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If $q=\sqrt{2}+\sqrt{3}$ were a rational number, then $q^2=5+2\sqrt{6}$ would make $\sqrt{6}$ rational. Write $\sqrt{6}=a/b$ in lowest terms. Then $a^2=6b^2$, which is impossible by unique factorization: the exponent of the prime $2$ is even on the left and odd on the right. Therefore

If ${\log }_{2}3=a/b$ were a rational number, choose positive integers $a,b$ because ${\log }_{2}3>0$. Exponentiation gives $2^a=3^b$, but the left side is even and the right side is odd. ThusThis is the basic irrational logarithm criterion: a rational logarithm would force positive integer powers of its base and argument to coincide.

The integers are countable: $j(z)=2z+1$ for $z\ge 0$ and $j(z)=-2z$ for $z<0$ is an injection $\mathbb{Z}\to \mathbb{N}$. Hence $\mathbb{Z}\times \mathbb{N}$ is a countable set. The map $(a,b)\mapsto a/b$ is a surjection onto the rational numbers, so the image result already proved gives $\mathbb{Q}$ is countable. Unique reduced fractions are not required for this surjection argument.

For the final unheaded geometry request, suppose for contradiction that no two distinct circles in the collection intersect. Restrict to the subcollection ${\mathcal{C}}_0$ of circles tangent to the $x$-axis. Each such nondegenerate circle has a unique tangency point, and the tangency-point map ${\mathcal{C}}_0\to \mathbb{R}$ is a surjection by the hypothesis.

The closed disk of a circle tangent at $(a,0)$ meets the $x$-axis only at $(a,0)$. Two distinct members of ${\mathcal{C}}_0$ cannot have nested disks: if one disk were inside the other, its tangency point would belong to the larger disk, so both tangency points would be equal and both circle boundaries would pass through it. That already contradicts nonintersection. Circles with disjoint boundaries and no nesting have disjoint disks. Thus their open disks are pairwise disjoint.

By the density of the rational numbers, every nonempty open disk contains a point of ${\mathbb{Q}}^2$, which is a countable set. Fix an enumeration of ${\mathbb{Q}}^2$ and assign to each disk the first enumerated point inside it. Disjointness makes this assignment an injection. This is the countability of pairwise disjoint open disks, and proves that ${\mathcal{C}}_0$ is a countable set. Its surjection onto $\mathbb{R}$ would make the real numbers countable, contrary to the Cantor diagonal argument. ThereforeThe proof allows circles on either side of the axis and counts tangential contact as intersection; it does not assume that all circles in the original collection are tangent to the axis.

For each positive integer $n$, the positive elements of $A_n$ form an infinite set, and the preceding increasing enumeration gives a bijection $f_n:\mathbb{N}\to A_n$ on these positive elements. The positive integers have a unique decomposition $10^{n-1}q$ with $10\nmid q$, so these sets are disjoint and exhaust the positive integers.

The printed natural numbers include $0$. The trailing-zero count of the representation of $0$ is not needed: shift the positive partition down by one and putUniqueness of the trailing-zero count proves the injective function property, and the decomposition of $N+1$ proves the surjective function property. An explicit decimal trailing-zero pairing function, using the increasing enumeration from the previous part, isIndeed, the bracket enumerates $1,2,\dots ,9,11,12,\dots$, exactly the positive integers not divisible by $10$. This handles both the positive index $n$ and the element $0$ without silently changing the PDF's convention.

To enumerate the rational numbers, first enumerate the integers by $e(0)=0$, $e(2r-1)=r$, $e(2r)=-r$ for $r\ge 1$. The function $(i,j)\mapsto e(i)/(j+1)$ maps ${\mathbb{N}}^2$ onto $\mathbb{Q}$, and composing with $g^{-1}$ gives a surjective function from $\mathbb{N}$. Choosing the first index representing each rational number gives an injective function $\mathbb{Q}\to \mathbb{N}$. Thus $\mathbb{Q}$ is countably infinite; its infinitude follows because it contains all integers.

Use the standard definition of a partition regular matrix: for every finite coloring of the positive integers, there is a positive solution whose coordinates all have one color. Coordinates may repeat. Clearing denominators lets us assume $a_i\in \mathbb{Z}\setminus \{0\}$; multiplication of the row by a nonzero rational number changes neither its zero-sum subsets nor its solutions. We prove both directions of the Rado theorem for one equation directly, without invoking any form of Rado's theorem.

For necessity, choose a prime number $p>{\sum }_i|a_i|$ and use the last nonzero digit coloring: write $x=p^{v_p(x)}u$, with $p\nmid u$, and assign color $u\mathrm{mod}p$. Given a monochromatic solution, let $v={\min }_i{v}_p(x_i)$, let $I=\{i:v_p(x_i)=v\}$, and let $r\ne 0$ be the common color. Divide ${\sum }_i{a}_i{x}_i=0$ by $p^v$ and reduce modulo $p$. Terms outside $I$ vanish; the others giveThe nonzero residue $r$ is invertible modulo the prime number $p$, so $p$ divides ${\sum }_{i\in I}{a}_i$. This sum has absolute value less than $p$, forcing it to be zero. The set $I$ is nonempty by its definition.

For sufficiency we first derive the needed Brauer progression theorem from the permitted Van der Waerden theorem. Fix positive integers $s,\ell$. We claim that for every number of colors $r$, a finite integer interval $[B(r,s,\ell )]$ forces a monochromatic setThe case $\ell =1$ is immediate by taking $a=s$ and $d=1$. For $r=1$ and $\ell \ge 2$, take $a=d=1$ in an interval of length at least $\max (s,\ell )$. Suppose the claim holds for $r-1$ and write $M=B(r-1,s,\ell )$. By the Van der Waerden theorem, some $R$ forces an arithmetic progression of length $(\ell -1)M+1$ in $[R]$ with one color $q$. Work in $[\max (R,sMR)]$ so that all needed multiples also lie in the coloring's domain. If a subprogression of length $\ell$ and step $td$ has $std$ of color $q$, we are done. Otherwise, for every $1\le t\le M$, the initial long arithmetic progression contains such a subprogression, and $std$ avoids color $q$. The induced finite coloring $t\mapsto \chi (sdt)$ of $[M]$ therefore uses at most $r-1$ colors. The induction hypothesis gives a monochromatic set $\{se,b,b+e,\dots ,b+(\ell -1)e\}$ for this induced finite coloring. Multiplying by $sd$ gives the desired configuration in the original coloring, with initial term $sdb$ and step $sde$. This proves the claim for $\ell \ge 2$; the only use below has $\ell \ge 3$.

Now suppose ${\sum }_{i\in I}{a}_i=0$ for a nonempty $I$. Choose $i_0\in I$, put $B={\sum }_{i\notin I}{a}_i$, and put $s=|a_{i_0}|$. If $B=0$, then ${\sum }_i{a}_i=0$ and any constant positive vector is already a monochromatic solution. Otherwise take $L=|B|$ and apply the proved Brauer progression theorem with $\ell =2L+1$. All of $sd$ and $a+jd$ for $0\le j\le 2L$ have one color. DefineEvery coordinate is a positive integer in that monochromatic set, since $L-\mathrm{sgn}(a_{i_0})B$ is either $0$ or $2L$. Using the zero sum on $I$ givesThus both directions are proved:For $n=1$ the right side is impossible, agreeing with the absence of positive solutions for a single nonzero coefficient.

The Liouville approximation theorem states that an irrational algebraic number $\alpha$ of degree $d\ge 2$ satisfies $|\alpha -a/b|\ge C_{\alpha }{b}^{-d}$ for all integers $a$ and $b\ge 1$, for some $C_{\alpha }>0$. The constant is independent of the rational approximation.

The required condition is infinitely many of the coefficients equal one. If only finitely many do, the series is a finite sum of rational numbers. Conversely let $\alpha$ be the sum with infinitely many nonzero terms. Its truncation $s_N$ has denominator dividing $q_N=10^{N!}$. For $N\ge 2$, its strictly positive tail satisfiesIf $\alpha =a/b$ were rational, the distinct fractions $\alpha ,s_N$ would have distance at least $1/(b{q}_N)$, contradicting the tail bound for large $N$. Thus $\alpha$ is irrational. If it were algebraic of fixed degree $d$, apply the Liouville approximation theorem to $s_N$. Even if its reduced denominator is smaller than $q_N$, the lower bound is at least $C_{\alpha }{q}_N^{-d}$. The displayed upper bound contradicts this as $N\to \infty$. HenceThe coincidence $0!=1!$ changes only the first rational contribution and has no effect on this criterion.

Rado's criterion. Let $A$ be a matrix over the rational numbers, with columns $a_1,\dots ,a_q$. A partition regular matrix is one for which every finite coloring of the positive integers gives a monochromatic positive vector $x\in {\mathbb{N}}^q$ with $Ax=0$; repeated coordinates are allowed. The columns condition is the existence of an ordered set partition $B_1,\dots ,B_s$ of $[q]$ into nonempty blocks such thatRado's theorem states

Necessity. Clear denominators so the columns are integer vectors. For each pair of disjoint index sets $I,B$, with $B\ne \varnothing$, for which $v_B={\sum }_{i\in B}{a}_i$ is outside the linear span of the $I$-columns, choose an integer linear functional ${\varphi }_{I,B}$ that vanishes on those columns and satisfies ${\varphi }_{I,B}(v_B)\ne 0$. Such a functional exists by extending a basis of the span to one also containing $v_B$, defining a functional on that basis, and clearing its rational denominators.

There are only finitely many such pairs. Choose a prime number $p$ dividing none of these nonzero integer evaluations. If there are no such pairs, any prime will do. Use the last nonzero digit coloring: write $x=p^{v_p(x)}u$, with $p\nmid u$, and color $x$ by $u\mathrm{mod}p\in \{1,\dots ,p-1\}$. Since $A$ is a partition regular matrix, this coloring supplies a monochromatic positive solution, so all its unit parts have a common nonzero residue $d$.

Group the coordinates by increasing P-adic valuation, obtaining blocks $B_1,\dots ,B_s$ with valuations $e_1<\cdots <e_s$. Suppose for some $j$ the sum $v_{B_j}$ lay outside the linear span of the earlier columns, with earlier index set $I$. Apply ${\varphi }_{I,B_j}$ to $Ax=0$. Earlier terms vanish exactly. Divide the remaining integer equation by $p^{e_j}$ and reduce modulo $p$. Later terms vanish modulo $p$, while the current block givescontrary to the choice of $p$. Thus every block sum lies in the earlier linear span, and for the first block this means it is zero. This proves the columns condition by the finite separating-functional proof of the columns condition.

Sufficiency. Suppose the columns condition holds. For $j>1$, choose rational coefficients ${\lambda }_{ij}$, indexed by earlier columns, such thatTake a positive integer $c$ clearing their denominators and $p\ge 1$ bounding all $|c{\lambda }_{ij}|$. We use the allowed monochromatic m-p-c set theorem with $s$ levels. Its M-p-c set is the full setwhere the generators make every displayed expression positive. Positivity is part of this convention, rather than a rule that discards nonpositive expressions. For $i\in B_h$, setAll these numbers lie in the monochromatic set $D$. The coefficient of $z_j$ in $Ax$ iswith the first coefficient zero by ${\sum }_{i\in B_1}{a}_i=0$. Hence $Ax=0$. This Rado solution inside an m-p-c set proves sufficiency.

The equality-or-inequality dichotomy. If every finite coloring gives a monochromatic solution with $x_1=x_2$, the first alternative holds. Otherwise fix a finite coloring $\chi$ admitting no such solution. Given any other finite coloring $\psi$, use the refinement of a finite coloring $n\mapsto (\chi (n),\psi (n))$. Since $A$ is a partition regular matrix, it supplies a solution monochromatic for both colorings. Its first two coordinates cannot be equal, by the defining property of $\chi$, so it is the required unequal solution for $\psi$. Thus one of the two universal alternatives always holds. This is the partition regularity dichotomy under an extra constraint; the alternatives are not asserted to be mutually exclusive.

Both restrictions can fail. The Schur matrixis a partition regular matrix: blocks $\{1,3\}$ and $\{2\}$ satisfy the columns condition. Color $n$ by $v_2(n)\mathrm{mod}2$, using the 2-adic valuation. An equal-coordinate solution would have $x_3=2x_1$, and $v_2(2x_1)=v_2(x_1)+1$, so its coordinates cannot be monochromatic. This exhibits failure of the equality restriction.

Conversely, $A=(1-1)$ satisfies the columns condition in one block. Every solution has $x_1=x_2$, so even the constant one-color finite coloring admits no unequal-coordinate solution. Consequently

The Martingale convergence theorem says that if a discrete-time martingale $(M_n)$ satisfies $C={\sup }_n\mathbb{E}|M_n|<\infty$, then there is an integrable random variable $M_{\infty }$ such thatThe hypothesis does not by itself ensure convergence in L1.

For each pair of rational numbers $a<b$, the Doob upcrossing inequality gives $\mathbb{E}{U}_N[a,b]\le (C+|a|)/(b-a)$. The monotone convergence theorem therefore gives a finite expected value for $U_{\infty }[a,b]={\lim }_N{U}_N[a,b]$, so this upcrossing count is finite almost surely. There are only countably many such pairs, hence all their upcrossing counts are simultaneously finite outside one event of probability zero.

If ${\mathrm{liminf}}_n{M}_n<{\mathrm{limsup}}_n{M}_n$, the density of the rational numbers supplies $a<b$ strictly between them, forcing infinitely many upcrossings. Consequently $M_n$ has a limit in the extended real numbers almost surely. By the Fatou lemma,A limit of either $+\infty$ or $-\infty$ would make this lower limit infinite, so the limit is finite almost surely, and the same inequality proves its integrability.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers:

Fix an integer $k\ge 1$ and consider the independent incrementsFor fixed $k$ these are independent and identically distributed random variables with the normal distribution $N(0,1/k)$, so $p_k=\mathbb{P}(|{\Delta }_{j,k}|>1)>0$. The second Borel-Cantelli lemma gives $|{\Delta }_{j,k}|>1$ for infinitely many $j$, almost surely. Intersecting these probability-one events over the countably many $k$ preserves probability one.

On any path in this intersection, uniform continuity on $[0,\infty )$ would give some $\delta >0$ such that $|B_t-B_s|<1$ whenever $|t-s|<\delta$. Choosing $k$ with $1/k<\delta$ contradicts the existence of an increment $|{\Delta }_{j,k}|>1$. Therefore Brownian paths are not uniformly continuous on the half-line:The event is measurable: for continuous paths, uniform continuity can be tested by countably many pairs of nonnegative rational numbers and tolerances $1/m$. Brownian motion has uniform continuity on every compact time interval, but almost surely fails it on the whole half-line.

The rational numbers are countable, sois a countable union of closed sets and hence a Borel set; its Lebesgue measure is zero. Its complement is also Borel and has infinite measure:

The intervals are nested in the relevant directions, and the strongest lower and upper restrictions both occur at $n=2$:Thus this set is closed, hence Borel, and

Suppose first that a simple continued fraction is eventually periodic. Its periodic tail $\xi$ is fixed by the Möbius transformation represented by the continued-fraction matrix of one full period:ThusThe infinite tail is not rational, since a rational simple continued fraction terminates. Hence $\xi$ is a quadratic irrational, and the finite initial block expresses the original number as another rational Möbius transform of $\xi$. Therefore every eventually periodic simple continued fraction represents a quadratic irrational.

For the purely periodic fraction $x=[7,7,7,\dots ]$,so $x^2-7x-1=0$. Positivity selects

Apply the continued-fraction algorithm to $\sqrt{23}$:The complete quotient then repeats, givingThe convergents through one term before the final $8$ areThe last of these satisfies the Pell equationso a positive solution is