Which one is the right answer in this boolean expression?
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0
down vote
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(A+B)' (A'+B')' =(A'B')+(AB)= A'B'+AB (Ans)
OR
(A+B)' (A'+B')' =(A'+B')(A+B)
= A'A+ A'B+ AB'+ BB'
= 0+ A'B+ AB'+ 0
= A'B+ AB' (Ans)
I get two different answers. I think the procedures I used in both are right. So which one is the right solution or are both of them wrong? The solution to the expression in my lecture has been given as 0 which really doesn't add up.
boolean-algebra
add a comment |
up vote
0
down vote
favorite
(A+B)' (A'+B')' =(A'B')+(AB)= A'B'+AB (Ans)
OR
(A+B)' (A'+B')' =(A'+B')(A+B)
= A'A+ A'B+ AB'+ BB'
= 0+ A'B+ AB'+ 0
= A'B+ AB' (Ans)
I get two different answers. I think the procedures I used in both are right. So which one is the right solution or are both of them wrong? The solution to the expression in my lecture has been given as 0 which really doesn't add up.
boolean-algebra
It looks like you confused it with the method for doing transposed matrix of a product of matrices.
– user376343
Nov 23 at 21:30
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
(A+B)' (A'+B')' =(A'B')+(AB)= A'B'+AB (Ans)
OR
(A+B)' (A'+B')' =(A'+B')(A+B)
= A'A+ A'B+ AB'+ BB'
= 0+ A'B+ AB'+ 0
= A'B+ AB' (Ans)
I get two different answers. I think the procedures I used in both are right. So which one is the right solution or are both of them wrong? The solution to the expression in my lecture has been given as 0 which really doesn't add up.
boolean-algebra
(A+B)' (A'+B')' =(A'B')+(AB)= A'B'+AB (Ans)
OR
(A+B)' (A'+B')' =(A'+B')(A+B)
= A'A+ A'B+ AB'+ BB'
= 0+ A'B+ AB'+ 0
= A'B+ AB' (Ans)
I get two different answers. I think the procedures I used in both are right. So which one is the right solution or are both of them wrong? The solution to the expression in my lecture has been given as 0 which really doesn't add up.
boolean-algebra
boolean-algebra
asked Nov 16 at 19:40
Sam Chau
41
41
It looks like you confused it with the method for doing transposed matrix of a product of matrices.
– user376343
Nov 23 at 21:30
add a comment |
It looks like you confused it with the method for doing transposed matrix of a product of matrices.
– user376343
Nov 23 at 21:30
It looks like you confused it with the method for doing transposed matrix of a product of matrices.
– user376343
Nov 23 at 21:30
It looks like you confused it with the method for doing transposed matrix of a product of matrices.
– user376343
Nov 23 at 21:30
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
$(A+B)'=(A'B')$ and $(A'+B')'=(AB)$. This fact does not affect the multiplication between these two terms. So you should start
$$
(A+B)'(A'+B')'=(A'B')(AB)
$$
Okay, I see why this expression ends up as 0 now.
– Sam Chau
Nov 16 at 19:49
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$(A+B)'=(A'B')$ and $(A'+B')'=(AB)$. This fact does not affect the multiplication between these two terms. So you should start
$$
(A+B)'(A'+B')'=(A'B')(AB)
$$
Okay, I see why this expression ends up as 0 now.
– Sam Chau
Nov 16 at 19:49
add a comment |
up vote
1
down vote
$(A+B)'=(A'B')$ and $(A'+B')'=(AB)$. This fact does not affect the multiplication between these two terms. So you should start
$$
(A+B)'(A'+B')'=(A'B')(AB)
$$
Okay, I see why this expression ends up as 0 now.
– Sam Chau
Nov 16 at 19:49
add a comment |
up vote
1
down vote
up vote
1
down vote
$(A+B)'=(A'B')$ and $(A'+B')'=(AB)$. This fact does not affect the multiplication between these two terms. So you should start
$$
(A+B)'(A'+B')'=(A'B')(AB)
$$
$(A+B)'=(A'B')$ and $(A'+B')'=(AB)$. This fact does not affect the multiplication between these two terms. So you should start
$$
(A+B)'(A'+B')'=(A'B')(AB)
$$
answered Nov 16 at 19:44
Arthur
108k7103186
108k7103186
Okay, I see why this expression ends up as 0 now.
– Sam Chau
Nov 16 at 19:49
add a comment |
Okay, I see why this expression ends up as 0 now.
– Sam Chau
Nov 16 at 19:49
Okay, I see why this expression ends up as 0 now.
– Sam Chau
Nov 16 at 19:49
Okay, I see why this expression ends up as 0 now.
– Sam Chau
Nov 16 at 19:49
add a comment |
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It looks like you confused it with the method for doing transposed matrix of a product of matrices.
– user376343
Nov 23 at 21:30