**FLOATING POINT ADDITION AND SUBTRACTION**

**FLOATING POINT ADDITION**

To understand floating point addition, first we see addition of real numbers in decimal as same logic is applied in both cases.

**For**** example, **we have to add **1.1 * 10**^{3 }and **50.**

We cannot add these numbers directly. First, we need to align the exponent and then, we can add significand.

After aligning exponent, we get **50 = 0.05 * 10**^{3}

Now adding significand, **0.05 + 1.1 = 1.15**

So, finally we get **(1.1 * 10**^{3 }**+ 50) = 1.15 * 10**^{3}

Here, notice that we shifted **50** and made it **0.05** to add these numbers.

**Now let ****u****s take example of floating point number addition**

We follow these steps to add two numbers:

1. Align the significand

2. Add the significands

3. Normalize the result

**Let ****the ****two numbers ****be**

x = 9.75

y = 0.5625

Converting them into 32-bit floating point representation,

**9.75**’s representation in 32-bit format = **0 10000010 00111000000000000000000**

**0.5625**’s representation in 32-bit format = **0 01111110 00100000000000000000000**

Now we get the difference of exponents to know how much shifting is required.

(**10000010 – 01111110**)2 = (**4**)10

Now, we shift the mantissa of lesser number right side by 4 units.

Mantissa of **0.5625 = 1.00100000000000000000000**

(note that 1 before decimal point is understood in 32-bit representation)

Shifting right by **4** units, we get** 0.00010010000000000000000**

Mantissa of **9.75 **= **1. 00111000000000000000000**

Adding mantissa of both

**0. 00010010000000000000000**

**+ 1. 00111000000000000000000**

————————————————-

**1. 01001010000000000000000**

In final answer, we take exponent of bigger number

So, final answer consist of :

Sign bit = **0**

Exponent of bigger number = **10000010**

Mantissa = **01001010000000000000000**

32 bit representation of answer = **x + y** = **0 10000010 01001010000000000000000**

**FLOATING POINT SUBTRACTION**

Subtraction is similar to addition with some differences like we subtract mantissa unlike addition and in sign bit we put the sign of greater number.

**Let ****the ****two numbers ****be**

x = 9.75

y = – 0.5625

Converting them into 32-bit floating point representation

**9.75**’s representation in 32-bit format = **0 10000010 00111000000000000000000**

**– 0.5625**’s representation in 32-bit format = **1 01111110 00100000000000000000000**

Now, we find the difference of exponents to know how much shifting is required.

(**10000010 – 01111110**)2 = (**4**)10

Now, we shift the mantissa of lesser number right side by 4 units.

Mantissa of **–** **0.5625 = 1.00100000000000000000000**

(note that 1 before decimal point is understood in 32-bit representation)

Shifting right by **4** units,** 0.00010010000000000000000**

Mantissa of **9.75**= **1. 00111000000000000000000**

Subtracting mantissa of both

**0. 00010010000000000000000**

**– 1. 00111000000000000000000**

————————————————

** 1. 00100110000000000000000**

Sign bit of bigger number =** 0**

So, finally the answer = **x – y = 0 10000010 00100110000000000000000**

This article has been contributed by Anuj Batham.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Cache Memory in Computer Organization
- Computer Arithmetic | Set - 1
- Computer Organization and Architecture | Pipelining | Set 1 (Execution, Stages and Throughput)
- Computer Organization and Architecture | Pipelining | Set 3 (Types and Stalling)
- Computer Organization and Architecture | Pipelining | Set 2 (Dependencies and Data Hazard)
- Computer Organization | Amdahl's law and its proof
- Computer Organization | Hardwired v/s Micro-programmed Control Unit
- Computer Architecture | Flynn's taxonomy
- Clusters In Computer Organisation
- Generations of Computer
- Simplified Instructional Computer (SIC)
- Computer Organization | Micro-Operation
- Computer Organization | Different Instruction Cycles
- Computer Organization | Booth's Algorithm
- Computer Organization | Basic Computer Instructions
- Arithmetic instructions in 8085 microprocessor
- Computer Organization | Instruction Formats (Zero, One, Two and Three Address Instruction)
- Arithmetic instructions in 8086 microprocessor
- Computer Organization | Problem Solving on Instruction Format
- Computer Organization | Von Neumann architecture