1}}{1+x^2}.
299 \]
300 Integration gives the exact remainder formula. Its integrand is positive on $(0,1)$, so $G-E_N>0$. Since $(1-x^2)/2\leq1/2$,
301 \[
302 G-E_N\leq2^{-N-1}G<2^{-N-1}.
303 \]
304 This gives the stated enclosure. An alternative bound sums the omitted terms using $\Delta^ja_0\leq1$, with the same geometric tail $2^{-N-1}$.
305 \end{proof}
306 
307 The stronger intermediate inequality also gives
308 \begin{equation}
309 G\leq\frac{E_N}{1-2^{-N-1}}.
310 \label{eq:c01-euler-sharper}
311 \end{equation}
312 The denominator is positive for every $N\geq0$. The release computations use the simpler interval $[E_N,E_N+2^{-N-1}]$, because that interval has an immediate rational width and is already sufficient for the tested signs.
313 
314 \begin{example}[Small Euler intervals]
315 The first difference values are
316 \[
317 \Delta^0a_0=1,\qquad \Delta a_0=\frac89,\qquad
318 \Delta^2a_0=\frac{184}{225}.
319 \]
320 They give
321 \[
322 E_0=\frac12,\qquad E_1=\frac{13}{18},\qquad
323 E_2=\frac{371}{450}.
324 \]
325 Thus, for example,
326 \[
327 \frac{371}{450}<G<\frac{1709}{1800}.
328 \]
329 These short transformed sums are useful for checking conventions. Their principal advantage appears as $N$ grows, because the certified width decreases geometrically.
330 \end{example}
331 
332 The independent audit retains $j=0,\ldots,1000$. Its exact rational endpoints differ by $2^{-1001}$. The finite differences and endpoint numerators are recorded in the evidence bundle. The preceding theorem proves the tail bound for every $N$. The program computes the finite rational portion at one chosen $N$ and checks its internal arithmetic. These are complementary forms of evidence with different scopes.
333 
334 \section{What a certified interval can decide}
335 
336 Suppose $l\leq G\leq h$ is a proven rational enclosure, and suppose $a,b\in\Q$. Then the interval for $aG-b$ is
337 \begin{equation}
338 \begin{cases}
339 [al-b,ah-b],&a\geq0,\\
340 [ah-b,al-b],&a<0.
341 \end{cases}
342 \label{eq:c01-linear-interval}
343 \end{equation}
344 All endpoints are rational. If the entire interval is positive, the form is positive. If the entire interval is negative, the form is negative. If it contains zero, the enclosure has not decided the sign. That last outcome is a limitation of the chosen certificate, not evidence that the form actually vanishes.
345 
346 This procedure makes the finite sign checks reproducible. It also limits their interpretation. A finite family of certified nonzero forms proves nonvanishing for those forms. It does not prove a universal assertion about all indices. A long decimal expansion cannot replace the rational enclosure when a coefficient is large enough to amplify the final rounding error.
347 
348 The next chapter explains why even an infinite family of nonzero real errors is not enough. Their coefficients must become integers, and the errors after that arithmetic operation must still tend to zero.
349 
350 \chapter{The arithmetic threshold for irrationality}
351 \label{ch:lattice}
352 
353 The real approximation problem asks for rational numbers near $G$. The arithmetic problem asks for nonzero integer linear forms in $G$ whose absolute values tend to zero. These problems are related, but the passage between them has a cost. A coefficient denominator can absorb the analytic gain. A common divisor can recover some of that gain. Nonvanishing has to survive the whole construction.
354 
355 The threshold is elementary enough to prove in a few lines. Its applications are demanding because all its hypotheses must hold together. This chapter states the criterion precisely, gives rational counterexamples to weaker assertions, and sets up the rate accounting used for the later filters.
356 
357 \section{Rational numbers leave a fixed gap}
358 
359 \begin{lemma}[Rational lattice gap]
360 \label{lem:c02-gap}
361 Suppose $\alpha=a/b$ with $a\in\Z$ and $b\in\Z_{>0}$. For every pair $P,Q\in\Z$, either $Q\alpha-P=0$ or
362 \begin{equation}
363 \abs{Q\alpha-P}\geq\frac1b.
364 \label{eq:c02-gap}
365 \end{equation}
366 \end{lemma}
367 
368 \begin{proof}
369 Multiplication by $b$ gives
370 \[
371 b(Q\alpha-P)=Qa-Pb\in\Z.
372 \]
373 A nonzero integer has absolute value at least $1$. Division by the positive integer $b$ gives the result. No coprimality hypothesis is needed.
374 \end{proof}
375 
376 For a fixed rational $\alpha$, every integer form $Q\alpha-P$ lies in the discrete set $b^{-1}\Z$. Its nonzero values cannot enter the open interval $(-1/b,1/b)$. The denominator $b$ need not be known to use this obstruction. If a constructed infinite family has nonzero values tending to zero, it eventually violates the gap for any proposed $b$.
377 
378 \begin{theorem}[Integer form criterion]
379 \label{thm:c02-criterion}
380 Let $\alpha\in\R$. Suppose there are integer pairs $(P_j,Q_j)$ for $j\geq1$ such that
381 \begin{equation}
382 0<\abs{Q_j\alpha-P_j}\longrightarrow0.
383 \label{eq:c02-form-criterion}
384 \end{equation}
385 Then $\alpha$ is irrational. The same conclusion holds if these conditions are verified along any infinite subsequence.
386 \end{theorem}
387 
388 \begin{proof}
389 If $\alpha=a/b$ were rational with $b>0$, Lemma~\ref{lem:c02-gap} would bound every nonzero form below by $1/b$. This contradicts convergence to zero. Restricting to a subsequence does not change the argument.
390 \end{proof}
391 
392 \begin{figure}[tbp]
393 \centering
394 \includegraphics[width=0.94\linewidth]{../visual/figures/06_rationality_lattice.pdf}
395 \caption{Under the assumption $G=a/b$, integer linear forms lie on a lattice with spacing $1/b$. A sequence of nonzero forms approaching zero would contradict that spacing. The spacing is conceptual here because the hypothetical denominator is unknown.}
396 \label{fig:c02-lattice}
397 \end{figure}
398 
399 The theorem separates three obligations. The coefficients must be integers. The forms must be nonzero at the indices being used. Their absolute values must approach zero. It is possible to establish any two while the third remains open.
400 
401 For instance, setting $P_j=Q_j=0$ gives integer coefficients and zero error at every index, but proves nothing. Taking a rational approximation with a tiny error gives a small real form, but its coefficients may have large rational denominators. Clearing those denominators may keep the resulting integer form far from zero. These examples explain why the hypotheses are kept distinct throughout this book.
402 
403 \subsection{Why the criterion is also a characterization}
404 
405 The criterion is sufficient for irrationality. Conversely, every irrational real number admits such a family. This observation clarifies the role of the later constructions. The existence of small integer forms is not itself a novel definition of irrationality. The research problem is to construct and justify them for a specific constant.
406 
407 \begin{proposition}[Elementary approximation by integer forms]
408 \label{prop:c02-pigeonhole}
409 If $\alpha$ is irrational, then for every integer $H\geq1$ there are integers $P,Q$ with $1\leq Q\leq H$ satisfying
410 \begin{equation}
411 0<\abs{Q\alpha-P}<\frac1H.
412 \label{eq:c02-pigeonhole}
413 \end{equation}
414 In particular, an infinite family as in Theorem~\ref{thm:c02-criterion} exists.
415 \end{proposition}
416 
417 \begin{proof}
418 Consider the $H+1$ fractional parts of $0,\alpha,2\alpha,\ldots,H\alpha$. Partition $[0,1)$ into $H$ intervals of length $1/H$, using half open intervals. Two fractional parts lie in one interval. Their difference has absolute value strictly smaller than $1/H$. Subtracting the corresponding multiples of $\alpha$ gives $Q\alpha-P$ with $1\leq Q\leq H$ after choosing the order of the two indices.
419 
420 The form is nonzero because $\alpha$ is irrational and $Q\ne0$. Choosing such a pair for every $H$ gives nonzero forms tending to zero. The denominators $Q$ cannot remain bounded. For any fixed finite set of positive $Q$, the distances from $Q\alpha$ to the nearest integers have a positive minimum. That minimum would contradict the bound as $H\to\infty$.
421 \end{proof}
422 
423 This proof assumes irrationality in order to ensure nonvanishing. It cannot be used to prove irrationality of $G$ by inserting $\alpha=G$ before that fact is known. The lattice criterion runs in the other direction: first establish small nonzero forms by independent mathematics, then deduce irrationality.
424 
425 \section{Approximation error and denominator are coupled}
426 
427 Let $P_j/Q_j$ be rational approximants written in lowest terms, with $Q_j>0$. Their ordinary errors and their integer form errors are related by
428 \begin{equation}
429 \abs{Q_j\alpha-P_j}
430 =Q_j\abs{\alpha-\frac{P_j}{Q_j}}.
431 \label{eq:c02-error-conversion}
432 \end{equation}
433 Thus a plot of the ordinary error omits one factor from the arithmetic test. A rapid decrease in that plot can coexist with a denominator that grows just as rapidly.
434 
435 \begin{corollary}[Reduced rational approximation criterion]
436 \label{cor:c02-reduced}
437 Suppose $P_j/Q_j$ is in lowest terms, $Q_j>0$, and $P_j/Q_j\ne\alpha$ at all indices used. If
438 \begin{equation}
439 Q_j\abs{\alpha-\frac{P_j}{Q_j}}\longrightarrow0,
440 \label{eq:c02-reduced-condition}
441 \end{equation}
442 then $\alpha$ is irrational.
443 \end{corollary}
444 
445 \begin{proof}
446 The left side equals the absolute value of the nonzero integer form $Q_j\alpha-P_j$. Apply Theorem~\ref{thm:c02-criterion}.
447 \end{proof}
448 
449 Reduction to lowest terms is not necessary for the theorem, but it gives the smallest denominator belonging to that rational number. If a common factor is left in $P_j,Q_j$, the form is multiplied by that factor. Keeping a redundant factor can conceal an available decay margin. Removing it cannot create nonvanishing if the original form was zero.
450 
451 \begin{proposition}[Separation from a rational target]
452 \label{prop:c02-rational-distance}
453 If $\alpha=a/b$ with $b>0$ and $P/Q\ne\alpha$ with $P\in\Z$, $Q\in\Z_{>0}$, then
454 \begin{equation}
455 \abs{\alpha-\frac PQ}\geq\frac1{bQ}.
456 \label{eq:c02-rational-distance}
457 \end{equation}
458 \end{proposition}
459 
460 \begin{proof}
461 The numerator of the difference is the nonzero integer $aQ-bP$:
462 \[
463 \abs{\frac ab-\frac PQ}=\frac{\abs{aQ-bP}}{bQ}\geq\frac1{bQ}.
464 \]
465 \end{proof}
466 
467 This lower bound explains a familiar sufficient exponent. If distinct approximants have $Q_j\to\infty$ and
468 \begin{equation}
469 \abs{\alpha-\frac{P_j}{Q_j}}\leq C Q_j^{-\theta}
470 \label{eq:c02-exponent-bound}
471 \end{equation}
472 for a fixed constant $C>0$ and an exponent $\theta>1$, then multiplying by $Q_j$ gives a quantity tending to zero. The criterion applies. An upper bound with $\theta\leq1$ does not supply that conclusion on its own. It may still describe useful approximation, and another argument may still prove irrationality, but this particular inequality does not cross the required threshold.
473 
474 \subsection{Two rational counterexamples}
475 
476 \begin{example}[A convergent rational sequence with a fixed form error]
477 \label{ex:c02-linear-rational}
478 Let $\alpha=1/2$ and take
479 \[
480 \frac{P_n}{Q_n}=\frac{n}{2n+1},\qquad n\geq1.
481 \]
482 The numerator and denominator are coprime because any common divisor of $n$ and $2n+1$ divides $1$. The ordinary error is
483 \[
484 \frac12-\frac{n}{2n+1}=\frac1{2(2n+1)}\longrightarrow0.
485 \]
486 Nevertheless,
487 \[
488 Q_n\alpha-P_n=\frac12
489 \]
490 for every $n$. The approximants are distinct from the target, all coefficients are integers, and the approximations converge. The integer form does not tend to zero.
491 \end{example}
492 
493 \begin{example}[Exponential approximation to a rational number]
494 \label{ex:c02-exponential-rational}
495 Again let $\alpha=1/2$. For $n\geq2$, set
496 \[
497 Q_n=2^n,\qquad P_n=2^{n-1}+1.
498 \]
499 The numerator is odd, so $P_n/Q_n$ is reduced. We have
500 \begin{equation}
501 \frac{P_n}{Q_n}=\frac12+2^{-n},\qquad
502 \abs{\alpha-\frac{P_n}{Q_n}}=2^{-n}.
503 \label{eq:c02-exponential-rational}
504 \end{equation}
505 The real error decreases exponentially. Yet
506 \[
507 Q_n\alpha-P_n=-1
508 \]
509 at every index. Exponential convergence of the ordinary error is therefore insufficient without the denominator comparison.
510 \end{example}
511 
512 The second example can also be written as a failure of denominator clearing. The rational coefficient form
513 \[
514 \ell_n=\alpha-\left(\frac12+2^{-n}\right)=-2^{-n}
515 \]
516 is nonzero and tends to zero. Its second coefficient is not an integer. Multiplication by $2^n$ makes both coefficients integral and gives the constant form $-1$. The analytic gain has been exactly consumed by the arithmetic scale.
517 
518 These counterexamples are deliberately elementary. They prevent a seductive but invalid inference: a known recurrence producing rapidly convergent rational approximants cannot settle irrationality until the actual coefficient denominators have been controlled.
519 
520 \section{Clearing rational coefficients}
521 
522 In the later chapters the starting form is
523 \begin{equation}
524 \ell_j=u_j\alpha-v_j,\qquad u_j,v_j\in\Q.
525 \label{eq:c02-rational-form}
526 \end{equation}
527 Choose a positive integer $D_j$ for which $D_ju_j$ and $D_jv_j$ are integers. Then
528 \begin{equation}
529 L_j=D_j\ell_j=(D_ju_j)\alpha-D_jv_j
530 \label{eq:c02-cleared-form}
531 \end{equation}
532 has integer coefficients. Its size is $D_j\abs{\ell_j}$, so the clearance factor must be included in any decay estimate.
533 
534 Such a $D_j$ always exists for one finite rational pair. Write $u_j=a_j/b_j$ and $v_j=c_j/d_j$ in lowest terms with positive denominators. The smallest positive common clearing factor is
535 \begin{equation}
536 D_j^{\min}=\operatorname{lcm}(b_j,d_j).
537 \label{eq:c02-minimal-clearance}
538 \end{equation}
539 Indeed, integrality of $D_ja_j/b_j$ forces $b_j\mid D_j$ by coprimality, and integrality of $D_jc_j/d_j$ forces $d_j\mid D_j$. Conversely, their least common multiple clears both denominators.
540 
541 A theorem may provide a larger convenient scale than the minimal one. A least common multiple indexed by the construction is often easier to control than the exact denominator of every term. That is a valid way to obtain integer coefficients, but any redundant scale should remain visible. It may account for part of an observed common divisor.
542 
543 \subsection{The primitive integer pair}
544 
545 Suppose a cleared form is $A_j\alpha-B_j$ with $A_j,B_j\in\Z$ and $(A_j,B_j)\ne(0,0)$. Define its common content by
546 \begin{equation}
547 h_j=\gcd(\abs{A_j},\abs{B_j})>0.
548 \label{eq:c02-content}
549 \end{equation}
550 The primitive pair is
551 \[
552 \left(\frac{A_j}{h_j},\frac{B_j}{h_j}\right),
553 \]
554 and the associated form has size
555 \begin{equation}
556 \frac{\abs{A_j\alpha-B_j}}{h_j}.
557 \label{eq:c02-primitive-form}
558 \end{equation}
559 Both coefficients remain integers. Thus a proved lower bound on common content may improve an irrationality argument. The word ``proved'' is essential when the improvement is used at infinitely many indices.
560 
561 If $A_j\ne0$, reducing the rational approximant $B_j/A_j$ gives denominator $\abs{A_j}/h_j$. This denominator controls the ordinary approximation criterion. The common content controls the primitive linear form. These are two descriptions of the same cancellation, not two independent gains that can be counted twice.
562 
563 There is also a useful invariance. Multiplying a cleared pair by a nonzero integer $c$ multiplies its common content by $\abs c$. After primitive reduction, the resulting pair is unchanged up to a common sign. A large gcd produced solely by an unnecessary clearance multiplier cannot improve the primitive form. The later gcd decomposition makes this observation exact for the historical filter data.
564 
565 \section{A strict rate margin is enough}
566 
567 We now state the rate criterion without assuming that the analytic estimate or denominator bound is sharp. A usable upper bound on the error and a usable upper bound on clearance are enough if their product is strictly below the threshold.
568 
569 \begin{theorem}[Exponential clearance criterion]
570 \label{thm:c02-rate}
571 Let $\alpha\in\R$, and let $u_j,v_j\in\Q$ for $j\geq1$. Let $D_j$ be positive integers such that $D_ju_j,D_jv_j\in\Z$. Assume $u_j\alpha-v_j\ne0$ for all sufficiently large $j$. Suppose
572 \begin{equation}
573 \limsup_{j\to\infty}\abs{u_j\alpha-v_j}^{1/j}\leq\lambda,
574 \qquad
575 \limsup_{j\to\infty}D_j^{1/j}\leq\delta,
576 \label{eq:c02-root-bounds}
577 \end{equation}
578 where $\lambda\geq0$, $\delta<\infty$ and $\delta\lambda<1$. Then $\alpha$ is irrational.
579 \end{theorem}
580 
581 \begin{proof}
582 Since $D_j\geq1$, the second hypothesis implies $\delta\geq1$. Choose positive numbers $\lambda'>\lambda$ and $\delta'>\delta$ with $\lambda'\delta'<1$. Such a choice is possible by the strict product inequality, including when $\lambda=0$.
583 
584 By the definitions of the two upper limits, for all sufficiently large $j$,
585 \[
586 \abs{u_j\alpha-v_j}\leq(\lambda')^j,
587 \qquad D_j\leq(\delta')^j.
588 \]
589 Consequently,
590 \[
591 0<\abs{D_j(u_j\alpha-v_j)}
592 \leq(\delta'\lambda')^j\longrightarrow0.
593 \]
594 The form has integer coefficients by hypothesis. Apply Theorem~\ref{thm:c02-criterion}.
595 \end{proof}
596 
597 The same proof applies along a specified infinite set of indices if the integrality, nonvanishing and rate hypotheses all hold on that set. An even filter index will later be useful precisely because it supplies a universal nonvanishing argument. The remaining rate inequalities still have to be proved along the chosen family.
598 
599 \begin{corollary}[Logarithmic rate accounting]
600 \label{cor:c02-log-rate}
601 Under the integrality and eventual nonvanishing hypotheses above, suppose
602 \begin{equation}
603 \limsup_{j\to\infty}\frac{\log D_j}{j}\leq\sigma,
604 \qquad
605 \limsup_{j\to\infty}\frac{\log\abs{u_j\alpha-v_j}}{j}\leq-\tau,
606 \label{eq:c02-log-bounds}
607 \end{equation}
608 with finite $\sigma$ and $\tau>\sigma$. Then $\alpha$ is irrational.
609 \end{corollary}
610 
611 \begin{proof}
612 Choose $\eps>0$ with $\sigma-\tau+2\eps<0$. Eventually $D_j\leq e^{(\sigma+\eps)j}$ and $\abs{u_j\alpha-v_j}\leq e^{(-\tau+\eps)j}$. Their product tends to zero. Apply the integer form criterion.
613 \end{proof}
614 
615 \begin{proposition}[Content can recover a proved rate]
616 \label{prop:c02-content-rate}
617 Assume the setup of Corollary~\ref{cor:c02-log-rate}, except that $\tau>\sigma$ is not required. Discard the finite initial set of indices before nonvanishing holds. For the remaining sufficiently large $j$, define
618 \[
619 h_j=\gcd(\abs{D_ju_j},\abs{D_jv_j}),
620 \]
621 which is positive because the form is nonzero at those indices. If
622 \begin{equation}
623 \liminf_{j\to\infty}\frac{\log h_j}{j}\geq\kappa
624 \quad\hbox{and}\quad \tau+\kappa>\sigma,
625 \label{eq:c02-content-rate}
626 \end{equation}
627 then $\alpha$ is irrational.
628 \end{proposition}
629 
630 \begin{proof}
631 The primitive form $D_j(u_j\alpha-v_j)/h_j$ has integer coefficients and remains nonzero. Choose $\eps>0$ so that $\sigma-\tau-\kappa+3\eps<0$. The upper limits for $D_j$ and the error, and the lower limit for $h_j$, give eventually
632 \[
633 \frac{D_j\abs{u_j\alpha-v_j}}{h_j}
634 \leq\exp\bigl((\sigma-\tau-\kappa+3\eps)j\bigr).
635 \]
636 The right side tends to zero. Apply Theorem~\ref{thm:c02-criterion} to the primitive pair.
637 \end{proof}
638 
639 The signs of the inequalities encode their roles. We need an upper bound on clearance growth, an upper bound on error size, and a lower bound on useful common content. A finite upper bound on a valuation cannot be relabeled as the lower bound on gcd growth that this proposition requires. It may instead help prove that a proposed cancellation mechanism cannot supply the missing gain, but that conclusion also requires a universal argument.
640 
641 \subsection{What happens at equality}
642 
643 The strictness of the rate comparison is not cosmetic. In Example~\ref{ex:c02-exponential-rational}, the rational coefficient form has root error rate $\lambda=1/2$ and clearance rate $\delta=2$. The product is exactly $1$. The cleared form is constantly $-1$, and the target is rational. Thus replacing $\delta\lambda<1$ by $\delta\lambda\leq1$ would make Theorem~\ref{thm:c02-rate} false.
644 
645 Equality in a root or logarithmic rate can still leave room for a more delicate argument. For example, an additional factor tending to zero can matter even when its logarithm divided by $j$ tends to zero. But such a factor needs to be proved. Rate equality alone leaves the behavior undetermined. A positive upper exponent likewise says that the displayed bound does not force decay. It does not prove that the actual forms grow.
646 
647 \section{Using the criterion without assuming the conclusion}
648 
649 For Catalan's constant, the procedure has a definite order. First construct rational coefficient pairs by a recurrence or integral identity. Then justify a universal integer clearance. Then establish nonvanishing of the chosen forms without assuming $G$ irrational. Finally compare the analytic decay with the full arithmetic scale, including any rigorously controlled common content.
650 
651 An exact integral of fixed sign is one possible source of nonvanishing. A parity bracket around a limit is another, once that limit has independently been identified with $G$. An interval computation can certify a particular finite form, but it cannot supply the universal step by simply repeating the test many times.
652 
653 The separation is especially important in a recurrence setting. Distinct adjacent rational approximants prove that the two sequences are independent. They do not by themselves prove that their limit is irrational. A rational number can be the limit of a strictly monotone sequence of distinct rationals, as the examples above show. Identifying the limit and classifying its arithmetic nature are therefore recorded as different claims.
654 
655 The later chapters will produce exact identities, integer coefficients and nonzero filtered forms for a useful even subfamily. They will also display the present rate obstruction. The criterion tells us exactly what a future successful step must change: it must create a strict decay margin for nonzero integer forms along an infinite justified family.

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