32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -98 765.432 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -98 765.432(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. Start with the positive version of the number:

|-98 765.432| = 98 765.432

2. First, convert to binary (in base 2) the integer part: 98 765.
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;
  • 98 765 ÷ 2 = 49 382 + 1;
  • 49 382 ÷ 2 = 24 691 + 0;
  • 24 691 ÷ 2 = 12 345 + 1;
  • 12 345 ÷ 2 = 6 172 + 1;
  • 6 172 ÷ 2 = 3 086 + 0;
  • 3 086 ÷ 2 = 1 543 + 0;
  • 1 543 ÷ 2 = 771 + 1;
  • 771 ÷ 2 = 385 + 1;
  • 385 ÷ 2 = 192 + 1;
  • 192 ÷ 2 = 96 + 0;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the integer part of the number.

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


98 765(10) =


1 1000 0001 1100 1101(2)


4. Convert to binary (base 2) the fractional part: 0.432.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.432 × 2 = 0 + 0.864;
  • 2) 0.864 × 2 = 1 + 0.728;
  • 3) 0.728 × 2 = 1 + 0.456;
  • 4) 0.456 × 2 = 0 + 0.912;
  • 5) 0.912 × 2 = 1 + 0.824;
  • 6) 0.824 × 2 = 1 + 0.648;
  • 7) 0.648 × 2 = 1 + 0.296;
  • 8) 0.296 × 2 = 0 + 0.592;
  • 9) 0.592 × 2 = 1 + 0.184;
  • 10) 0.184 × 2 = 0 + 0.368;
  • 11) 0.368 × 2 = 0 + 0.736;
  • 12) 0.736 × 2 = 1 + 0.472;
  • 13) 0.472 × 2 = 0 + 0.944;
  • 14) 0.944 × 2 = 1 + 0.888;
  • 15) 0.888 × 2 = 1 + 0.776;
  • 16) 0.776 × 2 = 1 + 0.552;
  • 17) 0.552 × 2 = 1 + 0.104;
  • 18) 0.104 × 2 = 0 + 0.208;
  • 19) 0.208 × 2 = 0 + 0.416;
  • 20) 0.416 × 2 = 0 + 0.832;
  • 21) 0.832 × 2 = 1 + 0.664;
  • 22) 0.664 × 2 = 1 + 0.328;
  • 23) 0.328 × 2 = 0 + 0.656;
  • 24) 0.656 × 2 = 1 + 0.312;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.432(10) =


0.0110 1110 1001 0111 1000 1101(2)


6. Positive number before normalization:

98 765.432(10) =


1 1000 0001 1100 1101.0110 1110 1001 0111 1000 1101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the left, so that only one non zero digit remains to the left of it:


98 765.432(10) =


1 1000 0001 1100 1101.0110 1110 1001 0111 1000 1101(2) =


1 1000 0001 1100 1101.0110 1110 1001 0111 1000 1101(2) × 20 =


1.1000 0001 1100 1101 0110 1110 1001 0111 1000 1101(2) × 216


8. 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 1 (a negative number)


Exponent (unadjusted): 16


Mantissa (not normalized):
1.1000 0001 1100 1101 0110 1110 1001 0111 1000 1101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


16 + 2(8-1) - 1 =


(16 + 127)(10) =


143(10)


10. 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;
  • 143 ÷ 2 = 71 + 1;
  • 71 ÷ 2 = 35 + 1;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


143(10) =


1000 1111(2)


12. 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. 100 0000 1110 0110 1011 0111 0 1001 0111 1000 1101 =


100 0000 1110 0110 1011 0111


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

Sign (1 bit) =
1 (a negative number)


Exponent (8 bits) =
1000 1111


Mantissa (23 bits) =
100 0000 1110 0110 1011 0111


The base ten decimal number -98 765.432 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 1000 1111 - 100 0000 1110 0110 1011 0111

The latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation