This note is based on my talk *Introduction to Diophantine Approximation* at Carnegie Mellon University on November 4, 2014. Due to limit of time, details of Thue’s method were not fully presented in the talk. The note is supposed to serve as a complement to that talk.

### Diophantine Approximation

Diophantine approximation deals with the approximation of real numbers by rational numbers. In other words, given a real number \alpha, we are interested in finding a rational approximation p/q such that the error |\alpha - p/q| is “small”. The smaller the error, the better the approximation. Since the set of rationals is dense, there is no best rational approximation of an irrational number (apparently, the best rational approximation of a rational number is itself). However, if we compare the error with the complexity of the rational, i.e., the denominator of the rational, it is natural to expect that the more complicated the rational approximation, the smaller the error. In particular, we can make sense of this in the following way.

**Theorem (Dirichlet).** Given an irrational \alpha, we have |\alpha - p/q| < 1/q^2 for infinitely many rationals p/q.

*Proof.* Think of C = \mathbb{R}/\mathbb{Z} as a circle with circumference 1. Fix a natural number N and mark nq on C for n=1,\ldots,N. By the pigeonhole principle, there are 1\leq n_1 \leq n_2\leq N such that n_1\alpha and n_2\alpha is \leq 1/N apart from each other on C. Let q = n_2 - n_1 < N. Then there is p such that |q\alpha - p| \leq 1/N, and so |\alpha - p/q| \leq 1/(Nq) < 1/q^2. Taking N large enough yields better approximation p/q with the desired property.

By means of continued fraction, one can push the result a bit further by a constant factor.

**Theorem (Hurwitz 1891). **Given an irrational \alpha, we have |\alpha - p/q| < 1/\sqrt{5}q^2 for infinitely many rationals p/q. Moreover, the constant \sqrt{5} is the best possible for \alpha = (1+\sqrt{5})/2.

These show us how “well” we can approximate reals by rationals. On the other hand, we would also like to say that we cannot approximate “too well”, i.e., given a real number \alpha, we have |\alpha - p/q| > 1/q^n for all but finite rationals p/q, where n may depend on \alpha. Unfortunately, this is not so because of the existence of Liouville numbers.

**Definition. **A Liouville number is an irrational number \alpha with the property that, for every natural number n, there exists a rational p/q and such that |\alpha - p/q| < q^{-n}.

For example, \alpha = \sum_{k=1}^\infty 2^{-k!} is a Liouville number.

However, we indeed can say something along those lines for algebraic numbers.

### Diophantine Approximation for Algebraic Numbers

**Definition. **An algebraic number is a number that is a root of a non-zero polynomial in one variable with rational coefficients (or equivalently integer coefficients). Given an algebraic number, there is a monic polynomial (with rational coefficients) of least degree that has the number as a root. This polynomial is called its minimal polynomial. If its minimal polynomial has degree d, then the algebraic number is said to be of degree d.

**Theorem (Liouville 1844). **Given an algebraic irrational number \alpha of degree d, there exists A>0 such that for all rationals p/q, |\alpha - p/q|>Aq^{-d}.

A immediate consequence of the theorem is that Liouville numbers are not algebraic.

*Proof.* Let f(x) be a polynomial with integer coefficients of degree d that has \alpha as a root. Note that f'(\alpha)\neq 0 otherwise f'(x) is a polynomial of degree <d that has \alpha as a root. As f'(x) is continuous, there is r>0 such that |f'(x)| < 1 / A for all |x-\alpha|<r. Without loss of generality, we may assume p/q is pretty close, i.e., within distance r, to \alpha by taking A < r. Note that also f(p/q)\neq 0 otherwise f(x) / (x-p/q) is a polynomial of degree <d that has \alpha as a root. Since q^df(p/q) is an integer, |f(p/q)| \geq q^{-d}. On the other hand, by mean value theorem, there is c between p/q and \alpha such that |f(\alpha)-f(p/q)| = |f'(c)||\alpha - p/q|. Therefore |\alpha - p/q| = |f(p/q)|/f'(c) > Aq^{-n} since |c-\alpha| < r.[qed]

Liouville’s theorem (on Diophantine approximation) can be phrased in an alternative way.

**Corollary.** Given an irrational algebraic number \alpha of degree d. If the equation |\alpha - p/q| < 1/q^n has infinitely many rational solutions p/q, then n\leq d.

The continuing working was established by Axel Thue (1909), Carl Ludwig Siegle (1921), Freeman Dyson (1947), and culminated at Klaus Roth’s result (1955):

**Theorem (Roth 1955). **Given an irrational algebraic number \alpha. If the equation |\alpha-p/q|<1/q^n has infinitely many rational solutions p/q, then n\leq 2.

The constant 2 cannot be improved due to Dirichlet’s theorem. Another way to interpret Roth’s theorem is as follows.

**Corollary. **Any irrational algebraic number \alpha has approximation exponent equal to 2, i.e., for any \epsilon>0, the inequality |\alpha-p/q|<1/q^{2+\epsilon} can only have finitely many solutions p/q.

This amounts to say that we cannot approximate irrational algebraic numbers “too well” in the sense that if we slightly restrict the error from q^{-2} to q^{-(2+\epsilon)}, the number of “good” approximation changes from infinite to finite.

### Thue’s Method

Although Roth’s proof has merit in itself, part of his idea dates back to Thue’s work. Indeed, Thue used a method, now known as the polynomial method, in a special way to prove the following theorem.

**Theorem (Thue 1909). **Given an irrational algebraic number \alpha. If the equation |\alpha-p/q|<1/q^n has infinitely many rational solutions p/q, then n\leq d/2+1, where d is the degree of \alpha.

*Proof.* Suppose that the irrational algebraic number \alpha has two “good” rational approximations r_1 = p_1/q_1, r_2 = p_2/q_2, i.e., |\alpha - p_i/q_i| < q_i^{-n} for i = 1,2. Since there are infinitely many “good” rational approximations, we get to choose later how large we want q_i to be.

The rest of the proof can be broken into three parts. Let m\in\mathbb{N} and 0 < \epsilon \ll 1 be two constants to be determined. The constant \epsilon is fixed before we send it to 0.

We are going to use C_0, \ldots, C_6 to indicate positive constants that may depend on \epsilon, d and c, a constant determined by \alpha (defined later). We may always assume \alpha, d, c, C_0,\ldots, C_6 \ll m.

**Part I.** Find a non-zero polynomial f(x,y) = P(x) + Q(x)y with integer coefficients that satisfies the following properties:

- it vanishes at (\alpha, \alpha) to order (m, 1), i.e., \partial_x^k f(\alpha, \alpha) = 0 for all k<m;
- its degree in x is \leq D := \frac{1}{2}(1+\epsilon)dm.
- its size, i.e., the maximum of all its coefficients’ absolute values, denoted by |f|, is \leq C_1^m for some constant C_1.

*Proof of Part I.* The proof is a classical “parameter counting” argument. Consider a polynomial f(x,y)=P(x)+Q(x)y = \sum_{i<D}a_ix^i + \sum_{i<D}b_ix^iy with degree \leq D in x and think of its 2D coefficients as undetermined. We can interpret the first property as a homogeneous system of m linear equations with coefficients in \mathbb{Z}[\alpha] because \frac{1}{k!}\partial_x^k f(\alpha, \alpha) = 0 \text{ if and only if } \sum_{i < D}{i\choose k}\alpha^{i-k}a_i + \sum_{i < D}{i\choose k}\alpha^{i-k+1}b_i = 0.

Fix a (non-zero) polynomial c_dx^d + \ldots + c_0 with integer coefficients that has \alpha as a root and c the size of this polynomial. Notice that c_d\alpha^d = -(c_{d-1}\alpha^{d-1} + \ldots + c_0). For each of the linear equation above of the form L_0 + L_1\alpha + \ldots + L_t\alpha^t = 0, for some t \geq d where L_k are linear combination of a_i, b_i with integer coefficients, we can replace it by c_d(L_0 + L_1\alpha + \ldots + L_{t-1}\alpha^{t-1}) - L_t(c_0\alpha^{t-d} + \ldots + c_{d-1}\alpha^{t-1}) = 0, i.e., L'_0 + L_1'\alpha + \ldots + L_{t-1}\alpha^{t-1} = 0 with L'_k = \begin{cases} c_dL_k & 0\leq k < t-d \\ c_dL_k - c_{k-t+d}L_t & t-d \leq k < t\end{cases} and we can repeat this until its degree in \alpha becomes smaller than d. Let |L_k|_1 be the sum of the absolute values of all coefficients in L_k. Note that each replacement increases the \max |L_k|_1 to \max |L_k'|_1 \leq 2c \max|L_k|. Note that the initially \max |L_k| < 2^D. After at most D replacements, the equation satisfies \max |L_k| < 2^D(2c)^D = C_0^m for some constant C_0.

Finally, for each of the linear equation, after the replacements, of the form L_0 + L_1\alpha + \ldots + L_{d-1}\alpha^{d-1} = 0, it is equivalent to have L_0 = L_1 = \ldots = L_{d-1} because 1, \alpha, \ldots, \alpha^{d-1} are linearly independent over \mathbb{Q}. Therefore each linear equation with coefficient in \mathbb{Z}[\alpha] corresponding to \partial_x^k f(\alpha, \alpha) = 0 is equivalent to d linear equations whose |\cdot|_1 are all bounded by C_0^m. In total, we have dm linear equations with 2D undetermined.

One can view this homogeneous system of linear equations as a linear transformation L = (L_1, \ldots, L_{dm}): \mathbb{Z}^{2D} \to \mathbb{Z}^{dm}. Consider all integral points in [-M, M]^{2D} and their image under L. Because |L_k|_1 \leq C_0^m, their image is contained in [-C_0^mM, C_0^mM]^{dm}. By the pigeonhole principle, as long as (2M+1)^{2D} > (2C_0^mM+1)^{dm}, there are two integral points v', v'' in [-M,M]^{2D} such that Lv'=Lv'', hence there is an integral point v=v'-v''\in [-2M, 2M]^{2D} such that Lv=0 and v\neq 0, hence a non-zero solution to the linear system. To guarantee the above inequality involving M, it is enough to have 2M > C_0^{m/\epsilon}, and so 2M = C_1^m, for some constant C_1, works.

**Part II. **Find l < L := \epsilon m such that g(x,y) = \frac{1}{l!}\partial_x^l f(x,y) satisfies the following properties:

- it vanishes at (\alpha, \alpha) to order (m-L, 1);
- its degree in x is \leq D (=\frac{1}{2}(1+\epsilon)dm);
- its size is |g| \leq {D\choose l}C_1^m < C_2^m for some constant C_2;
- and g(r_1, r_2)\neq 0.

*Proof of Part II.* The first three properties follow immediately from the properties of f. Henceforth, we only have to show there exists l < L such that the last property holds. For the sake of contradiction, assume that \partial_x^l f(r_1, r_2) = 0 for all l < L. Define D(x) := \begin{vmatrix}P(x) & Q(x) \\ P'(x) & Q'(x)\end{vmatrix} = P(x)Q'(x) - P'(x)Q(x). Because f(x,y) vanishes at (r_1, r_2) to order (L,1), by Leibniz rule, D(x) vanishes at r_1 to order L-1. Note that D(x) is in \mathbb{Z}[x]. By Gauss’ lemma, (q_1x-p_1)^{L-1} divides D(x). Therefore either (1) |D(x)| \geq q_1^{L-1}; or (2) D(x) = 0 as a polynomial.

If it is the first case, because |D(x)| = |P(x)Q'(x) - P'(x)Q(x)| < 2D|f|^2, we have 2D|f|^2 > q_1^{L-1}. Because |f| < C_1^m, we get a contradiction as long as q_1 > C_3, for some constant C_3.

If it is the second case, P(x) and Q(x) are linearly dependent over \mathbb{Q}. Therefore either (1) P(x) = -r_2'Q(x) and Q(x) vanishes at r_1 to order < L or (2) Q(x) vanishes at r_1 to order L. If it is the first case, because f(x,y) = (y-r_2')Q(x) vanishes at (r_1, r_2) to order (L, 1) and Q(x) vanish at r_1 to order <L, we know that r_2' = r_2, hence by Gauss’ lemma, f(x,y) is a multiple of q_2x - p_2, and so |f| \geq q_2. If it is the second case, because Q vanishes at r_1 to order L, by Gauss’ lemma, Q(x) is a multiple of (q_1x - p_1)^L, and so |f| \geq |Q| \geq q_1^L. To get a contradiction in both cases, we need q_1 > C_4\text{ and }q_2 > C_1^m.

Therefore, it is enough to have q_1 > C_5\text{ and }q_2 > q_1^m for some constant C_5.

**Part III. **We claim that |g(r_1, r_2)| < C_6^m\left(q_1^{-(1-\epsilon)m}+q_2^{-1}\right).

*Proof of Part III.* We know that |g(r_1, r_2)-g(r_1, \alpha)| = |Q(r_1)||\alpha - r_2| \leq D|g|(|\alpha|+1)^D|\alpha-r_2|. Because g vanishes at (\alpha_1, \alpha_1) to order (m-L, 1), using Taylor expansion, we have

\begin{aligned}|g(r_1, \alpha) - g(\alpha, \alpha)| & \leq \sup_{|x| < |\alpha|+1}\frac{1}{(m-L)!}\partial_x^{m-L}g(x,\alpha) |\alpha_1-r_1|^{m-L} \\ & < D2^D|g|(|\alpha|+1)^D||\alpha_1-r_1|^{m-L}.\end{aligned}

As r_1, r_2 are two “good” approximations of \alpha, we obtain \begin{aligned}|g(r_1, r_2)| & \leq D|g|(|\alpha|+1)^D|\alpha-r_2| + D2^D|g|(|\alpha|+1)^D||\alpha_1-r_1|^{m-L} \\ & < C_6^m\left(q_2^{-n} + q_1^{(-m+L)n}\right),\end{aligned} for some constant C_6

From Part II, we know that g(r_1, r_2)\neq 0, and so |g(r_1, r_2)| \geq q_1^{-D}q_2^{-1}. From Part III, we know that |g(r_1, r_2)| < C_6^m(q_1^{(-m+L)n}+q_2^{-n}). Therefore, we obtain q_1^{-D}q_2^{-1} < C_6^m\left(q_1^{(-m+L)n}+q_2^{-n}\right). Set m to be the biggest integer less than \log{q_2} / \log{q_1}, and so q_1^m < q_2 \leq q_1^{m+1}. To ensure Part II to work, we only need to take q_1 > C_5. In addition, we require q_1^\epsilon > C_6 and so C_6^m < q_1^{\epsilon m} < q_2^\epsilon. The choice of m implies that

q_2^{-\frac{1}{2}(1+\epsilon)d - 1} = q_2^{-D/m}q_2^{-1} < q_2^\epsilon\left(q_2^{-\frac{m-L}{m+1}n}+q_2^{-n}\right) < 2q_2^{-(1-2\epsilon)n+\epsilon}.

In order to have this inequality hold for arbitrarily large q_2, we need \frac{1}{2}(1+\epsilon)d + 1 > (1-2\epsilon)n-\epsilon. Let \epsilon go to 0 and obtain n\leq \frac{1}{2}d + 1. [qed]

### Acknowledgement

The proof largely relies on Larry Guth‘s notes on polynomial method.