# SOLUTIONS TO HOMEWORK 6

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1 SOLUTIONS TO HOMEWORK 6 1. Suppose that Q(x, y) = Ax + Bxy + Cy is a binary integral quadratic form. Recall from class that the discriminant of Q is defined to be D = B 4AC. The form Q is said to represent zero if there are integers x and y, not both zero, such that Q(x, y) = 0. 1(a). Suppose that either A or C is zero. Show that then Q represents zero, and the discriminant is a perfect square. Ans: If A = 0, then Q(x, 0) = 0 for all x. If C = 0, then Q(0, y) = 0 for all y. In either case, the discriminant is B. 1(b). Suppose that A and C are both non-zero, (x, y) (0, 0), but Q(x, y) = 0. Explain why x and y are both non-zero. Ans: Assume x = 0. Then Q(0, y) = Cy and, since C 0, we must have y = 0, a contradiction, hence x 0. The same argument shows y 0: just switch the roles of x and y and A and C. 1(c). Suppose that A and C are both non-zero, (x, y) (0, 0), but Q(x, y) = 0. Prove that (Ax + By)/y is a (rational) square root of D. Explain why it follows that D must be a perfect square. Ans: We calculate We are given 0 = Ax + Bxy + Cy, which implies y = ( Ax Bxy)/C. ( ) Ax + By = 4A x + 4ABxy + B y y y = (4A x + 4ABxy)C Ax + B Bxy = 4AC(Ax + Bxy) ( Ax + B Bxy) = B 4AC = D Hence, the given term is a rational square root of D. By unique factorization into primes, the square of a prime factor in the denominator can only be killed by the square of the same prime factor in the numerator. Our rational number squared having one in the denominator thus implies it was an integer to begin with. 1(d). Suppose that A and C are both non-zero, and that D is a perfect square. Prove that Q represents zero. 1

2 SOLUTIONS TO HOMEWORK 6 Ans: form Fixing a value of y, by the quadratic equation, the solutions for x have the x = By ± B y 4ACy. A Of course, we are only interested in non-zero integer solutions. Let y = A (a non-zero integer). Then x = B ± D. So, since D is a perfect square, we have found two solutions (unless D = 0, in which case we ve found one solution): ( B ± D, A). 1(e). Suppose that Q represents zero. Prove that there is a new coordinate system u = px + qy, v = rx + sy (with p, q, r, s integers satisfying ps qr = 1) so that in the new coordinates Q(u, v) = A u + B uv, with B a square root of D. (Hint: you can quote results from the exercises for section 5.7. This problem is still a bit more than I should really expect you to be able to do, so don t lose sleep over it.) Ans: Let (r, s) be a zero of Q(x, y). We can assume that r and s are relatively prime, because Q(kx, ky) = k Q(x, y), and so (x/ gcd(x, y), y/ gcd(x, y)) is also a zero of Q. Because r and s are relatively prime, we can use the Euclidean algorithm to find p and q such that ps qr = 1. Now, change coordinates to u = px + qy and v = rx + sy. Say the quadratic form corresponding to this new basis is Q (u, v) = A u + B uv + C v. We can determine C by evaluating Q at (0, 1), which is the same as evaluating our original quadratic form Q at (r, s), and we find by construction that C = 0. Now let s determine the discriminant of Q. In general, assume Q is the quadratic form Q after a basis change sending v to M 1 v where M is an invertible two-by-two matrix with integer terms. Recall the correspondence between quadratic forms and certain matrices, given on page 89 of the textbook. Let N and N denote the matrices associated to Q and Q respectively. Write v T for the transpose of the vector (or matrix) v. Then we have Hence, N = M T NM, and so Q (v) = Q(Mv) = (Mv) T NMv = v T M T NMv. det(n ) = det(m T ) det(n) det(m) = det(m) det(n). As our matrix M was invertible, had integer terms, and had an inverse with integer terms, det(m) = ±1. Thus, det(n) = det(n ). Finally, note that 4 det(n) is exactly the discriminant of Q. Hence, our quadratic forms have the same discriminant. Because C = 0, there is no 4A C term in the formula for the discriminant for Q, and so (B ) = D.. I proved in class Dirichlet s theorem that if ξ is any irrational number, then there are infinitely many rational numbers p/q so that ξ p/q < 1/q.

3 SOLUTIONS TO HOMEWORK 6 3 The point of this problem is to show that you can t do much better than this in general. Define ξ 0 = (1 + 5)/, the larger root of the equation f(x) = x x 1. This is the Golden Ratio, about which you can read in art history classes as well as in mathematics. I ll talk about this equation using the binary quadratic form Q(p, q) = p pq q. (a). Calculate the discriminant of the quadratic form Q. Explain why Q is indefinite and does not represent zero. Ans: The discriminant is B 4AC = 1 4( 1) = 5. By part (c) of problem 1, Q does not represent zero. Q(0, 1) = 1 and Q(1, 0) = 1, hence the form is indefinite. (The definition of indefinite is given on p. 96 of the textbook.) (b). List the pairs of numbers appearing on successive edges of John Conway s river for the quadratic form Q. (The sequence of pairs eventually repeats; you can list the pairs in one period, then put a bar over it.) Ans: The river is particularly simple in this case, with 1 always appearing immediately to the south side, and 1 always appearing immediately to the north side. (See the.jpg image linked to on the website.) In this case, the pairs of numbers are ( 1, 1). (Note: That has period 1, not period, because associated to each line segment is a pair of numbers, and in our case, every pair is ( 1, 1).) (c). Show that if p and q are any integers with q not zero, then f(p/q) 1/q. Ans: We have f(p/q) = p pq q q, and as we know that the quadratic form Q does not represent zero, we have that the numerator is 1. (d). Suppose that Prove that Deduce that 1/ x ξ 0. f(x) (ξ 0 x)(ξ 0 1) = (ξ 0 x)( 5). ξ 0 x f(x) / 5. Ans: We know the roots of f, and hence we know its factorization into linear pieces: f(x) = (x ξ 0 )(x 1 5 ). In the specified range for x, we have (x ξ 0 ) = (ξ 0 x), and so to prove the first inequality it suffices to show that x 1 5 (ξ 0 1) = 5, for x in the given range. It s clear that the left hand side attains its maximum when x attains its maximum, and this maximum is 5, as desired. Note that for x in the given range, ξ 0 x = ξ 0 x. Hence, to deduce the bottom inequality, we can simply divide by 5. (e). Suppose that p/q is a rational approximation to ξ 0, and that p/q < ξ 0. Prove that ξ 0 p/q > 1/ 5q.

4 4 SOLUTIONS TO HOMEWORK 6 Ans: First, note that for any (rational or not) x < 1/, ξ 0 x > 5/ > 1/ 5q for any non-zero integer q. Hence, the assertion is automatically true for p/q < 1/. For 1/ p/q < ξ 0, we can combine the results of c and d: ξ 0 p/q f(p/q) / 5 1/ 5q. It remains only to move from to >. In the course of proving d, we saw that for x in the given range, f(x) = (ξ 0 x)( 5) if and only if (x 1 5 ) = 5. But this does not happen in our case because (worse reason) the problem specified x < ξ 0 or (better reason) ξ 0 is not rational. Remark. This problem says that ξ 0 can t be very well approximated by rational numbers from below; for example, you can t get something like Dirichlet s theorem with 1/3q in place of 1/q. (It s also true that you can t get better-than-1/ 5q approximations from above, but the inequalities are not quite so simple as in (d) above.) The number 1/ 5 is best possible: a theorem of Hurwitz says that if ξ is any irrational, then there are infinitely many rationals p/q with ξ p/q < 1/ 5q. Do you see why this doesn t contradict (e)? 3. Write down a specific irrational number ξ 1 with the property that for every positive integer k, there are infinitely many rational numbers p/q such that ξ 1 p/q < 1/q k. (Hint: taking a finite number of terms in the decimal expansion of ξ gives a rational approximation to ξ. Usually the error in this approximation is terrible (like 1/q), but sometimes it s much smaller.) Ans: Warning: The solution I present here doesn t lead to the simplest possible ξ 1 ; although the ideas behind my solution are exactly the same, there is another ξ 1 which looks a little less intimidating. I ll describe it briefly in the last paragraph. Note it suffices to find an irrational number ξ 1 with the property that for every positive integer k, there is at least one rational number p/q such that ξ 1 p/q < 1/q k, because knowing one such rational approximation for some k, we also know a rational approximation for each smaller k 1, because ξ 1 p/q < 1/q k < 1/q k1. (Of course, it could happen that for k 1 < k we find the same rational approximation, and thus cannot count these as distinct rational approximations relative to k 1. But this cannot happen for infinitely many values of k, because ξ 1 p/q cannot be less than 1/q k for infinitely many values of k, because ξ 1 is irrational and hence ξ 1 p/q is non-zero.) We re about to write down an ugly formula for ξ 1 and then prove that it works. In this paragraph, though, we ll give the motivation for that ugly formula. We want a decimal expansion.a 1 a a 3 a 4... so that the first n 1 digits approximate it to better than n 1 digits of accuracy, so that the first n digits approximate it to better than 3 n digits of accuracy, and so on. (How we choose the value n i for each i is a vital step.) Start with.1 = 1/10. We want that approximation to be better than.01, so our number must start.10a 3 a Next take a 3 = 1. So we want.101 to approximate our number better than 1/ So, put a 10 = 1, and

5 SOLUTIONS TO HOMEWORK 6 5 a i = 0 for 4 i 9. What is the pattern? The first non-zero digit is the first digit. The next non-zero digit is the third. The next is the tenth. This can be figured out inductively. If ϕ(k) denotes the place of the kth non-zero digit, then ϕ(k) = kϕ(k 1) + 1. An explicit expression is ϕ(k) = k i=1 k! i!, because this satisfies the case k = 1 and the recursive relationship. Now we re ready to actually write down ξ 1. Put ξ 1 = 10 ϕ(j). Of course, that s exactly the number.a 1 a... that we were describing above. We claim that the rational approximation k 1 10 ϕ(j) =: p/10 ϕ(k 1) satisfies ξ 1 p/10 ϕ(k 1) < 1/(10 ϕ(k 1) ) k. From the construction of ξ 1, we see that the error is strictly between 1/10 ϕ(k) and 1/10 ϕ(k) 1. So, it remains to show that 1/10 ϕ(k) 1 (the upper-bound for the error) is less than 1/(10 ϕ(k 1) ) k. Is (10 ϕ(k 1) ) k 10 ϕ(k) 1? Is kϕ(k 1) ϕ(k) 1? Yes, they are in fact equal, by the inductive definition of ϕ. (Equality is fine, because 1/10 ϕ(k) 1 was a strict upper bound for the error.) By the remarks in the first paragraph, having found a single rational approximation for each k, we re done! Alternate Answer: I will leave out the details. Above we found a decimal expansion.a 1 a a 3 a 4... so that the first n 1 digits approximate it to better than n 1 digits of accuracy, so that the first n digits approximate it to better than 3 n digits of accuracy, and so on. Instead, we could ask that the first n 1 digits approximate it to better than 1 n 1 digits of accuracy, the first n to better than n digits of accuracy, and so on. It s easy to see that n k = k! works. So, another answer is ξ 1 = 10 j! =

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