Monday, April 19, 2010

how to prove it - ch2, sec2.1(quantifiers) ex

This section introduces the two well known quantifiers, universal($\forall$) meaning "for all" and existential($\exists$) meaning "there exists".

It also says that quantifier "binds" a variable, that is, a variable that is bound by a quantifier can always be replaced with a new variable without changing the meaning of the statement. For example $\forall x P(x)$ = $\forall y P(y)$

One statement can also contain more than one quantifier, for example, $\forall x \exists y (x + y = 5)$ is a valid statement that means, for all values of x there exists at lease one y such that (x + y = 5).
It may be best in these cases to think about the quantifiers one at a time , in order.

Quantifiers have higher precedence than conditional/biconditional which means $\forall x P(x) \rightarrow Q(x)$ means $[\forall x P(x)] \rightarrow Q(x)$ and not $\forall x [P(x) \rightarrow Q(x)]$

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Ex-1(a) $\forall x [\exists y F(x,y) \rightarrow S(x)]$
where F(x,y) means x has forgiven y
S(x) means x is a saint

Ex-1(b) $\lnot \exists x C(x) \land [\forall y D(y) \land S(x,y)]$
C(x) means, x is in Calculus class
D(y) means, y is in discrete math class
S(x,y) means, x is smarter than y

Ex-1(c) $\forall x L(x, Mary) \land \lnot(x = Mary)$
L(x,Mary) means, x likes Mary
x = Mary means, x is Mary

Ex-1(d) $ [\exists x P(x) \land S(Jane, x)] \land [\exists y P(y) \land S(Roger,y)]$
P(x) means, x is a police officer
S(x,y) means, x saw y

Ex-1(e) $ \exists x P(x) \land S(Jane, x) \land S(Roger, x)$


Ex-2(a) $ \forall x [B(x,Cash) \rightarrow \exists y R(y) \land U(y,x)]$
B(x,y) means, x bought a Rolls Royce with y
R(y) means, y is Rich
U(y,x) means, y is Uncle of x

Ex-2(b) This statement says, "If anyone in the dorm has the measles, then everyone who has a friend in the dorm will have to be quarantined."

$\exists x (D(x) \land M(x)) \rightarrow \forall z [ \exists y (F(z,y) \land D(y)) \rightarrow Q(z) ]$

D(x) means, x lives in the form
F(x,y) means, x and y are friends
M(x) means, x has measles
Q(x) means, x has to be quarantined

Ex-2(c) This statement says, "If there does not exists anyone who failed the test, then everybody with grade A will tutor someone with grade D"

$\lnot \exists x F(x) \rightarrow \forall y (A(y) \rightarrow \exists z D(z) \land T(y,z))$

F(x) means, x has failed the test
A(y) means, y got grade A
D(z) means, z got grade D
T(y,z) means, y will tutor z

Ex-2(d) $ \exists x D(x) \rightarrow D(Jones)$
D(x) means, x can do it

Ex-2(e) $ D(Jones) \rightarrow \forall x D(x)$


Ex-3(a) $\forall z [(z>x) \rightarrow (z > y)]$

this statements is about x and y, that is values of x,y fix the True/False value of the statement.. x,y are the free variables

Ex-3(b) $\forall a \exists x [(ax^2 + 4x -2 = 0) \leftrightarrow (a \geq -2)]$

clearly, there is no free variable.

Ex-3(c) $\forall x [(x^3 - 3x < 3) \rightarrow x < 10]$
no free variables

Ex-3(d) $[\exists x (x^2 + 5x = w)] \land [\exists y (4 - y^2 = w)] \rightarrow -10 < w < 10$
w is the free variable


Ex-4(a) Every unmarried man is unhappy.

Ex-4(b) y is a sibling(sister) of one of x's parent.


Ex-5(a) Every prime number except 2 is odd.

Ex-5(b) There exists a perfect number such that all the other perfect numbers are either less or equal to it.


Ex-6(a) It means, there exists atleast one person who is parent of everyone. clearly its False.

Ex-6(b) It means, Everyone is a parent. clearly its False.

Ex-6(c) It means, There does not exist anyone who is parent of someone. clearly its False.

Ex-6(d) It means, There exists someone who is parent of none. its True.

Ex-6(e) It means, There exists someone who is not parent of someone. its True.


Ex-7(a) using $y = 2x$ we can find a y for any x, clearly its True.

Ex-7(b) It means, there exists a y such that for all x, $2x - y = 0$. clearly its False.

Ex-7(c) Using y = x/2, for odd x there would not be any y which is a natural number. so its False. This would be true if universe of discourse was real numbers.

Ex-7(d) True

Ex-7(e) holds true for y = z = 50; y = 40 , z = 60 ..etc. Clearly its True.

Ex-7(f) This statement will fail for x = 100, there is no y which is greater than 100 for which there exists some (natural number)z such that $y + z = 100$. Clearly its False.


Ex-8(a) True

Ex-8(b) False

Ex-8(c) True

Ex-8(d) False

Ex-8(e) True

Ex-8(f) True


Ex-9(a) True

Ex-9(b) False

Ex-9(c) False

Ex-9(d) True

Ex-9(e) True

Ex-9(f) True

12 comments:

  1. This comment has been removed by the author.

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  2. This comment has been removed by the author.

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  3. Ex-2(b)
    I think that the correct answer is:
    ∃x(D(x) ∧ M(x)) → ∀z[∃y(F(z,y) ∧ D(y)) → Q(z)]

    Ex-6(b)
    ∀x∃yP(x,y): Everyone *is* a parent.
    The statement is False.

    Ex-8(d)
    Let x = 9.9 < 10 and y = 9.8 < x. Then y > 9. Thus, the statement is False.

    Ex-9
    The universe of discourse is Z, not C (complex numbers).
    (a) True
    (b) False
    (c) False
    (d) True
    (e) True
    (f) True

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  4. This is my take on 2 (b). I think this right is because

    ∀x[(∃yD(y)∧F(x,y))∧(∃zD(z)∧M(z))→Q(x)]
    for all x ,if there exists a y in the dorm and y friends with x and a person z in the dorm with measles, then x needs to be quarantined. This fits the single if then statement in the question.

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    Replies
    1. In this statement the word "anyone" meant "someone". Your answer is wrong. Turn to page 60 and read the example 4 again,

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    2. Yes, I agree I was wrong. Maybe you could explain why the next one 2(c) is:

      ¬∃xF(x)→∀y(A(y)→∃zD(z)∧T(y,z))

      instead of:

      ¬∃xF(x)→∀y(A(y)→∀zD(z)∧T(y,z))

      I was thinking that the statement doesn't work because it is saying if no one failed then for every person with an A there exists someone with a D and y will tutor them. But maybe one person getting an A does not imply that someone else got a D? thanks, sry I find these tricky.

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    3. The word "someone" tips us off that we should use ∃ instead of ∀.

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    4. I had ¬∃xF(x)→∀y(A(y)→∃zD(z)∧T(y,z)) initially for my answer for 2(c), but I changed it to:

      ¬∃xF(x)→∀y∃z[(A(y)∧D(z)]→T(y,z))

      I found that the initial statement, which said "If y got an A then at least one person z got a D and y will tutor z, didn't sound right. How does someone who received an A cause someone to get a D?

      This new statement says "For every person y ∃z[(A(y)∧D(z)]→T(y,z)) is true. In other words, for every person y, there is at least one person z in this statement. The statement then reads "If y got an A and z got a D, then y will tutor z."

      I think this sounds more logical. Tell me if I'm wrong.

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  5. This comment has been removed by the author.

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  7. 6(b) I agree is false. if we take all the people in the world, there exist at least one person such that x is y's parent. let x be a 4 year old child, counterexample.

    and what Stavros Mekesis said for 8,9.

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  8. I guess Stavros is correct in
    2(b), 6(b), 8(d) and 9.

    updating all the answers in the original post. Thanks.

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