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For $P(x)=x^3+4x-7$, the derivative $P^{\prime }(x)=3x^2+4$ is positive everywhere, so the mean value theorem makes $P$ strictly increasing. Since a polynomial is a continuous function, the intermediate value theorem and the values $P(1)=-2$, $P(2)=9$ locate its unique real root in $(1,2)$. By the rational root theorem, a rational root of a monic integer polynomial must be an integer. There is no integer in $(1,2)$, so the unique real root is an irrational number. The denominator part of that theorem also follows directly: if $a/b$ in lowest terms were a root, $a^3+4a{b}^2-7b^3=0$ would force $b\mid a^3$, hence $b=1$.

Choose a positive common denominator $d$ so that $x=a/d$ and $y=b/d$ with nonzero integers $a,b$. Put $h=\gcd (a,b)>0$. Every element of the generated subgroup has the form $(ma+nb)/d$, so it lies in $(h/d)\mathbb{Z}$. Conversely, Bezout identity gives integers $u,v$ with $ua+vb=h$, so $h/d$ belongs to the generated subgroup. ConsequentlyThis map is a bijective additive group homomorphism. It is the two-generator instance of a finitely generated subgroup of the rational additive group being an infinite cyclic group.

For the real example, take $x=1$ and $y=\sqrt{2}$. The group homomorphism $(m,n)\mapsto m+n\sqrt{2}$ from ${\mathbb{Z}}^2$ onto the generated subgroup is injective because $\sqrt{2}$ is an irrational number. Thus this subgroup is isomorphic to ${\mathbb{Z}}^2$. More directly, if a single real number $t$ generated it, then $1=at$ and $\sqrt{2}=bt$ for integers $a\ne 0,b$, giving $\sqrt{2}=b/a$, a contradiction. The choice $1,\sqrt{2}$ generates a noncyclic additive subgroup.

Let $D_n=p_{n-1}{q}_{n-2}-p_{n-2}{q}_{n-1}=(-1{)}^{n-2}$. Solving the given Möbius transformation for the complete quotient and rationalizing givesThereforeThe denominator $D_n$ is $\pm 1$, so both are integers. The algebraic norm denominator is nonzero because $\sqrt{d}$ is an irrational number.

For $x>0$, $N(x)=\lfloor 1/x\rfloor$ and $T(x)=1/x-\lfloor 1/x\rfloor$. Iterating this Gauss continued-fraction map on an irrational number never reaches zero, and its continued fraction isThe branches $T:(1/(n+1),1/n)\to (0,1)$ have inverse $x_n(y)=1/(n+y)$, for $n\ge 1$. The change of variables formula for a probability density function gives, almost everywhere,The endpoints of these branches are a countable set of probability zero. Thus $T(X)$ and $X$ have the same distribution; this Gauss measure is an invariant measure for the Gauss continued-fraction map.

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.

Let ${\beta }_{n+1}>0$ be the complete quotient after $a_n$. Part (a) writes the irrational number $\theta$ asso $\theta$ lies between the adjacent continued fraction convergents $p_n/q_n$ and $p_{n-1}/q_{n-1}$. Their separation isBecause all $a_j$ are positive integers, the recurrence $q_n=a_n{q}_{n-1}+q_{n-2}$ implies $q_n\to \infty$. Hence this separation tends to zero, and the squeeze theorem gives

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.

Take an irrational number $a\in (0,1)$. The density of the rational numbers provides a sequence in $\mathbb{Q}\cap [0,1]$ converging to $a$, which is outside that subset. Thus the subset is not closed in $\mathbb{R}$. By the Heine-Borel theorem, it is not compact, despite being bounded.

False, with “disjoint union” understood as a topological disjoint union. Take the rational numbers $\mathbb{Q}$ with the usual subspace topology from the real line.

Every connected subset of $\mathbb{Q}$ has at most one point. Indeed, if it contained $r<s$, choose an irrational number $t\in (r,s)$. The two relatively open setspartition $\mathbb{Q}$ and separate any subset containing both $r$ and $s$. Hence the connected components of $\mathbb{Q}$ are singletons.

If a topological disjoint union of nonempty connected spaces were homeomorphic to $\mathbb{Q}$, the image of each summand would be connected, hence a singleton. Every summand would therefore be a one-point topological space. Their topological disjoint union is a discrete space, since each summand is open. But $\mathbb{Q}$ is not a discrete space: for each $r\in \mathbb{Q}$, the distinct rational numbers $r+1/n$ approach $r$, so $r$ is not an isolated point. This is a contradiction. The set decomposition into connected components always exists; giving those pieces the topological disjoint union topology need not recover the original topology.

True. Choose the irrational number $\alpha =\sqrt{2}$, and partition the rational numbers intoNo endpoint is rational, so these disjoint sets cover $\mathbb{Q}$. Each is nonempty by the density of the rational numbers, and each is clopen in the subspace topology on $\mathbb{Q}$.

Let two rational numbers be equivalent exactly when they belong to the same $C_n$. The bijectionis well-defined. For every subset $B\subseteq \mathbb{Z}$, the preimage of $h^{-1}(B)$ under the canonical projection of a quotient set is ${\bigcup }_{n\in B}{C}_n$, which is open in $\mathbb{Q}$. Thus every subset of $\mathbb{Q}/\sim$ is open: its quotient topology is the discrete topology. The usual subspace topology on the integers is also the discrete topology, because $\{n\}=\mathbb{Z}\cap (n-1/2,n+1/2)$. Consequently $h$ is a homeomorphism.

For a proof by contradiction, suppose $\sqrt{3}=a/b$ for positive integers $a,b$ in lowest terms, so $\gcd (a,b)=1$. Squaring gives $a^2=3b^2$, and part (a) implies $3\mid a$. Write $a=3c$; then $9c^2=3b^2$, hence $b^2=3c^2$, and part (a) again implies $3\mid b$. This contradicts the coprimality of $a,b$. Therefore $\sqrt{3}$ is an irrational number.

Assuming $\pi =a/b$ and integrating a suitably normalized polynomial times $\sin x$ produces a positive integer that tends to zero as the polynomial degree grows. This contradiction proves that $\pi$ is an irrational number.

On $[0,1]$, use reduced representations of rational numbers with positive denominator. This reverses the nonzero values of the Thomae function, while leaving its values at irrational numbers zero. Every interval of positive length contains rational numbers of arbitrarily large reduced denominator: there are only finitely many fractions in $[0,1]$ with denominator at most a fixed $M$, whereas any open interval contains infinitely many rational numbers. Thus this function taking finite real number values is unbounded on every nondegenerate subinterval.

Fix a real angle fraction $\alpha$, and repeatedly multiply by $g=e^{2\pi i\alpha }$ in the circle group. The generated cyclic subgroup is $H_{\alpha }=\{g^k:k\in \mathbb{Z}\}$. Equivalently, the angle fractions are the fractional parts of $k\alpha$, with one full turn identified with zero.

If $\alpha =p/q$ is a rational number in lowest terms, with $q>0$ and $p,q$ coprime, then $g^q=1$ and no smaller positive power is $1$. Hence $H_{\alpha }$ is a finite cyclic group of order $q$, consisting of equally spaced points. For $\alpha =2/7$, the successive angle fractions, starting at $k=0$, areThe subgroup is finite, so it is a closed subgroup of the circle group; it is also a discrete subgroup. It is not an open subgroup, since every ambient neighbourhood of any of its points contains an arc, rather than only these seven points.

If $\alpha$ is an irrational number, no nonzero integer $k$ satisfies $k\alpha \in \mathbb{Z}$, so the powers never repeat. The stronger statement is that $H_{\alpha }$ is a dense set in the circle group: every nonempty open arc contains a power $g^k$. Thus its closure is all of $\mathbb{T}$, while $H_{\alpha }$ itself is a countable proper subgroup and is not a closed subgroup or a discrete subgroup. See irrational rotation and everywhere interval frequency under an irrational rotation for the corresponding motion and its long-term frequencies.

For $\alpha =\sqrt{2}-1$, direct computation of the fractional parts of $k\alpha$ gives, to six decimal places,For example, $\{3\alpha \}=3\sqrt{2}-4\approx 0.242641$ and $\{5\alpha \}=5\sqrt{2}-7\approx 0.071068$. These calculations specify exactly which points the diagram displays.

This example shows why algebraic generation and closure must be distinguished: one irrational rotation generates a countable cyclic subgroup, yet its closure is the whole uncountable compact group.