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Need Help in Boolean Algebra

  1. Sep 28, 2005 #1
    I would like you guys to check over my Boolean Algebra equations to see if I did them right.

    My teacher uses the same symbols as the computer science books

    Legend:
    [​IMG]= XOR

    XOR = a[​IMG]b = a * b' + a' * B

    + = OR
    * = And
    ' = Not

    a) X[​IMG]1 = X

    proof:

    X [​IMG] + 1 = Def
    X * 1' + X' * 1 = P5a
    X * 0 + X' 1 = T9a, T10a
    0 + X' = T9b, L6b

    Conclusion X [​IMG] 1 ≠ X

    b) X [​IMG] 1 = X’

    proof:

    X [​IMG] 1 = Def

    X* 1' + X' * 1 = P5a
    X * 0 + X' * 1 = T9a,T10a
    0 + X’ = L6b, T9b
    X' =

    Conclusion X [​IMG] 1 = X’

    c)
    Proof:

    X [​IMG] 0 = Def

    X * 0' + X' * 0 = P5b
    X * 1 + X' * 0 = T10a
    X + X' * 0 = T9a
    X + 0 = T9b
    X=

    Conclusion X [​IMG] 0 = X

    d)
    X [​IMG] ) = X'

    Proof:

    X [​IMG] 0 = Def
    X * 0' + X' * 0 = P5b
    X * 1 + X' * 0 = T10a
    X + X' * 0 = T9a
    X + 0 = T9b
    X =

    Conclusion X [​IMG] 0 ≠ X'

    e)
    X [​IMG] X' = 0

    Proof:

    X [​IMG] X' = Def

    X * X'' + X' * X' = T13a
    X * X + X' * X' = T11a
    X + X’ * X' = T12a
    X + 0 = T9b
    X =

    Conclusion X [​IMG] X' ≠ 0

    --------------------------------------------------------

    Postulates:

    P1a X = 1 if X ≠ 0 ----- P1b X = 0 if X ≠ 1
    P2a 0*0 = 0 ----- P2b 0+0 = 0
    P3a 1*1 = 1 ---- P3b 1+1 = 1
    P4a 1*0 = 0 ----- P4b 1+0 = 1
    P5a 1’ = 0 ----- P5b 0’ = 1

    Algebraic Laws

    Commutative laws

    L6a x*y = y*x ----- L6b x+y = y+x

    Associative laws

    L7a x*(y*z) = (x*y)*z ----- L7b x+(y+z) = (x+y)+z

    Distributive laws

    L8a x*(y+z) = x*y+x*z ----- L8b x+y*z = (x+y)*(x+z)

    Thermos

    T9a x*0 = 0 ------ T9b x+0 = 0
    T10a x*1 = x ------ T10b x+1 = 1
    T11a x*x = x ------ T11b x+x = x
    T12a x*x’ = 0 ------ T12b x+x’ = 1
    T13a x’’ = x ------ T13b x = x’’

    Absorption Theorems

    T14a x+x*y = x -------- T14b x*(x+y)=x
    T14c X*(x’ + y) = x*y ------- T14d x+x’*y= x+y

    De Morgan’s Theorems

    T15a (x*y*z)’ = x’ + y’ + z’
    T15b (x+y+z)’ = x’ * y’ * z’
     
  2. jcsd
  3. Sep 29, 2005 #2

    CarlB

    User Avatar
    Science Advisor
    Homework Helper

    This was amazingly unreadable.

    Carl
     
  4. Sep 29, 2005 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, I understand that your link is supposed to be the symbol [itex]\oplus[/itex]. (Click on the image to see what you can type to generate that image) So, for example, the thing at the top is supposed to say [itex]a \oplus b = a \cdot b' + a' \cdot b[/itex]. (Or are you using [itex]\bar{a}[/itex]?)


    Oh, I think I get what you're trying to ask here -- each "x) _____" is the problem you were given, and what's below is your proof or disproof, and you're looking for us to check over your work.

    I will say one thing: you have the idea of a disproof wrong. To disprove something, what you should be doing is to come up with a particular example for which the statement is incorrect. E.G. because [itex]0 \oplus 1 = 1[/itex], that is sufficient for disproving [itex]x \oplus 1 = x[/itex], because this counterexample shows that it fails when x = 0.

    Now, if you can prove a statement is always wrong, that is enough to disprove that statement, but many times it is impossible to do that. For example, [itex]x + 0 = 1[/itex] is not a tautology, but it is impossible to prove that it is always wrong.


    Now, the format of your proofs are very confusing. :frown: For (a), for example, it looks like you're trying to say:

    [tex]x \oplus 1 = x \cdot 1' + x' \cdot 1 = x \cdot 0 + x' \cdot 1 = 0 + x'[/tex]

    And giving a justification for each equality. Aside from the typos you made, the way you've actually written that is extremely confusing. Normally, when you spread an equation out over multiple lines, you put the equals signs on the left. E.G.

    x [itex]\oplus[/itex] 1 = x 1' + x' 1
    = x 0 + x' 1
    = 0 + x'
    = x'

    (Normally you would align the equals signs, but that's difficult in this text format)

    And when you're giving justifications for steps, you shouldn't make them look like part of the equation. In a text format like this forum, I might have written it something like:

    x [itex]\oplus[/itex] 1 = x 1' + x' 1 (by definition)
    = x 0 + x' 1 (by P5a)
    ...

    I suppose you can write it in a two-column format when you turn in your homework, though. In text, though, you really need some sort way to visually separate the justifications from the equations.


    Anyways, the steps of your work generally look right, though I haven't tried matching up the steps with the justification.
     
  5. Sep 29, 2005 #4
    Yes I did not know that this forum did have that symbol since another fourm I posted it did not have the symbol and it was just a copy and paste from it.

    As far as you saying I don't understand how to disprove stuff or whatever. It maybe true but from the examples given from my teacher that is they way she writes the conlusions up so I am just following the way she does it.

    And the format I use is the same as my teacher.

    So I rather keep it the way she does since she seems to like it like that.
     
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