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Thread: Propositional functions/ D as domain . Ill try and solve but help would be good.

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    Propositional functions/ D as domain . Ill try and solve but help would be good.

    Suppose P(x) and Q(x) are propositional functions and D is their domain.
    Let A = {xD :P(x) is true} and B = {xD: Q(x) is true}.

    (a) Give an example for a domain D and functions P(x) and Q(x) such that A B = .

    (b) Give an example for a domain D and functions P(x) and Q(x) such that A B butA B.

    (c) Given that x A - B, what is the truth value of Q(x)?

    (d) Given that x A - B, what is the truth value of P(x)V Q(x)?
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    Re: Propositional functions/ D as domain . Ill try and solve but help would be good.

    Quote Originally Posted by bee77 View Post
    Suppose P(x) and Q(x) are propositional functions and D is their domain.
    Let A = {x∈D :P(x) is true} and B = {x∈D: Q(x) is true}.
    (a) Give an example for a domain D and functions P(x) and Q(x) such that A ∩ B = .
    (b) Give an example for a domain D and functions P(x) and Q(x) such that A ⊆ B butA ≠ B.
    (c) Given that x ∈ A - B, what is the truth value of Q(x)?
    (d) Given that x ∈ A - B, what is the truth value of P(x)V Q(x)?
    First let $\mathcal{I}$ be the set of integers.
    A) let $D=\mathcal{I}$ & $P(x): x\text{ is even and }Q(x): x\text{ is odd. }$ Show that this works for part A.

    B) Use the same in part A) except $P(x)$ means that $x=4j+1,~j\in\mathcal{I}$ YOU show that works for part B).
    Thanks from bee77
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    Re: Propositional functions/ D as domain . Ill try and solve but help would be good.

    Ok Ill try ...
    (a) Let I be a set of integers{....-1,0,1...}
    let D = I and P(x) : x is even and Q(x): x is odd
    so P(x) ={...-2,0,2,4,6...} set of all even numbers
    and Q(x) = {...-1,0,1,3..} set of all odd numbers

    there fore A ∩ B =

    (b) to prove A ⊆ B but A ≠ B
    j∈I (being a set of integers) {...,-1,0,1...}
    we use an element of set P(x) being even and the same element in j and substitute into the equation x=4j+1 also substituting an element
    P(x)= 4*j+1
    so P(2) = 2 =4*(2)+1
    which shows 2=9
    so there fore A is a subset of B but it does not = B .
    I hope this is ok ...
    I had another way I was trying to do before I saw your answer ..as I am a beginner with this kind of stuff does this work too?
    (a) So I'll make a domain with P (x)= {1,2,3,4}
    and Q(x) = {1,2,3}
    if I make a function P(x) +Q(x) >0 which proves true
    then A ∩ B = holds true by subbing in the set values of P(x) and Q(x)

    so therefore it satisfies A ∩ B = .
    Last edited by bee77; Aug 9th 2017 at 05:07 PM.
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    Re: Propositional functions/ D as domain . Ill try and solve but help would be good.

    (b) I'll make a domain once again with all integers D(x)= {....,1,2,3,4.....]
    Q(x) x^2<4
    P(x) x^2>4
    by subbing the values in we can see that P(x) and Q(x) such that A ⊆ B but A ≠ B
    =is this correct I've probably written it wrong ...??
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    Re: Propositional functions/ D as domain . Ill try and solve but help would be good.

    Sorry Plato ,
    Before I go to bed
    I'm still kind of not sure how to answer these ...I know that a domain is a set of integers for example and the functions are P(x) and Q(x) and we make an equation to suit it answering A ∩ B = and A ⊆ B butA ≠ B.
    c)x ∈ A - B, what is the truth value of Q(x) means the truth table is drawn up with P , Q and P ∩ Q ?
    and
    d) x ∈ A - B, what is the truth value of P(x)V Q(x)? means the truth table is drawn up with P, Q ,Q and P/Q ?
    Is that correct for my truth tables..?
    I better go to bed too much math has killed my brain thanks
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    Re: Propositional functions/ D as domain . Ill try and solve but help would be good.

    Quote Originally Posted by bee77 View Post
    Sorry Plato ,
    Before I go to bed
    I'm still kind of not sure how to answer these ...I know that a domain is a set of integers for example and the functions are P(x) and Q(x) and we make an equation to suit it answering A ∩ B = and A ⊆ B butA ≠ B.
    c)x ∈ A - B, what is the truth value of Q(x) means the truth table is drawn up with P , Q and P ∩ Q ?
    and
    d) x ∈ A - B, what is the truth value of P(x)V Q(x)? means the truth table is drawn up with P, Q ,Q and P/Q ?
    Is that correct for my truth tables..?
    I better go to bed too much math has killed my brain thanks
    This posting shows that you are profoundly confused by this material. The material introduces you to propositional functions. Both in notation and operation they are similar to the functions with which you have already worked. However, with propositional functions the domain of discourse has far more importance.
    Suppose that $D$ is the set of positive integers and $P(x)$ means that x is a positive integer the square of which is less than $10^4$. Now if one sees $P(t)$ we know that $t\in\mathcal{I}^+$, i.e. a positive integer and $t^2<10^4$. We can be sure that $P(50)$ is a true proposition because fifty is a positive integer and its square is less than ten thousand.

    Is $P(100)$ true or false? Is $\neg P(200)$ true or false?
    If $A=\{n: P(n)\text{ is true}\}$, can you describe the set $A~?$
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    Re: Propositional functions/ D as domain . Ill try and solve but help would be good.

    Set A is 0>A <1000 for true
    I think thtas right ..i think symbols for me are a bit confusing ...
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    Re: Propositional functions/ D as domain . Ill try and solve but help would be good.

    Part c and d am i right in making a truth table with the previous mentioned?
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    Re: Propositional functions/ D as domain . Ill try and solve but help would be good.

    Is $P(100)$ true or false? Is $\neg P(200)$ true or false?t
    If $A=\{n: P(n)\text{ is true}\}$, can you describe the set $A~?$
    Quote Originally Posted by bee77 View Post
    Set A is 0>A <1000 for true
    I think that's right ..i think symbols for me are a bit confusing ...
    Again, you understand nothing about the basic prerequisites for the material in the course you are now pursuing.
    Come on guy, do you not that $\large\color{red}{0>A <1000}$ is absolute nonsense?
    What can $0>A$ possibly mean. $A$ is a set and so how is a set comparable to zero?

    Now actually $P(100)$ is false because $100^2\not<10^4$. If you don't follow that the you are in real need of help.
    Thus $A=\{1,2,3,\cdots\,98,99\}$, all positive integers less than $100$. You may not even understand that $(10^2)^2=10^4~? $
    You will be expected to understand all basic mathematics such as that above.

    Add that $Q(x)$ means that $x$ is prime. Now you see we already know that $x$ is a positive integer because of the domain of definition.
    Now suppose that $B=\{x: Q(x)\text{ is true }\}$. Do you see that $97\in A\cap B~?$ If not then that means more confusion on your part.

    Is it clear to you that $101\in(\neg A\cap B)~?$ If not why not?
    Thanks from bee77
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    Re: Propositional functions/ D as domain . Ill try and solve but help would be good.

    There's a few things in there like (10^2)^2 = 10^4 that I understand ...sorry It was 1:30 A.M in the morning and I overloaded my brain with alot of discrete math ...such that simple things were looking difficult for me..
    yes you are right saying 0>A <1000 is absolute nonsense ...P(100) is false ,
    100^2 is not less than 10^4 but is = to it so therefore it is false .
    97 is an element of A as it is an integer of set A {1,2,3....98,99} and is also an intersection of prime numbers being {2,3,.....,89,97} so it is true ie 97 is in set A as an integer and in set B as a prime number...
    101 is an element of (not A intersection B ) because any integer that is positive in set A{1,2,3....98,99} and has a symbol as you wrote 101 ∈ (A ∩ B) means it has to be an integer that is 100 or over ... set B are all Prime numbers so to be an intersection it can't be 100 as 100 is a not a prime number ...ie it can be divided by more than just 1 ie 50 and 2 etc ...101 however is a prime number thus 101 is an element of both B and A.

    Thanks for your description ...the strictness is really making me work hard at it ...

    Unfortunately I haven't done math in about 15 years and am at university with young students coming fresh out of highschool with extension1 and extension2 math .(advanced math in Australia)

    so new concepts look like this to me 阪, 熊, 奈, 岡, 鹿, 梨, 阜 .....(wouldn't surprise me if you could read that too ) ;p well what I mean is a foreign language ...

    The frustrations of this you'll be pleased to know or not.....that in the weekly quizzes with these I am keeping up with them and on some occassions have bettered them in the results ..thanks to the help from you guys on this forum yes I am a frustrating student but there's a good rocky bow bower click i like to watch https://www.youtube.com/watch?v=D_Vg4uyYwEk
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