Difference between revisions of "Boolean Algebra"
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+ | =Boolean Algebra Precedence= | ||
+ | the order of precedence for boolean algebra is: | ||
+ | # Brackets | ||
+ | # Not | ||
+ | # And | ||
+ | # Or | ||
− | + | =Boolean Identities= | |
+ | ===TRC Video=== | ||
+ | <youtube>https://www.youtube.com/watch?v=ym73-rgnrOQ</youtube> | ||
− | + | https://www.youtube.com/watch?v=ym73-rgnrOQ | |
− | + | ===Using AND=== | |
− | + | <math> A.1 = A </math> | |
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− | |||
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− | |||
− | |||
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This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa. | This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa. | ||
Line 37: | Line 28: | ||
<math> A.\overline{A}=0 </math> | <math> A.\overline{A}=0 </math> | ||
− | + | Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0. | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | This | ||
− | + | ===Using OR=== | |
− | |||
− | ==OR | ||
<math> 0+A = A </math> | <math> 0+A = A </math> | ||
− | + | 0 or A can be simplified as just A. | |
<math> 1+A = 1 </math> | <math> 1+A = 1 </math> | ||
− | + | 1 or A can be simplified as just 1. | |
− | <math> | + | <math> A+A=A</math> |
− | + | A or A can be simplified as just A. | |
<math> \overline{A}+A=1</math> | <math> \overline{A}+A=1</math> | ||
− | + | NOT A or A can be simplified as just 1. | |
+ | |||
+ | =Boolean Laws= | ||
+ | ===TRC Video=== | ||
+ | <youtube>https://www.youtube.com/watch?v=Cdqj4XDsUVY</youtube> | ||
+ | |||
+ | https://www.youtube.com/watch?v=Cdqj4XDsUVY | ||
− | |||
==Commutative Law== | ==Commutative Law== | ||
The Commutative Law is where equations are the same no matter what way around the letters are written. For example | The Commutative Law is where equations are the same no matter what way around the letters are written. For example | ||
− | + | ||
+ | <math> A+B = B+A </math> | ||
+ | |||
or | or | ||
− | |||
− | = | + | <math> A.B = B.A </math> |
+ | |||
==Associate Law== | ==Associate Law== | ||
+ | If all of the symbols are the same it doesn't matter which order the equation is evaluated. | ||
+ | |||
+ | <math> A+(B+C) = B + (A+C) </math> | ||
+ | |||
+ | <math> A+(B+C) = B + (A+C) </math> | ||
+ | |||
+ | <math> A+(B+C) = C + (A+B) </math> | ||
+ | |||
+ | So: | ||
+ | |||
+ | <math> A.(B.C) = B . (A.C) </math> | ||
+ | |||
+ | <math> A.(B.C) = B . (A.C) </math> | ||
+ | |||
+ | <math> A.(B.C) = C . (A.B) </math> | ||
==Distributive Law== | ==Distributive Law== | ||
Line 82: | Line 85: | ||
<math> A+(B.C) = (A+B).(A+C) </math> | <math> A+(B.C) = (A+B).(A+C) </math> | ||
+ | |||
+ | This is essentially factorising or expanding the brackets, but you can also remove the common factor: | ||
+ | |||
+ | <math> A.B + A.C = A.(B+C)</math> | ||
+ | |||
+ | <math> A+B.A+C = A+(B.C) </math> | ||
+ | |||
+ | You can also remove the common factor if you only have 1 term on one side: | ||
+ | |||
+ | <math> | ||
+ | A.(A + B) = (0+A) . (A + B) | ||
+ | </math> | ||
+ | |||
+ | <math> | ||
+ | A+(A . B) = (1.A) + (A . B) | ||
+ | </math> | ||
+ | |||
+ | if the symbol inside the brackets is a '+' you can add '+0' or if the symbol inside the brackets is '.' you can add '.1'. Doing this will not change the nature of the brackets because 'A' is the same as 'A+0' and is the same as 'A.1'. | ||
==Redundancy Law== | ==Redundancy Law== | ||
− | <math> \overline{A} = \overline{A} </math> | + | ===Law 1 :=== |
− | + | <math> A + (\overline{A}. B) = A + B </math> | |
− | <math> \overline{\overline{A}} = A </math> | + | |
+ | Proof : | ||
+ | |||
+ | <math>= A + (\overline{A}. B) = A + B \\ | ||
+ | = (A + \overline{A})(A + B) \\ | ||
+ | = 1 . (A + B) \\ | ||
+ | = A + B </math> | ||
+ | |||
+ | <hr> | ||
+ | |||
+ | ===Law 2:=== | ||
+ | <math> A.(\overline{A} + B) = A.B</math> | ||
+ | |||
+ | Proof : | ||
+ | |||
+ | <math>= A.(\overline{A} + B) \\ | ||
+ | = A.\overline{A} + A.B \\ | ||
+ | = 0 + A.B \\ | ||
+ | = A.B </math> | ||
+ | |||
+ | <hr> | ||
+ | ===Law 3:=== | ||
+ | <math> A.(A + B) = A</math> | ||
+ | |||
+ | Proof using distributive law: | ||
+ | |||
+ | <math> | ||
+ | A.(A + B) = (0+A) . (A + B) | ||
+ | </math> | ||
+ | |||
+ | So: | ||
+ | <math> | ||
+ | A + (0 . B) | ||
+ | </math> | ||
+ | |||
+ | So: | ||
+ | <math> | ||
+ | A + 0 = A | ||
+ | </math> | ||
+ | |||
+ | <hr> | ||
+ | ===Law 4:=== | ||
+ | <math> A+(A . B) = A</math> | ||
+ | |||
+ | Proof using distributive law: | ||
+ | |||
+ | <math> | ||
+ | A+(A . B) = (1 . A) + (A . B) | ||
+ | </math> | ||
+ | |||
+ | So: | ||
+ | <math> | ||
+ | A . (1 + B) | ||
+ | </math> | ||
+ | |||
+ | So: | ||
+ | <math> | ||
+ | A . 1 = A | ||
+ | </math> | ||
==Identity Law== | ==Identity Law== | ||
+ | This is also in the identities section: | ||
+ | |||
+ | <math> A.A = A </math> | ||
+ | |||
<math> A+A = A </math> | <math> A+A = A </math> | ||
==Negation Law== | ==Negation Law== | ||
+ | Just like in any other logic negating a negative is a positive so: | ||
+ | |||
+ | <math> \overline{ \overline{A} } = A </math> | ||
− | =Equations= | + | =Solving Boolean Equations= |
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find: | Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find: | ||
+ | ===TRC Video=== | ||
+ | <youtube>https://www.youtube.com/watch?v=N1r1D__NMGg</youtube> | ||
+ | |||
+ | https://www.youtube.com/watch?v=N1r1D__NMGg | ||
+ | |||
+ | ===Example 1=== | ||
+ | <math> | ||
+ | 𝐶+(𝐶.𝐷) | ||
+ | </math> | ||
+ | |||
+ | ------------------------------------ | ||
+ | |||
+ | Take out the common factor C: | ||
+ | |||
+ | <math>(C.D)+(C.1)=C.(D+1)</math>, | ||
+ | |||
+ | We know that <math>1+A=1</math>, | ||
+ | |||
+ | Therefore, <math>C.1</math>, | ||
+ | |||
+ | Use identity <math>A.1=A</math>, | ||
+ | |||
+ | Answer = <math>C</math> | ||
+ | |||
+ | ------------------------------------ | ||
+ | |||
+ | ===Example 2=== | ||
+ | '''A.(C+A)''' | ||
+ | |||
+ | |||
+ | ------------------------------------ | ||
+ | |Use Distributive Law| | ||
+ | |||
+ | ->'''(A.C)+(A.A)''' | ||
+ | |||
+ | |Use Identity| | ||
+ | '''A.A=A''' | ||
+ | |||
+ | ->'''(A.C)+A''' | ||
+ | |||
+ | |This is the same as writing (could straight apply redundancy rule here)| | ||
+ | |||
+ | ->'''(A.C)+(A.1)''' | ||
+ | |||
+ | |Take out the common factor| | ||
+ | |||
+ | ->'''A.(C+1)''' | ||
+ | |||
+ | |Use Identity| | ||
+ | '''C+1 = 1''' | ||
+ | |||
+ | ->'''A''' | ||
+ | |||
+ | ===Example 3=== | ||
+ | <math> | ||
+ | 𝐵.(𝐴+\overline{𝐵}) | ||
+ | </math> | ||
+ | |||
+ | ------------------------------------ | ||
+ | B.(A+ NOT B) | ||
+ | REDUNDANCY | ||
+ | (A + NOT B) | ||
+ | REDUNDANCY | ||
+ | |||
+ | ANSWER = NOT B | ||
+ | |||
+ | ===Example 4=== | ||
+ | <math> | ||
+ | 𝑋.(\overline{𝑋}+𝑌) | ||
+ | </math> | ||
+ | |||
+ | ------------------------------------ | ||
+ | <math> | ||
+ | 𝑋.\overline{𝑋} = 0 | ||
+ | </math> | ||
+ | |||
+ | <math> | ||
+ | 0+𝑌 = 𝑌 | ||
+ | </math> | ||
+ | |||
+ | ===Example 5=== | ||
+ | <math> | ||
+ | 𝑋.(X+\overline{Y}) | ||
+ | </math> | ||
+ | |||
+ | ------------------------------------ | ||
+ | <math> | ||
+ | (0+𝑋).(X+\overline{Y}) | ||
+ | </math> | ||
+ | |||
+ | <math> | ||
+ | 𝑋+(0.\overline{Y}) | ||
+ | </math> | ||
+ | |||
+ | <math> | ||
+ | 𝑋+(0) | ||
+ | </math> | ||
+ | |||
+ | <math> | ||
+ | 𝑋 | ||
+ | </math> | ||
+ | |||
+ | ===Example 8=== | ||
+ | <math> | ||
+ | 𝐷.𝐸+𝐸.\overline{𝐷} | ||
+ | </math> | ||
+ | ------------------------------------ | ||
+ | D.E+E.D | ||
+ | Distributivetive Law | ||
+ | D.(E+D) | ||
+ | Redundancy Law | ||
+ | D | ||
+ | |||
+ | ===Example 13=== | ||
+ | <math> | ||
+ | (\overline {A}+\overline {B}).B | ||
+ | </math> | ||
+ | |||
+ | ------------------------------------ | ||
+ | Expand the brackets: | ||
+ | <math> | ||
+ | (\overline {A} . B) + (\overline{B} . B) | ||
+ | </math> | ||
+ | |||
+ | Not B AND B = 0: | ||
+ | <math> | ||
+ | \overline {A}.B) + (0) | ||
+ | </math> | ||
+ | |||
+ | Something OR 0 is Something: | ||
+ | <math> | ||
+ | \overline {A}.B | ||
+ | </math> | ||
+ | |||
+ | ===Example 14=== | ||
+ | <math> | ||
+ | \overline{B} + (A.B) | ||
+ | </math> | ||
+ | |||
+ | ------------------------------------ | ||
+ | (B) + (A.B) | ||
+ | Distributive Law. | ||
+ | (B + A) . (B + B) | ||
+ | Not B cancels out. | ||
+ | B + A . 1 | ||
+ | = B+A | ||
+ | |||
+ | |||
+ | ===Example 19=== | ||
+ | <math>(X + Y) . (X + \overline{Y})</math> | ||
− | == | + | ------------------------------------ |
+ | <br>Distributive: | ||
+ | <br><math>X . (Y + \overline{Y})</math> | ||
+ | <br>Identity laws: | ||
+ | <br><math>Y + \overline{Y} = 1</math> | ||
+ | <p><math> X.1 = X</math> | ||
+ | ====Alternative==== | ||
+ | <math> X.X + X.\overline{Y} + Y.X + Y.\overline{Y} </math> Expanding the brackets | ||
− | == | + | <math> X + X.\overline{Y} + Y.X + 0 </math> Use of <math> X.X = X </math> and <math> Y.\overline{Y} = 0 </math> |
− | + | <math> X + X(\overline{Y}+Y) </math> Taking X out of the brackets | |
− | = | + | <math> X + X(1) </math> Use of <math> Y + \overline{Y} = 1 </math> |
− | + | <math> X(1) </math> | |
− | + | <math> X </math> | |
− | == | + | =====End of Page===== |
Latest revision as of 08:10, 23 August 2023
Contents
Boolean Algebra Precedence
the order of precedence for boolean algebra is:
- Brackets
- Not
- And
- Or
Boolean Identities
TRC Video
https://www.youtube.com/watch?v=ym73-rgnrOQ
Using AND
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.
The output is determined by A alone in this equation. This can be simplified to just "A".
Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0.
Using OR
0 or A can be simplified as just A.
1 or A can be simplified as just 1.
A or A can be simplified as just A.
NOT A or A can be simplified as just 1.
Boolean Laws
TRC Video
https://www.youtube.com/watch?v=Cdqj4XDsUVY
Commutative Law
The Commutative Law is where equations are the same no matter what way around the letters are written. For example
or
Associate Law
If all of the symbols are the same it doesn't matter which order the equation is evaluated.
So:
Distributive Law
The distributive law is these two equations.
This is essentially factorising or expanding the brackets, but you can also remove the common factor:
You can also remove the common factor if you only have 1 term on one side:
if the symbol inside the brackets is a '+' you can add '+0' or if the symbol inside the brackets is '.' you can add '.1'. Doing this will not change the nature of the brackets because 'A' is the same as 'A+0' and is the same as 'A.1'.
Redundancy Law
Law 1 :
Proof :
Law 2:
Proof :
Law 3:
Proof using distributive law:
So:
So:
Law 4:
Proof using distributive law:
So:
So:
Identity Law
This is also in the identities section:
Negation Law
Just like in any other logic negating a negative is a positive so:
Solving Boolean Equations
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:
TRC Video
https://www.youtube.com/watch?v=N1r1D__NMGg
Example 1
Take out the common factor C:
,
We know that
,Therefore,
,Use identity
,Answer =
Example 2
A.(C+A)
|Use Distributive Law|
->(A.C)+(A.A)
|Use Identity| A.A=A
->(A.C)+A
|This is the same as writing (could straight apply redundancy rule here)|
->(A.C)+(A.1)
|Take out the common factor|
->A.(C+1)
|Use Identity| C+1 = 1
->A
Example 3
B.(A+ NOT B) REDUNDANCY (A + NOT B) REDUNDANCY
ANSWER = NOT B
Example 4
Example 5
Example 8
D.E+E.D Distributivetive Law D.(E+D) Redundancy Law D
Example 13
Expand the brackets:
Not B AND B = 0:
Something OR 0 is Something:
Example 14
(B) + (A.B) Distributive Law. (B + A) . (B + B) Not B cancels out. B + A . 1 = B+A
Example 19
Distributive:
Identity laws:
Alternative
Expanding the brackets
Use of and
Taking X out of the brackets
Use of