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LetAssume for a contradiction that $S=a/b$ is rational. For every $N\ge b$, both $N!S$ andare integers, because the $m_n$ are integers. Their positive differencemust therefore be a positive integer.

On the other hand, concavity of $x^{\alpha }$ for $0\le \alpha <1$ gives $(N+k{)}^{\alpha }\le N^{\alpha }+k^{\alpha }\le N^{\alpha }+k$. Since $(N+1)\cdots (N+k)\ge (N+1{)}^k$,Eventually $0<R_N<1$, contradicting its integrality. Hence

The integrality assumption is essential. For example, take $\alpha =0$, $A=2$, andThen $1<m_n<2$, but the exponential series givesThus the result does not remain true for arbitrary real $m_n$.

For a partition of an interval $P:0=x_0<\cdots <x_m=1$, putThe upper Darboux sum and lower Darboux sum areThe upper and lower integrals areThe bounded function is Riemann integrable when these values agree, and their common value is its Riemann integral.

For the indicator of the rational numbers, every nondegenerate interval contains both a rational number and an irrational number. Hence every upper sum is one and every lower sum is zero. The function is not Riemann integrable.

The Cauchy functional equation first gives $g(0)=0$ and $g(-x)=-g(x)$. If $h\to 0$, then continuity at the given point $z$ givesso $g$ is continuous at zero. For every $x$,so $g$ is continuous everywhere.

Put $c=g(1)$. Additivity gives $g(n)=nc$ for integers $n$ and then $g(m/n)=(m/n)c$ for rational numbers. For any real $x$, choose rationals $q_j\to x$; continuity gives

For the multiplicative equation, $h(0)=h(0{)}^2$. If $h(0)=0$, then $h(x)=h(x)h(0)=0$ for every $x$. Otherwise $h(0)=1$. In that caseand $h(x)$ cannot vanish because $1=h(0)=h(x)h(-x)$. Hence $h$ is everywhere positive. The continuous function $g=\log h$ satisfies the Cauchy functional equation, so $g(x)=cx$. Therefore the complete family is

The first series is telescoping, sinceIts $N$th partial sum is $1-1/(N+1)$, soa rational number.

The second series converges absolutely, for example by the ratio test, and is the odd part of the exponential series:This value is irrational. Indeed, if $a=\sinh 1$ were algebraic, then $e$ would satisfymaking $e$ algebraic over the algebraic numbers and hence algebraic, contrary to the Hermite theorem on the transcendence of e. Thus the second sum is in fact transcendental.

A set is countable if it is finite or admits an injection into the natural numbers; equivalently, its elements can be listed by a finite list or a sequence.

The product of two countable sets is countable: after choosing enumerations, its elements are indexed by pairs $(m,n)\in {\mathbb{N}}^2$, and these pairs can be listed diagonally. Thus an injection $X\to A\times B$ followed by an injection $A\times B\to \mathbb{N}$ proves that $X$ is countable.

The integers are listed as $0,1,-1,2,-2,\dots$, so $\mathbb{Z}$ is countable. Every rational number is represented by a pair $(p,q)\in \mathbb{Z}\times {\mathbb{N}}_{>0}$, so $\mathbb{Q}$ is countable as the image of a countable set.

Finally, if $A_0,A_1,\dots$ are countable, choose listings $A_n=\{a_{n,0},a_{n,1},\dots \}$, allowing repetitions for finite sets. The diagonal listing of the pairs $(n,m)$ lists every $a_{n,m}$, proving that a countable union of countable sets is countable.

LetThe tail satisfiesFor every fixed positive integer $d$, this is smaller than $q_N^{-d}$ for all sufficiently large $N$.

The binary expansion has $1$ digits at factorial positions and arbitrarily long blocks of zeros, but it is neither eventually zero nor eventually periodic. Since a rational number has an eventually periodic binary expansion, $\alpha$ is irrational. If it were an algebraic number of degree $d$, part (b), or equivalently the Liouville approximation theorem, would give a fixed positive multiple of $q^{-d}$ as a lower bound for all rational approximations. The fractions $p_N/q_N$ violate that bound. Thus $\alpha$ is a transcendental number.

Let $\varphi \in Z^2(G,\mathbb{Q})$. Part iii gives $d\psi =|G|\varphi$. Since division by $|G|$ is possible in the rational numbers,is a two-coboundary. HenceThis is a degree-two instance of the vanishing of finite-group cohomology when the group order is invertible on the coefficient module.

Let $f:[a,b]\to \mathbb{R}$ be continuous. If it were unbounded, one could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x\in [a,b]$, but continuity would then give $f(x_{n_j})\to f(x)$, contradicting unboundedness.

Now let $M=\sup f([a,b])$. Choose $x_n$ with $f(x_n)>M-1/n$. A convergent subsequence and continuity give $f(x)=M$ at its limit. Applying the same argument to $-f$ shows that the infimum is attained. This proves the extreme value theorem on a closed bounded interval.

The functionis continuous and bounded, but attains neither its infimum $0$ nor its supremum $1$. For the second example, enumerate the rationals in $[0,1]$ as $(q_n)$ and setEvery nondegenerate interval contains infinitely many rational numbers, hence some $q_n$ with arbitrarily large $n$; therefore $\psi$ is unbounded on every such interval.

For the running extrema, compactness lets us writeThe function $f$ is uniformly continuous on $[a,b]$. Given $\epsilon >0$, choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon$ whenever $|u-v|<\delta$. If $x<y$ and $y-x<\delta$, every new value $f(\xi )$ with $x\le \xi \le y$ is at most $f(x)+\epsilon \le M(x)+\epsilon$. Since $M(y)\ge M(x)$,Interchanging $x,y$ handles the other order. Applying this argument to $-f$ proves continuity of $m$ as well.

Finally fix $T>0$ and put $h(x)=g(x+T)-g(x)$. For each positive integer $n$, there must be some $x_n>n$ with $|h(x_n)|<1/n$. Otherwise, for some $n$ the continuous function $h$ would satisfy $|h(x)|\ge 1/n$ on the interval $(n,\infty )$. By the intermediate value theorem, $h$ would have a constant sign there. The valueswould then increase or decrease by at least $1/n$ at every step, contradicting boundedness of $g$. Thus $x_n\to \infty$ and

LetIf $x$ were rational, thenwould make $\sqrt[3]{6}=(x^3-5)/(3x)$ rational. But if $\sqrt[3]{6}=a/b$ in lowest terms, then $a^3=6b^3$, which is impossible by unique prime factorization: the exponent of $2$ on the two sides is respectively a multiple of three and one more than a multiple of three. Therefore

Use the given convergent seriesSuppose this number were $a/b\in \mathbb{Q}$. Choose $N$ large enough that $b\mid (2N+1)!$. Multiplication by $(2N+1)!$ makes both the rational number and the partial sum through $n=N$ integers. Their differencewould therefore be an integer. It is positive, whilefor sufficiently large $N$, a contradiction. Hence

A transcendental number is a complex number that is not a root of any nonzero polynomial with rational, equivalently integer, coefficients. LetIf $a\ne 0$ and $y$ were an algebraic number, then $e$ would satisfyover the algebraic extension $\mathbb{Q}(y)$. By transitivity of algebraic extensions, $e$ would be algebraic over $\mathbb{Q}$, contradicting its transcendence. If $a=0$, then $b\ne 0$ and $e=b/y$ would again be algebraic. Thus

The integers are countable under the enumerationPart (i) then shows that $\mathbb{Z}\times \mathbb{N}$ is countable. The mapis a surjection from $\mathbb{Z}\times \mathbb{N}$ onto the rational numbers. Selecting, for example, the representation in lowest terms with positive denominator gives an injection in the other direction. Thus $\mathbb{Q}$ is countable.

Start two registers in $|0⟩|0⟩$. Apply ${\mathrm{QFT}}_N$ to the first and then reversibly evaluate $f$ into the second, obtainingThe evaluation can be implemented by repeated squaring and reversible modular multiplication using a number of elementary arithmetic operations polynomial in $\log N$.

Measure the second register. Because $f$ is injective within one period and $r\mid N$, the first register becomes a uniformly weighted periodic cosetThe quantum Fourier transform of a periodic coset state followed by measurement returnsReduce the rational number $c/N$ to lowest terms. Sinceits denominator is $q=r/\gcd (s,r)$. Test whether $a^q\equiv 1(\mathrm{mod}N)$ by repeated squaring. The promised injectivity means that $r$ is the least positive period, and $q\mid r$, so this congruence holds exactly when $q=r$. The test therefore certifies whether the run succeeded; this is heralded exact quantum period finding when the period divides the register size.

One run succeeds with probabilitybecause success is equivalent to $s$ being coprime to $r$. A standard estimate for the Euler totient function gives $\phi (r)/r\ge C/\log \log r$ for all sufficiently large $r$, with the finitely many smaller cases absorbed by changing the positive constant $C$. Repeating independently $O(\log \log N)$ times therefore makes the probability that every run fails at most $1/2$. Every step, including the repetitions and the classical verification, takes time polynomial in $\log N$.

For a real boundary point $x\ne 0$, the Boundary-point Bessel flow for SLE says that $V_t^x/\sqrt{\kappa }$ is a Bessel process of dimensionWhen $\kappa \le 4$, one has $d\ge 2$, so part 2(b), including its logarithmic $d=2$ case, shows that no fixed boundary point is swallowed.

If the trace touched the real line away from its starting point, or if a later segment crossed an earlier segment, the resulting hull would disconnect from infinity a nonempty real interval, which contains a rational number. Applying the preceding argument after every rational time and using the Conformal Markov property of SLE rules this out on a countable probability-one event. Continuity of the trace then shows that no two distinct times have the same image. Thus ${\mathrm{SLE}}_{\kappa }$ is simple for $0<\kappa \le 4$.

If $q\ne 0$, divide the cubic equation by $q^3$ and define the rational numbersThen every integer anomaly-free assignment supplies a rational solution of

Assume ${\sup }_n\mathbb{E}|M_n|<\infty$. For every pair of rationals $a<b$, Doob upcrossing inequality givesby monotone convergence theorem. Thus every rational interval is upcrossed only finitely often almost surely. If a real sequence has distinct limit inferior and limit superior, it completes infinitely many upcrossings of some rational interval between them. Hence $M_n$ converges in the extended real line almost surely.

Fatou lemma givesso the limit is finite almost surely and integrable. This is the $L^1$-bounded form of the Martingale convergence theorem.

Suppose the columns $a_i$ of a matrix over the rational numbers satisfy the columns condition with blocks $B_1,\dots ,B_s$. Choose rational coefficients ${\lambda }_{ij}$ such thatChoose a positive integer $c$ clearing all denominators and $p\ge \max |c{\lambda }_{ij}|$. For generators of any full positive M-p-c set, the coordinateslie in that set and satisfy $Ax=0$. Thus the monochromatic m-p-c set theorem supplies the sufficiency direction of Rado's theorem.

The additive group ${\mathbb{Q}}_{\mathrm{sq}}$ consists of the rational numbers whose reduced denominators are square-free integers. If $p_0,p_1,\dots$ lists the primes without repetition, then

The rational homotopy group is obtained by tensoring a homotopy group with the rational numbers. It retains the free part and discards torsion.

The indicator of the rational numbers in the real numbers is discontinuous everywhere because both the rationals and irrationals are dense. The discontinuity theorem for Baire class one functions therefore shows that it is not Baire class one.

A rational matrix is a matrix whose entries are rational numbers. Multiplying by a common denominator produces an integer matrix and preserves its homogeneous solution set.

The square class of $a\in K^{\ast }$ is its image in the square-class group of a field $K^{\ast }/K^{\ast 2}$. Two nonzero elements have the same square class exactly when their quotient is a square. Every element of this quotient has order dividing $2$. Over the rational numbers, a square class has a unique signed squarefree integer representative.