Convert the Number 2 464 477 979 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number. Detailed Explanations

Number 2 464 477 979(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

The first steps we'll go through to make the conversion:

Convert to binary (base 2) the integer number.


1. Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 2 464 477 979 ÷ 2 = 1 232 238 989 + 1;
  • 1 232 238 989 ÷ 2 = 616 119 494 + 1;
  • 616 119 494 ÷ 2 = 308 059 747 + 0;
  • 308 059 747 ÷ 2 = 154 029 873 + 1;
  • 154 029 873 ÷ 2 = 77 014 936 + 1;
  • 77 014 936 ÷ 2 = 38 507 468 + 0;
  • 38 507 468 ÷ 2 = 19 253 734 + 0;
  • 19 253 734 ÷ 2 = 9 626 867 + 0;
  • 9 626 867 ÷ 2 = 4 813 433 + 1;
  • 4 813 433 ÷ 2 = 2 406 716 + 1;
  • 2 406 716 ÷ 2 = 1 203 358 + 0;
  • 1 203 358 ÷ 2 = 601 679 + 0;
  • 601 679 ÷ 2 = 300 839 + 1;
  • 300 839 ÷ 2 = 150 419 + 1;
  • 150 419 ÷ 2 = 75 209 + 1;
  • 75 209 ÷ 2 = 37 604 + 1;
  • 37 604 ÷ 2 = 18 802 + 0;
  • 18 802 ÷ 2 = 9 401 + 0;
  • 9 401 ÷ 2 = 4 700 + 1;
  • 4 700 ÷ 2 = 2 350 + 0;
  • 2 350 ÷ 2 = 1 175 + 0;
  • 1 175 ÷ 2 = 587 + 1;
  • 587 ÷ 2 = 293 + 1;
  • 293 ÷ 2 = 146 + 1;
  • 146 ÷ 2 = 73 + 0;
  • 73 ÷ 2 = 36 + 1;
  • 36 ÷ 2 = 18 + 0;
  • 18 ÷ 2 = 9 + 0;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

Take all the remainders starting from the bottom of the list constructed above.


2 464 477 979(10) =


1001 0010 1110 0100 1111 0011 0001 1011(2)



The last steps we'll go through to make the conversion:

Normalize the binary representation of the number.

Adjust the exponent.

Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Normalize the mantissa.


3. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the left, so that only one non zero digit remains to the left of it:


2 464 477 979(10) =


1001 0010 1110 0100 1111 0011 0001 1011(2) =


1001 0010 1110 0100 1111 0011 0001 1011(2) × 20 =


1.0010 0101 1100 1001 1110 0110 0011 011(2) × 231


4. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)


Exponent (unadjusted): 31


Mantissa (not normalized):
1.0010 0101 1100 1001 1110 0110 0011 011


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


31 + 2(8-1) - 1 =


(31 + 127)(10) =


158(10)


6. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 158 ÷ 2 = 79 + 0;
  • 79 ÷ 2 = 39 + 1;
  • 39 ÷ 2 = 19 + 1;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

7. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


158(10) =


1001 1110(2)


8. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.


b) Adjust its length to 23 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 001 0010 1110 0100 1111 0011 0001 1011 =


001 0010 1110 0100 1111 0011


9. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (8 bits) =
1001 1110


Mantissa (23 bits) =
001 0010 1110 0100 1111 0011


The base ten decimal number 2 464 477 979 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 1110 - 001 0010 1110 0100 1111 0011

(32 bits IEEE 754)

Number 2 464 477 978 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

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Convert to 32 bit single precision IEEE 754 binary floating point representation standard

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

Available Base Conversions Between Decimal and Binary Systems

Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):


1. Integer -> Binary

2. Decimal -> Binary

3. Binary -> Integer

4. Binary -> Decimal