-0.000 000 001 89 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 001 89(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
-0.000 000 001 89(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. Start with the positive version of the number:

|-0.000 000 001 89| = 0.000 000 001 89


2. 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;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 001 89.

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) 0.000 000 001 89 × 2 = 0 + 0.000 000 003 78;
  • 2) 0.000 000 003 78 × 2 = 0 + 0.000 000 007 56;
  • 3) 0.000 000 007 56 × 2 = 0 + 0.000 000 015 12;
  • 4) 0.000 000 015 12 × 2 = 0 + 0.000 000 030 24;
  • 5) 0.000 000 030 24 × 2 = 0 + 0.000 000 060 48;
  • 6) 0.000 000 060 48 × 2 = 0 + 0.000 000 120 96;
  • 7) 0.000 000 120 96 × 2 = 0 + 0.000 000 241 92;
  • 8) 0.000 000 241 92 × 2 = 0 + 0.000 000 483 84;
  • 9) 0.000 000 483 84 × 2 = 0 + 0.000 000 967 68;
  • 10) 0.000 000 967 68 × 2 = 0 + 0.000 001 935 36;
  • 11) 0.000 001 935 36 × 2 = 0 + 0.000 003 870 72;
  • 12) 0.000 003 870 72 × 2 = 0 + 0.000 007 741 44;
  • 13) 0.000 007 741 44 × 2 = 0 + 0.000 015 482 88;
  • 14) 0.000 015 482 88 × 2 = 0 + 0.000 030 965 76;
  • 15) 0.000 030 965 76 × 2 = 0 + 0.000 061 931 52;
  • 16) 0.000 061 931 52 × 2 = 0 + 0.000 123 863 04;
  • 17) 0.000 123 863 04 × 2 = 0 + 0.000 247 726 08;
  • 18) 0.000 247 726 08 × 2 = 0 + 0.000 495 452 16;
  • 19) 0.000 495 452 16 × 2 = 0 + 0.000 990 904 32;
  • 20) 0.000 990 904 32 × 2 = 0 + 0.001 981 808 64;
  • 21) 0.001 981 808 64 × 2 = 0 + 0.003 963 617 28;
  • 22) 0.003 963 617 28 × 2 = 0 + 0.007 927 234 56;
  • 23) 0.007 927 234 56 × 2 = 0 + 0.015 854 469 12;
  • 24) 0.015 854 469 12 × 2 = 0 + 0.031 708 938 24;
  • 25) 0.031 708 938 24 × 2 = 0 + 0.063 417 876 48;
  • 26) 0.063 417 876 48 × 2 = 0 + 0.126 835 752 96;
  • 27) 0.126 835 752 96 × 2 = 0 + 0.253 671 505 92;
  • 28) 0.253 671 505 92 × 2 = 0 + 0.507 343 011 84;
  • 29) 0.507 343 011 84 × 2 = 1 + 0.014 686 023 68;
  • 30) 0.014 686 023 68 × 2 = 0 + 0.029 372 047 36;
  • 31) 0.029 372 047 36 × 2 = 0 + 0.058 744 094 72;
  • 32) 0.058 744 094 72 × 2 = 0 + 0.117 488 189 44;
  • 33) 0.117 488 189 44 × 2 = 0 + 0.234 976 378 88;
  • 34) 0.234 976 378 88 × 2 = 0 + 0.469 952 757 76;
  • 35) 0.469 952 757 76 × 2 = 0 + 0.939 905 515 52;
  • 36) 0.939 905 515 52 × 2 = 1 + 0.879 811 031 04;
  • 37) 0.879 811 031 04 × 2 = 1 + 0.759 622 062 08;
  • 38) 0.759 622 062 08 × 2 = 1 + 0.519 244 124 16;
  • 39) 0.519 244 124 16 × 2 = 1 + 0.038 488 248 32;
  • 40) 0.038 488 248 32 × 2 = 0 + 0.076 976 496 64;
  • 41) 0.076 976 496 64 × 2 = 0 + 0.153 952 993 28;
  • 42) 0.153 952 993 28 × 2 = 0 + 0.307 905 986 56;
  • 43) 0.307 905 986 56 × 2 = 0 + 0.615 811 973 12;
  • 44) 0.615 811 973 12 × 2 = 1 + 0.231 623 946 24;
  • 45) 0.231 623 946 24 × 2 = 0 + 0.463 247 892 48;
  • 46) 0.463 247 892 48 × 2 = 0 + 0.926 495 784 96;
  • 47) 0.926 495 784 96 × 2 = 1 + 0.852 991 569 92;
  • 48) 0.852 991 569 92 × 2 = 1 + 0.705 983 139 84;
  • 49) 0.705 983 139 84 × 2 = 1 + 0.411 966 279 68;
  • 50) 0.411 966 279 68 × 2 = 0 + 0.823 932 559 36;
  • 51) 0.823 932 559 36 × 2 = 1 + 0.647 865 118 72;
  • 52) 0.647 865 118 72 × 2 = 1 + 0.295 730 237 44;

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).


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:


0.000 000 001 89(10) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0001 1110 0001 0011 1011(2)

6. Positive number before normalization:

0.000 000 001 89(10) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0001 1110 0001 0011 1011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 001 89(10) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0001 1110 0001 0011 1011(2) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0001 1110 0001 0011 1011(2) × 20 =

1.0000 0011 1100 0010 0111 011(2) × 2-29


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): -29

Mantissa (not normalized):
1.0000 0011 1100 0010 0111 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-29 + 2(8-1) - 1 =

(-29 + 127)(10) =

98(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;
  • 98 ÷ 2 = 49 + 0;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

98(10) =

0110 0010(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 000 0001 1110 0001 0011 1011 =

000 0001 1110 0001 0011 1011


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) =
0110 0010

Mantissa (23 bits) =
000 0001 1110 0001 0011 1011


Decimal number -0.000 000 001 89 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0010 - 000 0001 1110 0001 0011 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