Introduction to Mathematical Rigor, MTH 307
Exercises
10 Express each of the following statements as a conditional statement in “if-then” form
or as a universally quantified statement. Also write the negation.
(a) Every odd number is prime.
– (∀x ∈ odd integers ) x is prime.
– Negation: (∃x ∈ odd integers ) x is not prime.
(b) The sum of the angles of a triangle is 180 degrees.
– If x, y, z are the angles of a triangle, then their sum is 180 degrees.
– Negation: x, y, z are the angles of a triangle and their sum is not 180
degrees.
(c) Passing the test requires solving all the problems.
– If you passed the test then you solved all the problems.
– Negation: You passed the test and you did not solve all the problems.
(d) Being first in line guarantees getting a good seat.
– If you are first in line, then you will get a good seat.
– Negation: You are first in line and you do not get a good seat.
(e) Lockers must be turned in by the last day of class.
– (∀x ∈ lockers) x is turned in by last day of class.
– Negation: (∃x ∈ lockers ) x is not turned in by the last day of class.
(f) Haste makes waste.
– (∀x ∈ haste) x makes waste
– Negation: (∃x ∈ haste ) x does not make waste.
(g) I get mad whenever you do that.
– If you do that then I get mad.
– Sometimes you do that and I do not get mad.
(h) I won’t say that unless I mean it.
– If I mean that then I will say that.
– Sometimes I won’t say that and I mean that.
31 Which of these statements are believable?
(a)
(b)
(c)
(d)
“All of my 5-legged dogs can fly.” true
“I have no 5-legged dog that cannot fly.” true
“Some of my 5-legged dogs cannot fly.” false
“I have a 5-legged dog that cannot fly.” false
35 Prove that if x and y are distinct real numbers, then (x + 1)2 = (y + 1)2 if and only if
x + y = −2. How does the conclusion change if we allow x = y?
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Proof: We start with (x + 1)2 = (y + 1)2 . Expanding both sides, we get x2 + 2x + 1 =
y 2 + 2y + 1. Cancelling the 1’s and rearranging terms, we get x2 − y 2 = 2y − 2x. We
factor each side to get (x + y)(x − y) = 2(y − x) = −2(x − y). As we are given x 6= y, we
can divide both sides by x − y to get x + y = −2. As equations preserve equality and
these steps are reversible, x + y = −2 implies (x + 1)2 = (y + 1)2 , so the dependence
holds both ways. If x can equal y, the statement only holds true when x and y are
distinct as before or when x = y = −1.
37 Given a real number x, let A be the statement 1/2 < x < 5/2, let B be the statement
x ∈ Z, let C be the statement x2 = 1, and let D be the statement x = 2. Which
statements below are true for all x ∈ R.
(a)
(b)
(c)
(d)
A ⇒ C. False. For a counterexample, let x = 2.
B ⇒ C. False. For a counterexample, let x = 2.
(A ∧ B) ⇒ C. False. For a counterexample, let x = 2.
(A ∧ B) ⇒ (C ∨ D). True. If A and B are true, then x = 1 or x = 2. If x = 1
then C is true. If x = 2 then D is true. So in either case, either C or D is
true.
(e) C ⇒ (A ∧ B). False. For a counterexample let x = −1.
(f) D ⇒ [A ∧ B ∧ (¬C)]. True. If x = 2 then A is true, B is true, and x2 6= 1 so
¬C is true.
(g) (A ∨ C) ⇒ B. True. A ∨ C means that x = 1, 2, or −1. So, B is true.
38 Let x, y be integers. Determine the truth value of each statement below.
(a) xy is odd if and only if x and y are odd. TRUE. If x is odd and y is odd then
x = 2a + 1 for some a and y = 2b + 1 for some b. Then xy = (2a + 1)(2b + 1) =
4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1 which is odd. Suppose it is not true that
both x and y are odd. Then at least one is even. Without loss of generality,
suppose x is even. Then x = 2a for some a. xy = 2ay which is even.
(b) xy is even if and only if x and y are even. FALSE. For a counterexample to
the “only if” part, let x = 3 and y = 4. Then, xy = 12 which is even.
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