**LAB 7**
**Answer 1**
Converting 0.1 to binary involves representing its fractional part in binary.
However, representing 0.1 precisely in binary is not possible as it is a repeating decimal in base 2.
We can only represent it approximately using a finite number of bits.

Using the process of multiplication by 2 and extracting the integer part of the result,
we can approximate the binary representation of 0.1.
1. 0.1 * 2 = 0.2 ⟶ Bit: 0
2. 0.2 * 2 = 0.4 ⟶ Bit: 0
3. 0.4 * 2 = 0.8 ⟶ Bit: 0
4. 0.8 * 2 = 1.6 ⟶ Bit: 1
5. 0.6 * 2 = 1.2 ⟶ Bit: 1
6. 0.2 * 2 = 0.4 ⟶ Bit: 0

This process repeats indefinitely because 0.1 in binary is a repeating fraction, similar to how 1/3 is represented as 0.333... in decimal.

So, a typical approximation of 0.1 in binary would be 0.0001100110011..
Comment

**Answer 2**
now it's exponent in scientific form is -4 so it's exponent is 123 since it's bias for precesion is 127

now 0.1/0.0625 is equal to 1.6

now representing 0.6 in binary form:

1. Start with 0.6.
2. Multiply 0.6 by 2:
- 0.6 * 2 = 1.2 ⟶ Bit: 1
3. Take the fractional part (0.2) and repeat:
- 0.2 * 2 = 0.4 ⟶ Bit: 0
- 0.4 * 2 = 0.8 ⟶ Bit: 0
- 0.8 * 2 = 1.6 ⟶ Bit: 1
- 0.6 * 2 = 1.2 ⟶ Bit: 1
- 0.2 * 2 = 0.4 ⟶ Bit: 0
- 0.4 * 2 = 0.8 ⟶ Bit: 0
- 0.8 * 2 = 1.6 ⟶ Bit: 1
- 0.6 * 2 = 1.2 ⟶ Bit: 1
- ... and so on.

The pattern repeats: 1001. Note that this pattern repeats every 4 bits after the binary point. To get 23 bits after the binary point, we can extend this pattern:

0.6 in binary with 23 bits after the binary point:
0.10011001100110011001100

**Answer 3**
sign bit of 0.2 is 0

the exponent of 0.2 in scientific notation is -3 so it's value is 124

representation of 124 in binary is 01111100

when we divide 0.2/0.125 we get 1.6 , represent 0.6 in 23 bit binary number:

now representing 0.6 in binary form:

1. Start with 0.6.
2. Multiply 0.6 by 2:
- 0.6 * 2 = 1.2 ⟶ Bit: 1
3. Take the fractional part (0.2) and repeat:
- 0.2 * 2 = 0.4 ⟶ Bit: 0
- 0.4 * 2 = 0.8 ⟶ Bit: 0
- 0.8 * 2 = 1.6 ⟶ Bit: 1
- 0.6 * 2 = 1.2 ⟶ Bit: 1
- 0.2 * 2 = 0.4 ⟶ Bit: 0
- 0.4 * 2 = 0.8 ⟶ Bit: 0
- 0.8 * 2 = 1.6 ⟶ Bit: 1
- 0.6 * 2 = 1.2 ⟶ Bit: 1
- ... and so on.

The pattern repeats: 1001. Note that this pattern repeats every 4 bits after the binary point. To get 23 bits after the binary point, we can extend this pattern:

0.6 in binary with 23 bits after the binary point:
0.10011001100110011001100

now entire representation:

0 01111100 10011001100110011001100

now reverting back

sign bit 0

exponent is 124 so the exponent in scientific notation is -3

now we can see in binary form it is

1x2-1+0x2-2+…

so we approximately get 0.6

so answer is 1.6*(0.125)

so value is 0.

**Answer 4**
To represent floating-point numbers and perform addition, let's use a simple representation where each floating-point number is split into two parts: one for the integer part and another for the fractional part. We'll store these parts in separate registers.

For this example, let's use two registers to store each floating-point number: one for the integer part and another for the fractional part.

Here's how we can represent and perform addition for the given example of adding 5.391 with 3.012:

1. **Representation**:
- Register 1: Store the integer part of the floating-point number.
- Register 2: Store the fractional part of the floating-point number.

For 5.391:
- Register 1: Integer part = 5
- Register 2: Fractional part = 391

For 3.012:
- Register 3: Integer part = 3
- Register 4: Fractional part = 012

2. **Performing Addition**:
- Add the fractional parts together. If the sum exceeds 999 (or the maximum value of the fractional part), carry over to the integer part.
- Add the integer parts together along with the carry, if any.

Let's perform the addition:
- Fractional part: 391 + 012 = 403
- Since 403 is within the range of the fractional part (0 to 999), no carryover is needed.
- Integer part: 5 + 3 (no carryover) = 8

So, the result is:
- Register 5: Integer part = 8
- Register 6: Fractional part = 403

Therefore, the result of adding 5.391 and 3.012 is 8.403.

**Answer 5**
.section .data
# Define the two floating-point numbers in IEEE 754 format
num1: .float 3.14159 # Example: 3.14159
num2: .float -3.14159 # Example: -3.14159

# Define the memory location to store the result
result: .float 0.0 # Initialize to 0.0

.section .text
.globl _start

_start:
# Load the address of num1 into a register
la t0, num1

# Load the address of num2 into another register
la t1, num2

# Load the first floating-point number into a floating-point register (f0)
flw f0, 0(t0)

# Load the second floating-point number into another floating-point register (f1)
flw f1, 0(t1)

# Perform floating-point addition
fadd.s f2, f0, f1

# Store the result in the memory location
la t2, result
fsw f2, 0(t2)

# Exit the program
li a0, 0 # Exit status
li a7, 93 # Exit system call number
ecall # Invoke system call