Download Math 401: Homework 2 Due September 12, 2016 Solutions Exercise

Math 401: Homework 2
Due September 12, 2016
Solutions
Exercise 1.4.1: Recall that I stands for the set of irrational numbers.
(a) Show that if a, b ∈ Q then ab and a + b ∈ Q as well.
(b) Show that if a ∈ Q and t ∈ I then a + t ∈ I and if a , 0 then at ∈ I as well.
(c) Part (a) says that the rational numbers are closed under multiplication and addition.
What can be said about st and s + t when s, t ∈ I?
Solution:
(a) Suppose a, t ∈ Q. Then a = m/n and b = p/q where m, n, p, q ∈ Z and n, q , 0. But
then a + b = (mq + np)/(nq) since mq + np ∈ Z and nq ∈ Z we conclude that a + b ∈ Q.
Similarly, at = (mp)/(nq) so at ∈ Q.
(b) Suppose a + t = c where a and c are rational. Then t = c − a which, by part a, is
rational. Hence if a is rational and t is irrational, then c is irrational.
Suppose that at = c where a and c are rational and a , 0. Then t = c/a ∈ Q. So if
a ∈ Q and t ∈ I with a , 0, then at ∈ I
(c) Not much√can be√ said about
√ the product and sum of
√ irrational
√ numbers. For example,
the
sum
2
+
2
=
2
2
∈
I
by
part
b).
But
2
+
(−
2) = √
0 ∈ Q. Similarly,
√
√
√
2 · (1/
√ 2) = 1. Note that 1/ 2 is irrational, for otherwise if 1/ 2 = c is rational,
then 2 = 1/c and is also rational. So the √product
√ irrational number can be
√ of two
2,
3
and
6 are all irrational (emulate
rational. On the
other
hand,
it
is
known
that
√
√ √
√
the proof for 2). Since 6 = 2 3 it is also possible for the product of irrational
numbers to be irrational.
Exercise 1.4.2: Let A ⊆ R be nonempty and bounded above. Let s ∈ R have the property
that for all n ∈ N, s + (1/n) is an upper bound for A but s − (1/n) is not an upper bound for
A. Show that s = sup A.
Solution:
We first show that if x < s, then x is not an upper bound for A. Indeed, suppose x < s
and pick n ∈ N such that 1/n < s − x. Observe that x < s − (1/n). Since s − (1/n) is not an
upper bound for A there exists a ∈ A such that s − (1/n) < a. But then x < a as well, so x is
not an upper bound. We conclude, therefore, that if x is an upper bound for A then s ≤ a.
To show that s = sup A it remains to show that s is an upper bound for A. Indeed,
suppose x > s and pick n ∈ N such that 1/n < x − s. Hence s + (1/n) < x. Now s + (1/n)
is an upper bound for A and therefore x < A. We have therefore shown that if x > s then
x < A. Equivalently, if x ∈ A then x ≤ s.
Exercise 1.4.3: Show that ∩∞
n=1 (0, 1/n) = ∅.
Math 401: Homework 2
Due September 12, 2016
Solutions
Solution:
We need show that 0 is a lower bound, and that if b is any other lower bound, then b ≤ 0.
Clearly 0 is a lower bound since 1/n > 0 for every n ∈ N. Suppose b is any other lower
bound. We cannot have b > 0, for otherwise (by Theorem 1.4.2) there exists n ∈ N such
that 1/n < b and b is not a lower bound. Hence b ≤ 0 as required.
Exercise 1.4.4: (W) (Hand this one in to David.)
Let a < b be real numbers and let T = [a, b] ∩ Q. Show that sup T = b.
Solution:
We observe trivially that b is an upper bound for T . Thus, we need only show that if c is an
upper bound for T , then b ≤ c. Proceeding via the contrapositive, suppose c < b; we will
show that c is not an upper bound for T . Indeed, consider d = max(a, c). Since d < b, there
is a rational number q ∈ (d, b). Since (d, b) ⊆ [c, d] we know that q ∈ T . Since c ≤ d < q,
c is not an upper bound for T .
Exercise 1.4.5: Use
√ Exercise√1.4.1 to provide a proof of Corollary 1.4.4 by considering
real numbers a − 2 and b − 2.
Solution:
Suppose a < √b. We wish
√ to find an irrational number c with a < c < b. Consider the
number r such that
interval
√ (a − 2, b −
√ 2). By Theorem√1.4.3 we can find a rational
√
a − 2 < r < b − 2. Hence a < r + 2 < b. Let c = r + 2. By exercise 1.4.2, c is
the sum of a rational number and an irrational number, and is therefore irrational. Thus c
satisfies the desired properties.
Exercise Supplemental 1: Show that the sets [0, 1) and (0, 1) have the same cardinality.
Solution:
Let xn = 1/(n + 1).
Define f : [0, 1) → (0, 1) by



xn+1




f (x) = 
x1




x
x = xn
x=0
otherwise.
To show that f is surjective consider x ∈ (0, 1). Then either x = xn for some n or not. If
x , xn for any n, then (noting that x ∈ (0, 1) and is in the domain of f ) we have f (x) = x.
Otherwise either x = xn for some n > 1, or x = x1 . If x = x1 , then f (0) = x. Otherwise,
f (xn−1 ) = xn = x.
We now show that f is injective. Let A = {xn : n ∈ N}, B = {0} and C = (0, 1) \ A.
Suppose a, b ∈ A and f (a) = f (b). Then a = xn and b = xm for some n, m and
f (a) = xn+1 = 1/(n + 2) and f (b) = xm+1 = 1/(m + 2). So n + 2 = m + 2 and n = m. That
is, a = b.
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Math 401: Homework 2
Due September 12, 2016
Solutions
Suppose a, b ∈ B and f (a) = f (b). Then a, b = 0. So a = b.
Suppose a, b ∈ C and f (a) = f (b). Then f (a) = a and f (b) = b. So a = f (a) = f (b) =
b.
We have now shown that if f (a) = f (b), and if a and b are both in A or B or C, then
a = b. On the other hand, if a ∈ A and b ∈ B, then f (a) = xn for some n > 1 while
f (b) = x1 . So a , b. If a ∈ A and b ∈ C then f (a) = xn for some n > 1 and f (b) = b.
Since b ∈ C, b , xn for any n, so a , b. Finally, if a ∈ B and b ∈ C then f (a) = x1 and
f (b) = b , x1 .
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