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Homework
Exercises
- Mark each statement True or False. Justify each answer. (a) The symbol ∀ means ”for every.” True, this is one of the replacements for ∀. (b) The negation of a universal statement is another universal statement. False the negation of ∀ is ∃ and the negation of ∃ is ∀.
- Mark each statement True or False. Justify each answer. (a) The symbol ”∃” means ”there exist several” False, ”∃” means ”there exist at least one” (b) If a variable is used in the antecedent of an implication without being quantified, then the universal quantifier is assumed to apply. True, if you are working with a proposition that has variables and the an- tecedent has no quantifier then you apply the universal quantifier. (c) The order in which quantifiers are used affects the truth value. True, since the ∀ and ∃ quantifiers have different definitions the order in which they are used affects the truth value. Example: ∀x ∈R ∃y ∈R such that ∀ > 0, |x − y| < , True ∃y ∈R such that ∀x ∈R, ∀ > 0 |x − y| < , False
- Write the negation of each statement. (a) Some pencils are red. Statement in mathematics: At least one pencil is red. All pencils are not red. (b) All chairs have four legs. There exist a chair that does not have four legs. (c) No one one the basketball team is over 6 feet 4 inches tall. Statement in mathematics: For all members of the basketball team no one is 6 feet 4 inches tall. There exist a member of the basketball team such that he/she is 6 feet 4 inches
tall. (d) ∃ x > 2 such that f (x) = 7. ∀ x> 2, f (x)=/ 7. (e) ∀ x ∈ A, ∃ y > 2 such that 0 < f(y) < f(x). ∃ x ∈ A such that ∀ y > 2 f(y) ≥ 0 or f(y) ≥ f(x). (f) If x > 3, then ∃ E > 0 such that x^2 > 9+ E. x > 3, and ∀ E > 0, x^2 ≤ 9+ E.
- Write the negation of each statement. (a) Everyone likes Robert. Statement in mathematics: All members like Robert. At least one student does not like Robert. (b) Some students work part time. Statement in mathematics: At least one student works part time All Students do not work part time. (c) No square matrices are triangular. Statement in mathematics: All square matrices are not triangular. There exist a square matrix that is triangular. (d) ∃ x in B such that f(x) > k. ∀ x ∈ B, f(x) ≤ k. (e) If x > 5, then f(x) < 3 or f(x) > 7. x > 5 and f(x) ≥ 3 and f(x) ≤ 7. (f) x is in A, then ∃ y in B such that f(x) < f(y). if x ∈ A, and ∀ y ∈ B, f(x) ≥ f(y).
- Determine the truth value of each statement, assuming x is a real number. Justify your answer. (a) ∃ x in the interval [2, 4] such that x < 7. What is this statement saying in English? There is at least one x in the interval [2,4] and x < 7. True, Take x:=3, x:=3 satisfies x ∈ [2, 4] and x < 7. (b) ∀ x in the interval [2, 4], x < 7. What is this saying in English? For all x’s in the interval [2,4] x < 7 True, Let x in the interval [2, 4] 2 ≤ x ≤ 4 < 7, x < 7 (c) ∃ x ∈ R such that x^2 = 5. What is this statement saying in English? There exist a least one x such that if you solve for x in x^2 = 5, x ∈ R. True, Take x:=
5 , x:=
5 satisfies x∈ R and x^2 = 5 (d) ∀ x ∈ R , x^2 = 5. What is this statement saying in English? For all x, you can solve x^2 = 5, x ∈ R. False, take x:=2, x^2 =4, x:=2 satisfies x ∈ R and x^2 6 = 5.
There is at least one element in th R that makes x − x = 0 true True, Take x:=5, x:=5 satisfies x ∈ R and x − x= (h) ∀ x ∈ R , x-x=0. English translation: For all x’s in the R, this equation is true. True, Let x ∈ R, whenever x > 0, x < 0, or x=0, x − x = 0
- Below are two strategies for determining the truth value of a statement involving a positive number x and another statement P(x). For each statement below, indicate which strategy is appropriate. (i) Find some x>0 such that P(x) is true. (ii) Let x be the name for any number greater than 0 and show that P(x) is true. (a) ∀x > 0, P(x). (ii) (b) ∃ x > 0 such that P(x). (i) (c) ∃ x > 0 such that ¬ P(x). (ii) (d) ∀x > 0, ¬ P(x). (i)
- Which of the following best identifies f as a constant function where x and y are real numbers. Constant Function: The output value is the same no matter what the input value is. (a) ∃x ∈ R such that ∀y ∈ R f (x) = y No This is telling us that the image of x of R, which is not compatible with the definition of a map from R (b) ∀x ∈ R ∃y ∈ R such that f (x) = y No This is just telling you that f is a map from R to R (c) ∃y ∈ R such that ∀x ∈ R, f (x) = y Yes (d) ∀y ∈ R ∃x ∈ R such that f (x) = y No This is the definition of subjectivity.
- Determine the truth value of each statement, assuming that x and y are real num- bers. Justify your answer. (a) ∀ x,y ∈ R y, x ≤ y. False, since the original statement is false we will prove the negation of the original statement. The negation is: ∃ x,y ∈ R y, x > y. Take: x:= Take y:= x and y both satisfy that x ∈ R and that x > y. This is contradictory to the original statement. (b) ∃ x and y ∈ R such that x ≤ y. True
Take x:= Take y:= x and y both satisfy that x,y ∈ R and that x ≤ y (c) ∀ x R, ∃ y R such that x ≤ y. True Draft: Let x ∈ R We can find a y ∈ R such that x ≤ y Proof: Let x ∈ R Take y:=x+ Therefore x ≤ y. Notice how the above problem could’ve been solved by taking y:=x because both values ca be equal. It is easier to equate the variables when possible instead of finding another example. The difference between this problem and the problems we have done before is that this one involves two different quantifiers. As a result, it is easy to get confused on what you are evaluating. It is essential to do a draft before the proof. (d) ∃ x ∈ R such that ∀ y ∈ R x ≤ y. False, the original statement is false because x is independent and this is not true for all y’s if x is set beforehand. Since the original statement is false we will prove the negation of the original statement. The negation is: ∀ x ∈ R ∃ y ∈ R such that x > y Proof: Let x ∈ R Take y:=x- Then x > y Which is contradictory to the original statement we started with.
- Determine the truth value of each statement, assuming that x and y are real num- bers. Justify your answer. (a) ∀ x ∈ R, ∃ y ∈ R such that xy = 0 True Draft: For all x ∈ R We can find a y ∈ R such that xy= Proof: Let x ∈ R Take y:= then clearly xy=
will prove the negation of the original statement. The Negation is: ∀ x and y ∈ R ∃ z ∈ R , x+y 6 = z Proof: True Proof: Let x,y ∈ R Take x:= x+y+ then clearly x+y 6 = z (c) ∀ x and y ∈ R, ∃ z ∈ R such that y − z = x Let x,y ∈ R Take z:=-x+y Then clearly y-z=x (d) ∀ x and y ∈ R, ∃ z ∈ R such that xz=y False, The original statement is false. Since the original statement is false so we will prove the negation of the original statement. The negation is: ∃ x and y ∈ R such that ∀ z ∈ R xz 6 = y Proof Take x:= Take y:= Let z ∈ R Then clearly xz 6 = y (e) ∃ x ∈ R such that ∀ y,z ∈ R, z > y implies that z > x+y. True, Draft: We want to find an x R such that For all y,z ∈ R Suppose z > y We want to prove that z > x+y Proof: Take x:= 0 Let x, y ∈ R Suppose z > y Then clearly z > x+y. (f)∀ x ∈ R , ∃ y,z ∈ R such that z > y implies that z > x+y. True, Draft: For all x ∈ R We want to find a y, z ∈ R Suppose z > y We want to prove that z > x+y Proof: Let x ∈ R
Take y:= Take z:= x+ Suppose z > y Then clearly z > x+y.
- Determine the truth value of each statement, assuming that x,y, and z are real numbers. Justify your answer. (a) ∀ x and y ∈ R, ∃ z ∈ R such that ∈ R x + y = z True, Let x, y ∈ R Take z:=x+y Then clearly x+y=z (b) ∀ x ∈ R ∃ y ∈ R such that ∀ z ∈ R , x + y = z False, The original statement is false. Since the original statement is false we will prove by negating the original statement. The negation is: ∃ x ∈ R such that ∀ y ∈ R ∃ z ∈ R such that x+y 6 = z Take x:= Let y ∈ R Take z:=y+ then clearly x+y 6 = z (c) ∃ x ∈ R such that ∀ y ∈ R , ∃ z ∈ R such that xz = y True, Take x:= Let y ∈ R Take z:=y Then clearly xz=y (d) ∀ x and y ∈ R, ∃ z ∈ R such that yz = x False, the original statement is false. Since the original statement is false we will prove the negation of the original statement. The negation is: ∃ x and y ∈ R such that ∀ z ∈ R yz= 6 = x Proof: Take x:= Take y:= Let z ∈ R Then clearly yz= 6 = x (e) ∀ x ∈ R , ∃ y ∈ R such that ∀ z ∈ R, z > y implies that z > x+y. False, the original statement is false. Since the original statement is false we will prove the negation of the original statement. The negation is: ∃ x ∈ R such that ∀ y ∈ R ∃ z ∈ R, z > y and z ≤ x+y. Proof: Take x:= Let y ∈ R Take z:=y+
that |f (x) − f (y)| < E whenever x and y are in S and |x − y| < δ (a) ∀ E> 0 ∃ δ > 0 such that |f (x) − f (y)| < E ⇒ x,y ∈S and |x − y| < δ (b) ∃ E > 0 such that ∀ δ > 0 such that |f (x) − f (y)| < E and x,y ∈/S or |x − y| ≥ δ (21) The real number L is the limit of the function f: D → R at the point c if for each E> 0 there exists a δ > 0 such that |f (x) − L| < E whenever x ∈ D and 0 < |x − c| < δ (a) ∀ E> 0 , ∃ δ > 0 such that |f (x) − L| < E ⇒ x ∈ D and 0 < |x − c| < δ (b) ∃ E> 0 such that ∀ δ > 0 |f (x) − L| < E and x ∈/ D or |x − c| ≥ 0 or |x − c| ≥ δ ∗ Using negation to understand mathematical definitions
Part II
Find the logic connector between the sentences (a) All men are mortal ∀ x ∈ M, x ∈/ I (b) All men are immortal ∀ x ∈ M x ∈ I (c) None of the men are mortal ∀ x ∈ M, x ∈ I (d) None of the men are immortal ∀ x ∈ M, x ∈/ I (e) There exist a man that is immortal ∃ x ∈ M x ∈ I (f) There exist a man that is mortal ∃ x ∈ M x ∈/ I a ⇔ d, b ⇔ c
Let C be the set of cats, c ∈ C M(c) be the proposition ”c has whiskers” P(c) be the proposition ”c likes fish” S(c) be the proposition ”c fears mice”
Write the sentence with quantifiers and the set defined (1) The cats with whiskers always likes fish. ∀ c ∈ C, M(c) ⇒ P(c) (2) It is wrong that the cats which likes fish has whiskers. ∃ c ∈ C, P(c) ∧ ¬ M(c) (3) None of the cats which like fish fear the mice. ∀ c ∈ C, ¬ P(c) ⇒ ¬ S(c) (4) The cats that have whiskers fear the mice. ∀ c ∈ C, M(c) ⇒, S(c) (5) The cats which fear the mice have no whiskers. ∀ c ∈ C, S(c) ⇒ ¬ M(c)
Suppose 1, 2, 3 is true. Make a picture presenting the set of cats. The sentences (1, 2, 3) are true. What can you say about statements 4 and 5? Statements 4 and 5 are inverses of each other.
Let H be the set of men. We propose the two math translations for the sentence. ”All men are happy and quiet.” (1) ∀ x ∈ H. (x is happy ∧ x is quiet) All men are happy and quiet. Is these statement the same as the first: Yes
(2) (∀ x ∈ H, x is happy) ∧ (∀ x ∈ H, x is quiet) All men are happy and all men are quiet. Is these statement the same as the first: Yes (-) Are the 3 propositions the same: Yes
”Happy men are quiet” (1) ∀ x ∈ H, (x is happy ⇒ x is quiet) Happy men are quiet Is these statement the same as the first: Yes (2) (∀ x ∈ H, x is happy) ⇒ (∀ x ∈ H, x is quiet) All men are happy then all men are quiet, which has a different meaning than all men are quiet. Is these statement the same as the first: No (-) Are the 3 propositions the same: No
”There exist a man happy and quiet.” (1) (∃ x ∈ H, x is happy) ∧ (∃ x ∈ H, x is quiet) There exists a man that is happy and there exist a man that is quiet. Is these statement the same as the first: No (2) ∃ x ∈ H,[ (x is happy) ∧ (q is quiet)] There exist a man that is happy and quiet. Is these statement the same as the first: Yes (-) Are the 3 propositions the same: No
”All men are not happy” (1) ∀ x ∈ H, ¬ (x is happy) All men are not happy Is these statement the same as the first: Yes (2) ∃ x ∈ H, ¬ (x is not happy) There exist a man that is happy Is these statement the same as the first: No (3) ¬ (∀ x ∈ H, x is happy) There exist a man that is not happy. Is these statement the same as the first: No (-) Are the 3 propositions the same: No
Write the negation of the following proposition: (a) p: All men are kind. There exist a man that is not kind. (b) p: All interval of R contains an element of the interval [0,1]. There exists an interval of R that does not contains an element of the interval [0,1]
Proof: Take i= f (i) = − 1 < 0 (b) ∃ i ∈ { 0 , 1 , 2 }, f(i) ≥ 0 Negation: ∀ i ∈ { 0 , 1 , 2 }, f(i) < 0 The negated statement is false so we will prove the original statement. Proof: True, Take i=0, f (i) = 3 ≥ 0 (c) ∀ j ∈ {− 1 , 3 , 5 }, ∃ i ∈ { 0 , 1 , 2 } f(i)=j Negation: ∃ j ∈ {− 1 , 3 , 5 } such that ∀ i ∈ { 0 , 1 , 2 } f(i) 6 = j The negated statement is false so we will prove the original statement. Proof: Let j ∈ {− 1 , 3 , 5 }. If j= -1 then i=1, f (1) = − 1 , If j=3 then i=2, f (2) = 3, If j=5 then i=2, f (2) = 5 (d) ∀ j ∈ {− 1 , 3 }, ∃ i ∈ { 0,1,2 } f(i) =j Negation: ∃ j ∈ {− 1 , 3 } such that ∀ i ∈ { 0,1,2 } f(i) 6 = j Proof: Let j ∈ {− 1 , 3 , 5 } If j=-1, Take i:=1 therefore f(1)=- If j=3, Take i:=0 therefore f(0)= If j=5, Take i:=2 therefore f(2)= (e) ∃ j ∈ {− 1 , 3 , 5 } such that ∀ i ∈ { 0 , 1 , 2 } f (i) = j. Negation: ∀ j ∈ {− 1 , 3 , 5 } ∃ i ∈ { 0 , 1 , 2 } such that f(i) 6 = j , The negated statement is true so we will prove it. Proof: If j=-1, Take i=0 then f(0) 6 = - If j=3, Take i=2 then f(2) 6 = 3 If j- 5, Take i=1, then f(1) 6 = 5
For all the propositions below. Give the negation and say if the statement is false. Justify your answer. (a) ∀ x ∈ R, x^2 > 0 Negation: ∃ x ∈ R such x^2 ≤ 0 True, The negated statement is true so we will prove it. Take x=0, x=0 satis- fies that x ∈ R and x^2 = 0 ≤ 0 (b) ∃ x ∈ R, x^2 > 0 Negation: ∀ x ∈ R, x^2 ≤ 0 False, This negated statement is false so we will prove the original statement. Take x:=2, x^2 = 4, x:=2 satisfies that x ∈ R and x^2 = 4 > 0 (c) ∀ x ∈ R,
x^2 = x Negation:∃ x ∈ R such that
x^2 =/ x True, The negated statement is true so we will prove it. Take x=-5, x=- satisfies that x ∈ R and
x^2 = 5=/ − 5
(d) (∀ x ∈ R), (∃ y ∈ R), x + y = 0 Negation:(∃ x ∈ R) such that (∀ y ∈ R), x + y=/ 0 False, This negated statement is false so we will prove the original statement. Let x ∈ R Take: y = −x Then clearly x + y = 0 (e) (∃ y ∈ R), (∀ x ∈ R), x + y = 0 Negation: (∀ y ∈ R), (∃ x ∈ R) such that x + y=/ 0 The negated statement is True so we will prove it. Let y ∈ R Take x=-y+ then clearly x + y=/ 0 Explain if there is a difference between e and f and why? Yes, there is a difference between both statements because we intertwine the quantifier, in the first statement y can be chosen in terms of z while in the second statement y is independent of x.
Write with quantifiers. (a) All natural numbers have a real square root. ∀ n ∈ N, ∃ x ∈ R such that x^2 = m (b) All natural numbers have a positive real number greater than them. ∀ n ∈ N ∃ x ∈ R (positive) such that n < x (c) ∃ a real number smaller than all the integers. ∃ x ∈ R such that ∀ n ∈ Z x < n (d) The interval I is included in [1,2] ∀ x ∈ I, x ∈ [1,2]
Let F be the of French. We denote, ∀ ∈ F, P(x), the proposition ”x has brown hats” Q(x), the proposition ”x is tall” Answer the following questions (1) In a picture, represent in F the set of elements of F such that P(x) is true then the set of elements such that Q is true. (2) Consider the following sentence: (∀ x ∈ F) [P-[(x) or Q(x)] and (∀ x ∈ F, P(x)) or (∀ x ∈ F, Q(x)) Tell if the propositions are true or fall in the case of the picture. (∀ x ∈ F) [P(x) or Q(x)] True (∀ x ∈ F, P(x)) or (∀ x ∈ F, Q(x)) True
