Boolean algebra - class-XII

By De Morgan's laws and logical implication rules, p and q is equivalent to not (p implies not q). p -> not q is equivalent to not p or not q. The negation is p and q.

Using Boolean algebra, the expression P + P_bar Q can be simplified using the distributive law. Since P + P_bar = 1, the expression simplifies to P + Q, which represents the logical OR operation. Therefore, the circuit is an OR gate.

Person who use boolean algebra for describing the operation of logic gates first was Claude Shannon. 

In binary number system only $0$'s and $1$'s are used to built the whole number system.
Hence,
$0 = 0$
$1 = 1$
Start back at $0$ again, but add $1$ on the left,
$2 = 10$
$3 = 11$
Start back at $0$ again, and add one to the number on the left, but that number is already at $1$, so it also goes back to $0$ and $1$ is added to the next position on the left. Hence,
$4 = 100$ .... and so on.

Boolean algebra is essentially based on logic. It is also known as logical algebra.

The Boolean algebra uses binary number system i.e. two digits $0$ and $1$.

$\bar 1 + \bar 1 = 0 + 0 = 0$

When $A = 1,$ then $A + \bar{A} = 1 + 0 =1$
and when $A = 0$, then $A + \bar{A} = 0 + 1 = 1$

$(1000) _2 = 0 \times  2^0 + 0 \times 2^1 + 0 \times 2^2 + 1 \times  2^3 = 8$

Boolean algebra is based on logic.

$A=1$, $B=0$

Here,

$A+\bar{B}=1+0=1=A$.

$\uparrow$ read up the remainders
So, $(16)$$ _{10}$ $= (10000)$$ _2$

$(9.25) _{10} = (9) _{10} + (0.25) _{10} = (1001) _2 + (0.01) _2 = (1001.01) _2$

$100010 + 11011 = (111101) _2$

The binary representation of $13$ is $1101$. The $1's$ complement of $13$ is $0010$. The $2's$ complement of $13$ is $0010 + 1 = 0011$.

In standard BCD (Binary Coded Decimal) or common character encoding representations, 'A' is often represented as 1010 in specific 4-bit contexts or extended codes.