# I need help in boolean algebra?

Discussion in 'Math' started by HMHQ8, Jan 30, 2008.

1. ### HMHQ8 Thread Starter Member

Jan 29, 2008
10
0
Hi

I need help in solving these problems, for my homework!!

I really need to understand the solution

2. Assuming that the Boolean equation on the left is valid, write several Boolean expressions that are equivalent to the expression on the right.

a. A = A . A A . B + A . C

b. A . B = B . A (c . d) . (f . e) . (b . A)

c. (A . B) = A + B ((X1 . X2) . (X3/X4))

2. ### HMHQ8 Thread Starter Member

Jan 29, 2008
10
0
Sorry, these are at the correct way:

a. A = A . A A <====> B + A . C

b. A . B = B . A <====> (c . d). (f . e) . (b . A)

c. (A . B)’ = A’ + B’ <===> ((X1 . X2)’ . (X’3/X’4))’

3. ### HMHQ8 Thread Starter Member

Jan 29, 2008
10
0
sorry again, I dont know what happened to me
(a) should be : A = A . A <=====> A . B + A . C

4. ### Dave Retired Moderator

Nov 17, 2003
6,960
144
I assume the trick is to work out what the example on the left is doing to produce a valid solution and apply it to the right.

This one looks like simplification by common factors, therefore A.B + A.C = A(B + C)

This one is an example of the Commutative Property of Addition whose principles are equally applicable to the Commutative Property of Multiplication. Take your pick on the order of the brackets.

The example on the left is DeMorgans theorems. Since I don't know what the "/" operator is I cannot help.

Dave

5. ### HMHQ8 Thread Starter Member

Jan 29, 2008
10
0
Thank you Dave, u r the man

6. ### Dave Retired Moderator

Nov 17, 2003
6,960
144
Can you tell us what the "/" operator is? Perhaps you could scan in the question paper.

I'd say the answer is: ((X1 . X2)’ . (X’3/X’4))’ = (X1 . X2) + (X'3/X'4)'

The (X'3/X'4)' term could possibly be simplified further using DeMorgans Theorem, but we need to know the meaning of the "/" operator.

Dave

7. ### HMHQ8 Thread Starter Member

Jan 29, 2008
10
0
I check with my Proff. about ''/'' operater and he says its a trick problem, because there is not a ''/'' operater in boolean algebra!!

8. ### Dave Retired Moderator

Nov 17, 2003
6,960
144
Yes, and we established that here

Sometimes people use all sorts of symbols to denote Boolean operators on the forums - in this case it was actually a division operator. You can use the LaTex feature of the forums to script Boolean equations (on the Post Reply screen press the SIGMA button in the upper right-hand corner and select "Above and Below").

$\overline{A.B}$$=$$\overline{A}$$+$$\overline{B}$

Is one of DeMorgans theorems!

Dave