SOLUTIONS FOR THE FIRST PROBLEM SET
1. Show that the equation x2 + 2y 2 = 1 has infinitely many solutions where x, y ∈ Q. Give
a parametric description of the set of solutions.
SOLUTION: We will use the secant-line method. The point Po = (−1, 0) is on the curve
C : x2 + 2y 2 = 1. Let m ∈ Q. The line Lm of slope m going through the point Po is given
by the equation y = mx + m. To find the intersection between Lm and C, we substitute
y = mx + m into the equation for C, obtaining an equation for the x-coordinates of the
points of intersection. This equation is x2 + 2(mx + m)2 = 1. It simplifies to
(1 + 2m2 )x2 + 4m2 x + 2m2 − 1 = 0
This gives the equation x2 +
4m2
x
1+2m2
+
2m2 −1
2m2 +1
= 0.
This quadratic equation has two roots, say r1 and r2 . One root of this quadratic equation
2 −1
. Thus the
must be r1 = −1. The product of the two roots is given by r1 r2 = 2m
2m2 +1
1−2m2
second root must be x = r2 = 1+2m2 . For this x, the corresponding value of y is given by
1−2m2
2m
y = m 1+2m
2 + m = 1+2m2 . Hence, the corresponding point of intersection between Lm and
1−2m2
2m
C, which we will denote by Pm , is given by Pm = 1+2m2 , 1+2m2 .
Note that this point Pm is indeed in C(Q) (since m ∈ Q) and that Pm cannot be the same
2m
as Po = (−1, 0). Indeed, if Pm = Po , then 1+2m
2 = 0 and this implies that m = 0. But then
Pm = P0 = (1, 0) 6= Po .
Suppose that m, m0 ∈ Q and m0 6= m. It is clear that the points Pm and Pm0 will be
different simply because the two lines Lm and Lm0 are different and hence have only one
point in common, namely Po . Since Q is infinite, it follows that we must have infinitely
many different points in C(Q).
The parametric description for C(Q) − {(−1, 0)} is given by a function F defined by
2m
1 − 2m2
,
.
F (m) =
1 + 2m2 1 + 2m2
We have already explained why F is injective. The fact that F is surjective follows by noting
that if P is any point in C(Q) − {(−1, 0)}, then there will be a unique line containing both
P and Po . Since both P and Po have rational coordinates, the slope of this line must be
a rational number m. Thus, that line must be Lm and P will just be the second point of
intersection between Lm and C. That is, P = Pm = F (m) for some m ∈ Q.
1
2. By using the “secant line method,” we have given a description of the set of solutions to
the equation x2 − 2y 2 = 1 in terms of a parameter m. Describe the values of m which give a
solution satisfying x > 0 and y > 0. Answer the same question for solutions satisfying x < 0
and y > 0.
SOLUTION: The set of solutions to the equation x2 − 2y 2 = 1, where x, y ∈ Q, was derived
in class. Here is a description of that set
2
2m
2m + 1
,
m∈Q
∪ { (1, 0) }
S=
2m2 − 1 2m2 − 1 To find the set of points where x > 0, y > 0, we use this description. The point (1, 0) is not
in that set. Note that 2m2 + 1 > 0 for all m ∈ Q. The following inequalities are equivalent.
1
1
2m2 + 1
> 0 ⇐⇒ 2m2 − 1 > 0 ⇐⇒ 2m2 > 1 ⇐⇒ m2 >
⇐⇒ |m| > √
2
2m − 1
2
2
Assume that these equivalent inequalities hold. Then
2m
> 0 ⇐⇒ m > 0
2m2 − 1
Therefore, the points in S satisfying x > 0, y > 0 are the points corresponding the values
of m satisfying m > √12 .
To answer the same question for points in S satisfying x < 0, y > 0, first note that
1
2m2 + 1
< 0 ⇐⇒ 2m2 − 1 < 0 ⇐⇒ 2m2 < 1 ⇐⇒ m2 <
2
2m − 1
2
1
⇐⇒ |m| < √
2
Assuming that these equivalent inequalities hold, we have
2m
> 0 ⇐⇒ m < 0.
2m2 − 1
Therefore, the points in S satisfying x < 0, y > 0 are the points corresponding the values
of m satisfying − √12 < m < 0.
3. Assume that x, y ∈ Z satisfy the equation x2 − 2y 2 = 1 and that x > 0, y > 0. Prove the
following inequalities.
√
1
x
2 − < √1
√
<
y
(2 2 + 12 )y 2
2 2y 2
2
√
√
SOLUTION: The given equation can be rewritten in factored form (x − 2y)(x + 2y) = 1.
Dividing both sides by y 2 gives
x √
x √
1
− 2
+ 2
= 2
y
y
y
This gives
(a)
x √
− 2 =
y
x
y
1
√ + 2 y2
Note that since x > 0 and y > 0, the right
√ hand side of (a) is clearly positive. Hence so is
x
the left hand side. This implies that y > 2. Therefore
√
x √
+ 2 > 2 2
y
It follows that
(b)
x
y
1
1
√ < √
+ 2 y2
2 2y 2
Combining (a) and (b), we obtain one of the stated inequalities, namely
√
√
2 − x = x − 2 < √1
y
y
2 2y 2
To get the other inequality (the lower bound), first note that the inequality just proved
implies that
√
x
1
< 2+ √
y
2 2y 2
√ 2
√
But y ≥ 1 implies that 2 2y ≥ 2 2 > 2 and hence that 2√12y2 < 12 . It follows that
√
x
1
< 2+
y
2
This implies that
√
2 < 2 2 + 12 . If we use this together with (a), we obtain
√
√
1
x
1
2− = x − 2 =
√ √
>
x
1
2
y
y
+
2
y
2
2
+
y2
y
2
x
y
+
√
3
√
That is, | 2 − xy | >
inequalities.
√ 1
,
(2 2+ 12 )y 2
which is the other stated inequality. We have proved both
4. Find a rational number r such that |
√
2 − r |< 10−6 without using a calculator.
SOLUTION: As in class, let T = { (x, y) | x2 −2y 2 = 1, x, y ∈ Z }. According to √
problem 3, it
suffices to let r = xy , where (x, y) ∈ T , x, y > 0, and y satisfies the inequality 2 2y 2 > 106 ,.
In fact, it is clearly sufficient that y satisfy the weaker inequality y 2 > 106 . In class, we
described the following method for generating infinitely many points in T : If (a, b) is such
a point, we obtain another point (3a + 4b, 2a + 3b). Thus, we obtain the following points
starting with Q1 = (3, 2).
Q2 = (17, 12),
Q3 = (99, 70),
Q4 = (477, 408),
Q5 = (3363, 2378)
The calculations can easily be done without a calculator. Thus, since 2378 > 103 , we can
.
take r = 3363
2378
5. Let s be a fixed rational number. Prove that there is a constant c > 0, depending only
on s, such that
x
s − > 1
y
cy
x
for all rational numbers y 6= s. (Here x, y ∈ Z and y > 0.)
SOLUTION: Suppose that xy ∈ Q and that xy
assume that b ≥ 1. Now
x
s − = a −
b
y
6= s. Write s in reduced form s = ab . We can
ay − bx x = y
by Since xy 6= s by assumption, the above quantities are nonzero. Hence ay − bx 6= 0. Also,
clearly ay − bx ∈ Z. Therefore, | ay − bx | ≥ 1. It follows that
s − x ≥ 1 = 1 > 1
by y
by
cy
where c is any number greater than b. For example, we could take c = b + 1.
6. Let s be a fixed rational number. Prove that there are only finitely many rational numbers
x
(in reduced form) such that
y
s − x < 1
y
y2
4
SOLUTION: We may assume that xy 6= s since that assumption excludes only one rational
number. We may therefore use the inequality in problem 5. Thus, if xy satisfies the inequality
in this question, then we have
x 1
1
< s − < 2
cy
y
y
where c is some positive number depending only on s. This implies that
1
1
< 2
cy
y
and therefore that y 2 < cy. Since y > 0, this gives y < c. Hence we have 0 < y < c. Only
finitely many integers y exist satisfying this inequality.
For each possible value for y, the stated inequality asserts that | s − xy | < y12 . Note that
1
≤ 1. Hence xy must be in the interval s − 1 < xy < s + 1. Hence, (s − 1)y < x < (s + 1)y.
y2
Such an interval can contain only finitely many integers x. Thus, for each possible value of
y, there can only be finitely many values of x such that xy satisfies the inequality stated in
the problem. Since we have already shown that there are only finitely many possible values
of y, we have proved that there are only finitely many rational numbers xy satisfying the
inequality. This is what we wanted to prove.
(NOTE: With a slightly more careful argument, one can easily show that for each possible
y > 1, there is at most one possible value of x. The inequality asserts that xy is in a certain
open interval of length y22 .)
7. The infinite series
number.
P∞
1
n=1 23n
converges to some real number α. Prove that α is an irrational
SOLUTION: We will use the result from problem 6. Let α be a real number. The contrapositive of the statement proved in problem 6 can be stated as follows:
If there exist infinitely many rational numbers
then α is an irrational number.
x
y
(in reduced form) such that | α −
x
y
|<
1
y2
,
This gives a possible approach to proving that certain real numbers are irrational. We will
apply this approach to the number α defined in this problem.
Pk
1
For each k ≥ 1, let rk =
n=1 23n . Since the terms in the infinite series defining α are
all positive, it is clear that r1 < r2 < ... < rk < rk+1 < .... That is, the sequence {rk } is
5
strictly increasing and so consists of distinct numbers. Also, note that for each k ≥ 1, rk is
a finite sum of rational numbers and so rk is a rational number. It is clear that a common
k
denominator for all the terms 231n where 1 ≤ n ≤ k is 23 . We can thus write rk = abkk , where
k
ak ∈ Z and bk = 23 . (This turns out to be the reduced form for rk , but we won’t need that
fact and so won’t bother to prove it.)
We want to find an upper bound for the quantity
∞ 3n
X
1
1
α − rk =
n =
3
2
2
n=k+1
n=k+1
∞
X
The terms of this series are of the form ( 12 )m , where m is a power of 3, starting with m = 3k+1 .
We can obtain an upper bound on the series by including all integral values of m ≥ 3k+1 .
That is, we have
α − rk = α − rk =
∞ 3n
X
1
<
2
n=k+1
m
∞
X
1
2
k+1
m=3
Using the summation formula for a geometric series, we obtain
m
3k+1 X
3k+1
3k+1 −1
∞
∞ m
X
1
1
1
1
1
=
·
=
·2 =
2
2
2
2
2
k+1
m=0
m=3
We have proved that
3k+1 −1
1
a
k
α − <
bk
2
The following inequality will be useful.
3k+1 − 1 = 3 · 3k − 1 = 2 · 3k + 3k − 1 > 2 · 3k
for any k ≥ 1. We have used the fact that 3k − 1 > 0 for k ≥ 1. It follows that
3k+1 −1
2·3k
1
1
1
<
= 2
2
2
bk
k
k
Here we have used the fact that bk = 23 and so b2k = 22·3 .
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In summary, we have proven that
α − rk = α − ak < 1
bk b2k
Since we haven’t verified that rk = abkk is in reduced form, we can only say that when we
write rk in reduced form, the denominator will be at most bk . That is, suppose that rk = xykk
is in reduced form. Then 1 ≤ yk ≤ bk . Hence b2k ≥ yk2 . Therefore, we have the inequality
x
k
α − rk = α − < 1 ≤ 1
yk b2k
yk2
for all values of k ≥ 1. The rational numbers rk =
xk
yk
are distinct and we have
α − xk < 1
yk yk2
Therefore, using the contrapositive form of problem 6, we can now conclude that α is indeed
an irrational number.
√
8. Compute N (α) for the following elements α ∈ Q[ 2]:
1
,
2
√
2,
1 2√
−
2,
3 5
1+
√
2,
√
5 + 3 2,
√
1+2 2
SOLUTION:
N ( 21 ) =
1
2
·
1
2
=
1
4
√
√
√
N ( 2) = 2 · (− 2) = −2
√
√
√
N ( 31 − 25 2) = ( 13 − 25 2)( 13 + 52 2) = ( 13 )2 − 2( 52 )2 =
√
√
2) = (1 + 2)(1 − 2) = 1 − 2 = −1.
√
√
√
N (5 + 3 2) = (5 + 3 2)(5 − 3 2) = 25 − 9 · 2 = 7
√
√
√
N (1 + 2 2) = (1 + 2 2)(1 − 2 2) = 1 − 4 · 2 = −7
N (1 +
√
7
1
9
−
8
25
47
= − 225
√
9. Assume that n ∈ Z and that there exists
an
element
α
∈
Z[
2] such that N (α) = n.
√
Prove that there exists
√ an element β ∈ Z[ 2] such that N (β) = −n. Prove that there exists
an element γ ∈ Z[ 2] such that N (γ) = 2n.
√
SOLUTION: We will use some√of the results from√problem 8. Suppose that α ∈ Z[ 2] and
that N (α) = n. Let β = (1 + 2)α. Then β ∈ Z[ 2] and
√
N (β) = N (1 + 2)N (α) = (−1)n = −n
√
√
Also, let γ = 2β. Then γ ∈ Z[ 2] and
√
N (γ) = N ( 2)N (β) = (−2)(−n) = 2n.
8