Mathematics 220
Homework for Week 4
Due February 1 or 2
1. Question 2.68. State the negation of the following quantified statements:
(a) For every rational number r, the number 1/r is rational.
(b) There exists a rational number r such that r 2 = 2.
Solution:
(a) There exists a rational number r such that 1/r is not rational.
(b) For all rational numbers r, r 2 6= 2.
2. Let P be the statement ‘∀x ∈ R ∃y ∈ R, y 2 = x’.
(a) State this in words (no notation allowed).
Solution: For every real number x there is a real number y such that the square
of y is x.
(b) Is this statement true?
Solution: No, it is false. If we take x = −1 then for any real number y,
y 2 ≥ 0 > −1 and so y 2 6= −1.
(c) State the negation in symbols.
Solution: ∃x ∈ R s.t. ∀y ∈ R, y 2 6= x.
3. Modified Question 2.70 Determine the truth value for each of the following statements.
(a) ∃x ∈ R, x2 − x = 0.
√
(b) ∀x ∈ R, x2 = x.
(c) ∃x ∈ R, ∃y ∈ R, x + y + 3 = 8.
(d) ∀x, y ∈ R, x + y + 3 = 8.
(e) ∀x ∈ R, ∃y ∈ R, x + y + 3 = 8.
(f) ∃x ∈ R, ∀y ∈ R, x + y + 3 = 8.
(g) ∃m, n ∈ N, n2 + m2 = 25.
(h) ∀m ∈ N ∃n ∈ N, n2 + m2 = 25.
Solution: Pay attention to the situations when we use a counterexample to prove
that a statement about all numbers is false, and when we use an example to prove
that the statement that there exists such-and-such number is true.
Page 1 of 6
Mathematics 220
Homework for Week 4
Due February 1 or 2
(a) True — for example, x = 0.
(b) False — for example, take x = −1, then
√
x2 = 1 6= x.
(c) True — for example x = 0, y = 5.
(d) False — pick x = 0, y = 0 — this example proves the negation is true. ∃x ∈
R s.t. ∃y ∈ R s.t. x + y + 3 6= 8.
(e) True – but here we need a general proof, an example would not work. Proof:
suppose x ∈ R is given. Then take y = 5−x. Then x+y +3 = x+(5−x)+3 = 8,
and so we have shown that for every x there exists a suitable y (that depends on
x) which makes x + y + 3 = 8 true.
(f) False – this statement is saying that there exists such a magical number x ∈ R
that no matter what y we add to it, we always get x + y = 5. Clearly, such
a magical number cannot exist. Better, let us prove that the negation is true:
∀x ∈ R, ∃y ∈ R, x + y + 3 6= 8. This is proved almost the same way as (e):
suppose x ∈ R is given. Now take y = 6 − x, for example (almost anything would
do here). Then x + y + 3 = x + (6 − x) + 3 = 9 6= 8, and so we have shown that for
every x there exists a suitable y (that depends on x) which makes x + y + 3 6= 8.
(g) True – take, for example, m = 3 and n = 4.
(h) False – take, for example, m = 1. Then for this m, we would need to find n ∈ N
such that n2 + 1 = 25, that is, n2 = 24. But we know that 24 is not a square of
a natural number: 44 < 24 but 52 > 24, so for m = 1 such an n does not exist,
which makes the statement False.
4. Question 2.72. Let P (x) and Q(x) be open sentences where the domain of the variable
x is a set S. Which of the following implies that ∼ P (x) ⇒ Q(x) is false for some x ∈ S?
(a)
(b)
(c)
(d)
P (x) ∧ Q(x) is false for all x ∈ S.
P (x) is true for all x ∈ S.
Q(x) is true for all x ∈ S.
P (x) ∨ Q(x) is false for all x ∈ S.
Solution: First, let R be the statement “∼ P (x) ⇒ Q(x) is false for some x ∈ S”.
We need to simplify R before we can proceed. First, recall that P ⇒ Q ≡ ∼ P ∨ Q;
then ∼ P ⇒ Q ≡ P ∨ Q. Note that R says:
∃x ∈ S s.t.
∼ (∼ P (x) ⇒ Q(x)).
The statement R contains the negation of an implication, and replacing it by its
equivalent as discussed above, we get they R is logically equivalent to:
∃x ∈ S s.t. ∼ (P (x) ∨ Q(x)) ≡ ∃x ∈ S s.t. (∼ P (x)) ∧ (∼ Q(x)).
Page 2 of 6
Mathematics 220
Homework for Week 4
Due February 1 or 2
Now, we are asked which of the listed statements imply R. We get:
(a) “P (x) ∧ Q(x) is false for all x ∈ S.”
This says: ∀x ∈ S ∼ (P (x) ∧ Q(x)), which is equivalent to
∀x ∈ S(∼ P (x)) ∨ (∼ Q(x)), which does not imply R.
To see this, we just need to give a counterexample: an example of open sentences
P (x) and Q(x) and a set S such that “∀x ∈ S(∼ P (x)) ∨ (∼ Q(x))” is True, but
“∃x ∈ S s.t. (∼ P (x)) ∧ (∼ Q(x))” is False. Consider, for example, S = {1, 2, 3},
and P (x) : x 6= 1, Q(x) : x ∈
/ {2, 3}. Then for all x ∈ S, (∼ P (x))∨(∼ Q(x)) says
that x = 1 or x ∈ {2, 3}, which is true. At the same time, (∼ P (x)) ∧ (∼ Q(x))
says x = 1 and x ∈ {2, 3}, which is false for all x.
(b) P (x) is true for all x ∈ S.
This says, ∀x ∈ S P (x). This actually makes R false, so it does not imply R (it
implies ∼ R, in fact).
(c) Q(x) is true for all x ∈ S. This also implies ∼ R, so does not imply R.
(d) P (x) ∨ Q(x) is false for all x ∈ S. This says: “∀x ∈ S, ∼ (P (x) ∨ Q(x))”, which is
logically equivalent to “∀x ∈ S s.t. (∼ P (x)) ∧ (∼ Q(x))”, which, in particular,
implies that ∃x ∈ S s.t. (∼ P (x)) ∧ (∼ Q(x)), so this statement does imply R.
5. Let S = [1, 2] and T = (3, ∞). Determine the truth value of the following statements,
and state each of them in words. Prove your answers— no proof = no marks
(a)
(b)
(c)
(d)
∃x ∈ S s.t. ∃y ∈ T s.t. |x − y| > 3
∃x ∈ S s.t. ∀y ∈ T, |x − y| > 3
∀x ∈ S, ∃y ∈ T s.t. |x − y| > 3
∀x ∈ S, ∀y ∈ T, |x − y| > 3
Solution:
(a) True — Let x = 1, y = 5 then |x − y| = 4 > 3.
(b) False — Consider the negation: ∀x ∈ S, ∃y ∈ T s.t. |x − y| ≤ 3. Pick any
x ∈ S and let y = 3.1. Since 1 ≤ x ≤ 2 we have −2.1 ≤ x − y ≤ −1.1. Hence
|x − y| ≤ 2.1 ≤ 3. Since the negation is true, the original statement must be
false.
(c) True — Pick any x ∈ S and let y = 100. Then since x ≤ 2 then x − y ≤ −98
and hence |x − y| ≥ 98 > 3.
(d) False — Consider the negation of the statement: ∃x ∈ S s.t. ∃y ∈ T s.t. |x−y| ≤
3. Pick x = 2, y = 4, so |x − y| = 2 ≤ 3. Since the negation is true, the original
statement must be false.
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Mathematics 220
Homework for Week 4
Due February 1 or 2
6. Let I = {n2 | n ∈ Z}. Let P be the statement
[
[k, 2k] = R.
k∈I
(a) Reformulate P using quantifiers.
Solution: ∀x ∈ R ∃k ∈ I s.t. x ∈ [k, 2k].
Even more concretely,
∀x ∈ R ∃n ∈ Z s.t. n2 ≤ x ≤ 2n2 .
(b) Is P true or false? Prove or disprove.
Solution: P is false. We claim
[
k∈I
[k, 2k] ⊂ [1, ∞),
(1)
and therefore this union of intervals is not the whole real line.
S
To see this let y ∈ k∈I [k, 2k]. Then for some k ∈ I, y ∈ [k, 2k], where k ≥ 1
since it is the square of an integer. Therefore y ≥ k ≥ 1. This proves y ∈ [1, ∞)
and so establishes (1).
S
We maySconclude that −1 ∈
/ k∈I [k, 2k] since it is not in the larger set [1, ∞),
and so k∈I [k, 2k] is not the set of all real numbers. That is, P is false.
7. Let A = {x | ∀n ≥ 3, 1/n < x < 1 − 1/n}.
(a) Represent the set A as an intersection of an indexed collection of sets.
Solution: A = ∩∞
n=3 (1/n, 1 − 1/n).
(b) Find A (that is, represent it in the simplest possible form, using the notation for
intervals in R). No proof required.
Solution: A = ( 31 , 23 ).
8. Suppose you have the following information about the population of the planet QE220:
• Among the inhabitants of QE220 who can watch TV, not all have antennae on their
head.
• The inhabitants of QE220 that are green and do not have antennae, cannot watch
TV.
Page 4 of 6
Mathematics 220
Homework for Week 4
Due February 1 or 2
Does it follow that not all the inhabitants of QE220 that can watch TV are green? Justify
your answer.
Hint. Start by writing down the statement you are asked about.
Solution: Let Q be the universal set – the set of all inhabitants of QE220. Let A
be the set of those who have antennae, let G be the set of those who are green, and
let T V be the set of those who can watch TV. The the question is asking, is it true
that T V − G is not empty?
We are given: T V − A 6= ∅, and (G − A) ∩ T V = ∅. Let b ∈ T V − A – such an
element b exists by the first condition. (In words, this says that there is an alien on
QE220 who can watch TV but has no antenna. Let us call him Billy.) We claim that
Billy cannot be green. Indeed, by the second statement, since Billy can watch TV,
he has to either not be green, or have an antenna. But we know that he does not
have an antenna, so he has to be not green. So the set of QE220-iers who can watch
TV but are not green contains Billy, and thus this set is not empty.
This is how some mathematicians might write the same solution: (Check for
yourself that these solutions are in fact equivalent solutions, just stated differently).
Let Q be the set of inhabitants of QE220. Introduce the following open statements:
G(x) : x is green; A(x) : x has antennae; T V (x) : x can watch TV.
The statement we are asked about is:
∼ (∀x ∈ Q T V (x) ⇒ G(x)).
(2)
We claim this is true. To prove this we can show the logically equivalent (recall
∼ (P ⇒ Q) ≡ P ∧ ∼ Q):
∃x ∈ Q s.t. T V (x)∧ ∼ G(x).
(3)
The second given bit of information is that
∀x ∈ Q G(x)∧ ∼ A(x) ⇒ ∼ T V (x),
or equivalently (using the contrapositive),
∀x ∈ Q T V (x) ⇒ ∼ G(x) ∨ A(x).
(4)
So lets prove the existence of an inhabitant who can watch TV and is not green, as is
required in (3). By the first bit of given information there is an x0 ∈ Q who can watch
TV and does not have an antenna. By (4), either x0 is not green or has an antenna.
The latter is impossible by our choice of x0 , so x0 cannot be green. Therefore x0 is
the required inhabitant who can watch TV and is not green. We have verified (3)
and are done.
9. Question 3.8. Prove that if x is an odd integer, then 9x + 5 is even.
Page 5 of 6
Mathematics 220
Homework for Week 4
Due February 1 or 2
Solution: We assume that the hypothesis is true and work our way to the conclusion
— just like in class.
Proof. Asume that x is odd. Hence there is some k ∈ Z such that x = 2k + 1. So
9x + 5 = 9(2k + 1) + 5 = 18k + 9 + 5 = 18k + 14 = 2(9k + 7)
Since 9k + 7 is an integer, it follows that 9x + 5 is even.
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