32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 37 286.008 300 4 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 37 286.008 300 4(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. First, convert to binary (in base 2) the integer part: 37 286.
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;
  • 37 286 ÷ 2 = 18 643 + 0;
  • 18 643 ÷ 2 = 9 321 + 1;
  • 9 321 ÷ 2 = 4 660 + 1;
  • 4 660 ÷ 2 = 2 330 + 0;
  • 2 330 ÷ 2 = 1 165 + 0;
  • 1 165 ÷ 2 = 582 + 1;
  • 582 ÷ 2 = 291 + 0;
  • 291 ÷ 2 = 145 + 1;
  • 145 ÷ 2 = 72 + 1;
  • 72 ÷ 2 = 36 + 0;
  • 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 integer part of the number.

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


37 286(10) =


1001 0001 1010 0110(2)


3. Convert to binary (base 2) the fractional part: 0.008 300 4.

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) 0.008 300 4 × 2 = 0 + 0.016 600 8;
  • 2) 0.016 600 8 × 2 = 0 + 0.033 201 6;
  • 3) 0.033 201 6 × 2 = 0 + 0.066 403 2;
  • 4) 0.066 403 2 × 2 = 0 + 0.132 806 4;
  • 5) 0.132 806 4 × 2 = 0 + 0.265 612 8;
  • 6) 0.265 612 8 × 2 = 0 + 0.531 225 6;
  • 7) 0.531 225 6 × 2 = 1 + 0.062 451 2;
  • 8) 0.062 451 2 × 2 = 0 + 0.124 902 4;
  • 9) 0.124 902 4 × 2 = 0 + 0.249 804 8;
  • 10) 0.249 804 8 × 2 = 0 + 0.499 609 6;
  • 11) 0.499 609 6 × 2 = 0 + 0.999 219 2;
  • 12) 0.999 219 2 × 2 = 1 + 0.998 438 4;
  • 13) 0.998 438 4 × 2 = 1 + 0.996 876 8;
  • 14) 0.996 876 8 × 2 = 1 + 0.993 753 6;
  • 15) 0.993 753 6 × 2 = 1 + 0.987 507 2;
  • 16) 0.987 507 2 × 2 = 1 + 0.975 014 4;
  • 17) 0.975 014 4 × 2 = 1 + 0.950 028 8;
  • 18) 0.950 028 8 × 2 = 1 + 0.900 057 6;
  • 19) 0.900 057 6 × 2 = 1 + 0.800 115 2;
  • 20) 0.800 115 2 × 2 = 1 + 0.600 230 4;
  • 21) 0.600 230 4 × 2 = 1 + 0.200 460 8;
  • 22) 0.200 460 8 × 2 = 0 + 0.400 921 6;
  • 23) 0.400 921 6 × 2 = 0 + 0.801 843 2;
  • 24) 0.801 843 2 × 2 = 1 + 0.603 686 4;

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...)


4. 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:


0.008 300 4(10) =


0.0000 0010 0001 1111 1111 1001(2)


5. Positive number before normalization:

37 286.008 300 4(10) =


1001 0001 1010 0110.0000 0010 0001 1111 1111 1001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the left, so that only one non zero digit remains to the left of it:


37 286.008 300 4(10) =


1001 0001 1010 0110.0000 0010 0001 1111 1111 1001(2) =


1001 0001 1010 0110.0000 0010 0001 1111 1111 1001(2) × 20 =


1.0010 0011 0100 1100 0000 0100 0011 1111 1111 001(2) × 215


7. 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): 15


Mantissa (not normalized):
1.0010 0011 0100 1100 0000 0100 0011 1111 1111 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


15 + 2(8-1) - 1 =


(15 + 127)(10) =


142(10)


9. 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;
  • 142 ÷ 2 = 71 + 0;
  • 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;

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

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


Exponent (adjusted) =


142(10) =


1000 1110(2)


11. 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 0001 1010 0110 0000 0010 0001 1111 1111 1001 =


001 0001 1010 0110 0000 0010


12. 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) =
1000 1110


Mantissa (23 bits) =
001 0001 1010 0110 0000 0010


The base ten decimal number 37 286.008 300 4 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 1110 - 001 0001 1010 0110 0000 0010

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