0.000 003 98 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 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
0.000 003 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: 0.000 003 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) 0.000 003 98 × 2 = 0 + 0.000 007 96;
  • 2) 0.000 007 96 × 2 = 0 + 0.000 015 92;
  • 3) 0.000 015 92 × 2 = 0 + 0.000 031 84;
  • 4) 0.000 031 84 × 2 = 0 + 0.000 063 68;
  • 5) 0.000 063 68 × 2 = 0 + 0.000 127 36;
  • 6) 0.000 127 36 × 2 = 0 + 0.000 254 72;
  • 7) 0.000 254 72 × 2 = 0 + 0.000 509 44;
  • 8) 0.000 509 44 × 2 = 0 + 0.001 018 88;
  • 9) 0.001 018 88 × 2 = 0 + 0.002 037 76;
  • 10) 0.002 037 76 × 2 = 0 + 0.004 075 52;
  • 11) 0.004 075 52 × 2 = 0 + 0.008 151 04;
  • 12) 0.008 151 04 × 2 = 0 + 0.016 302 08;
  • 13) 0.016 302 08 × 2 = 0 + 0.032 604 16;
  • 14) 0.032 604 16 × 2 = 0 + 0.065 208 32;
  • 15) 0.065 208 32 × 2 = 0 + 0.130 416 64;
  • 16) 0.130 416 64 × 2 = 0 + 0.260 833 28;
  • 17) 0.260 833 28 × 2 = 0 + 0.521 666 56;
  • 18) 0.521 666 56 × 2 = 1 + 0.043 333 12;
  • 19) 0.043 333 12 × 2 = 0 + 0.086 666 24;
  • 20) 0.086 666 24 × 2 = 0 + 0.173 332 48;
  • 21) 0.173 332 48 × 2 = 0 + 0.346 664 96;
  • 22) 0.346 664 96 × 2 = 0 + 0.693 329 92;
  • 23) 0.693 329 92 × 2 = 1 + 0.386 659 84;
  • 24) 0.386 659 84 × 2 = 0 + 0.773 319 68;
  • 25) 0.773 319 68 × 2 = 1 + 0.546 639 36;
  • 26) 0.546 639 36 × 2 = 1 + 0.093 278 72;
  • 27) 0.093 278 72 × 2 = 0 + 0.186 557 44;
  • 28) 0.186 557 44 × 2 = 0 + 0.373 114 88;
  • 29) 0.373 114 88 × 2 = 0 + 0.746 229 76;
  • 30) 0.746 229 76 × 2 = 1 + 0.492 459 52;
  • 31) 0.492 459 52 × 2 = 0 + 0.984 919 04;
  • 32) 0.984 919 04 × 2 = 1 + 0.969 838 08;
  • 33) 0.969 838 08 × 2 = 1 + 0.939 676 16;
  • 34) 0.939 676 16 × 2 = 1 + 0.879 352 32;
  • 35) 0.879 352 32 × 2 = 1 + 0.758 704 64;
  • 36) 0.758 704 64 × 2 = 1 + 0.517 409 28;
  • 37) 0.517 409 28 × 2 = 1 + 0.034 818 56;
  • 38) 0.034 818 56 × 2 = 0 + 0.069 637 12;
  • 39) 0.069 637 12 × 2 = 0 + 0.139 274 24;
  • 40) 0.139 274 24 × 2 = 0 + 0.278 548 48;
  • 41) 0.278 548 48 × 2 = 0 + 0.557 096 96;

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 98(10) =

0.0000 0000 0000 0000 0100 0010 1100 0101 1111 1000 0(2)

5. Positive number before normalization:

0.000 003 98(10) =

0.0000 0000 0000 0000 0100 0010 1100 0101 1111 1000 0(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 98(10) =

0.0000 0000 0000 0000 0100 0010 1100 0101 1111 1000 0(2) =

0.0000 0000 0000 0000 0100 0010 1100 0101 1111 1000 0(2) × 20 =

1.0000 1011 0001 0111 1110 000(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 1011 0001 0111 1110 000


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 0101 1000 1011 1111 0000 =

000 0101 1000 1011 1111 0000


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 0101 1000 1011 1111 0000


Decimal number 0.000 003 98 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1101 - 000 0101 1000 1011 1111 0000

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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