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If Euler's number were $a/b$, choose $n\ge \max (b,2)$ and multiply its factorial-series tail by $n!$. This produces an integer, but the tail is positive and smaller than ${\sum }_{j\ge 1}(n+1{)}^{-j}=1/n<1$. The contradiction proves that the number is an irrational number. The argument illustrates how a rapidly convergent rational series can force a hypothetical rational remainder into an impossible integer interval.

For integers $a,b>1$, rationality of ${\log }_{a}b$ is equivalent to equality of some positive integer powers $a^p=b^q$. Exponentiation proves one direction and taking logarithms proves the other. Thus multiplicatively independent integers give an irrational number as their logarithm ratio. For distinct prime bases and arguments, unique factorization supplies that independence.

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.