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Truth Set for the given Equivalence

  1. May 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the truth set of the given equivalence. Assume U=ℝ
    #56. (x2=1)↔[(x=1)∨(x=-1)]
    Source: Principles of Mathematics by Allendoefer and Oakley section 1.10

    2. Relevant equations
    {x I px↔qx}=(P∩Q)∪(P'∩Q')

    3. The attempt at a solution
    P={x I px}={x I x2=1}={x I x=1 or -1)
    P'={x I x≠1 or -1}
    Q={x I qx is true}={x I (x=1)∨(x=-1)}
    {x I ax∨bx}=A∪B
    Let ax be x=1 and bx be x=-1
    A={x I ax} and B={x I bx}
    A={1} and B={-1}
    Q={x I x=-1 or 1}
    Q'={x I x≠-1 or 1}
    (P∩Q)={x I x=1 or -1}∩{x I x=1 or -1}={x I x=1 or -1}
    (P'∩Q')={x I x≠1 or -1}∩{x I x≠1 or -1}={x I x≠1 or -1}
    (P∩Q)∪(P'∩Q')={x I x=1 or -1}∪{x I x≠1 or -1}= U

    Just wanted to know if someone could go over my work and verify it. Thank you for your time.
  2. jcsd
  3. May 28, 2015 #2


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    The result is right, I guess there are shorter ways to prove it but it looks reasonable.
    That notation is problematic, as it does not work with a mathematical "or" here (-1 is not meant as logical statement). Same for the = before, but there it is clearer because there is no "not" involved.
  4. May 28, 2015 #3
    What do you suggest in changing? In the statement that you quoted, I'm trying to express that x cannot take both of those values. Which is a negation of the statement before in which x, could take those values. Thank you for your time.
  5. May 28, 2015 #4


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    You could write, ##P' = \{ x \in \mathbb{R} \ |\ x \ne 1 \text{ and } x \ne -1\}##. Both conditions have to be true for an element to be in P'.
  6. May 28, 2015 #5
    I'm also uploading a picture of the author using the "or" inside the logical statement. Maybe someone could explain to me how to interpret the author correctly or whether he is wrong about this.

    Attached Files:

  7. May 30, 2015 #6


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    Now you broke your proof. When you write ##Q = \{x\ |\ x=-1\text{ and }x=1\}##, you end up with Q being empty because there's no value of ##x## equal to 1 and -1 simultaneously.

    You're interpreting the author's intent correctly. I just don't like the way the author writes that kind of expression. The words "and" and "or" have a precise meaning in the context of logic, but the author is using the human language interpretation of the word "or", which doesn't align with the mathematical interpretation. So the word caught my eye and initially I thought you made a mistake until I realized what you were going for there. I wouldn't say it's technically wrong, but it's not as clear as it could be.
  8. May 30, 2015 #7
    What would be the best way to express these kinds of sets? Thank you for your help.
  9. May 30, 2015 #8


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    The way I wrote it in post #4, for example.
  10. May 30, 2015 #9
    The way you wrote it applies to the set P' where x cannot equal 1 and cannot equal -1. My question would be, how can I write a set where x is allowed to take 2 values, either one or the other. In this case, set Q. I want to express that Q can take either the value of positive one or negative one.
  11. May 30, 2015 #10


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    ##Q = \{q | q = 1 ∨ q = -1\}##
    Or you could write ##q = \pm 1##
    Last edited: May 30, 2015
  12. May 30, 2015 #11
    I think I will go the plus or minus route. Thank you again Mark.
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