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Boolean Algebra Questions and Answers
Boolean Algebra:
Boolean logic, originally developed by George Boole in the mid 1800s. The outcomes of the Boolean quantities are either true (1) or false (0).
Boolean Algebra Questions and Answers :
Boolean Algebra is a digital electronic subjects, involves simplification of logical expression. By practicing a sets of problems on Boolean helps to understand the logical relationship between the variables. The equation is combination of sum of product and product of sum's
Laws of Boolean Algebraic Expression are:
We have given some of the Boolean algebra question answer for self study and understanding of the laws.
Simplify the Equation:
Laws of Boolean Algebraic Expression are:
Sl.No  Law  Expression 
1  Commutative Law  A + B = B + A A B = B A 
2  Associate Law  (A + B) + C = A + (B + C) (A B) C = A (B C) 
3  Distributive Law  A (B + C) = A B + A C A + (B C) = (A + B) (A + C) 
4  Identity Law  A + A = A A A = A 
5  Redundant Law  A + A B = A A (A + B) = A 
0 + A = A 0 A = 0 1 + A = 1 1 A = A $\bar{A}$ + A = 1 $\bar{A}$.A =0 A + AB = A + B A($\bar{A}$+ B) = AB 

6  De Morgan's Theorem  $(\overline{A+B})=\bar{A}\bar{B}$ $(\overline{AB})=\bar{A}+\bar{B}$ 
We have given some of the Boolean algebra question answer for self study and understanding of the laws.
Simplify the Equation:
1. (A + C)(AD + AD) + AC + C
(A + C)A(D + D) + AC + C // Distributive rule //
(A + C)A + AC + C // Complementary Identity. //
A((A + C) + C) + C // Commutative & Distributive //
A(A + C) + C
AA + AC + C // simplify //
A+(A+1)C // A+1 = 1 //
A+C
2. Prove that LHS = RHS
$\bar{A}$BC + A$\bar{B}$C + AB$\bar{C}$+ABC = AB+BC+AC
Take the LHS ABC + ABC + ABC + ABC
= $\bar{A}$ BC+A$\bar{B}$C+AB$\bar{C}$+ABC // TABE AB as common term //
= $\bar{A}$ BC+A$\bar{B}$C+AB($\bar{C}$+C ) // $\bar{C}$+C = 1 //
= $\bar{A}$ BC+A$\bar{B}$C+AB // take A as common term //
= $\bar{A}$ BC+A($\bar{B}$C+B) // B+BC = B+C //
= $\bar{A}$BC+A($\bar{B}$+C)
= $\bar{A}$BC+AB+AC
= C(AB+A)+AB+AC
= C(B+A)+AB+AC
= CB+AC+AB+AC // AC+AC = AC //
= AB+BC+AC
= RHS
LHS = RHS
3. Give an example for sum of products form of equation.
AB + CD
4. Applying DeMorgan's theorem to the expression $\overline{\overline{(X+Y)}+\overline{Z}}$
$\overline{\overline{(X+Y)}+\overline{Z}}$ \\ $\overline{(\bar{A})}$ = A //
= (X + Y)+Z
Best Results From Yahoo Answers
From Yahoo Answers
Question:The question is to simplify ab+b`c+ac`d
My question is Is it valid to put parenthesis here: (ab+b`)(c+ac`d)?
if so that would make the question quite simple but if it cannot then the most simple expression I can come up with is
a(b+c`d)+b`c
Answers:a(b+c'd)+b'c is correct. if u had eg. a+bc+d u cant do (a+b)(c+d) cause otherwise that would mean that if u open brackets u wld get the same answer but u dont.. u wld get ac+ad+bc+bd = a(c+d)+b(c+d) hope this helped :)
Answers:a(b+c'd)+b'c is correct. if u had eg. a+bc+d u cant do (a+b)(c+d) cause otherwise that would mean that if u open brackets u wld get the same answer but u dont.. u wld get ac+ad+bc+bd = a(c+d)+b(c+d) hope this helped :)
Question:The problem is to prove that the statement is true through algebraic manipulation:
AD' + A'B + C'D + B'C = (A'+B'+C'+D')(A+B+C+D)
Or in another format:
(A AND NOT D) OR (NOT A AND B) OR (NOT C AND D) OR (NOT B AND C) equals (NOT A OR NOT B OR NOT C OR NOT D) AND (A OR B OR C OR D)
I've tried several different approaches but have gotten nothing near the RHS of the equation. On my most recent attempt through usage of distributive property/consensus theorem I arrived at: (C'+D')(D+A) + (B+C)(A+B'). Any tips on which expressions to combine, where to multiply by various identities, partial solutions, etc are greatly appreciated. Thanks for helping! Edit: Solved the problem by using consensus theorem to expand items rather than condensing them
Answers:I haven't solved it, but I have one insight that may help you. I'll admit that I'm not using strict algebraic operations. Looking at the righthand side of the equation, we can see that it is only TRUE if the set of values ABCD contains at least one TRUE and at least one FALSE. The "all TRUEs" and "all FALSEs" cases will both make the right side FALSE. (Don't use pencil and paper; just look at it.) So, another equality would be NOT((ABCD)(A'B'C'D')), right? Maybe that helps. Or start writing "at least one TRUE and at least one FALSE" expressions: A'(B+C+D) + B'(A+C+D) + C'(A+B+D) + D'(A+B+C) Which should be easy to simplify into a bunch of twovariable terms, many of which may be identical. There's a lot of gut feel, here. But I haven't done anything invalid. Interesting problem!
Answers:I haven't solved it, but I have one insight that may help you. I'll admit that I'm not using strict algebraic operations. Looking at the righthand side of the equation, we can see that it is only TRUE if the set of values ABCD contains at least one TRUE and at least one FALSE. The "all TRUEs" and "all FALSEs" cases will both make the right side FALSE. (Don't use pencil and paper; just look at it.) So, another equality would be NOT((ABCD)(A'B'C'D')), right? Maybe that helps. Or start writing "at least one TRUE and at least one FALSE" expressions: A'(B+C+D) + B'(A+C+D) + C'(A+B+D) + D'(A+B+C) Which should be easy to simplify into a bunch of twovariable terms, many of which may be identical. There's a lot of gut feel, here. But I haven't done anything invalid. Interesting problem!
Question:Hello, I have been working on solving these canonical sum of minterm:
A'B'C'D + A'B'CD + AB'C'D' + AB'C'D + AB'CD + ABC'D' + ABC'D + ABCD
I know the answer is
A'C + AD + B'D
But I have not simplified it successfully.
Please help me solve this and show your work.
I would appreciate it a lot.
Thak you!! heidegger_001:
Yes, also used a Karnaugh map to know the answer. Unfortunately when I typed my question I put the term A'C which should have been AC'.
What I need is to show the boolean algebra involved.
Thank you for the time you took to work the problem!
Answers:I think Karnaugh mapping is a faster way than Boolean algebra when dealing with this type of equation. Simplifying this with Boolean algebra is too long for me. Using Karnaugh mapping i came up with C'D' C'D CD CD' A'B' 0 1 1 0 A'B 0 0 0 0 AB 1 1 1 0 AB' 1 1 1 0 According to this kmap I came up with the answer AC' + AD + B'D.
Answers:I think Karnaugh mapping is a faster way than Boolean algebra when dealing with this type of equation. Simplifying this with Boolean algebra is too long for me. Using Karnaugh mapping i came up with C'D' C'D CD CD' A'B' 0 1 1 0 A'B 0 0 0 0 AB 1 1 1 0 AB' 1 1 1 0 According to this kmap I came up with the answer AC' + AD + B'D.
Question:Well I have a quick issue, its my first time getting my hands on some boolean algebra and I need to reduce some equations down to the bare minimum, I have some basic identities to use but I dont quite understand the feel for it just yet...
Take for example if an equation is A+A'B+AB how would one go about reducing this thing? I look at the basic identities and I'm not too sure if someone could do this but I see WITHIN this equation an A+A', am I able to use the identity on the A+A' even though the A' is paired up with the B?
Answers:You can use Karnaugh maps, which may be beyond the scope of what you're suppose to do. You cannot do A + A', because its paired with an AND But there is an identity that says: A'B + AB = B Therefore your example would be simplified to A + B. an easy check for this is to see that whenever you plug in a 1 for B you get 1 for the equation.
Answers:You can use Karnaugh maps, which may be beyond the scope of what you're suppose to do. You cannot do A + A', because its paired with an AND But there is an identity that says: A'B + AB = B Therefore your example would be simplified to A + B. an easy check for this is to see that whenever you plug in a 1 for B you get 1 for the equation.