Boolean algebra is a form of algebra that deals with single digit binary

values and variables.

Values and variables can indicate some of the following binary pairs of

values:

• ON / OFF
• TRUE / FALSE
• HIGH / LOW
• CLOSED / OPEN
• 1 / 0

There are three fundamental operators in Boolean algebra :

• NOT: unary operator that complements represented as A’
• AND: binary operator which performs logical multiplication
• example: A ANDed with B would be represented as A.B or AB
• OR: binary operator which performs logical addition
• example: A ORed with B would be represented as A+B

Boolean expressions must be evaluated with the following order of operator precedence

• parentheses
• NOT
• AND
• OR

FUNCTION MANIPULATION

Example: Simplify the following expression

F =BC +BC’ + BA

• Simplification
F = B(C + C’) +BA
F = B.1 + BA

F = (1 + A)

F= B

SUM OF PRODUCT (SOP) & PRODUCT OF SUM (POS)

Boolean expressions can be manipulated into many forms.

• Some standardized forms are required for Boolean expressions to simplify
communication of the expressions.

Combination of input value that produce 1 s is convert into equivalent variables, ANDed together then ORed with other combination variables with same output.
SOP is easier to derive from truth table.

-EXAMPLE.

F = A’B’ + ABC’D

The truth table :

PRODUCT TERM : A’B’C’D’

A’B’C’D

A’B’CD’

A’B’CD

SOP EXPRESSION :

F = (A’B’C’D’) + (A’B’C’D) + (A’B’ CD’) + (A’B’C’D) + (ABC’D)

Note that the method is reversible. You can find the SOP expression from the truth table or build the truth table from the expression.

Input combinations that produce 0 in sum terms (ORed variables) are ANDed together.
Convert input values that produce 0s into equivalent variables, ORed the variables, then ANDed with other ORed forms.
Usually use if more 1s produce in output function.

EXAMPLE :

F = (A+B+C)(A+B+C’)(A+B’+C)(A’+B+C’)

The truth table :

PRODUCT TERM : A + B + C

A + B + C’

A + B’ + C

A’ + B + C’

POS EXPRESSION:

F = (A + B + C)(A + B + C’)(A + B’+C)(A’ + B + C’)

Prepared by :

RkUsman

BOOLEAN EQUATION FORM

Hello everyone ,my name is RkUsman

let me explain to all of you about Arihmetics for Addition,

BINARY ADDITION

Rules of Binary Addition

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0, and carry 1 to the next more significant bit

For Example,

00011010 + 00001100 = 00100110	1 1		carries
	0 0 0 1 1 0 1 0	=	26(base 10)
	+ 0 0 0 0 1 1 0 0	=	12(base 10)
	0 0 1 0 0 1 1 0	=	38(base 10)

2. DECIMAL ADDITION

To add two multiple-digit decimal numbers:

you first need to know how to add two single-digit decimal numbers.
This requires the memorization of 100 facts, or 55 facts if you exclude the commutative or “turnaround” facts.
Also, because of carries, you need to know ten additional facts: 10 + 0 = 10, 10 + 1 = 11,…, 10 + 9 = 19.
The latter apply when there’s a carry (always 1) and the “top” digit is 9.

For Example,

let’s add 19.7 and 12.8 ,

3.HEXADECIMAL ADDITION

Just think of it like this:

A = 10
B = 11
C = 12
D = 13
E = 14
F = 15

So, you can either wrap your head around the math in a straight way, and realize that a carry only happens when the addtion results in a single column equaling 16 or greater:

AE9
+ F
——
AF8

Or you can do the conversion to base 10 then back:
A = 10
E = 14

Next you multiply each of those by 16 raised to the place minues 1. Since A is in the third place, it’s raised by 2. Since E is in the second place, it’s raised by 1. 8 is in the first place, so it’s raised by 0.

10*16^2 = 2560
14*16^1 = 224
9*16^0 = 9

Now add those together:
AE9 = 2560 + 224 + 9 = 2793

Now, you can add the other number:
F = 15
2793 + 15 = 2808

Now you convert back by dividing the number by the base 16 powers and saving the remainders:

Third Place:
16^2 = 256

2808 / 256 = 10 (remainder 248) = A

Seconed Place:
16^1 = 16

247 / 16 = 15 (remainder 8) = F

Ones Place:
16^0 = 1

8 / 1 = 8

Answer:
AF8

I perfer just getting used to the idea of adding in base 16 since it takes less time.

thanks for reading i'm RkUsman

Subtraction is generally simpler than addition since only two numbers are involved and the upper value representation is greater than the lower value representation. The problem of “borrow” is similar in binary subtraction to that in decimal. We can contruct a subtraction table that has two parts – the three cases of subtracting without borrow, and the one case of the involvement of a borrow digit, no matter how far to the left is the next available binary digit.

Next, there are four rules we should follow in binary subtraction as in table 1.1

BINARY RULES	SUM	BORROW
0 – 0 = 0	0	0
0 – 1 = 1	1	10
1 – 0 = 1	1	0
1 – 1 = 1	0	1

Table 1.1

The first three are the same as in decimal. The fourth fact is the only new one; it is the borrow case. It applies when the “top” digit in a column is 0 and the “bottom” digit is 1. (Remember: in binary, 10 are pronounced “one-zero” or “two.”)

Now let’s subtract 1011.11 from 10101.101, following the same algorithm I used for decimal numbers:

Step 1: 1 – 0 = 1.
Step 2: Borrow to make 10 – 1 = 1.
Step 3: Borrow to make 10 – 1 = 1.
Step 4: Cascaded borrow to make 10 – 1 = 1.
Step 5: 1 – 1 = 0.
Step 6: 0 – 0 = 0.
Step 7: Borrow to make 10 – 1 = 1.

Since there are lots of 0s in binary numbers, there can be lots of borrows — and lots of messy looking cross-outs.

There is another example for subtraction:

• 1011011 − 10010 = 1001001:


	1	0	1	1	0	1	1
−			1	0	0	1	0
	1	0	0	1	0	0	1

created by RkUsman

Binary multiplication is much easier because each digit we multiply by is either zero or one.F
Multiplication is achieved by adding a list of shifted multiplicands according to the digits of the multiplier.
There are four rules we should follow in binary multiplication as in table below: