0.000 003 18 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 18(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 18(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 18.

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 18 × 2 = 0 + 0.000 006 36;
  • 2) 0.000 006 36 × 2 = 0 + 0.000 012 72;
  • 3) 0.000 012 72 × 2 = 0 + 0.000 025 44;
  • 4) 0.000 025 44 × 2 = 0 + 0.000 050 88;
  • 5) 0.000 050 88 × 2 = 0 + 0.000 101 76;
  • 6) 0.000 101 76 × 2 = 0 + 0.000 203 52;
  • 7) 0.000 203 52 × 2 = 0 + 0.000 407 04;
  • 8) 0.000 407 04 × 2 = 0 + 0.000 814 08;
  • 9) 0.000 814 08 × 2 = 0 + 0.001 628 16;
  • 10) 0.001 628 16 × 2 = 0 + 0.003 256 32;
  • 11) 0.003 256 32 × 2 = 0 + 0.006 512 64;
  • 12) 0.006 512 64 × 2 = 0 + 0.013 025 28;
  • 13) 0.013 025 28 × 2 = 0 + 0.026 050 56;
  • 14) 0.026 050 56 × 2 = 0 + 0.052 101 12;
  • 15) 0.052 101 12 × 2 = 0 + 0.104 202 24;
  • 16) 0.104 202 24 × 2 = 0 + 0.208 404 48;
  • 17) 0.208 404 48 × 2 = 0 + 0.416 808 96;
  • 18) 0.416 808 96 × 2 = 0 + 0.833 617 92;
  • 19) 0.833 617 92 × 2 = 1 + 0.667 235 84;
  • 20) 0.667 235 84 × 2 = 1 + 0.334 471 68;
  • 21) 0.334 471 68 × 2 = 0 + 0.668 943 36;
  • 22) 0.668 943 36 × 2 = 1 + 0.337 886 72;
  • 23) 0.337 886 72 × 2 = 0 + 0.675 773 44;
  • 24) 0.675 773 44 × 2 = 1 + 0.351 546 88;
  • 25) 0.351 546 88 × 2 = 0 + 0.703 093 76;
  • 26) 0.703 093 76 × 2 = 1 + 0.406 187 52;
  • 27) 0.406 187 52 × 2 = 0 + 0.812 375 04;
  • 28) 0.812 375 04 × 2 = 1 + 0.624 750 08;
  • 29) 0.624 750 08 × 2 = 1 + 0.249 500 16;
  • 30) 0.249 500 16 × 2 = 0 + 0.499 000 32;
  • 31) 0.499 000 32 × 2 = 0 + 0.998 000 64;
  • 32) 0.998 000 64 × 2 = 1 + 0.996 001 28;
  • 33) 0.996 001 28 × 2 = 1 + 0.992 002 56;
  • 34) 0.992 002 56 × 2 = 1 + 0.984 005 12;
  • 35) 0.984 005 12 × 2 = 1 + 0.968 010 24;
  • 36) 0.968 010 24 × 2 = 1 + 0.936 020 48;
  • 37) 0.936 020 48 × 2 = 1 + 0.872 040 96;
  • 38) 0.872 040 96 × 2 = 1 + 0.744 081 92;
  • 39) 0.744 081 92 × 2 = 1 + 0.488 163 84;
  • 40) 0.488 163 84 × 2 = 0 + 0.976 327 68;
  • 41) 0.976 327 68 × 2 = 1 + 0.952 655 36;
  • 42) 0.952 655 36 × 2 = 1 + 0.905 310 72;

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 18(10) =

0.0000 0000 0000 0000 0011 0101 0101 1001 1111 1110 11(2)

5. Positive number before normalization:

0.000 003 18(10) =

0.0000 0000 0000 0000 0011 0101 0101 1001 1111 1110 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 003 18(10) =

0.0000 0000 0000 0000 0011 0101 0101 1001 1111 1110 11(2) =

0.0000 0000 0000 0000 0011 0101 0101 1001 1111 1110 11(2) × 20 =

1.1010 1010 1100 1111 1111 011(2) × 2-19


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): -19

Mantissa (not normalized):
1.1010 1010 1100 1111 1111 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(8-1) - 1 =

(-19 + 127)(10) =

108(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;
  • 108 ÷ 2 = 54 + 0;
  • 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) =

108(10) =

0110 1100(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. 101 0101 0110 0111 1111 1011 =

101 0101 0110 0111 1111 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 1100

Mantissa (23 bits) =
101 0101 0110 0111 1111 1011


Decimal number 0.000 003 18 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1100 - 101 0101 0110 0111 1111 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