Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Callisto Group Title

Boolean algebra How can I get \(a+b+\bar{a}\) from \(a+b+\bar{a}\bar{b}\)?

  • one year ago
  • one year ago

  • This Question is Closed
  1. modphysnoob Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Try DeMorgan

    • one year ago
  2. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[a+b+\bar{a}\bar{b}\]\[=\overline{\bar{a}\bar{b}} + \bar{a}\bar{b}\]\[=1\]*cough* I just get the answer directly!?

    • one year ago
  3. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    what is bar ?

    • one year ago
  4. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Does \(+\) mean \(\wedge\) and \(\bar{a}\) =\(\neg{a}\)

    • one year ago
  5. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\overline{a}\] = complement of a

    • one year ago
  6. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay, so \(\neg a\) indeed.

    • one year ago
  7. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[a + b + \bar{a} \iff a \wedge b \wedge \neg a\]It's zero . . .

    • one year ago
  8. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    The answer is 1 :|

    • one year ago
  9. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    So does \(+\) stand for the OR operator?

    • one year ago
  10. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes.

    • one year ago
  11. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    then it is 1 indeed since either a or neg(a) is 1 :-)

    • one year ago
  12. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    What does \(ab\) mean? And?

    • one year ago
  13. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes :| Don't you write in this way?

    • one year ago
  14. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    You can write this both ways :-)

    • one year ago
  15. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[a\lor b\lor \neg a\iff a\lor b\lor\neg a\land\neg b\] ?

    • one year ago
  16. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes, they're pretty much equivalent using truth tables.

    • one year ago
  17. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I can't read (don't understand) all those symbols!! :(

    • one year ago
  18. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Eh, it's easy once you are used to it.

    • one year ago
  19. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Apparently we're not even caring about \(c\) in that expression.

    • one year ago
  20. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    truth table ?

    • one year ago
  21. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    If ∨ = OR ¬ = NOT ∧ = AND Then, the way to get a∨b∨¬a⟺a∨b∨¬a∧¬b is what I want know :|

    • one year ago
  22. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Well, manipulation. I thought you wanted a proof.

    • one year ago
  23. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It's just that if b is positive then you have the whole thing true, and if it's false then definitely ¬a∧¬b is true. If it's not true, then definitely a is true, so the whole expression will be true. So we have both expressions equivalent.

    • one year ago
  24. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\begin{array}{|c|c|c|c|}\hline a&\bar a&b&\bar b&\bar a\bar b&a+b+\bar a& a+b+\bar a \equiv \bar a\bar b \\\hline\ \end{array}\]

    • one year ago
  25. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Dangy, that's just a truth table. I was waiting to see the manipulations @UnkleRhaukus does.

    • one year ago
  26. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I can get a∨b∨¬a = 1 and a∨b∨¬a∧¬b =1 (so they must be equal), but I don't know how to get a∨b∨¬a from a∨b∨¬a∧¬b

    • one year ago
  27. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Because they have the same boolean values for each case.

    • one year ago
  28. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ...... I know they are equal, but how to get a∨b∨¬a from a∨b∨¬a∧¬b without showing that they are both equal to 1 and without using truth table (I'm sorry :( )?

    • one year ago
  29. UnkleRhaukus Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    is \(a\)= 1 , \(\bar a\)= 0 ?

    • one year ago
  30. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes.

    • one year ago
  31. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @UnkleRhaukus \(\neg\) lol

    • one year ago
  32. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    If a = 1, then \(\bar{a}\)=0.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.