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

Convert decimal -0.000 000 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 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 89| = 0.000 000 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 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 89 × 2 = 0 + 0.000 001 78;
  • 2) 0.000 001 78 × 2 = 0 + 0.000 003 56;
  • 3) 0.000 003 56 × 2 = 0 + 0.000 007 12;
  • 4) 0.000 007 12 × 2 = 0 + 0.000 014 24;
  • 5) 0.000 014 24 × 2 = 0 + 0.000 028 48;
  • 6) 0.000 028 48 × 2 = 0 + 0.000 056 96;
  • 7) 0.000 056 96 × 2 = 0 + 0.000 113 92;
  • 8) 0.000 113 92 × 2 = 0 + 0.000 227 84;
  • 9) 0.000 227 84 × 2 = 0 + 0.000 455 68;
  • 10) 0.000 455 68 × 2 = 0 + 0.000 911 36;
  • 11) 0.000 911 36 × 2 = 0 + 0.001 822 72;
  • 12) 0.001 822 72 × 2 = 0 + 0.003 645 44;
  • 13) 0.003 645 44 × 2 = 0 + 0.007 290 88;
  • 14) 0.007 290 88 × 2 = 0 + 0.014 581 76;
  • 15) 0.014 581 76 × 2 = 0 + 0.029 163 52;
  • 16) 0.029 163 52 × 2 = 0 + 0.058 327 04;
  • 17) 0.058 327 04 × 2 = 0 + 0.116 654 08;
  • 18) 0.116 654 08 × 2 = 0 + 0.233 308 16;
  • 19) 0.233 308 16 × 2 = 0 + 0.466 616 32;
  • 20) 0.466 616 32 × 2 = 0 + 0.933 232 64;
  • 21) 0.933 232 64 × 2 = 1 + 0.866 465 28;
  • 22) 0.866 465 28 × 2 = 1 + 0.732 930 56;
  • 23) 0.732 930 56 × 2 = 1 + 0.465 861 12;
  • 24) 0.465 861 12 × 2 = 0 + 0.931 722 24;
  • 25) 0.931 722 24 × 2 = 1 + 0.863 444 48;
  • 26) 0.863 444 48 × 2 = 1 + 0.726 888 96;
  • 27) 0.726 888 96 × 2 = 1 + 0.453 777 92;
  • 28) 0.453 777 92 × 2 = 0 + 0.907 555 84;
  • 29) 0.907 555 84 × 2 = 1 + 0.815 111 68;
  • 30) 0.815 111 68 × 2 = 1 + 0.630 223 36;
  • 31) 0.630 223 36 × 2 = 1 + 0.260 446 72;
  • 32) 0.260 446 72 × 2 = 0 + 0.520 893 44;
  • 33) 0.520 893 44 × 2 = 1 + 0.041 786 88;
  • 34) 0.041 786 88 × 2 = 0 + 0.083 573 76;
  • 35) 0.083 573 76 × 2 = 0 + 0.167 147 52;
  • 36) 0.167 147 52 × 2 = 0 + 0.334 295 04;
  • 37) 0.334 295 04 × 2 = 0 + 0.668 590 08;
  • 38) 0.668 590 08 × 2 = 1 + 0.337 180 16;
  • 39) 0.337 180 16 × 2 = 0 + 0.674 360 32;
  • 40) 0.674 360 32 × 2 = 1 + 0.348 720 64;
  • 41) 0.348 720 64 × 2 = 0 + 0.697 441 28;
  • 42) 0.697 441 28 × 2 = 1 + 0.394 882 56;
  • 43) 0.394 882 56 × 2 = 0 + 0.789 765 12;
  • 44) 0.789 765 12 × 2 = 1 + 0.579 530 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).


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 89(10) =

0.0000 0000 0000 0000 0000 1110 1110 1110 1000 0101 0101(2)

6. Positive number before normalization:

0.000 000 89(10) =

0.0000 0000 0000 0000 0000 1110 1110 1110 1000 0101 0101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 89(10) =

0.0000 0000 0000 0000 0000 1110 1110 1110 1000 0101 0101(2) =

0.0000 0000 0000 0000 0000 1110 1110 1110 1000 0101 0101(2) × 20 =

1.1101 1101 1101 0000 1010 101(2) × 2-21


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): -21

Mantissa (not normalized):
1.1101 1101 1101 0000 1010 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

106(10) =

0110 1010(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. 110 1110 1110 1000 0101 0101 =

110 1110 1110 1000 0101 0101


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 1010

Mantissa (23 bits) =
110 1110 1110 1000 0101 0101


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

1 - 0110 1010 - 110 1110 1110 1000 0101 0101

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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