32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 35 192 852 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 35 192 852(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;
  • 35 192 852 ÷ 2 = 17 596 426 + 0;
  • 17 596 426 ÷ 2 = 8 798 213 + 0;
  • 8 798 213 ÷ 2 = 4 399 106 + 1;
  • 4 399 106 ÷ 2 = 2 199 553 + 0;
  • 2 199 553 ÷ 2 = 1 099 776 + 1;
  • 1 099 776 ÷ 2 = 549 888 + 0;
  • 549 888 ÷ 2 = 274 944 + 0;
  • 274 944 ÷ 2 = 137 472 + 0;
  • 137 472 ÷ 2 = 68 736 + 0;
  • 68 736 ÷ 2 = 34 368 + 0;
  • 34 368 ÷ 2 = 17 184 + 0;
  • 17 184 ÷ 2 = 8 592 + 0;
  • 8 592 ÷ 2 = 4 296 + 0;
  • 4 296 ÷ 2 = 2 148 + 0;
  • 2 148 ÷ 2 = 1 074 + 0;
  • 1 074 ÷ 2 = 537 + 0;
  • 537 ÷ 2 = 268 + 1;
  • 268 ÷ 2 = 134 + 0;
  • 134 ÷ 2 = 67 + 0;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 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.


35 192 852(10) =


10 0001 1001 0000 0000 0001 0100(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the left, so that only one non zero digit remains to the left of it:


35 192 852(10) =


10 0001 1001 0000 0000 0001 0100(2) =


10 0001 1001 0000 0000 0001 0100(2) × 20 =


1.0000 1100 1000 0000 0000 1010 0(2) × 225


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): 25


Mantissa (not normalized):
1.0000 1100 1000 0000 0000 1010 0


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


25 + 2(8-1) - 1 =


(25 + 127)(10) =


152(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;
  • 152 ÷ 2 = 76 + 0;
  • 76 ÷ 2 = 38 + 0;
  • 38 ÷ 2 = 19 + 0;
  • 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) =


152(10) =


1001 1000(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. 000 0110 0100 0000 0000 0101 00 =


000 0110 0100 0000 0000 0101


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 1000


Mantissa (23 bits) =
000 0110 0100 0000 0000 0101


The base ten decimal number 35 192 852 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1001 1000 - 000 0110 0100 0000 0000 0101

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