Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 4 / 7E (Numbers and Sets)

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.

For $M=\max \{|s|,|t|,1\}$, factorization of a difference of powers givesConsequently the polynomial $f$ obeys the Lipschitz bound

Since $p/q$ is not a root, $q^{d}f(p/q)$ is a nonzero integer, and henceMoreover $0<\theta -p/q<1$ implies $|p/q|<|\theta |+1$. Applying part (i) to $s=\theta$, $t=p/q$, and using $f(\theta )=0$ givesRearrangement provesThis is the core estimate behind the Liouville approximation theorem.

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.

This can be false. For any nonzero transcendental number $\beta$, the number $\gamma =1/\beta$ is also transcendental, since otherwise its reciprocal would be algebraic. Yet $\beta \gamma =1$ is algebraic.

This is always true for every positive integer $n$. If ${\beta }^n$ were algebraic, then $\beta$ would be a root of the polynomial $X^n-{\beta }^n$ over the field of algebraic numbers. By transitivity of algebraic extensions, $\beta$ would then be algebraic over $\mathbb{Q}$, contradicting its transcendence.