0.000 002 71 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 002 71(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
0.000 002 71(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: 0.000 002 71.

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) 0.000 002 71 × 2 = 0 + 0.000 005 42;
  • 2) 0.000 005 42 × 2 = 0 + 0.000 010 84;
  • 3) 0.000 010 84 × 2 = 0 + 0.000 021 68;
  • 4) 0.000 021 68 × 2 = 0 + 0.000 043 36;
  • 5) 0.000 043 36 × 2 = 0 + 0.000 086 72;
  • 6) 0.000 086 72 × 2 = 0 + 0.000 173 44;
  • 7) 0.000 173 44 × 2 = 0 + 0.000 346 88;
  • 8) 0.000 346 88 × 2 = 0 + 0.000 693 76;
  • 9) 0.000 693 76 × 2 = 0 + 0.001 387 52;
  • 10) 0.001 387 52 × 2 = 0 + 0.002 775 04;
  • 11) 0.002 775 04 × 2 = 0 + 0.005 550 08;
  • 12) 0.005 550 08 × 2 = 0 + 0.011 100 16;
  • 13) 0.011 100 16 × 2 = 0 + 0.022 200 32;
  • 14) 0.022 200 32 × 2 = 0 + 0.044 400 64;
  • 15) 0.044 400 64 × 2 = 0 + 0.088 801 28;
  • 16) 0.088 801 28 × 2 = 0 + 0.177 602 56;
  • 17) 0.177 602 56 × 2 = 0 + 0.355 205 12;
  • 18) 0.355 205 12 × 2 = 0 + 0.710 410 24;
  • 19) 0.710 410 24 × 2 = 1 + 0.420 820 48;
  • 20) 0.420 820 48 × 2 = 0 + 0.841 640 96;
  • 21) 0.841 640 96 × 2 = 1 + 0.683 281 92;
  • 22) 0.683 281 92 × 2 = 1 + 0.366 563 84;
  • 23) 0.366 563 84 × 2 = 0 + 0.733 127 68;
  • 24) 0.733 127 68 × 2 = 1 + 0.466 255 36;
  • 25) 0.466 255 36 × 2 = 0 + 0.932 510 72;
  • 26) 0.932 510 72 × 2 = 1 + 0.865 021 44;
  • 27) 0.865 021 44 × 2 = 1 + 0.730 042 88;
  • 28) 0.730 042 88 × 2 = 1 + 0.460 085 76;
  • 29) 0.460 085 76 × 2 = 0 + 0.920 171 52;
  • 30) 0.920 171 52 × 2 = 1 + 0.840 343 04;
  • 31) 0.840 343 04 × 2 = 1 + 0.680 686 08;
  • 32) 0.680 686 08 × 2 = 1 + 0.361 372 16;
  • 33) 0.361 372 16 × 2 = 0 + 0.722 744 32;
  • 34) 0.722 744 32 × 2 = 1 + 0.445 488 64;
  • 35) 0.445 488 64 × 2 = 0 + 0.890 977 28;
  • 36) 0.890 977 28 × 2 = 1 + 0.781 954 56;
  • 37) 0.781 954 56 × 2 = 1 + 0.563 909 12;
  • 38) 0.563 909 12 × 2 = 1 + 0.127 818 24;
  • 39) 0.127 818 24 × 2 = 0 + 0.255 636 48;
  • 40) 0.255 636 48 × 2 = 0 + 0.511 272 96;
  • 41) 0.511 272 96 × 2 = 1 + 0.022 545 92;
  • 42) 0.022 545 92 × 2 = 0 + 0.045 091 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:


0.000 002 71(10) =

0.0000 0000 0000 0000 0010 1101 0111 0111 0101 1100 10(2)

5. Positive number before normalization:

0.000 002 71(10) =

0.0000 0000 0000 0000 0010 1101 0111 0111 0101 1100 10(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:


0.000 002 71(10) =

0.0000 0000 0000 0000 0010 1101 0111 0111 0101 1100 10(2) =

0.0000 0000 0000 0000 0010 1101 0111 0111 0101 1100 10(2) × 20 =

1.0110 1011 1011 1010 1110 010(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):
1.0110 1011 1011 1010 1110 010


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. 011 0101 1101 1101 0111 0010 =

011 0101 1101 1101 0111 0010


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) =
011 0101 1101 1101 0111 0010


Decimal number 0.000 002 71 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1100 - 011 0101 1101 1101 0111 0010

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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