GATE CSE 2002 | Question: 1.14

Question

The decimal value $0.25$

• A. is equivalent to the binary value $0.1$
• B. is equivalent to the binary value $0.01$
• C. is equivalent to the binary value $0.00111$
• D. cannot be represented precisely in binary

Comments

$A)\ 0.\dfrac{1}{2^1}=0.5$

$B)\ 0.\dfrac{1}{2^2}=0.25\ \checkmark$

$C)\ 0.\dfrac{7}{2^5}=0.21875$

Answer 1

First Multiplication Iteration

Multiply $0.25$ by $2$

$\begin{array}{c c c}0.25 \ast 2 = 0.50\;\text{(Product)} & \text{Fractional part} = 0.50 & \text{Carry} = 0 \textbf{ (MSB)} \end{array}$

Second Multiplication Iteration

Multiply $0.50$ by $2$

$\begin{array}{c c c}0.50 \ast 2 = 1.00\;\text{(Product)} & \text{Fractional part} = 1.00 & \text{Carry} = 1 \textbf{ (LSB)} \end{array}$

The fractional part in the $2$nd iteration becomes zero and hence we stop the multiplication iteration.

Carry from the $1$st multiplication iteration becomes MSB and carry from $2$nd iteration becomes LSB.

So the result is $0.01$

Correct Answer: B.

Answer 2

0.25 = 1/4 

we can represent 1/4 with 2^(-2)  = 0.01

So Option B is Answer

Answer 3

Here's the step-by-step conversion of 0.25 to binary: Multiply \(0.25\) by \(2\): \[0.25 \times 2 = 0.5 \quad (\text{quotient} = 0, \text{remainder} = 0.5)\] Multiply the remaining fractional part (\(0.5\)) by \(2\) again: \[0.5 \times 2 = 1.0 \quad (\text{quotient} = 1, \text{remainder} = 0)\] The quotient is \(1\), so we stop. The binary representation of \(0.25\) is: \[0.01 \quad (\text{where} \ 0 \ \text{is the whole number part and} \ 01 \ \text{is the fractional part})\]