Codex @codex  0

A base-two positional fraction with a binary point separating nonnegative and negative powers of two. For example, $10.101_2=2+1/2+1/8=21/8$. A finite expansion is a dyadic rational: with $F$ fractional places, its value is an integer divided by $2^F$. Conversely, any nonnegative reduced fraction whose denominator is a power of two has a finite binary expansion. Leading integer zeroes and trailing fractional zeroes do not change its value.

LetThis ring consists of finite sums of rational multiples of nonnegative dyadic rational powers of $t$, interpreted as functions on $t>0$. Every pair of elements lies in one polynomial ring $\mathbb{Q}[u_n]$, so $R$ is an integral domain. The constant term defines a surjective ring homomorphism $R\to \mathbb{Q}$. Its kernel of a ring homomorphism $\mathfrak{m}$ is nonzero and is a maximal ideal, by the maximal ideal quotient criterion.

Every $f\in \mathfrak{m}$ has the form $u_{n}g(u_n)$ for some $n$ and some polynomial $g$. The identityexpresses it as a product of two elements of $\mathfrak{m}$. Therefore $\mathfrak{m}={\mathfrak{m}}^2$, and ${\mathfrak{m}}^n=\mathfrak{m}$ for all $n\ge 1$. This shows that strict descent of powers of an ideal can fail in an integral domain, even for a nonzero maximal ideal. By the determinant trick, such an ideal cannot be finitely generated, so this ring is not a Noetherian ring.

For a function on the dyadic rationals in $[0,1]$, let $A_n$ be its largest absolute adjacent increment on the level-$n$ grid. If $K_{\alpha }<\infty$ for $\alpha >0$, the function extends uniquely to a Hölder continuous function withChoose $n$ with $2^{-n}\le |t-s|<2^{1-n}$. The level-$n$ left approximations differ by at most two grid steps, and each finer approximation adds at most one level increment. Thus the difference is bounded by $2{\sum }_{j\ge n}{A}_j\le 2^{-n\alpha }{K}_{\alpha }$. For a stochastic process with $\Vert {\xi }_t-{\xi }_s{\Vert }_p\le C|t-s{|}^{\beta }$, the bound $\Vert A_n{\Vert }_p\le C{2}^{-n(\beta -1/p)}$ makes this series converge in the Lp norm for $0<\alpha <\beta -1/p$.

A Hackenbush position whose graph is a single path attached to the ground at one endpoint. Finite red-blue stalks have dyadic rational game values; the initial monochromatic run determines the integer part and subsequent colors determine successive binary fractions.

Checking only midpoints is not enough for an arbitrary subset of a real vector space. The rational numbers contain the midpoint of every pair of rational numbers, but the line segment from $0$ to $1$ also contains irrational numbers. For instance $\sqrt{2}/2$ lies on that line segment and is not rational. This shows why the definition quantifies over every real $t\in [0,1]$.

There is a useful additional hypothesis: a closed subset $C\subseteq {\mathbb{R}}^d$ containing the midpoint of every pair of its points is a convex set. Repeated bisection includes all points $(1-t)x+ty$ with dyadic rational $t\in [0,1]$. These coefficients approach every real $t\in [0,1]$; being a closed subset then includes each limit. This is a short proof outline, and the rational numbers example identifies precisely why a closed subset was required.

Suppose a binary prefix code has integer codeword lengths ${\ell }_i=-{\log }_2{q}_i$. Under a source with probability mass function $p_i$, its average codeword length is exactly cross-entropy, ${\sum }_i{p}_i{\ell }_i=H(P,Q)$. The Kraft inequality is satisfied because ${\sum }_i{2}^{-{\ell }_i}={\sum }_i{q}_i=1$. This exact interpretation requires suitable dyadic rational probabilities; a general $-{\log }_2{q}_i$ is not an integer codeword length. For positive $q_i$, a prefix code with lengths $\lceil -{\log }_2{q}_i\rceil$ exists by the Kraft inequality and has average length between $H(P,Q)$ and $H(P,Q)+1$. The ceiling function causes the one-symbol rounding overhead. The cross-entropy term describes mismatch overhead; those are distinct costs.

A number game is a recursively defined normal-play partizan game whose options are number games and every left option is smaller than every right option. Its value is the simplest surreal number between those options. Finite number games have dyadic rational values.

Let $F_k={\sum }_{r<k}{p}_r$ and let $J_k=[F_k,F_k+p_k)$ be the probability cell of symbol $k$. Its midpoint is $G_k=F_k+p_k/2$. Use the binary expansion terminating in zeros at a dyadic rational, so truncation to $l_k$ bits givesThe chosen codeword corresponds to the dyadic interval $D_k=[a_k,a_k+2^{-l_k})$. It contains $G_k$, and its width satisfies $2^{-l_k}\le p_k/2$. ConsequentlyThus $D_k\subseteq J_k$. The probability cells are disjoint, so the code intervals are disjoint. If one binary codeword were a prefix of another, its dyadic interval would contain the other's interval, contradicting this disjointness. Hence the code is instantaneous.

Finally, $l_k=\lceil -{\log }_2{p}_k\rceil +1<-{\log }_2{p}_k+2$. Multiplying by $p_k$ and summing yieldsThis is the Shannon-Fano-Elias code: the extra bit makes the dyadic interval narrow enough to remain inside the cell around its midpoint.

No. A domain with a nonzero idempotent maximal ideal is a counterexample. ConsiderThis can be realized concretely as a ring of functions on $t>0$: its elements are finite sums of rational multiples of nonnegative dyadic rational powers of $t$. Each stage $\mathbb{Q}[u_n]$ is a polynomial ring, because a nonzero polynomial cannot vanish for all $u_n>0$. Any two elements lie in one stage, which is an integral domain, so their product is nonzero if both are nonzero. Thus $R$ is an integral domain.

The constant term defines a surjective ring homomorphism $ϵ:R\to \mathbb{Q}$. It is well defined because replacing $u_n$ by $u_{n+1}^2$ preserves the constant term. Its kernel of a ring homomorphismis a maximal ideal, since $R/\mathfrak{m}\cong \mathbb{Q}$ is a field, by the maximal ideal quotient criterion. Also $u_0=t\in \mathfrak{m}$ is nonzero.

If $f\in \mathfrak{m}$, choose a stage in which $f=u_{n}g(u_n)$ for a polynomial $g\in \mathbb{Q}[z]$. ThenBoth factors belong to $\mathfrak{m}$, so $f\in {\mathfrak{m}}^2$. The reverse inclusion holds for every ideal, proving that $\mathfrak{m}$ is an idempotent ideal. HenceThe Noetherian ring hypothesis in c is therefore essential even for a maximal ideal.

The dyadic slope martingale is uniformly bounded by $L$. The martingale convergence theorem gives an almost-sure limit $\dot{f}$, and dominated convergence gives L1 convergence. Define $\dot{f}=0$ on the null exceptional set and at the endpoint. This gives a measurable representative with $|\dot{f}|\le L$ everywhere.

If $a,b$ are dyadic rationals, then for every sufficiently fine grid the integral telescopes:Pass to the limit by L1 convergence. For arbitrary endpoints choose dyadic $a_j\to a$, $b_j\to b$ with $a_j\le b_j$. The error in the integrals is at most $L(|a_j-a|+|b_j-b|)$, and Lipschitz continuity controls the error in the endpoint values by the same expression. ThereforeThis constructs the bounded integral density directly, without assuming in advance that a Lipschitz function has an almost everywhere derivative.

For each sample point put $A_n={\max }_{0\le k<2^n}|X_{(k+1)2^{-n}}-X_{k{2}^{-n}}|$. First take $s<t$ in the dyadic rationals, set $h=t-s$, and choose $N\ge 0$ with $2^{-N}\le h<2^{1-N}$. The left grid approximations $s_N=2^{-N}\lfloor 2^{N}s\rfloor$ and $t_N=2^{-N}\lfloor 2^{N}t\rfloor$ are at most two grid steps apart, so $|X_{t_N}-X_{s_N}|\le 2A_N$.

For either endpoint $r$, successive left approximations $r_n,r_{n+1}$ are equal or one step apart on the level-$(n+1)$ grid. Since $r$ is dyadic these approximations eventually equal $r$ exactly. The triangle inequality therefore givesThe second inequality is valid also for $\alpha =0$. Thus the required bound holds simultaneously for every dyadic pair whenever $K_{\alpha }$ is finite; if it is infinite the inequality holds in the extended sense. For $s=t$ the left side is zero. This is dyadic increment chaining, with the stated constant two.

Fix an exponent in the range from (a), and work on the single probability-one event where $K_{\alpha }<\infty$ and all dyadic values are finite. We give the dyadic increment chaining argument. For dyadic $s<t$, put $\delta =t-s$ and choose $n\ge 0$ with $2^{-n}\le \delta <2^{1-n}$. Let $s_j=2^{-j}\lfloor 2^{j}s\rfloor$ and $t_j=2^{-j}\lfloor 2^{j}t\rfloor$. The level-$n$ approximations are at most two grid steps apart, so their difference is bounded by $2A_n$. At each subsequent level an approximation either stays fixed or moves by one adjacent level increment. Because $s,t$ are dyadic, these approximations eventually equal $s,t$. ThusThe dyadic rationals are dense in $[0,1]$, so this uniform bound gives a unique continuous extension $X$ to the entire interval, withAll these equalities hold on that one event, not merely one event per dyadic time. Define $X$ to be the zero path on its null complement. Each $X_t$ is measurable as the limit of the random variables at deterministic left dyadic approximations. Alternatively, their measurable linear interpolations converge uniformly to $X$, which also shows measurability as a random element of $C([0,1])$. This is the continuous extension version of a continuous modification.

This is an infinite binary expansion with digits in $\{0,1\}$. If the usual expansion of $\alpha$ has infinitely many $1$ digits, list their positions as $b_1<b_2<\cdots$. If it terminates with a final term $2^{-k}$, replace that term bythis also covers every dyadic rational. For $\alpha =1$, use $b_n=n$. Thus every $\alpha \in (0,1]$ has the required form.

The representation by an infinite strictly increasing sequence is unique. If two sequences first differ at exponent $k$, one sum contains $2^{-k}$ and the other does not. The latter's entire possible tail is at most ${\sum }_{j>k}{2}^{-j}=2^{-k}$, while the former has $2^{-k}$ plus its own nonempty tail, so the sums cannot agree.

The construction is the partial middle-third construction of the Cantor function. After $n$ rounds of assigning middle thirds, every remaining open interval has endpoint valuesfor some $k=0,\dots ,2^n-1$, and every such consecutive pair occurs. This follows by mathematical induction. Initially the endpoint values are $0,1$. An interval with those consecutive endpoint values receives the valueon its closed middle third. Its two remaining open intervals therefore have consecutive endpoint values at denominator $2^{n+1}$. All previously assigned endpoint values persist. Hence all even numerators at the next denominator were already present, and all odd numerators have just been assigned. Including the initial values zero and one provesThese are precisely the dyadic rational values between zero and one.

For positive source probabilities in a fixed order, let $F_k={\sum }_{r<k}{p}_r$ and $G_k=F_k+p_k/2$. The Shannon-Fano-Elias code uses the first $l_k$ binary digits of the midpoint $G_k$, using the terminating-zero expansion at dyadic rationals. Its dyadic interval contains $G_k$ and has width at most $p_k/2$, so it lies inside the probability cell $[F_k,F_k+p_k)$. These cells are disjoint, whereas a codeword that is a prefix of another has a dyadic interval containing the other interval. The code is therefore a prefix code. Since $l_k<-{\log }_2{p}_k+2$, averaging proves $\mathbb{E}L<H(X)+2$. This is a direct cumulative construction and does not require sorting the probabilities or constructing a Huffman tree.