# Help needed with Boolean Algebra and DeMorgan

Discussion in 'Homework Help' started by randb, Mar 27, 2014.

1. ### randb Thread Starter New Member

Dec 29, 2012
16
0
Demorgan problem

(( AB') (A'C))' One broke bar recd (AB')' + (A'C)'

then broke the bar again and recd A'+B+A+C' A'+a =1 1+ other variables =1?

Book states answer is (A+B') ( A'+C)

Where did I go wrong?

2. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,785
771
What's wrong with your title ?

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3. ### amilton542 Active Member

Nov 13, 2010
494
64
Why are you applying boolean operators to expressions alone?

4. ### jjw Member

Dec 24, 2013
173
31
The answer in the book is wrong if the original expression is correct.

5. ### amilton542 Active Member

Nov 13, 2010
494
64
$Y = [(AB')(A'C)]'$

$Y = (AB')' + (A'C)'$

$Y = (A'+B)(A+C')$

Which is the same as you've got.

You're operating on expressions alone though, which is O.K, but could you type this "question" in it's entirety because something's missing.

Last edited: Mar 29, 2014
6. ### jjw Member

Dec 24, 2013
173
31
Last expression is missing + should be (A'+B)+(A+C')

7. ### randb Thread Starter New Member

Dec 29, 2012
16
0
Applying DeMorgan's theorem and Boolean algebra to the expression results in _____.

8. ### randb Thread Starter New Member

Dec 29, 2012
16
0
Thank you for your responses and help

9. ### amilton542 Active Member

Nov 13, 2010
494
64
Oops, yes you're right. I missed that one somehow.

In all fairness, I really don't know. Maybe someone else could help you.

10. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
Did you go wrong?

Let's check it out. You are claiming that the expression

((AB') (A'C))' = 1

This is of the form

(FG)' = 1

which is

FG = 0

F = AB'
G = A'C

If A is False, then F is False and FG is False.
If A' is False, the G is False and FG is False.

What does that tell you?

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