32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 4 999 999 999 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 4 999 999 999(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)

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;
  • 4 999 999 999 ÷ 2 = 2 499 999 999 + 1;
  • 2 499 999 999 ÷ 2 = 1 249 999 999 + 1;
  • 1 249 999 999 ÷ 2 = 624 999 999 + 1;
  • 624 999 999 ÷ 2 = 312 499 999 + 1;
  • 312 499 999 ÷ 2 = 156 249 999 + 1;
  • 156 249 999 ÷ 2 = 78 124 999 + 1;
  • 78 124 999 ÷ 2 = 39 062 499 + 1;
  • 39 062 499 ÷ 2 = 19 531 249 + 1;
  • 19 531 249 ÷ 2 = 9 765 624 + 1;
  • 9 765 624 ÷ 2 = 4 882 812 + 0;
  • 4 882 812 ÷ 2 = 2 441 406 + 0;
  • 2 441 406 ÷ 2 = 1 220 703 + 0;
  • 1 220 703 ÷ 2 = 610 351 + 1;
  • 610 351 ÷ 2 = 305 175 + 1;
  • 305 175 ÷ 2 = 152 587 + 1;
  • 152 587 ÷ 2 = 76 293 + 1;
  • 76 293 ÷ 2 = 38 146 + 1;
  • 38 146 ÷ 2 = 19 073 + 0;
  • 19 073 ÷ 2 = 9 536 + 1;
  • 9 536 ÷ 2 = 4 768 + 0;
  • 4 768 ÷ 2 = 2 384 + 0;
  • 2 384 ÷ 2 = 1 192 + 0;
  • 1 192 ÷ 2 = 596 + 0;
  • 596 ÷ 2 = 298 + 0;
  • 298 ÷ 2 = 149 + 0;
  • 149 ÷ 2 = 74 + 1;
  • 74 ÷ 2 = 37 + 0;
  • 37 ÷ 2 = 18 + 1;
  • 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.


4 999 999 999(10) =

1 0010 1010 0000 0101 1111 0001 1111 1111(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the left, so that only one non zero digit remains to the left of it:


4 999 999 999(10) =

1 0010 1010 0000 0101 1111 0001 1111 1111(2) =

1 0010 1010 0000 0101 1111 0001 1111 1111(2) × 20 =

1.0010 1010 0000 0101 1111 0001 1111 1111(2) × 232


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): 32

Mantissa (not normalized):
1.0010 1010 0000 0101 1111 0001 1111 1111


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

32 + 2(8-1) - 1 =

(32 + 127)(10) =

159(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;
  • 159 ÷ 2 = 79 + 1;
  • 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) =

159(10) =

1001 1111(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 0101 0000 0010 1111 1000 1 1111 1111 =

001 0101 0000 0010 1111 1000


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 1111

Mantissa (23 bits) =
001 0101 0000 0010 1111 1000


The base ten decimal number 4 999 999 999 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 1111 - 001 0101 0000 0010 1111 1000

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