Results 1 to 3 of 3

Math Help - Sorry everyone!!

  1. #1
    Newbie
    Joined
    Aug 2006
    Posts
    12

    Sorry everyone!!

    I'm sorry that i'm asking so many q's it's just that im repeating my discrete maths exam nxt wk.Im studying Cs and need to pass it to get to nx yr..So here is another Q.The nand operator is important in logic circuit theory, and is defined by

    p nand q=(p^q)
    By means of truth tables or otherwise show that;
    (i)p=(p nand p)
    (ii)pVq|=(p nand p) nand (q nand p)

    preferably truth tables...
    Thanks,
    Kod
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by kodirl
    (i)p=(p nand p)
    I believe that this is wrong.
    p nand q= -(p^q)=-p v -q (By de Morgan's laws).
    You cannot conclude that,
    -p v -q = -p
    Unless -q is false, i.e. q is true.

    Quote Originally Posted by kodirl
    (ii)pVq=(p nand p) nand (q nand p)
    (p nand p) nand (q nand q) = -[(p nand p) ^ (q nand q)]
    =-(p nand p) v -(q nand q)=-[-(p ^ p)] v -[-(q ^ q)]
    =(p^p) v (q^q)=p v q
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    778
    Hello, Kid!

    Here's the first one . . .


    By means of truth tables or otherwise show that:
    . . . (1)\;\sim p \;\longleftrightarrow \;(p\text{ nand }p)

    The truth values for nand are:
    Code:
        p | q | p nand q
        - + - + - - - - - 
        T | T |    F
        T | F |    T
        F | T |    T
        F | F |    T

    For (1) we have:
    Code:
        p | ~p  <--->  (p nand p)
        - + - - - - - - - - - - - 
        T |  F    T     T  F   T
        F |  T    T     F  T   F
                  ↑
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum