SOLUTIONS TO HOMEWORK 6
1. Suppose that
Q(x, y) = Ax2 + Bxy + Cy 2
is a binary integral quadratic form. Recall from class that the discriminant of Q is
defined to be
D = B 2 − 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 2 .
1(b). Suppose that A and C are both non-zero, (x, y) 6= (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 2 and, since C 6= 0, we must have y = 0,
a contradiction, hence x 6= 0. The same argument shows y 6= 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) 6= (0, 0), but Q(x, y) = 0.
Prove that (2Ax + By)/y is a (rational) square root of D. Explain why it follows
that D must be a perfect square.
Ans: We are given 0 = Ax2 + Bxy + Cy 2 , which implies
y 2 = (−Ax2 − Bxy)/C.
We calculate
2Ax + By
y
2
4A2 x2 + 4ABxy + B 2 y 2
y2
(4A2 x2 + 4ABxy)C
=
+ B2
−Ax2 − Bxy
4AC(Ax2 + Bxy)
+ B2
=
(−Ax2 − Bxy)
=
= B 2 − 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: Fixing a value of y, by the quadratic equation, the solutions for x have the
form
p
−By ± B 2 y 2 − 4ACy 2
x=
.
2A
Of course, we are only interested in non-zero integer solutions. Let y = 2A (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, 2A).
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) = A0 u2 + B 0 uv,
with B 0 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 2 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 Q0 (u, v) = A0 u2 + B 0 uv + C 0 v 2 . We can determine C 0 by evaluating Q0 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 = 0.
Now let’s determine the discriminant of Q0 . In general, assume Q0 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 0
denote the matrices associated to Q and Q0 respectively. Write v T for the transpose
of the vector (or matrix) v. Then we have
Q0 (v) = Q(M v) = (M v)T N M v = v T M T N M v.
Hence, N 0 = M T N M , and so
det(N 0 ) = det(M T ) det(N ) det(M ) = det(M )2 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 0 ). Finally, note that −4 det(N ) is
exactly the discriminant of Q. Hence, our quadratic forms have the same discriminant. Because C 0 = 0, there is no 4A0 C 0 term in the formula for the discriminant
for Q0 , and so (B 0 )2 = D.
2. 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 2 .
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
√
ξ0 = (1 + 5)/2,
the larger root of the equation
f (x) = x2 − 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) = p2 − pq − q 2 .
2(a). Calculate the discriminant of the quadratic form Q. Explain why Q is
indefinite and does not represent zero.
Ans: The discriminant is B 2 − 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.)
2(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 2, because associated to each line
segment is a pair of numbers, and in our case, every pair is (−1, 1).)
2(c). Show that if p and q are any integers with q not zero, then
|f (p/q)| ≥ 1/q 2 .
2
2
|
Ans: We have |f (p/q)| = |p −pq−q
, and as we know that the quadratic form Q
q2
does not represent zero, we have that the numerator is ≥ 1.
2(d). Suppose that
1/2 ≤ x ≤ ξ0 .
Prove that
Deduce that
√
|f (x)| ≤ (ξ0 − x)(2ξ0 − 1) = (ξ0 − x)( 5).
√
|ξ0 − x| ≥ |f (x)|/ 5.
Ans: We know the roots of√f , and hence we know its factorization into linear
pieces: f (x) = (x − ξ0 )(x − 1−2 5 ). In the specified range for x, we have |(x − ξ0 )| =
√
(ξ0 − x), and so to prove the first inequality it suffices to show that |x − 1−2 5 | ≤
√
(2ξ0 − 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, |ξ0 −
√ x| = ξ0 − x. Hence, to deduce the
bottom inequality, we can simply divide by 5.
2(e). Suppose that p/q is a rational approximation to ξ0 , and that p/q < ξ0 . Prove
that
√
|ξ0 − p/q| > 1/ 5q 2 .
4
SOLUTIONS TO HOMEWORK 6
Ans: First, note that for any (rational or not) x < 1/2,
√
√
|ξ0 − x| > 5/2 > 1/ 5q 2
for any non-zero integer q. Hence, the assertion is automatically true for p/q < 1/2.
For 1/2 ≤ p/q < ξ0 , we can combine the results of 2c and 2d:
√
√
|ξ0 − p/q| ≥ |f (p/q)|/ 5 ≥ 1/ 5q 2 .
It remains only to move from ≥ to >. In the course of proving √
2d, we saw that for
√
√
x in the given range, |f (x)| = (ξ0 − x)( 5) if and only if (x − 1−2 5 ) = 5. But this
does not happen in our case because (worse reason) the problem specified x < ξ0
or (better reason) ξ0 is not rational.
Remark. This problem says that ξ0 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 2 in place of 1/q 2 . (It’s also true that you can’t get better-than-1/ 5q 2
approximations from above,
√ but the inequalities are not quite so simple as in 2(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 2 .
Do you see why this doesn’t contradict 2(e)?
3. Write down a specific irrational number ξ1 with the property that for every
positive integer k, there are infinitely many rational numbers p/q such that
|ξ1 − 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
ξ1 ; although the ideas behind my solution are exactly the same, there is another ξ1
which looks a little less intimidating. I’ll describe it briefly in the last paragraph.
Note it suffices to find an irrational number ξ1 with the property that for every
positive integer k, there is at least one rational number p/q such that
|ξ1 − p/q| < 1/q k ,
because knowing one such rational approximation for some k, we also know a rational approximation for each smaller k1 , because |ξ1 − p/q| < 1/q k < 1/q k1 . (Of
course, it could happen that for k1 < k we find the same rational approximation,
and thus cannot count these as distinct rational approximations relative to k1 . But
this cannot happen for infinitely many values of k, because |ξ1 − p/q| cannot be
less than 1/q k for infinitely many values of k, because ξ1 is irrational and hence
|ξ1 − p/q| is non-zero.)
We’re about to write down an ugly formula for ξ1 and then prove that it works.
In this paragraph, though, we’ll give the motivation for that ugly formula. We
want a decimal expansion .a1 a2 a3 a4 . . . so that the first n1 digits approximate it to
better than 2 ∗ n1 digits of accuracy, so that the first n2 digits approximate it to
better than 3 ∗ n2 digits of accuracy, and so on. (How we choose the value ni 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 .10a3 a4 . . . . Next take a3 = 1. So we
want .101 to approximate our number better than 1/103∗3 . So, put a10 = 1, and
SOLUTIONS TO HOMEWORK 6
5
ai = 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
X
k!
i=1
i!
,
because this satisfies the case k = 1 and the recursive relationship.
Now we’re ready to actually write down ξ1 . Put
∞
X
ξ1 =
10−ϕ(j) .
j=1
Of course, that’s exactly the number .a1 a2 . . . that we were describing above. We
claim that the rational approximation
k−1
X
10−ϕ(j) =: p/10ϕ(k−1)
j=1
satisfies
|ξ1 − p/10ϕ(k−1) | < 1/(10ϕ(k−1) )k .
From the construction of ξ1 , we see that the error is strictly between 1/10ϕ(k) and
1/10ϕ(k)−1 . So, it remains to show that 1/10ϕ(k)−1 (the upper-bound for the error)
is less than 1/(10ϕ(k−1) )k . Is (10ϕ(k−1) )k ≤ 10ϕ(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
1/10ϕ(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 .a1 a2 a3 a4 . . . so that the first n1 digits approximate it to better than
2 ∗ n1 digits of accuracy, so that the first n2 digits approximate it to better than
3 ∗ n2 digits of accuracy, and so on. Instead, we could ask that the first n1 digits
approximate it to better than 1 ∗ n1 digits of accuracy, the first n2 to better than
2 ∗ n2 digits of accuracy, and so on. It’s easy to see that nk = k! works. So, another
answer is
∞
X
ξ1 =
10−j! = .11000100 . . . .
j=1