0.000 000 000 86 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 86(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 000 86(10) to 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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 86.

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) 0.000 000 000 86 × 2 = 0 + 0.000 000 001 72;
  • 2) 0.000 000 001 72 × 2 = 0 + 0.000 000 003 44;
  • 3) 0.000 000 003 44 × 2 = 0 + 0.000 000 006 88;
  • 4) 0.000 000 006 88 × 2 = 0 + 0.000 000 013 76;
  • 5) 0.000 000 013 76 × 2 = 0 + 0.000 000 027 52;
  • 6) 0.000 000 027 52 × 2 = 0 + 0.000 000 055 04;
  • 7) 0.000 000 055 04 × 2 = 0 + 0.000 000 110 08;
  • 8) 0.000 000 110 08 × 2 = 0 + 0.000 000 220 16;
  • 9) 0.000 000 220 16 × 2 = 0 + 0.000 000 440 32;
  • 10) 0.000 000 440 32 × 2 = 0 + 0.000 000 880 64;
  • 11) 0.000 000 880 64 × 2 = 0 + 0.000 001 761 28;
  • 12) 0.000 001 761 28 × 2 = 0 + 0.000 003 522 56;
  • 13) 0.000 003 522 56 × 2 = 0 + 0.000 007 045 12;
  • 14) 0.000 007 045 12 × 2 = 0 + 0.000 014 090 24;
  • 15) 0.000 014 090 24 × 2 = 0 + 0.000 028 180 48;
  • 16) 0.000 028 180 48 × 2 = 0 + 0.000 056 360 96;
  • 17) 0.000 056 360 96 × 2 = 0 + 0.000 112 721 92;
  • 18) 0.000 112 721 92 × 2 = 0 + 0.000 225 443 84;
  • 19) 0.000 225 443 84 × 2 = 0 + 0.000 450 887 68;
  • 20) 0.000 450 887 68 × 2 = 0 + 0.000 901 775 36;
  • 21) 0.000 901 775 36 × 2 = 0 + 0.001 803 550 72;
  • 22) 0.001 803 550 72 × 2 = 0 + 0.003 607 101 44;
  • 23) 0.003 607 101 44 × 2 = 0 + 0.007 214 202 88;
  • 24) 0.007 214 202 88 × 2 = 0 + 0.014 428 405 76;
  • 25) 0.014 428 405 76 × 2 = 0 + 0.028 856 811 52;
  • 26) 0.028 856 811 52 × 2 = 0 + 0.057 713 623 04;
  • 27) 0.057 713 623 04 × 2 = 0 + 0.115 427 246 08;
  • 28) 0.115 427 246 08 × 2 = 0 + 0.230 854 492 16;
  • 29) 0.230 854 492 16 × 2 = 0 + 0.461 708 984 32;
  • 30) 0.461 708 984 32 × 2 = 0 + 0.923 417 968 64;
  • 31) 0.923 417 968 64 × 2 = 1 + 0.846 835 937 28;
  • 32) 0.846 835 937 28 × 2 = 1 + 0.693 671 874 56;
  • 33) 0.693 671 874 56 × 2 = 1 + 0.387 343 749 12;
  • 34) 0.387 343 749 12 × 2 = 0 + 0.774 687 498 24;
  • 35) 0.774 687 498 24 × 2 = 1 + 0.549 374 996 48;
  • 36) 0.549 374 996 48 × 2 = 1 + 0.098 749 992 96;
  • 37) 0.098 749 992 96 × 2 = 0 + 0.197 499 985 92;
  • 38) 0.197 499 985 92 × 2 = 0 + 0.394 999 971 84;
  • 39) 0.394 999 971 84 × 2 = 0 + 0.789 999 943 68;
  • 40) 0.789 999 943 68 × 2 = 1 + 0.579 999 887 36;
  • 41) 0.579 999 887 36 × 2 = 1 + 0.159 999 774 72;
  • 42) 0.159 999 774 72 × 2 = 0 + 0.319 999 549 44;
  • 43) 0.319 999 549 44 × 2 = 0 + 0.639 999 098 88;
  • 44) 0.639 999 098 88 × 2 = 1 + 0.279 998 197 76;
  • 45) 0.279 998 197 76 × 2 = 0 + 0.559 996 395 52;
  • 46) 0.559 996 395 52 × 2 = 1 + 0.119 992 791 04;
  • 47) 0.119 992 791 04 × 2 = 0 + 0.239 985 582 08;
  • 48) 0.239 985 582 08 × 2 = 0 + 0.479 971 164 16;
  • 49) 0.479 971 164 16 × 2 = 0 + 0.959 942 328 32;
  • 50) 0.959 942 328 32 × 2 = 1 + 0.919 884 656 64;
  • 51) 0.919 884 656 64 × 2 = 1 + 0.839 769 313 28;
  • 52) 0.839 769 313 28 × 2 = 1 + 0.679 538 626 56;
  • 53) 0.679 538 626 56 × 2 = 1 + 0.359 077 253 12;
  • 54) 0.359 077 253 12 × 2 = 0 + 0.718 154 506 24;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 000 000 86(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1011 0001 1001 0100 0111 10(2)

5. Positive number before normalization:

0.000 000 000 86(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1011 0001 1001 0100 0111 10(2)

6. Normalize the binary representation of the number.

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


0.000 000 000 86(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1011 0001 1001 0100 0111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1011 0001 1001 0100 0111 10(2) × 20 =

1.1101 1000 1100 1010 0011 110(2) × 2-31


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): -31

Mantissa (not normalized):
1.1101 1000 1100 1010 0011 110


8. 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) =

96(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;
  • 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;

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) =

96(10) =

0110 0000(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 110 1100 0110 0101 0001 1110 =

110 1100 0110 0101 0001 1110


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) =
0110 0000

Mantissa (23 bits) =
110 1100 0110 0101 0001 1110


Decimal number 0.000 000 000 86 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0000 - 110 1100 0110 0101 0001 1110

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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:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

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

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

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

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

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

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111