Math 5070 Spring 2006 Homework 2: Sets and Logic
Due 1/31/06
Jan Mandel
1. 1/1: If r 6= 0 is rational and x is irrational, prove that r + x and rx are
irrational.
Solution. Suppose r + x is rational, then x = (r + x) − r is rational
because rational numbers form a field, so the difference of two rational
numbers is rational.
Suppose rx is rational, then x = (rx)/r is rational because rational numbers form a field, so the quotient of two rational numbers is rational. The
division is possible because r 6= 0 by assumption. (-1 pt for not using
r 6= 0)
2. 1/4: Let E 6= ∅ be a subset of an ordered set. Suppose α is a lower bound
of E and β is an upper bound of E. Prove that α ≤ β.
Solution. Because α is a lower bound of E,
∀x ∈ E : α ≤ x
and because β is an upper bound of E,
∀x ∈ E : β ≥ x.
Since E 6= ∅, there exists x ∈ E. Then
α ≤ x and x ≤ β
so α ≤ β. (It is OK to write α ≤ x ≤ β. -1pt for not using E 6= ∅)
3. Let A be a nonempty set of real numbers which is bounded below. Let −A
be the set of all numbers −x, where x ∈ A. Prove that inf A = − sup(−A).
Solution. Let a ∈ R. We have
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
a = − sup(−A)
−a = sup(−A) (multiplying by -1)
(∀x ∈ −A : x ≤ −a) ∧ (∀y < −a∃x ∈ −A : x > y) (from the definition of sup )
(∀ − x ∈ A : x ≤ −a) ∧ (∀y < −a∃ − x ∈ A : x > y) (from the definition of − A)
(∀x ∈ A : −x ≤ −a) ∧ (∀ − y < −a∃x ∈ A : −x > −y) (substitute − x for x and − y for y)
(∀x ∈ A : x ≥ a) ∧ (∀y > a∃x ∈ A : x < y) (multiplying the all inequalities by − 1)
a = inf A (since this is the definition of infimum)
Since A is nonempty and bounded below, inf A exists. Hence inf A =
− sup(−A). (-1 pt for not using that inf A exists.)
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(a) The assumption that A is nonempty is not needed for the equivalences
above. But all the equivalences prove is that − sup(−A) exists if and
only if inf A does, and if they do, then they equal.
(b) I have of course derived this proof by writing the definitions of a =
− sup(−A) and a = inf A and comparing them; this should be the
starting point of any attempt at the solution.
(c) Another way to structure this proof is: Since A is nonempty and
bounded below, a = inf A exists. Then
a = − sup(−A) =⇒ −a = sup(−A) (multiplying by -1)
etc., replacing all ⇐⇒ above by =⇒.
(d) A more verbal way of writing the same proof: Since A is nonempty
and bounded below, a = inf A exists. Then a is lower bound on
A and no number y > a is a lower bound on A, etc., . . . hence
a = − sup(−A).
4. 1/11 If z ∈ C prove that there exists an r ≥ 0 and w ∈ C with |w| = 1
such that z = rw. Are w and r always uniquely determined by z?
Solution.
Suppose that
z = rw, r ≥ 0, |w| = 1.
Then
|z| = |rw| = |r| |w| = r,
(1)
so r is determined uniquely by r = |z|. If z 6= 0, then dividing z = rw by
r, we get
z |z|
r
z
= = 1.
w = , |w| = =
r
r
|r|
r
So, in conclusion, if z 6= 0, then
r = |z| , w =
z
, r > 0,
r
(2)
and r and w are unique. We need to verify that (2) implies z = rw:
rw = |z|
z
z
= r = z.
r
r
If z = 0, then z = rw with r = 0 and any w ∈ C with |w| = 1; so, in this
case, w is not unique, but r is still unique by (1). (-1 pt for missing the
uniqueness of r)
5. 1/13 If x, y are complex, prove that ||x| − |y|| ≤ |x − y|.
Solution. Using the triangle inequality,
|x| − |y| ≤ |x − y|
2
because |x| + |x − y| = |x| + |y − x| ≤ |x + (y − x)| = |y|. Switching the
notation x and y,
|y| − |x| ≤ |y − x| = |x − y|
which gives
− |x − y| ≤ |x| − |y|
hence
− |x − y| ≤ |x| − |y| ≤ |x − y| ,
so ||x| − |y|| ≤ |x − y|.
6. 11/15 Under what conditions does equality hold in Schwarz inequality?
Solution. From 1.35: With
X
X
X
2
2
A=
|aj | , B =
|bj | , C =
aj bj ,
it was proved that
X
2
2
|Baj − Cbj | = B AB − |C| .
2
(3)
2
Schwarz inequality is |C| ≤ AB. Suppose that AB = |C| . First, if B =
2
0, then (bj ) = 0. Now consider the case B 6= 0. Then, from (3), AB = |C|
P
2
if and only if
|Baj − Cbj | = 0, which implies that Baj − Cbj = 0 for
all j.Dividing by B 6= 0,
C
∀j : aj = bj .
B
So, we have proved
2
AB = |C| =⇒ (bj ) = 0 ∨ ∃t ∈ C : (aj ) = t (bj ) .
(4)
Now we need to show the converse implication. If (bj ) = 0, then B = C =
2
0, so AB = |C| = 0. If (aj ) = t (bj ), then
X
X 2
2
2
2
A =
|aj | =
|t| |bj | = |t| B
X
X
X
2
C =
aj bj =
tbj bj =
t |bj | = tB
so
AB
2
|C|
2
2
= |t| BB = |t| B 2
2
2
= |tB| = |t| B 2
2
and AB = |C| . (-1pt by somehow assuming that t is real)
Note that by switching aj and bj and noting that (bj ) = 0 can be written
as (bj ) = t (aj ) with t = 0, (4) can be equivalently expressed “equality
holds if and only if one of (aj ) or (bj ) is a (complex) multiple of the other.”
Let’s prove this. (Not required, unless the result is somehow written in
this form.)
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