# Boolean algebra

Discussion in 'Homework Help' started by lusito92, Feb 4, 2013.

1. ### lusito92 Thread Starter New Member

Jan 29, 2013
25
0
Hello,
Im trying to solve this with identities

xz+z /z + zxy = z
z ( x +/x + xy) <-- x +/x = 1 (identity)
z (xy) = z so now what? Im trying to find identity to get rid xy but I dont know how to do it ..

2. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,328
Where are you getting xz+z /z + zxy?

Be careful with your distribution. Once you distribute everything correctly, you'll see a common factor.

EDIT: Just realized you're using "/" as a "not" symbol. I suggest you use an apostrophe after the variable instead of a / before it.

Last edited: Feb 4, 2013
3. ### lusito92 Thread Starter New Member

Jan 29, 2013
25
0
Sorry..
xz + z(x'+xy) = z <-- x' + xy = x' + y
xz + z ( x' + y ) = z
xz + zx' + zy = z < -- distributive
z ( x + x' + y ) = z <- x + x' = 1
z ( 1 + y ) = z <-- Im stuck! what can I do ?

4. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
what is any variable ORed with 1?

5. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
Code ( (Unknown Language)):
1. x' + xy = x' + y
I've never seen this identity, if that was taught as one, great, if not, I think you need to expand this out...

6. ### lusito92 Thread Starter New Member

Jan 29, 2013
25
0
well in my notes that my teacher gave me says C' +CB = C+B

7. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
Well, then you're good. I've never seen given like that, but since your teacher gave that, use it!

8. ### lusito92 Thread Starter New Member

Jan 29, 2013
25
0
but Im still stuck because you can see I carry out
xz + z ( x' + y ) = z
xz + zx' + zy = z < -- I expanded everything, and then the common factor is z and I got..
z ( x + x' + y ) = z <- since I have another x + x' = 1 according to my teacher A + A' = 1
z ( 1 + y ) = z <-- now Im not sure, I found another A + 1 = 1 ? is that can work 1 + y what do you think?

9. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,328
If anything is "OR"ed with a 1, the outcome is 1. So z (1 + y) is the same as z (1). What is Z AND 1?

10. ### lusito92 Thread Starter New Member

Jan 29, 2013
25
0
z (1 ) = z is that the way I should do it ?

11. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,328
It works, doesn't it?

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12. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
...sounds like a logical progression

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13. ### lusito92 Thread Starter New Member

Jan 29, 2013
25
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Now I have another that Im stuck sorry
x'+y' + xyz'= x'+y'+ z'
No idea how to do it .. the only think I can find x' y' could be xy'' ( xy are not) but there are not common factors or something in this equation that can lead me to do something

14. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,328
Remember the rule A' + AB' is A' + B'. If you rearrange a bit, you will see that this can help you twice.

Matt

15. ### lusito92 Thread Starter New Member

Jan 29, 2013
25
0
Sorry I dont know that rule, but I have this one with only one not .. A + A'B, but iff I have the whole term together because is XYZ' and the rule says A' + AB' ? is that applies?

16. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,328
Hmm, I was thinking it worked with two NOTs just as it would with only 1. I could be wrong though. I haven't done this in a while

Perhaps I should let someone else jump in here, just in case my memory is failing me....

17. ### lusito92 Thread Starter New Member

Jan 29, 2013
25
0
No problem I would wait lol I really need help :/

18. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
Let:
A = x
B = y'

Code ( (Unknown Language)):
1.
2. x' + xy = x' + y
3. ^    ^^   ^    ^
4. |    ||   |    |
5. A'   A B' A'   B'
6.

Last edited: Feb 4, 2013
19. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
Try putting it in a CODE block. You probably will still see it as proportional spacing while you are typing, but it will show up as monospaced when posted (or previewed).

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20. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
This is confusing because it isn't an identity. It is valid, but you only see why when you go through the steps...
A + A'B : Expand A
A + AB + A'B : Factor B'
A + B(A + A') : Using Identity A + A' = 1
A + B(1) : Using identity A(1) = A
=> A + A'B = A + B

So, if we were to use another form of B, namely B':
A + A'B' : Expand A
A + AB' + A'B' : Factor B'
A + B'(A + A') : Using Identity A + A' = 1
A + B'(1) : Using identity A(1) = A
A + B'
=> A + A'B' = A + B'