Please help me for midterm ON GRADE 11 boolen algebra

[joker_tony]

i really need someone to me verify my answer. i will be having my exam next week and iam stuck . please iam really stuck on some of this question.

' is not


'' its not, not


1. F = B.(A''. B' . C')

2. F = A. (A''+ B' + C')

3. F = A . (B''.C') + A . (B + C')

4. F = (A'' + B'') + A . B

5. F = A'.B'+C'+A.(B'+C')

6. F= A'. B+ D.(C+D') + A. B' . C' . D

7. F = A. C'.D'+(A''.B'.C') + (A''.B')

8. F = A.B'.C' + A'.B.C.D + A'.B.C'+A.B'C.D


MY ANSWERS

1. F=BA'.BC'
2 F=A.AB.AC'
3. F= AB + AC
4. F=A.B

 

[berkeman]

i really need someone to me verify my answer. i will be having my exam next week and iam stuck . please iam really stuck on some of this question.

1. F = B.(A''. B' . C')

2. F = A. (A''+ B' + C')

3. F = A . (B''.C') + A . (B + C')

4. F = (A'' + B'') + A . B

5. F = A'.B'+C'+A.(B'+C')

6. F= A'. B+ D.(C+D') + A. B' . C' . D

7. F = A. C'.D'+(A''.B'.C') + (A''.B')

8. F = A.B'.C' + A'.B.C.D + A'.B.C'+A.B'C.D


MY ANSWERS

1. F=BA'.BC'
2 F=A.AB.AC'
3. F= AB + AC
4. F=A.B

Could you please explain your notation? What is the difference between ' and " ? Also, could you please explain what techniques you are using to derive your answers? We do not give out answers to homework/coursework questions here on the PF, but we can provide tutorial help so that you can figure out the answers yourself.

 

[joker_tony]

' is not

'' its not, not


I just need someone to check my answers and i would really apreciated

 

[berkeman]

' is not

'' its not, not

Um, okay. What would not not always equal?

And we really do need to see what you are doing to derive your answers. What distributive rules are you using, for example? Please show your step-by-step work for each of the answers, so that we can see how you worked them out. Alternately, show the Karnaugh maps you used, if that's how you solved the problems.

 

[berkeman]

I just need someone to check my answers and i would really apreciated

As I said, we do not give out answers here. We also do not just check answers. That would be cheating. Please show all of your work.

 

[joker_tony]

step by step solution

1. AB'+BB'+BC'
= AB'+BC'

2. AA.AB.AC'
=A.AB'.AC'

3. AB+AC+AB+AC
=AB+AC

4. (A.B)+A.B
=A.AB.AB.B
=A.B

i don't know if this is right, also iam working on the other answers

 

[berkeman]

That's much better, but it would still help us if you could say what rule you are using in each step. Like the "Monotone Laws" listed at wikipedia.org (with v = OR and ∧ = AND):

from http://en.wikipedia.org/wiki/Boolean_algebra_(introduction)

Monotone laws
Boolean algebra satisfies many of the same laws as ordinary algebra when we match up ∨ with addition and ∧ with multiplication. In particular the following laws are common to both kinds of algebra.

(Associativity of ∨) x∨(y∨z) = (x∨y)∨z
(Associativity of ∧) x∧(y∧z) = (x∧y)∧z
(Commutativity of ∨) x∨y = y∨x
(Commutativity of ∧) x∧y = y∧x
(Distributivity of ∧ over ∨) x∧(y∨z) = (x∧y)∨(x∧z)
(Identity for ∨) x∨0 = x
(Identity for ∧) x∧1 = x
(Annihilator for ∧) x∧0 = 0

Boolean algebra however obeys some additional laws, in particular the following.

(Idempotence of ∨) x∨x = x
(Idempotence of ∧) x∧x = x
(Absorption 1) x∧(x∨y) = x
(Absorption 2) x∨(x∧y) = x
(Distributivity of ∨ over ∧) x∨(y∧z) = (x∨y)∧(x∨z)
(Annihilator for ∨) x∨1 = 1

 

[joker_tony]

yes iam using those laws but i don't know if my answers are right or wrong

 

[berkeman]

yes iam using those laws but i don't know if my answers are right or wrong

Yes the laws are right. They were on wikipedia, after all

But seriously, what I mean is to annotate each line in your solution with a comment on what law you've used to make that step. Like

1. F = B.(A''. B' . C')
= AB'+BB'+BC' (which law or laws were used in this step?)
= AB'+BC' (what property of BB' did you use?)

 

[joker_tony]

step by step solution

1. AB'+BB'+BC' (use complements laws B.B=0)
= AB'+BC'

2. AA.AB'.AC' ( Use idempotency A.A= A)
=A.AB'.AC'

3. AB+AC+AB+AC(use idempotency A+A=A)
=AB+AC

4. (A.B)+A.B ( i think this one is wrong)
=A.AB.AB.B
=A.B

 

[berkeman]

Looks like you're getting the hang of it. On #4...

First of all, I've never seen not-not, but whatever. Anything not-not would be itself, I would think. So:

4. F = (A'' + B'') + A . B = (A + B) + A . B

But you can use the Absorption #2 listed from wikipedia, or just truth tables to simplify that further...