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Idempotency (Boolean Algebra Theorem)

  1. May 27, 2013 #1

    JJBladester

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    Gold Member

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

    The idempotency theorem X∙X=X is defined for one variable, but does it apply to logic expressions with multiple variables?

    Example:
    ABC∙ABC = ABC

    Is this valid?

    2. Relevant equations

    Switching algebra single-variable theorems:

    X+X = X
    X∙X = X

    3. The attempt at a solution

    My book gives single-variable and general theorems in its switching algebra section. The single-variable theorem is listed above. The general idempotency theorems are:

    X+...+X = X
    X∙X...∙X = X

    However, it says nothing about whether or not the expression ABC∙ABC = ABC can be made.
     
  2. jcsd
  3. May 28, 2013 #2

    NascentOxygen

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    Staff: Mentor

    Hi JJBladester. The dot is AND, I presume? Could you draw a truth table to test whether A.B.C = ❲A.B.C❳.❲A.B.C❳
     
  4. May 28, 2013 #3

    JJBladester

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    Gold Member

    Idempotency (Multiple Variables)

    I use the middle dot for AND and the plus symbol for OR. Some authors use the ampersand symbol or even a period to mean AND.

    I've drawn the truth table below and it appears that idempotency is valid for multi-variable terms because F3=F1&F2. Am I correct?

    idempotency.jpg

    My professor claims that "the theorem is for single input and single output logic" but I don't know if that's correct because of my table above...
     
  5. May 28, 2013 #4
    You know that [itex]X \land X = X[/itex] is true, so why not just let [itex]X = A \land B \land C[/itex]?
     
  6. May 28, 2013 #5

    JJBladester

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    I believe what you suggested is valid. This is why I am utterly confused why my professor said that the idempotency theorem is "for single input and single output logic".

    I could also re-write the original expression as:

    (A∙B∙C)∙(A∙B∙C)=(A∙A)(B∙B)∙(C∙C)=A∙B∙C

    Which uses the associativity theorem and then applies the idempotency theorem to single variables.

    My book says "In all of the [switching algebra] theorems, it is possible to replace each variable with an arbitrary logic expression" which seems to back up what you said (let X = A∙B∙C).
     
  7. May 28, 2013 #6
    Your theorem states that [itex]X \land X = X[/itex] is true for any X, where X is a variable that has a truth value. [itex]A \land B \land C[/itex] has a truth value so it follows that [itex](A \land B \land C) \land (A \land B \land C) = A \land B \land C[/itex].
     
  8. May 29, 2013 #7

    NascentOxygen

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    It looks like you have it all figured out.
     
  9. May 29, 2013 #8

    JJBladester

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    Yep!

    Yep, thanks to you and miles. You've helped me on various things in the past. As I work my way through my degree I am always pleasantly surprised at the helpfulness of PF members.
     
  10. May 29, 2013 #9
    I'm not entirely sure what your professor meant, but you were critical and did good work in proving what you thought to be true.
     
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