# Floating Point Representation of Binary Numbers

**Binary numbers floating-point representation**: Here, we are going to learn about the **floating point representation of binary numbers**.

Submitted by Saurabh Gupta, on October 24, 2019

**Prerequisite:** Number systems

We all very well know that very small and very large numbers in the decimal number system are represented using scientific notation form by stating a number (**mantissa**) and an exponent in the power of **10**. Some of the examples are **6.27 * 10 ^{-27} and 5.21 * 10^{34}**. Similarly, Binary numbers can also be represented in the same form by stating a number (

**mantissa**) and an exponent of

**2**. The format of this representation will be different for different machines.

The 16-bit machine consists of 10 bits as the **mantissa** and 6 bits for the **exponent** part whereas 24-bit machine consists of 15 bits for mantissa and 9 bits for exponent.

Format of the 16-bit machine can be represented as:

Mantissa Part | Exponent Part |
---|---|

0110011010 | 101010 |

The **mantissa** is written in 2's complement form, so the **MSB of the Mantissa** can be thought of as a sign bit. The binary point is assumed to be to the right of this sign bit. The 6-bit of the exponent can be used to represent **0 to 63**, however, to express negative exponents a number **(32) _{10}** or

**(100000)**is added to the desired exponent.

_{2}**Excess-32 Representation**: This is a common system to represent floating-point numbers. In this notation, to represent a negative exponent, we add **(32) _{10}** to the given exponent which are given by the 6 bits.

Given table illustrates representation of exponent part.

Desired Exponent | 2's complement notation | Excess-32 notation (in 6 bits) | Binary representation |
---|---|---|---|

-32 | 100000 | 100000 +100000 = 000000 | 000000 |

-31 | 100001 | 100001 +100000 = 000001 | 000001 |

-30 | 100010 | 100010 +100000 = 000010 | 000010 |

-15 | 110001 | 110001 +100000 = 010001 | 010001 |

0 | 000000 | 000000 +100000 = 100000 | 100000 |

+1 | 000001 | 000001 +100000 = 100001 | 100001 |

+15 | 001111 | 001111 +100000 = 101111 | 101111 |

+30 | 011110 | 011110 +100000 = 111110 | 111110 |

+31 | 011111 | 011111 +100000 = 111111 | 111111 |

Mantissa Part | Exponent Part |
---|---|

0110011010 | 101010 |

As given above, the **floating-point number** given in the above format is:

At the extreme left (**MSB**) is the **sign-bit '0'**, which represents it is a positive number. Also, just after the sign-bit, we assume a binary point. Thus,

In Mantissa Part: .110011010 In Exponent Part: 101010, In Excess-32 notation,32 is already added. So Subtracting 100000 001010 (i.e.,10 in decimal, so exponent part is 2^{10}) The number is N = +(.110011010)_{2}* 2^{10}= +(1100110100.00) = +(820)_{10}

**Example 1: Express the following decimal number into 16-bit floating point number (45365.125) _{10}**

**Solution:**

Binary equivalent of (45365.125)_{10}: 1011000100110101.001 Binary format: .1011000100110101 * 2^{16}Mantissa: + .101100010 Exponent: 010000 (Value of exponent is 16) Equivalent exponent: 010000 + 100000 = 110000

Since the number is a positive number an additional sign-bit '0' is added in the **MSB**.

So, the floating-point format will be **0101100010110000**

**Example 2: What floating point number do the given number 0100101001101011 represents?**

**Solution:**

At the extreme left (MSB) is the sign-bit '0' which represents it is a positive number. Also, just after the sign-bit we assume a binary point. Thus,

In Mantissa Part: .100101001 In Exponent Part: 10101, In Excess-32 notation,32 is already added. So Subtracting 100000 001011 (i.e.,11 in decimal, so exponent part is 2^{11}) The number is N = +(.100101001)_{2}* 2^{11}= +(10010100100.0) = +(1188)_{10}

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