0.000 000 98 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 98(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 98(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 98.

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 98 × 2 = 0 + 0.000 001 96;
  • 2) 0.000 001 96 × 2 = 0 + 0.000 003 92;
  • 3) 0.000 003 92 × 2 = 0 + 0.000 007 84;
  • 4) 0.000 007 84 × 2 = 0 + 0.000 015 68;
  • 5) 0.000 015 68 × 2 = 0 + 0.000 031 36;
  • 6) 0.000 031 36 × 2 = 0 + 0.000 062 72;
  • 7) 0.000 062 72 × 2 = 0 + 0.000 125 44;
  • 8) 0.000 125 44 × 2 = 0 + 0.000 250 88;
  • 9) 0.000 250 88 × 2 = 0 + 0.000 501 76;
  • 10) 0.000 501 76 × 2 = 0 + 0.001 003 52;
  • 11) 0.001 003 52 × 2 = 0 + 0.002 007 04;
  • 12) 0.002 007 04 × 2 = 0 + 0.004 014 08;
  • 13) 0.004 014 08 × 2 = 0 + 0.008 028 16;
  • 14) 0.008 028 16 × 2 = 0 + 0.016 056 32;
  • 15) 0.016 056 32 × 2 = 0 + 0.032 112 64;
  • 16) 0.032 112 64 × 2 = 0 + 0.064 225 28;
  • 17) 0.064 225 28 × 2 = 0 + 0.128 450 56;
  • 18) 0.128 450 56 × 2 = 0 + 0.256 901 12;
  • 19) 0.256 901 12 × 2 = 0 + 0.513 802 24;
  • 20) 0.513 802 24 × 2 = 1 + 0.027 604 48;
  • 21) 0.027 604 48 × 2 = 0 + 0.055 208 96;
  • 22) 0.055 208 96 × 2 = 0 + 0.110 417 92;
  • 23) 0.110 417 92 × 2 = 0 + 0.220 835 84;
  • 24) 0.220 835 84 × 2 = 0 + 0.441 671 68;
  • 25) 0.441 671 68 × 2 = 0 + 0.883 343 36;
  • 26) 0.883 343 36 × 2 = 1 + 0.766 686 72;
  • 27) 0.766 686 72 × 2 = 1 + 0.533 373 44;
  • 28) 0.533 373 44 × 2 = 1 + 0.066 746 88;
  • 29) 0.066 746 88 × 2 = 0 + 0.133 493 76;
  • 30) 0.133 493 76 × 2 = 0 + 0.266 987 52;
  • 31) 0.266 987 52 × 2 = 0 + 0.533 975 04;
  • 32) 0.533 975 04 × 2 = 1 + 0.067 950 08;
  • 33) 0.067 950 08 × 2 = 0 + 0.135 900 16;
  • 34) 0.135 900 16 × 2 = 0 + 0.271 800 32;
  • 35) 0.271 800 32 × 2 = 0 + 0.543 600 64;
  • 36) 0.543 600 64 × 2 = 1 + 0.087 201 28;
  • 37) 0.087 201 28 × 2 = 0 + 0.174 402 56;
  • 38) 0.174 402 56 × 2 = 0 + 0.348 805 12;
  • 39) 0.348 805 12 × 2 = 0 + 0.697 610 24;
  • 40) 0.697 610 24 × 2 = 1 + 0.395 220 48;
  • 41) 0.395 220 48 × 2 = 0 + 0.790 440 96;
  • 42) 0.790 440 96 × 2 = 1 + 0.580 881 92;
  • 43) 0.580 881 92 × 2 = 1 + 0.161 763 84;

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 98(10) =

0.0000 0000 0000 0000 0001 0000 0111 0001 0001 0001 011(2)

5. Positive number before normalization:

0.000 000 98(10) =

0.0000 0000 0000 0000 0001 0000 0111 0001 0001 0001 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 98(10) =

0.0000 0000 0000 0000 0001 0000 0111 0001 0001 0001 011(2) =

0.0000 0000 0000 0000 0001 0000 0111 0001 0001 0001 011(2) × 20 =

1.0000 0111 0001 0001 0001 011(2) × 2-20


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): -20

Mantissa (not normalized):
1.0000 0111 0001 0001 0001 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 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) =

107(10) =

0110 1011(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 0011 1000 1000 1000 1011 =

000 0011 1000 1000 1000 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 1011

Mantissa (23 bits) =
000 0011 1000 1000 1000 1011


Decimal number 0.000 000 98 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1011 - 000 0011 1000 1000 1000 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