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Math 784: algebraic NUMBER THEORY
(Instructor’s Notes)*
Algebraic Number Theory:
- What is it? The goals of the subject include: (i) to use algebraic concepts to deduce information about integers and other rational numbers and (ii) to investigate generaliza- tions of the integers and rational numbers and develop theorems of a more general nature. Although (ii) is certainly of interest, our main point of view for this course will be (i). The focus of this course then will be on the use of algebra as a tool for obtaining information about integers and rational numbers.
- A simple example. Here, we obtain as a consequence of some simple algebra the following:
Theorem 1. Let θ ∈ Q with 0 ≤ θ ≤ 1 / 2. Then sin^2 (πθ) ∈ Q if and only if θ ∈ { 0 , 1 / 6 , 1 / 4 , 1 / 3 , 1 / 2 }.
It is easy to check that for θ ∈ { 0 , 1 / 6 , 1 / 4 , 1 / 3 , 1 / 2 }, we have sin^2 (πθ) is rational; so we are left with establishing that if sin^2 (πθ) ∈ Q, then θ ∈ { 0 , 1 / 6 , 1 / 4 , 1 / 3 , 1 / 2 }. Observe that we can use Theorem 1 to immediately determine for what θ ∈ Q the value of sin(πθ) is rational (see the upcoming homework assignment). Before getting to the proof of this theorem, we give some background.
- Some definitions and preliminaries.
Definition. Let α be a complex number. Then α is algebraic if it is a root of some f (x) ∈ Z[x] with f (x) 6 ≡ 0. Otherwise, α is transcendental.
Examples and Comments:
(1) Rational numbers are algebraic. (2) The number i =
−1 is algebraic. (3) The numbers π, e, and eπ^ are transcendental. (4) The status of πe^ is unknown. (5) Almost all numbers are transcendental.
Definition. An algebraic number α is an algebraic integer if it is a root of some monic polynomial f (x) ∈ Z[x] (i.e., a polynomial f (x) with integer coefficients and leading coef- ficient one).
Examples and Comments: (1) Integers (sometimes called “rational integers”) are algebraic integers. (2) Rational numbers which are not rational integers are not algebraic integers. In other words, we have
*These notes are from a course taught by Michael Filaseta in the Spring of 1997 and 1999 but based on notes from previous semesters.
1
Theorem 2. If α is a rational number which is also an algebraic integer, then α ∈ Z.
Proof. Suppose f (a/b) = 0 where f (x) =
∑n j=0 aj^ x
j (^) with an = 1 and where a and b are
relatively prime integers with b > 0. It suffices to show b = 1. From f (a/b) = 0, it follows that an^ + an− 1 an−^1 b + · · · + a 1 abn−^1 + a 0 bn^ = 0.
It follows that an^ has b as a factor. Since gcd(a, b) = 1 and b > 0, we deduce that b = 1, completing the proof.
(3) The number i is an algebraic integer. (4) Transcendental numbers are not algebraic integers. (5) If u ∈ Q, then 2 cos(πu) is an algebraic integer. This requires an explanation which we supply next.
Lemma. For each positive integer m, there is a gm(x) =
∑m j=0 bj^ x
j (^) ∈ Z[x] satisfying: (i) cos(mθ) = gm(cos θ) (ii) bm = 2m−^1 (iii) 2 k−^1 |bk for k ∈ { 2 , 3 ,... , m}.
Proof. We do induction on m. The cases m = 1 and m = 2 are easily checked. Suppose the lemma holds for m ≤ n. Observe that
cos((n + 1)θ) + cos((n − 1)θ) = 2 cos(nθ) cos θ.
Then (i), (ii), and (iii) follow by considering gn+1(x) = 2xgn(x) − gn− 1 (x).
Write u = a/m with m a positive integer. By the lemma,
±1 = cos(mπu) = gm(cos(πu)) =
∑^ m
j=
bj (cos(πu))j^ ,
where bm = 2m−^1 and for some integers b′ j we have bj = 2j−^1 b′ j for j ∈ { 2 , 3 ,... , m − 1 }. Multiplying through by 2 and rearranging, we deduce that 2 cos(πu) is a root of
xm^ + b′ m− 1 xm−^1 + · · · + b′ 2 x^2 + b 1 x + (2b 0 ∓ 2).
It follows that 2 cos(πu) is an algebraic integer.
- An application. Before proving Theorem 1, we determine for what θ ∈ Q, the value of cos(πθ) is rational. By the last example above, if θ ∈ Q, then 2 cos(πθ) is an algebraic integer. If we also have cos(πθ) is rational, then Theorem 2 implies that 2 cos(πθ) ∈ Z. Since |2 cos(πθ)| ≤ 2, we deduce that 2 cos(πθ) ∈ {− 2 , − 1 , 0 , 1 , 2 } so that cos(πθ) ∈ {− 1 , − 1 / 2 , 0 , 1 / 2 , 1 }. It follows that both θ and cos(πθ) are rational if and only if θ ∈ {k/2 : k ∈ Z} ∪
k/3 : k ∈ Z
- Completing the proof of Theorem 1. Let u = 2θ, and suppose that sin^2 (πθ) ∈ Q. Then 2 cos(πu) = 2 − 4 sin^2 (πθ) is an algebraic integer which is also rational. By Theorem
the size of h to be (k 1 ,... , kn) where (k 1 ,... , kn) is the element of T with k 1 as large as possible, k 2 as large as possible given k 1 , etc. Since h(α 1 ,... , αn) is symmetric, it follows that (1 ,... ,n) ∈ T if and only if each permutation of (1 ,... ,n) is in T. This implies that k 1 ≥ k 2 ≥ · · · ≥ kn. Observe that we can use the notion of size to form an ordering on the elements of R[α 1 ,... , αn] in the sense that if h 1 has size (k 1 ,... , kn) and h 2 has size (k 1 ′,... , k n′), then h 1 > h 2 if there is an i ∈ { 0 , 1 ,... , n − 1 } such that k 1 = k′ 1 ,... , ki = k′ i, and ki+1 > k i′+1. Note that the elements of R[α 1 ,... , αn] which have size (0, 0 ,... , 0) are precisely the constants (the elements of R). Suppose now that (k 1 ,... , kn) represents the size of some symmetric g ∈ R[α 1 ,... , αn] with g 6 ∈ R. For non-negative integers d 1 ,... , dn, the size of h = σd 11 σ 2 d 2 · · · σd nn is (d 1 + d 2 +· · ·+dn, d 2 +· · ·+dn,... , dn− 1 +dn, dn). Taking d 1 = k 1 −k 2 , d 2 = k 2 −k 3 ,... , dn− 1 = kn− 1 − kn, and dn = kn, we get the size of h is (k 1 ,... , kn). The coefficient of αk 11 · · · αk nn in h is 1. It follows that there is an a ∈ R such that g − ah is of smaller size than g. The above implies that for any symmetric element f ∈ R[α 1 ,... , αn], there exist a 1 ,... , am ∈ R and h 1 ,... , hm ∈ R[σ 1 ,... , σn] such that f − a 1 h 1 − · · · − amhm has size (0, 0 ,... , 0). This implies the theorem.
- Elementary symmetric functions on roots of polynomials. Let f (x) =
∑n j=0 aj^ x
j (^) be
a non-zero polynomial in C[x] of degree n with not necessarily distinct roots α 1 ,... , αn. Then it is easy to see that
f (x) = an
∏n
j=
(x − αj ) = anxn^ − anσ 1 xn−^1 + anσ 2 xn−^2 + · · · + (−1)nanσn,
where now we view the σj as elementary symmetric functions in the numbers α 1 ,... , αn. It follows that
(∗) σ 1 = −
an− 1 an
, σ 2 =
an− 2 an
,... , σn = (−1)n^
a 0 an
- An example (almost Putnam Problem A-1 from 1976). Consider all lines which pass through the graph of y = 2x^4 +7x^3 +3x−5 in 4 distinct points, say (xj , yj ) for j = 1, 2 , 3 , 4. We will show that the average of the xj ’s is independent of the line and find its value. If y = mx + b intersects y = 2x^4 + 7x^3 + 3x − 5 as indicated, then x 1 ,... , x 4 must be the four distinct roots of 2x^4 + 7x^3 + (3 − m)x − (b + 5) = 0. From the previous section, we deduce that the sum of the xj ’s is σ 1 = − 7 /2. The average of the xj ’s is therefore − 7 /8.
Homework:
(1) Show that the average of the x^2 j ’s is independent of the line and find its value.
(2) Prove or disprove that the average of the yj ’s is independent of the line.
(3) In the proof of Theorem 3, we deduced that the size of g − ah is smaller than the size of g. By continuing the process, we claimed that eventually we would obtain an element of R[α 1 ,... , αn] of size (0, 0 ,... , 0). Prove this as follows. Explain why the claim is true if n = 1. Consider n ≥ 2. Let (k 1 ,... , kn) be the size of g with g 6 ∈ R, and let (k′ 1 ,... , k n′)
be the size of g − ah. Let b be an integer ≥ k 1. Associate the integer
∑n− 1 j=0 `n−j^ b
j (^) with an
n-tuple (1 ,... ,n). Show that the integer associated with (k 1 ,... , kn) is greater than the integer associated with (k′ 1 ,... , k′ n). Explain why (0, 0 ,... , 0) is obtained by continuing the process as claimed. (There are other approaches to establishing that (0, 0 ,... , 0) will be obtained, and you can feel free to establish this in a different manner.)
Algebraic Numbers and Algebraic Integers as Algebraic Structures:
- The main theorems we deal with here are as follows.
Theorem 4. The algebraic numbers form a field.
Theorem 5. The algebraic integers form a ring.
To prove these, we suppose that α and β are algebraic numbers or integers, and prove that −α, α + β, and αβ are likewise. In the case that α is a non-zero algebraic number, we show that 1/α is as well.
- The case for −α. If f (x) is a polynomial with integer coefficients having α as a root, then we consider ±f (−x). If f (x) is monic, then one of these will be as well. Hence, if α is an algebraic number, then so is −α; and if α is an algebraic integer, then so is −α.
- The case for α + β. Suppose α is a root of f (x) ∈ Z[x] and β is a root of g(x) ∈ Z[x]. Let α 1 = α, α 2 ,... , αn denote the complete set of roots of f (x) (counted to their multiplicity so that the degree of f (x) is n) and let β 1 = β, β 2 ,... , βm denote the complete set of roots of g(x). Consider the polynomial
F (x) =
∏^ n
i=
∏^ m
j=
x − (αi + βj )
Taking R = Z[β 1 ,... , βm] in Theorem 3, we see that the coefficients of F (x) are symmet- ric polynomials in α 1 ,... , αn. Thus, if σ 1 ,... , σn correspond to the elementary sym- metric functions in α 1 ,... , αn and A is some coefficient (of xk) in F (x), then A = B(σ 1 ,... , σn, β 1 ,... , βm) for some polynomial B with integer coefficients. Now, the coef- ficients of F (x) are also symmetric in β 1 ,... , βm. Taking R = Z[σ 1 ,... , σn] in Theorem 3 and σ 1 ′,... , σ′ m to be the elementary symmetric functions in β 1 ,... , βm, we get that A = B′(σ 1 ,... , σn, σ 1 ′,... , σ m′) for some polynomial B′^ with integer coefficients. On the other hand, (∗) implies that σ 1 ,... , σn, σ′ 1 ,... , σ m′ are all rational so that A ∈ Q. Thus, F (x) ∈ Q[x] and m′F (x) ∈ Z[x] for some integer m′. Since α + β is a root of m′F (x), we deduce that α + β is an algebraic number. If α and β are algebraic integers, then we can take the leading coefficients of f (x) and g(x) to be 1 so that (∗) implies that each of σ 1 ,... , σn, σ′ 1 ,... , σ m′ is in Z so that F (x) ∈ Z[x]. Since F (x) is monic, we obtain that in this case α + β is an algebraic integer.
- The case for αβ. The same idea as above works to show αβ is an algebraic number (or integer) by defining
F (x) =
∏^ n
i=
∏^ m
j=
x − αiβj
integers as small as possible such that p - ak and p - b; these exist since u(x) and v(x) are primitive. One checks that the coefficient of xk+^ is not divisible by p. It follows that the content of u(x)v(x) cannot be divisible by p, completing the proof.
Theorem 8 (Gauss’ Lemma). Let f (x) ∈ Z[x]. Suppose that there exist u 1 (x) and v 1 (x) in Q[x] such that f (x) = u 1 (x)v 1 (x). Then there exist u 2 (x) and v 2 (x) in Z[x] such that f (x) = u 2 (x)v 2 (x) and deg u 2 (x) = deg u 1 (x) and deg v 2 (x) = deg v 1 (x).
Comment: The theorem implies that if f (x) ∈ Z[x] has content 1, then a necessary and sufficient condition for f (x) to be irreducible over the rationals is for it be irreducible over the integers. Also, we note that the proof will show more, namely that one can take u 2 (x) and v 2 (x) to be rational numbers times u 1 (x) and v 1 (x), respectively.
Proof. Let d denote the content of f (x). Then there are positive rational integers a and b and primitive polynomials u(x) and v(x) in Z[x] with deg u(x) = deg u 1 (x) and deg v(x) = deg v 1 (x) satisfying u 1 (x)v 1 (x) = (a/b)u(x)v(x). Then there is a primitive g(x) ∈ Z[x] for which f (x) = dg(x) and bdg(x) = bf (x) = au(x)v(x). By the lemma, u(x)v(x) is primitive. It follows that the content of au(x)v(x) is a. Since g(x) is primitive, the content of bdg(x) is bd. Hence, a = bd. We set u 2 (x) = du(x) and v 2 (x) = v(x). Then f (x) = u 1 (x)v 1 (x) = du(x)v(x) = u 2 (x)v 2 (x), and we deduce the theorem.
- The proof of Theorem 7. It is clear that if the minimal polynomial for α is in Z[x], then α is an algebraic integer. Now, consider an algebraic integer α, and let f (x) ∈ Z[x] be monic with f (α) = 0. Let u 1 (x) be the minimal polynomial for α. We want to prove that u 1 (x) ∈ Z[x]. By the division algorithm for polynomials in Q[x], there exist v 1 (x) and r(x) in Q[x] such that f (x) = u 1 (x)v 1 (x) + r(x) and either r(x) ≡ 0 or 0 ≤ deg r(x) < deg u 1 (x). Note that r(α) = f (α) − u 1 (α)v 1 (α) = 0. Since u 1 (x) is the monic polynomial of smallest degree having α as a root, it follows that r(x) ≡ 0 (otherwise, there would be a k ∈ Z for which (1/k)r(x) ∈ Q[x] is monic, is of smaller degree than deg u 1 (x), and has α as a root). Thus, f (x) = u 1 (x)v 1 (x) is a factorization of f (x) in Q[x]. By Gauss’ Lemma and the comment after it, there exist u 2 (x) and v 2 (x) in Z[x] with f (x) = u 2 (x)v 2 (x) and with u 2 (x) = mu 1 (x) for some non-zero rational number m. By considering f (x) = (−u 2 (x))(−v 2 (x)) if necessary, we may suppose that the leading coefficient of u 2 (x) is positive. Since f (x) is monic, we deduce that u 2 (x) is monic. Comparing leading coefficients in u 2 (x) = mu 1 (x), we see that m = 1 so that u 1 (x) = u 2 (x) ∈ Z[x] as desired.
Algebraic Number Fields:
- The definition. If α is an algebraic number, then Q(α) is defined to be the smallest field containing both α and the rationals.
- Some simple observations. Let f (x) ∈ Q[x] be the minimal polynomial for α. By considering each integer j ≥ 0 successively and αj^ f (α) = 0, one shows that αn+j^ can be expressed as a polynomial in α with coefficients in Q and with degree ≤ n − 1. It follows that Q(α) is the set of all numbers of the form g(α)/h(α) where g(x) and h(x) are in Z[x], deg g(x) ≤ n − 1, deg h(x) ≤ n − 1, and h(α) 6 = 0. By Theorem 4, every element of Q(α) is an algebraic number. For this reason, we refer to Q(α) as an algebraic number field.
- The ring of algebraic integers in Q(α).
Theorem 9. The algebraic integers contained in an algebraic number field Q(α) form a ring.
Proof. If α and β are in Q(α), then so are αβ and α − β since Q(α) is a field. If also α and β are algebraic integers, then Theorem 5 implies αβ and α − β are algebraic integers. The result follows.
Homework:
(1) Prove that for every algebraic number α, the minimal polynomial for α exists and is unique.
(2) Prove that the minimal polynomial f (x) for an algebraic number α is irreducible over the rationals. In other words, prove that there do not exist g(x) and h(x) in Q[x] of degree ≥ 1 satisfying f (x) = g(x)h(x).
(3) With the notation in the section above, let α 1 , α 2 ,... , αn be the roots of f (x) with α 1 = α. Show that w = h(α 2 )h(α 3 ) · · · h(αn) ∈ Q[α] (i.e., w can be expressed as a polynomial in α with rational coefficients). Also, show that w 6 = 0. By considering (g(α)w)/(h(α)w), show that every element of Q(α) can be written uniquely in the form u(α) where u(x) ∈ Q[x] and deg u(x) ≤ n − 1. (There are other ways to establish this; we will in fact do this momentarily. The homework problem is to establish this result about Q(α) by showing that one can “rationalize the denominator” of g(α)/h(α).)
Quadratic Extensions:
- Definition. Let m ∈ Z with m not a square. Then Q(
m) is a quadratic extension of the rationals. Note that the minimal polynomial for
m is x^2 − m (see the first homework exercises, problem (5)).
m). We have discussed this in more generality already. If β ∈ Q(
m), then there are rational integers a, b, c, and d such that
β =
a + b
m c + d
m
a + b
m c + d
m
×
c − d
m c − d
m
(ac − bdm) + (bc − ad)
m c^2 − md^2
Observe that the denominator is non-zero since
m 6 ∈ Q. The above corresponds to what took place in the last homework problem; we have shown that each element of Q(
m) can be expressed as a linear polynomial in
m with coefficients in Q. Note that each element of Q(
m) has a unique representation of the form a + b
m with a and b rational.
- When does Q(α 1 ) = Q(α 2 )? One way to show two algebraic number fields are the same field is to show that α 1 ∈ Q(α 2 ) (so that Q(α 2 ) is a field containing α 1 ) and that α 2 ∈ Q(α 1 ) (so that Q(α 1 ) is a field containing α 2 ). Explain why this is enough.
- When does Q(
m) = Q(
m′)? Given the above, equality holds if there are positive integers k and such that k^2 m =^2 m′. It follows that all quadratic extensions are of the form Q(
m) with m a squarefree integer and m 6 = 1. Since m squarefree implies m is not
- Answering the question. We can view the a/b we constructed as a rational approxi- mation of α, but it is possible to have better approximations in the following sense. The above choice for a/b satisfies |α − (a/b)| ≤ 1 /(2b). Let’s prove now that there are a/b satis- fying |α − (a/b)| < 1 /b^2 (but we note here a difference: for infinitely many positive integers b, there is an a for which a/b is within 1/b^2 of being α; for every positive integer b, there is an a for which a/b is within 1/(2b) of being α). We use the Dirichlet drawer principle as Dirichlet himself did. Fix a positive integer N. For each b ∈ [0, N ], consider a = [bα] so that bα − a ∈ [0, 1). Two of these N + 1 values must be within 1/N of each other. More precisely, there are integers b 1 and b 2 in [0, N ] with (b 2 α − a 2 ) − (b 1 α − a 1 ) ∈ [0, 1 /N ) for some integers a 1 and a 2. By taking b = |b 2 − b 1 | ∈ [1, N ], and a = ±(a 2 − a 1 ), we deduce
|bα − a| <
N
∣α^ −^
a b
∣ <^
bN
b^2
- Avoiding the question further. What exactly “good” means is questionable. We will see later that for every real number α, there are infinitely many positive integers b such that |α − (a/b)| < 1 /(
5 b^2 ) for some integer a. Furthermore, the number
5 is best possible. Perhaps then such a/b should be considered good rational approximations of α. Or maybe those are great rational approximations and we should view any rational number a/b within 1/b^2 of α or something close to that as being a good rational approximation. I didn’t really intend to define good because what’s good in general tends to depend on the individual asking the question, and whatever I tell you is good you might not believe anyway.
- Units and an example. A unit in a ring R is an element of R that has a multiplicative inverse. Let’s consider R to be the ring of algebraic integers in Q(
2). By Theorem 10,
R = Z[
2]. Let β ∈ R, so there are integers a and b such that β = a + b
- We suppose β is a unit in R. Then 1 β
a + b
a − b
a^2 − 2 b^2
∈ Z[
2].
Thus, (a^2 − 2 b^2 )|a and (a^2 − 2 b^2 )|b (use the uniqueness of the representation x + y
2 in Q(
- where x and y are rational). Let d = gcd(a, b). Then d^2 |(a^2 − 2 b^2 ) implies that d^2 |a and d^2 |b. But this means d^2 ≤ d. We deduce that d = 1. On the other hand, a^2 − 2 b^2 is a common divisor of a and b. It follows that a^2 − 2 b^2 = ±1. Remember this for later. If β = a + b
2 is a unit in R, then a^2 − 2 b^2 = ±1. The converse is easily seen to be true as well. We consider now the case when a and b are positive (the other solutions of a^2 − 2 b^2 = ±1 can be obtained from these). We obtain that (^) ( a b
a b
b^2
so that (^) ∣ ∣ ∣ ∣
a b
∣ =^
a b
b^2
2 b^2
We see then that a/b is in some sense a good rational approximation of
- It is actually better than indicated here. To see this note that the above implies
a b
2 b^2
and we get
∣ ∣ ∣ ∣
a b
∣ =^
a b
b^2
b^2
- 12132 · · · × b^2
2 b^2
We can repeat the above to show that a/b is still better than this suggests. One more time around, in fact, gives that |
2 − (a/b)| < 1 /(
5 b^2 ), a good approximation indeed.
2]. We now show how one can determine the complete set of units in Z[
2]. We begin with a lemma that basically asserts that the units in any ring form a mulitplicative group.
Lemma. Let 1 and 2 be units in a ring R. Then 1 2 and 1 − 2 1 are also units in R.
Proof. Use that 1 2 and − 2 1 − 1 1 are in R and their product is 1, and 1 − 2 1 and 2 − 1 1 are in R and their product is 1.
Comment: Let u = 1 +
- Since u(−1 +
- = 1, u is a unit in R = Z[
2]. Clearly, −1 is a unit in R as well. By the lemma, ±un^ is a unit in R for every n ∈ Z. In fact, we show the following:
Theorem 11. The units in Z[
2] are precisely the numbers of the form ±(1+
2)n^ where n ∈ Z.
Proof. By the comment, it suffices to show that Z[
2] contains no more units than those indicated by the Theorem. Let u = 1 +
- We first show that u is the only unit in (1, u]. Let = a + b
2 be a unit in (1, u] where a and b are in Z. As seen before, we have a^2 − 2 b^2 = ±1. From a + b
2 > 1 and 1 = |a^2 − 2 b^2 | = |a − b
2 ||a + b
2 |, we deduce − 1 < a − b
2 < 1. Since 1 < a + b
2, we obtain 0 < 2 a ≤ 2 +
- Hence, a = 1. Now, 1 < 1 + b
2 implies b = 1, so = u. Now, suppose is an arbitrary unit in Z[
2]. Clearly ∈ R. Note that is a unit if and only if − is, so we may restrict our attention to > 0 and do so. Let n ∈ Z with ∈ (un−^1 , un]. By the lemma, u−(n−1)^ is a unit. Also, u−(n−1)^ is in (1, u]. Hence, by the above, u−(n−1)^ = u so that = un, completing the proof.
- A corollary. As a consequence of the above discussion, we have the
Corollary. The solutions of x^2 − 2 y^2 = ± 1 in integers x and y are determined by the equation x + y
2)n^ where n ∈ Z.
Observe that [q 0 , q 1 ,... , qk, qk+1] = [q 0 , q 1 ,... , qk− 1 , qk + (1/qk+1)], and the induction hy- pothesis applies to the latter. Hence, the latter is a′/b′. On the other hand,
ak+1 = akqk+1 + ak− 1 = (ak− 1 qk + ak− 2 )qk+1 + ak− 1
and the analogous result for bk+1 imply that a′^ = ak+1/qk+1 and b′^ = bk+1/qk+1. Hence, ak+1/bk+1 = a′/b′^ = [q 0 , q 1 ,... , qk, qk+1]. One checks directly that the last equation in the theorem holds for j = −1. Suppose it holds for some j = k ≥ −1. Then
ak+1bk − akbk+1 = (akqk+1 + ak− 1 )bk − ak(bkqk+1 + bk− 1 ) = −(akbk− 1 − ak− 1 bk) = (−1)k+2,
from which the result follows.
- Simple Continued Fractions. If q 0 is an integer and q 1 , q 2 ,... are positive integers, then the continued fraction [q 0 , q 1 ,... ] is called a simple continued fraction. Throughout this section, we make use of the notation made in Theorem 12. Our main goal here is to show that simple continued fractions with infinitely many partial quotients converge (see Theorem 18). In each of the results stated below, we clarify however if the statments hold for continued fractions in general or if the statments hold specifically for simple continued fractions.
Theorem 13. For simple continued fractions, the numbers aj and bj are relatively prime integers.
Proof. This follows from aj bj− 1 − aj− 1 bj = (−1)j+1^ for j ≥ −1.
Theorem 14. For simple continued fractions, the numbers bj satisfy bj ≥ j for all j ≥ 0.
Proof. One checks directly that bj ≥ j for j = 0 and j = 1. In fact, b 0 = 1. For j > 1, the result follows by induction since bj = bj− 1 qj + bj− 2 ≥ bj− 1 + 1.
Theorem 15. For continued fractions, we have
an bn
an− 1 bn− 1
(−1)n+ bnbn− 1
for all n ≥ 1
and an bn
an− 2 bn− 2
(−1)nqn bnbn− 2
for all n ≥ 2
Proof. The first of these follows immediately from anbn− 1 − an− 1 bn = (−1)n+1^ (see The- orem 12). Also, by definition, bn − bn− 2 = qnbn− 1. Thus,
an bn
an− 2 bn− 2
an bn
an− 1 bn− 1
an− 1 bn− 1
an− 2 bn− 2
(−1)n+ bnbn− 1
(−1)n bn− 1 bn− 2
(−1)n bn− 1
bn − bn− 2 bnbn− 2
(−1)nqnbn− 1 bnbn− 1 bn− 2
(−1)nqn bnbn− 2
Theorem 16. For continued fractions, the convergents c 2 n strictly increase for n ≥ 0 and the convergents c 2 n+1 strictly decrease for n ≥ 0.
Proof. This follows immediately from the second equation in Theorem 15.
Theorem 17. For continued fractions, if n and m are ≥ 0 , then c 2 m+1 > c 2 n.
Proof. The first equation in Theorem 15 implies that c 2 n− 1 > c 2 n if n ≥ 1 and that c 2 m+1 > c 2 m. If m ≤ n − 1 (so n ≥ 1), then we use Theorem 16 to obtain c 2 m+1 ≥ c 2 n− 1 > c 2 n. If m ≥ n, then we use Theorem 16 to obtain c 2 m+1 > c 2 m ≥ c 2 n.
Theorem 18. For simple continued fractions containing infinitely many partial quotients, lim n→∞ cn exists.
Proof. By Theorems 16 and 17, the convergents c 2 n are increasing and bounded above by c 1. Hence, lim n→∞
c 2 n exists. Call this limit L. Consider an arbitrary ε > 0. Let N be a
positive integer such that if k ≥ N , then |c 2 k − L| < ε/2 and 2k(2k + 1) > 2 /ε. If n is an integer ≥ 2 N , then either n = 2k with k ≥ N and |cn − L| < ε/ 2 < ε or n = 2k + 1 with k ≥ N and (using Theorems 15 and 14)
|cn − L| ≤ |c 2 k+1 − c 2 k| + |c 2 k − L| <
b 2 kb 2 k+
ε 2
2 k(2k + 1)
ε 2
ε 2
ε 2
= ε.
The result follows. (Alternatively, one can use that lim n→∞
c 2 n and lim n→∞
c 2 n+1 both exist
and that lim n→∞ (c 2 n+1 − c 2 n) = 0.)
Comment: It is apparent that the convergents c 2 n increase to L = lim n→∞ cn and that the
convergents c 2 n+1 decrease to L.
- Is every real number α the value of a simple continued fraction? The answer is, “Yes,” and in fact every real number that is not rational has a unique representation as a simple continued fraction. We establish this next as well as a clarification of what happens in the case that α ∈ Q. By Theorem 18, we know that every simple continued fraction [q 0 , q 1 ,... ] has some unique value α. We will write α = [q 0 , q 1 ,... ]. In what follows, qj denotes an integer with qj > 0 for j > 0; however, q′ j will denote real numbers (which are possibly not integral) with q j′ > 0 if j > 0.
Lemma 1. Suppose α = [q 0 , q 1 ,... ]. If α 6 ∈ Q, then q 0 = [α]. If α ∈ Q, then either q 0 = [α] or α = [q 0 , 1] (with no further partial quotients existing).
Proof. The situation is clear if only one partial quotient exists. Suppose there are more. We have already seen that c 2 n ≤ α ≤ c 2 n+1 for all n so that, in particular, q 0 ≤ α ≤ [q 0 , q 1 ]. Since q 1 ≥ 1, we have q 0 ≤ α ≤ q 0 + 1 with α = q 0 + 1 only in the case that q 1 = 1 and no further partial quotients exist. The result follows.
Lemma 2. Define aj and bj for j ∈ {− 2 , − 1 , 0 ,... , n − 1 } as in Theorem 12 with the partial quotients q 0 , q 1 ,... , qn− 1. Then
α = [q 0 , q 1 ,... , qn− 1 , q′ n] ⇐⇒ q n′ =
an− 2 − bn− 2 α bn− 1 α − an− 1
[q 0 , q 1 ,... , qn− 1 , m − 1 , 1]. Otherwise, write q′ n = a/b with a and b relatively prime positive integers and a > b. By the division algorithm, there exist a positive integer qn and a remainder r ∈ (0, b) such that a = bqn + r. We get a/b = [qn, q′ n+1] with q n′+1 = b/r > 1. We continue as before with the one difference that we stop when some q′ n is an integer. Note also, however, that we just showed that if q n′ = a/b 6 ∈ Z, then q′ n+1 can be expressed as a rational number with a positive denominator strictly less than b. This implies that eventually, for some n, we will have q′ n ∈ Z. As we have just seen, this will give us two representations of α as a simple continued fraction. The argument that these representations are the only such representations follows like the uniqueness argument in the case that α 6 ∈ Q (the difference being in the use of Lemma 1). Summarizing, we have the following two theorems:
Theorem 19. If α ∈ R − Q, then α has a unique representation as a simple continued fraction.
Theorem 20. The simple continued fraction representation for α ∈ R is finite if and only if α ∈ Q. If α ∈ Q, then there are unique integers q 0 , q 1 ,... , qn with qj > 0 for j > 0 such that α = [q 0 , q 1 ,... , qn] = [q 0 , q 1 ,... , qn − 1 , 1]. In particular, we may arrange for the simple continued fraction representation for α ∈ Q to have an even or an odd number of partial quotients (whichever we choose).
Comments and Examples: The numbers q n′ above are called the complete quotients for the simple continued fraction α = [q 0 , q 1 ,... ]. It is also appropriate here to consider q 0 ′ = α. As examples of the above material, derive the simple continued fraction representations for 10/7 and
Homework:
(1) Compute the simple continued fraction representation for
(2) Compute the simple continued fraction representation for
n^2 + 1 where n is a positive integer.
(3) Let α be the positive real root of x^3 − x − 1. There is only one such root by Descartes’ Rule of Signs. Calculate the first 3 partial quotients q 0 , q 1 , and q 2 of the simple continued fraction representation for α as follows (yes, you must do it this way to get credit). First, calculate q 0 by using the Intermediate Value Theorem. Then find a polynomial with q′ 1 as a root. Then calculate q 1 by using the Intermediate Value Theorem and find a polynomial with q 2 ′ as a root. Finally, use the Intermediate Value Theorem to obtain q 2. Show your work.
- Good approximations by simple continued fractions. Throughout this section a and b will denote integers. We will refer to “the” simple continued fraction for α somewhat inappropriately given the content of Theorem 20 (there are actually two simple continued fraction representations for α if α ∈ Q).
Theorem 21. Let α ∈ R. If a/b is a convergent of the simple continued fraction for α
with gcd(a, b) = 1, then (^) ∣ ∣ ∣α^ −^
a b
∣ ≤^
b^2
Proof. Let n ≥ 0 be such that an = a and bn = b. If α = an/bn, then the result is clear. Otherwise, there is a further convergent an+1/bn+1 of the simple continued fraction for α. Since an+1/bn+1 and an/bn are on opposite sides of α on the number line, we obtain from Theorem 15 that (^) ∣ ∣ ∣α^ −^
a b
an+ bn+
an bn
∣ ≤^
bnbn+
b^2 n
b^2
completing the proof.
Theorem 22. Let α ∈ R. For every two consecutive convergents of the simple continued fraction for α, one of the convergents, say a/b with gcd(a, b) = 1, satisfies
∣ ∣ ∣α^ −^
a b
∣ ≤^
2 b^2
Proof. Let an/bn and an+1/bn+1 be two consecutive convergents of the simple continued fraction for α. Assume that ∣ ∣ ∣α −
an bn
2 b^2 n
and
∣α −
an+ bn+
2 b^2 n+
Since α is between an/bn and an+1/bn+1, we get from Theorem 15 that
1 bnbn+
an+ bn+
an bn
an+ bn+
− α
∣α −
an bn
2 b^2 n+
2 b^2 n
It follows that 2bn+1bn > b^2 n + b^2 n+1 so that (bn − bn+1)^2 < 0, a contradiction. The theorem follows.
Theorem 23. Let α ∈ R. Suppose that
∣ ∣ ∣α^ −^
a b
∣ ≤^
2 b^2
Then a/b is a convergent of the simpled continued fraction for α.
Proof. We may suppose that gcd(a, b) = 1 and do so. Write a/b = [q 0 , q 1 ,... , qn] where by Theorem 20 we can choose n to be either even or odd. We take n so that
α −
a b
(−1)n b^2
θ with 0 ≤ θ ≤
Let ak/bk denote the convergents of [q 0 , q 1 ,... , qn]. Note that if α = a/b, then a/b = an/bn and we’re done. Suppose now that α 6 = a/b. Define β ∈ R so that
α =
βan + an− 1 βbn + bn− 1
Theorem 24. Let m be a squarefree integer > 1 , and let R be the ring of algebraic integers in Q(
m). Suppose x + y
m ∈ R. Then x + y
m is a unit in R if and only if x^2 − my^2 = ± 1.
Proof. If x^2 − my^2 = ±1, then it is easy to check that ±(x − y
m) is in R and is the inverse of x + y
m. Hence, x^2 − my^2 = ±1 implies that x + y
m is a unit in R. Now, suppose x + y
m is a unit in R, and we want to show x^2 − my^2 = ±1. There exist rational numbers x′^ and y′^ such that x′^ + y′
m ∈ R and
(∗) 1 = (x + y
m)(x′^ + y′
m) = (xx′^ + yy′m) + (xy′^ + x′y)
m.
We obtain xx′^ + yy′m = 1 and xy′^ + x′y = 0. Solving for x and y, we deduce
x =
x′ (x′)^2 − m(y′)^2
and y =
−y′ (x′)^2 − m(y′)^2
We obtain that (x^2 − my^2 )
(x′)^2 − m(y′)^2
= 1. Note that even if x and y are not integers, x + y
m ∈ R implies that x^2 − my^2 ∈ Z; similarly, (x′)^2 − m(y′)^2 ∈ Z. Therefore, (x^2 − my^2 )
(x′)^2 − m(y′)^2
= 1 implies that x^2 − my^2 = ±1, completing the proof.
Examples: (1) Recall that
2 = [1, 2]. The convergents are 1, 3 / 2 , 7 / 5 , 17 / 12 ,.... The units a + b
2 in Z[
2] correspond to solutions of a^2 − 2 b^2 = ±1. All the convergents a/b above satisfy this equation. In fact, if a and b are positive relatively prime integers with a/b a convergent of the simple continued fraction for
2, then a^2 − 2 b^2 = ±1. More precisely, if an and bn are as in Theorem 12 with [q 0 , q 1 ,... ] = [1, 2], then a^2 n − 2 b^2 n = (−1)n+1. This follows easily by induction and the defining recursion relations for an and bn. These comments should be considered with Theorem 11 and the Corollaries to Theorem 11 and Theorem 23.
(2) Let R be the ring of algebraic integers in Q(
13). By Theorem 24, the units x + y
in R are derived from solutions to x^2 − 13 y^2 = ±1. We may suppose x and y are positive as other solutions come from replacing x with ±x and y with ±y. If x^2 − 13 y^2 = ±1 with x and y positive, we obtain
∣ ∣ ∣
x y
∣ =^
x y
∣y^2
13 y^2
2 y^2
This would imply by Theorem 23 that x/y is a convergent of the simple continued fraction for
13 except that we do not know that x and y are integral. If x and y are integers, then x/y will be a convergent. Since 13 ≡ 1 (mod 4), we might have x = x′/2 and y = y′/ 2 for some odd integers x′^ and y′. Note in this case x′/y′^ might not be a sufficiently good approximation to ensure from Theorem 23 that it is a convergent. On the other hand, even if Theorem 23 does not apply, x′/y′^ is a somewhat good approximation to
13 and it might happen that it is a convergent of the simple continued fraction for
- How can we
tell? Observe that x^2 − 13 y^2 = ±1 with x = x′/2 and y = y′/2 implies (x′)^2 −13(y′)^2 = ±4. A computation gives
13 = [3, 1 , 1 , 1 , 1 , 6] with convergents
3 1
One checks these convergents by considering 3^2 − 13 × 12 = −4, 4^2 − 13 × 12 = 3, 7^2 − 13 × 22 = −3, 11^2 − 13 × 32 = 4, 18^2 − 13 × 52 = −1, 119^2 − 13 × 332 = 4, etc. This gives us the units
, and
Past experience would lead us to consider the possibility that the smallest unit above, namely u = (3 +
13)/2, may generate all the units in R. In fact, it can be shown using methods before that the units in R are precisely the numbers of the form ±un^ where n ∈ Z.
Comment: We could probably have obtained the unit (3 +
13)/2 by trial and er- ror (without simple continued fractions); but if you doubt the usefulness of simple con- tinued fractions for this purpose, try describing the units in Z[
94] using a trial and error approach. The units are the numbers of the form ±un^ where n ∈ Z and u = 2143295 + 221064
- Hurwitz Theorem and others. We will not need the material in this section for the remainder of the course, but it is worth discussing it now that we have gone thus far.
Theorem 25 (Hurwitz). Let α ∈ R − Q. Then there exist infinitely many distinct rational numbers a/b (with a and b integers) such that
∣α^ −^
a b
∣ <^
5 b^2
Furthermore, if c >
5 , then (∗) cannot be improved by replacing
5 with c.
Proof. We show that one of every three consecutive convergents a/b of the simple continued fraction for α satisfies (∗). Assume that (∗) does not hold with (a, b) = (an− 1 , bn− 1 ) and with (a, b) = (an, bn) (where n ≥ 1). We show
bn bn− 1
bn− 1 bn
(given the assumption). In fact, this follows since α is between an− 1 /bn− 1 and an/bn so that 1 bn− 1 bn
an bn
an− 1 bn− 1
∣α −
an bn
∣α −
an− 1 bn− 1
5 b^2 n
5 b^2 n− 1
This implies (∗∗) upon noting that equality cannot hold in (∗∗) given that one side of the inequality is rational and the other is irrational. Taking x = bn/bn− 1 in (∗∗), we obtain x + x−^1 <
5 and x ≥ 1. Hence, ( x −
x −
= x^2 −
5 x + 1 < 0 ,
