0.000 003 87 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 87(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
0.000 003 87(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: 0.000 003 87.

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) 0.000 003 87 × 2 = 0 + 0.000 007 74;
  • 2) 0.000 007 74 × 2 = 0 + 0.000 015 48;
  • 3) 0.000 015 48 × 2 = 0 + 0.000 030 96;
  • 4) 0.000 030 96 × 2 = 0 + 0.000 061 92;
  • 5) 0.000 061 92 × 2 = 0 + 0.000 123 84;
  • 6) 0.000 123 84 × 2 = 0 + 0.000 247 68;
  • 7) 0.000 247 68 × 2 = 0 + 0.000 495 36;
  • 8) 0.000 495 36 × 2 = 0 + 0.000 990 72;
  • 9) 0.000 990 72 × 2 = 0 + 0.001 981 44;
  • 10) 0.001 981 44 × 2 = 0 + 0.003 962 88;
  • 11) 0.003 962 88 × 2 = 0 + 0.007 925 76;
  • 12) 0.007 925 76 × 2 = 0 + 0.015 851 52;
  • 13) 0.015 851 52 × 2 = 0 + 0.031 703 04;
  • 14) 0.031 703 04 × 2 = 0 + 0.063 406 08;
  • 15) 0.063 406 08 × 2 = 0 + 0.126 812 16;
  • 16) 0.126 812 16 × 2 = 0 + 0.253 624 32;
  • 17) 0.253 624 32 × 2 = 0 + 0.507 248 64;
  • 18) 0.507 248 64 × 2 = 1 + 0.014 497 28;
  • 19) 0.014 497 28 × 2 = 0 + 0.028 994 56;
  • 20) 0.028 994 56 × 2 = 0 + 0.057 989 12;
  • 21) 0.057 989 12 × 2 = 0 + 0.115 978 24;
  • 22) 0.115 978 24 × 2 = 0 + 0.231 956 48;
  • 23) 0.231 956 48 × 2 = 0 + 0.463 912 96;
  • 24) 0.463 912 96 × 2 = 0 + 0.927 825 92;
  • 25) 0.927 825 92 × 2 = 1 + 0.855 651 84;
  • 26) 0.855 651 84 × 2 = 1 + 0.711 303 68;
  • 27) 0.711 303 68 × 2 = 1 + 0.422 607 36;
  • 28) 0.422 607 36 × 2 = 0 + 0.845 214 72;
  • 29) 0.845 214 72 × 2 = 1 + 0.690 429 44;
  • 30) 0.690 429 44 × 2 = 1 + 0.380 858 88;
  • 31) 0.380 858 88 × 2 = 0 + 0.761 717 76;
  • 32) 0.761 717 76 × 2 = 1 + 0.523 435 52;
  • 33) 0.523 435 52 × 2 = 1 + 0.046 871 04;
  • 34) 0.046 871 04 × 2 = 0 + 0.093 742 08;
  • 35) 0.093 742 08 × 2 = 0 + 0.187 484 16;
  • 36) 0.187 484 16 × 2 = 0 + 0.374 968 32;
  • 37) 0.374 968 32 × 2 = 0 + 0.749 936 64;
  • 38) 0.749 936 64 × 2 = 1 + 0.499 873 28;
  • 39) 0.499 873 28 × 2 = 0 + 0.999 746 56;
  • 40) 0.999 746 56 × 2 = 1 + 0.999 493 12;
  • 41) 0.999 493 12 × 2 = 1 + 0.998 986 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:


0.000 003 87(10) =

0.0000 0000 0000 0000 0100 0000 1110 1101 1000 0101 1(2)

5. Positive number before normalization:

0.000 003 87(10) =

0.0000 0000 0000 0000 0100 0000 1110 1101 1000 0101 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the right, so that only one non zero digit remains to the left of it:


0.000 003 87(10) =

0.0000 0000 0000 0000 0100 0000 1110 1101 1000 0101 1(2) =

0.0000 0000 0000 0000 0100 0000 1110 1101 1000 0101 1(2) × 20 =

1.0000 0011 1011 0110 0001 011(2) × 2-18


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): -18

Mantissa (not normalized):
1.0000 0011 1011 0110 0001 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-18 + 2(8-1) - 1 =

(-18 + 127)(10) =

109(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;
  • 109 ÷ 2 = 54 + 1;
  • 54 ÷ 2 = 27 + 0;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

109(10) =

0110 1101(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. 000 0001 1101 1011 0000 1011 =

000 0001 1101 1011 0000 1011


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 1101

Mantissa (23 bits) =
000 0001 1101 1011 0000 1011


Decimal number 0.000 003 87 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1101 - 000 0001 1101 1011 0000 1011

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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