Unformatted text preview: Jennifer Klarfeld  Homework #2 Section 1.21.3 1. Identify the antecedent and the consequent for each of the following conditional sentences. (b) The antecedent is “The moon is made of cheese” and the consequent is “8 is an irrational number”. (c) The antecedent is “b divides 9” and the consequent is “b divides 3”. 1 Jennifer Klarfeld  Homework #2 Section 1.21.3 5. Which of the following conditional sentences are true? → P Q T T T F T T T F F F F T (b) If hexagons have six sides, then the moon is made of cheese. Let P, the antecedent, be “Hexagons have six sides” and let Q, the consequent, be “the moon is made of cheese”. Based on the truth table, since the antecedent is true and the consequent is false, the conditional statement is false. (f) If Euclid’s birthday was April 2nd, then rectangles have four sides. Let P, the antecedent, be “Euclid’s birthday was April 2nd” and let Q, the consequent, be “rectangles have four sides”. Based on the truth table, since the consequent is true, no matter whether the antecedent is true or false, the conditional statement is true. 2 Jennifer Klarfeld  Homework #2 Section 1.21.3 6. Which of the following are true? ↔ P Q T T T F T F T F F F F T (b) 7+5=12 if and only if 1+1=2. Let P be “7+5=12” and let Q be “1+1=2”. Since P is true and Q is true, then based on the above truth table, the biconditional statement is true. (f) 10 + 13 < 11 + 12 if and only if 13 ‐ 12 < 11 ‐ 10 10 + 13 < 11 + 12 iff 10 + 13 ‐ 10 < 11 + 12 ‐ 10 iff 13 ‐ 12 < 11 + 12 ‐ 10 ‐ 12 iff 13 ‐ 12 < 11 ‐ 10 . Thus 10 + 13 < 11 + 12 if and only if 13 ‐ 12 < 11 ‐ 10. (g) * ≥ 0 if and only if ≥ 0 Let P be " * ≥ 0” and let Q be “ ≥ 0". For this biconditional statement to be true, both → and → must be true. So first we will look at → : “if ≥ 0 then * ≥ 0. If Q is true, then P would be true. And if Q is false (x < 0), P could still be true, since the square of a negative number is positive. Thus → true no matter the truth values. Now, let’s look at → : “if * ≥ 0 then ≥ 0”. If P is true, Q is false (since x could be negative) making the statement false. And P cannot be false (since the square of any number is either zero or positive). Thus, → is false. Therefore, the biconditional statement is false. 3 Jennifer Klarfeld  Homework #2 Section 1.21.3 8. Prove Theorem 1.2.2 by constructing truth tables for each equivalence. (c) ~( → ) and ∧ ~ are equivalent. → ∧ ~ P Q ~( → ) ~Q T T T F F F F T T F F F T F F T T T F F T F T F Because the fourth and sixth columns are the same, ~( → ) and ∧ ~ are equivalent. (d) ~(⋀) and → ~ are equivalent. ~ → ~ ⋀ ~ ⋀ P Q T T F F T F F T F T F T T F T T F T F F T T F T Because the fourth and sixth columns are the same, ~(⋀) and → ~ are equivalent. 4 Jennifer Klarfeld  Homework #2 Section 1.21.3 14. Give an example of a false conditional statement for which: (a) the converse is true. A false conditional statement will always have a true converse. To illustrate this, we will make a truth table for a conditional statement: P T F T F → T T F T Q T T F F As you can see, the only time the conditional statement is false is when P is true and Q is false. Thus, the converse would be → , with a false antecedent and true consequent, which makes the converse true. EX: Conditional: “If a shape is a rectangle, then it is a square.” Converse: “If a shape is a square, then it is a rectangle.” (b) the converse is false. A false conditional statement will never have a false converse. To illustrate this, we will make a truth table for a conditional statement and its converse: P T F T F Q T T F F → T T F T → T F T T As you can see, for the conditional statement to be false, P would need to be true and Q would need to be false. But for the converse to be false, P would need to be false and Q would need to be true. Thus, it is not possible to have a false converse with a false conditional statement. (c) the contrapositive is false. A false conditional statement will always have a false contrapositive. To illustrate this, we will make a truth table for a conditional statement and its contrapositive: P T F T F Q T T F F → T T F T ~ F F T T ~ F T F T ~ → ~ T T F T 5 Jennifer Klarfeld  Homework #2 Section 1.21.3 As you can see, the only time both the contrapositive is false is if the original conditional statement is false. EX: Conditional: “If this class is in spring, then the sun is a planet.” Contrapositive: “If the sun is not a planet, then this class is not in spring.” (d) the contrapositive is true. As explained above, the contrapositive cannot be true if the conditional statement is false. 6 Jennifer Klarfeld  Homework #2 Section 1.21.3 1. Translate the following English sentences into symbolic sentences with quantifiers. (c) Some isosceles triangle is a right triangle. (All triangles) (∃)( ⋀ ), where () is the set of all x that are isosceles triangles, and () is the set of all right triangles. (g) Some people are honest and some people are not honest. ∃ (∃)( ), where is the set of all people who are honest and is the set of all people who are not honest. 7 Jennifer Klarfeld  Homework #2 Section 1.21.3 7. (a) Complete the following proof of Theorem 1.3.1(b). Proof: Let U be any universe. The sentence ~ ∃ () is true in U iff ∃ () is false in U iff the truth set of A(x) is not in U iff the truth set of A(x) is empty iff ∀ ~() is true in U. 8 Jennifer Klarfeld  Homework #2 Section 1.21.3 9. Give an English translation for each. (b) (∃! )( ≥ 0 ≤ 0) (ℝ) “There exists a unique x that is both greater than or equal to zero and less than or equal to zero in the set of all real numbers” is the literal translation of this statement. It could be simplified by stating, “Only one real number is both nonnegative and nonpositive.” (c) (∀)( ≠ 2 → ) (ℕ) “For all x, if x is a prime number and x is not two, then x is odd, in the set of all natural numbers” is the literal translation of this statement. It could be simplified by stating, “All prime natural numbers are odd, except for 2.” 9 Jennifer Klarfeld  Homework #2 Section 1.21.3 10. Which of the following are true in the universe of all real numbers? (f) (∃)(∀)( ≤ ) “There exists an x such that for all of y, ≤ .” This statement is false, as there is no value of x that is less than or equal to every value of y. A counterexample would be x=2. The statement would be true for y=3, but false for y=1. (g) (∀)(∃) ( ≤ ) “For all y, there exists an x such that ≤ .” This statement is true. For all real numbers y, there will always be a number x less than or equal to y. 10 Jennifer Klarfeld  Homework #2 Section 1.21.3 11. (d) Prove that ∃! is equivalent to ∃ ⋀ ∀ → = . ∃! is in U. Therefore, the truth set of A(x) is nonempty, which means ∃ . Moreover, A(x) contains only one element, let’s say x. So x ∈ A(x). Now, let’s say y ∈ A(x). If x ∈ A(x) and y ∈ A(x), and there is only one element is in A(x), the x=y. Hence, ∀ → = . So, ∃ ⋀ ∀ → = . (e) Find a useful denial for ∃! . This statement could be translated as “There exists a unique x with properties A(x)”. A denial would be anything equivalent to ~ ∃! , or “There doesn’t exist a unique x with properties A(x)”. Thus, either there are no elements with properties A(x) or there is more than one element with properties A(x). So either all x ∈ ~A(x) or there are at least two elements in A(x). This can be written with quantifiers as such: ∀ ~ ∨ (∃)(∃)( ⋀ → ≠ ). 11 ...
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 Spring '16
 Math, Jennifer Klarfeld

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