Digital Logic Design $////1-1$: Converting a decimal number to binary involves dividing the decimal number by $2$ repeatedly and noting down the remainders until the quotient becomes zero. The remainders, read in reverse order, give the binary equivalent of the decimal number.

Question $1$: Can you demonstrate the conversion of the decimal number $25$ into binary?

Answer:

Divide $25$ by $2$: Quotient $=(),$ Remainder $=()$

Divide $12$ by $2$: Quotient $=(),$ Remainder $=()$

Divide $6$ by $2$: Quotient $=(),$ Remainder $=()$

Divide $3$ by $2$: Quotient $=(),$ Remainder $=()$

Divide $1$ by $2$: Quotient $=(),$ Remainder $=()$

Reading the remainders from bottom to top

Getting the binary equivalent of $25$ as $()$

$1-2$: Converting a binary number to hexadecimal involves grouping the binary digits into sets of four, starting from the right side. If the leftmost group has fewer than four bits, add zeroes to make it a complete group. Then, each group is converted separately into its equivalent hexadecimal digit.

Question $2$: Illustrate the conversion of the binary number $1011011$ into hexadecimal?

Answer: $()$

group the binary digits into sets of four, starting from the right. Add leading zeroes if necessary to make complete groups of four. Then, convert each group separately into its equivalent hexadecimal digit.

Grouped the binary number $1011011$ as $00101101.$

Converting each group into its corresponding hexadecimal digit.

$1-3$: Binary addition adds two binary digits, considering like base$-10$ addition, but with a base of $2.$ When adding binary numbers, summing two $1$s results in $0$ with a carry of $1($since $1+1$ in binary equals $10).$ Adding $1$ and $0$ or two $0$s results in the corresponding binary sum without a carry.

Question $3$: Demonstrate the addition of the binary numbers $1011$ and $1101?$

Answer: Add $1011$ and $1101$ in binary:

$1011$

$+1101$

$\_\_\_\_\_\_$

$()$

$1-4$: Binary subtraction subtracts one binary number from another following in base of $2.$ To subtract binary numbers, borrow borrowing might be needed, and while subtracting digits, borrowing $2$ from the adjacent higher$-$order bit is necessary if needed.

Answer: Subtract $101$ from $1101$ in binary:

$1101$

$-101$

$\_\_\_\_\_\_$

$()$

$1-5$: Binary multiplication multiplies binary digits following base of $2.$ To multiply binary numbers, perform multiplication for each digit of one binary number with every digit of the other number, shifting as necessary, and then sum up the products to get the final result.

Question: illustrate the multiplication of the binary number $10101$ by $11?$

Answer: Multiply $101$ from $1101$ in binary:

$1011($multiplicand$)$

x $110($multiplier$)$

$\_\_\_\_\_\_\_$

$0000($Partial product: $1011$ x $0,$ shifted$)$

$1011($Partial product: $1011$ x $1)$

$+1011($Partial product: $1011$ x $1,$ shifted two places left$)$

$\_\_\_\_\_\_\_\_$

$10010010($Result: $1011$ multiplied by $110$ in binary$)$

Performing multiplication for each digit of the multiplier $(110)$ with the multiplicand $(1011)$ and summing the partial products results in $10010010$ as the product of $1011$ and $110$ in binary.

$1-5$: Binary division divides one binary number $($dividend$)$ by another binary number $($divisor$)$ to obtain a quotient and a remainder.

Question $4$: illustrate the division of the binary number $10101$ by $11?$

Answer: Divide $10101$ by $11$ in binary:

$()($Quotient: $11)$

$\_\_\_\_\_\_\_\_\_\_\_$

$11|10101($Dividend$)$

$()($Subtract $11$ from $101($first division$))$

$-------------$

$()($Remainder$)$

$()($Subtract $11$ from $100($second division$))$

$--------$

$()($Remainder$)$

Therefore, the binary division of $10101$ by $11$ results in a quotient of $()$ and a remainder of $().$

To get the $2$s complement of a binary number, you can follow these steps:

$1)$ Find the $1$s complement of the binary number by flipping all the bits $($changing every $0$ to $1$ and every $1$ to $0).$

$2)$ Add $1$ to the least significant bit $($LSB$)$ of the $1$s complement.

For example, let$$s say we want to find the $2$s complement of the binary number $1101.$

The $1$s complement of $1101$ is $().$

Adding $1$ to the LSB of $0010$ gives us $(),$ which is the $2$s complement of $().$

$1-6$: $2$s Compliment Addition adds numbers using the $2\text{'}$s complement representation to handle both positive and negative binary numbers.

To add two numbers, convert both to their $2\text{'}$s complement form, perform binary addition, and discard any overflow beyond the designated number of bits.

The result may involve discarding the carry bit to maintain the appropriate bit length.

Question $5$: $2\text{'}$s complement addition of $-3$ and $5$ in a $4-$bit syste