1
Math 431 Homework 4
√
2.7. If p is a prime number, then p is not a rational number. (You may assume
the uniqueness of prime factorization.)
Suppose that p is prime. Assume we can write
√
p=
m
,
n
where m and n have no common factors. Then
m2 = pn2 ,
and m2 is divisible by p.
Suppose that the prime factorization (in increasing order) of m is
m = q1 q2 . . . qk .
Then
m2 = q 1 q 1 q 2 q 2 . . . q k q k .
Since the factorization into primes is unique, every prime factor of m2 is a prime factor
of m. In particular p is a prime factor of m, and p2 is a prime factor of m2 . Since
n2 = m2 /p,
n2 has at least one factor p. Of course this implies that n also has a factor p. We now
have a contradiction, since both m and n have the common prime factor p.
2.8. Consider the quadratic equation
x2 + bx + c = 0,
x ∈ R.
Suppose that b and c are rational, and b2 − 4c is prime. Show the equation has no
rational solutions.
By the quadratic formula, any solution of the equation has the form
√
−b ± b2 − 4c
.
x=
2
√
b2 − 4c is irrational, and it is easy to check that
The
previous
problem
shows
that
√
2
− b − 4c is irrational too. If x is rational, so is
p
2 · x + b = ± b2 − 4c.
Thus x is not rational.
2.9. Suppose that F is an ordered field. Show that if 0 < a < b, then 0 < a2 < b2 .
Since 0 < a < b, by axiom O.6 we have 0 < a2 , and since 0 < b − a, we also have
0 < a(b − a) = ab − a2 ,
2
or
0 < a2 < ab.
Since 0 < b − a, we also have
0 < b(b − a) = b2 − ab,
or
0 < a2 < ab < b2 .
2.10. a) Use the Archimedean Property O.7 to show that if a, b ∈ R and a < b,
then there is a rational number q satisfying a < q < b. (Hint: Consider the numbers
m/n where 1/n < b − a.)
By the Archimedean Property O.7 (see Lemma 2.2.4) there is a positive integer n
with 1/n < b − a. Similarly there are integers L and M such that
L/n < a < b < M/n.
Let (m − 1)/n be the largest multiple of 1/n which is less than or equal to a,
(m − 1)/n ≤ a,
Since
0<
we have
m/n > a.
1
m m−1
−
= < b − a,
n
n
n
m
m−1 1
1
=
+ ≤ a + < a + (b − a) = b.
n
n
n
n
Thus
m
< b.
n
To get strict inequality, pick another number k > 0 such that
a≤
m
1
<b− .
k
n
Then
a<
m 1
+ < b,
n
k
q=
m 1
+ <b
n
k
and
is the desired rational number.
b)Using the ideas of part a), show that for every real number x there is a sequence
of rational numbers {qk } such that
lim qk = x.
k→∞
3
Pick the numbers qk so that
x < qk < x + 1/k.
11. a) Find an example of a bounded sequence without a limit.
The sequence
xk = (−1)k
is bounded, since |xk | ≤ 1, but this sequence has no limit.
To carefully justify the last statement you can take the following approach.
If there were a limit L, then for any ǫ > 0 there would be an N such that
|xk − L| < ǫ,
k ≥ N.
This would mean, in particular, that for k ≥ N
|xk+1 − xk | = |(xk+1 − L) + (L − xk )| ≤ |xk+1 − L| + |L − xk | < 2ǫ,
k ≥ N.
But in this example we actually find that for all k
|xk+1 − xk | = 2,
so as soon as ǫ < 1 the sequence does not have the required property.
b) Find an example of a monotone sequence without a limit.
The sequence xk = k is unbounded, so cannot have a limit.
12. Find an example of a pair of sequences {xk } and {yk } such that the intervals
[xk , yk ] are nested, but there are two distinct numbers L1 and L2 satisfying xk ≤ L1 <
L2 ≤ y k .
One example is
xk = −1/k,
yk = 1 + 1/k,
k = 1, 2, 3, . . . .
Notice that for k = 1, 2, 3, . . .
xk < xk+1 < · · · < 0 < 1 < yk+1 < yk ,
so the intervals [xk , yk ] are nested. For this example we can take L1 = 0 and L2 = 1.
13. Find the least upper bound for the following sets:
a) S1 = {x ∈ R | − 2 < x < 1},
b) S2 = {x ∈ R | |x − 3| ≤ 5},
c) S3 is the set of rational numbers less than π.
d) S4 = {1 − 1/k, k = 2, 3, 4, . . . }.
I’m really only looking for the numerical answer here. The answers are: a) 1, b) 8,
c) π , d) 1.