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The word equvilalent

  1. Oct 5, 2011 #1
    The word "equvilalent"

    1. The problem statement, all variables and given/known data



    Yes I know I spelled equivlaence wrong.

    [PLAIN]http://img834.imageshack.us/img834/9285/unledkzo.png [Broken]



    3. The attempt at a solution

    What do they mean by equivlance? What is to show?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 5, 2011 #2

    SammyS

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    Re: The word "equvilalent"

    Statement (i) means the same thing as statement (ii) which means the same thing as statement (iii) .

    They're equivalent logically speaking.

    (A ⊆ B) iff (A ∩ B = A)

    etc.
     
  4. Oct 5, 2011 #3

    Mark44

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    Re: The word "equvilalent"

    Two statements P and Q are equivalent if P implies Q and Q implies P. To show that three statements, P, Q, and R, are equivalent you could show that P implies Q and Q implies P, then show that P implies R and R implies P, and finally, show that P implies R and R implies P.

    A simpler way to show that statements P, Q, and R are equivalent is to show that these implications are true.

    P ==> Q (that is, P implies Q)
    Q ==> R
    R ==> P


    See above.
    No, it's also true if A is a subset of B, which is what statement i says.
    The three statements don't stand alone. What you are supposed to do is to show that they are all equivalent; loosely speaking, that they all say the same thing.
     
    Last edited by a moderator: May 5, 2017
  5. Oct 5, 2011 #4
    Re: The word "equvilalent"

    Before I even attempt this (don't want to head a long direction), this is my idea.

    SHould I let some element x be in A and then deduct that since (i) A is a subset of B then x is also in B?

    Then to go from there, I say that x is indeed (from what I just said) is in the intersection of A and B, i.e. x is in A ∩ B, which also means that x is in A (I am not sure about the whole A ∩ B = A (being only A?)

    Then to go from there, I would say that from (i), since x is in A or in B, it follows that x is in either A or B?

    That equal sign is messing me up
     
  6. Oct 5, 2011 #5

    Mark44

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    Re: The word "equvilalent"

    No, you can assume that statement i is true, then deduce that statement ii is true.

    IOW, assume that A [itex]\subseteq[/itex] B. Now let x be an element of A ∩ B. Show that x is necessarily in A.

    And so on.
     
  7. Oct 5, 2011 #6
    Re: The word "equvilalent"

    But that doesn't say if ii) is true, then i) is true. What if you assume ii) is true first? (follow below)

    No that can't right

    [tex]x \in A \cap B \iff x \in A[/tex] and [tex]x \in B[/tex]


    [tex]\therefore x \in A[/tex]

    But that doesn't say that [tex]A \subseteq B[/tex]

    And I just realize I did nothing at all...


    Let's say now I do believe you (or the question) that [tex]A \subseteq B[/tex]

    (ii)
    [tex]x \in A \cap B \iff x \in A[/tex] and [tex]x \in B[/tex]


    [tex]\therefore x \in A[/tex]

    But (i) is true, so that [tex]x \in B[/tex] is unnecessary, and that only [tex]x \in A[/tex] because A is a subset of B

    (iii) [tex]x \in A \cup B \iff x \in A[/tex] or [tex]x\in B[/tex]

    Since (i) is true, i.e., then only [tex]x \in B [/tex] is necessary

    How do I link (ii) and (iii) together? They give off the same results from agreeing on (i), but they don't have relations
     
    Last edited: Oct 5, 2011
  8. Oct 5, 2011 #7

    Mark44

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    Re: The word "equvilalent"

    Assume i is true.
    Show that if i is true, then ii must be true.
    Show that if ii is true, then iii must be true.
    Show that if iii is true, then i must be true.

    If you have done all this, then you're finished. How does this show that ii ==> i? It does so in a roundabout way.
    Having done all the things I said in the first paragraph, we have
    ii ==> iii ==> i

    How do you link ii and iii together?
    Assume ii is true (i.e., that A ∩ B = A). Show that this being true implies that A U B = B.

    It would be helpful to your understanding to draw a simple Venn diagram, but you won't be able to use the diagram in your actual proof.
     
  9. Oct 5, 2011 #8
    Re: The word "equvilalent"

    My example in post 6 says it is true, so i know they are true, just how to prove it.

    So I just showed that

    Assume i) is true and that ii) is true

    Then I showed that

    Assume i) is true, then iii) is true

    Now I need to show that ii) and iii),


    [tex]x \in A \cap B = x\in A[/tex]


    I don't know how to go from here.

    My so far proof did

    i) -> ii)
    i -> iii)

    Does ii) -> iii) complete the circle?
     
  10. Oct 5, 2011 #9

    SammyS

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    Re: The word "equvilalent"

    For two sets, C & D : To show (prove) they are equal, you must show that C ⊆ D and that D ⊆ C .

    So for (i) implies (ii), show that A ∩ B = A is true by proving that A ∩ B ⊆ A and that A ⊆ A ∩ B , of course you prove these assuming that A ⊆ B is true.
     
  11. Oct 5, 2011 #10

    Mark44

    Staff: Mentor

    Re: The word "equvilalent"

    No, because you don't have a circle.
    You have i --> ii, ii --> iii, and i --> iii. You don't have anything that shows i is true.

    Did you not understand this?
     
  12. Oct 5, 2011 #11

    SammyS

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    Re: The word "equvilalent"

    fp,

    If i --> ii and ii --> iii then you have i --> iii by default.

    That along with iii --> i , gives i <--> iii .
     
  13. Oct 5, 2011 #12
    Re: The word "equvilalent"

    Show that if (i) is true, then (ii) is true


    This is a bit messy, but I think I did this part correctly.

    [tex]x \in A \cap B \iff x \in A[/tex] and [tex] x\in B[/tex]


    [tex]x \in A \iff x \in B \iff x \in A \cap B[/tex]

    Assume [tex]A \subseteq B[/tex] is true

    Then [tex]x \in A \iff x \in A \cap B[/tex]

    So I just showed

    phew...
     
  14. Oct 5, 2011 #13

    SammyS

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    Re: The word "equvilalent"

    What do you mean by your two way arrows?

    I know what Mark means by them, nut what do you mean?
     
  15. Oct 5, 2011 #14
    Re: The word "equvilalent"

    if and only if? That seems to be used correctly in this problem
     
  16. Oct 5, 2011 #15

    SammyS

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    Re: The word "equvilalent"

    Some of the iff's shouldn't be there.
     
  17. Oct 5, 2011 #16
    Re: The word "equvilalent"

    You mean this right...?

    [tex]x \in A \iff x \in B \iff x \in A \cap B[/tex]


    So everything after this is wrong...

    that is my guess because that looks the most suspicious.

    I am going nowhere with this...
     
  18. Oct 5, 2011 #17
    Re: The word "equvilalent"

    Actually wait, something hit me


    [tex]x \in A \iff x \in B \implies x \in A \cap B[/tex]

    I just have to change iff to an "implies"

    Which flows better
     
  19. Oct 5, 2011 #18

    Mark44

    Staff: Mentor

    Re: The word "equvilalent"

    The above makes no sense as you have written it.

    [tex]x \in A ~\text{and}~ x \in B \iff x \in A \cap B[/tex]

    Definitions are often written as iff statements. Otherwise, P <==> Q means P ==> Q and Q ==> P.
     
    Last edited: Oct 5, 2011
  20. Oct 5, 2011 #19
    Re: The word "equvilalent"

    Sigh, this is due tomorrow and i have another midterm tomorrow.

    But it does make sense because I assumed that [tex]A \subseteq B[/tex]
     
  21. Oct 5, 2011 #20
    Re: The word "equvilalent"

    I've attached a picture of where I was going with my algebra

    [PLAIN]http://img198.imageshack.us/img198/91/unledctr.png [Broken]
     
    Last edited by a moderator: May 5, 2017
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