0.000 000 000 42 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 42(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
0.000 000 000 42(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: 0.000 000 000 42.

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) 0.000 000 000 42 × 2 = 0 + 0.000 000 000 84;
  • 2) 0.000 000 000 84 × 2 = 0 + 0.000 000 001 68;
  • 3) 0.000 000 001 68 × 2 = 0 + 0.000 000 003 36;
  • 4) 0.000 000 003 36 × 2 = 0 + 0.000 000 006 72;
  • 5) 0.000 000 006 72 × 2 = 0 + 0.000 000 013 44;
  • 6) 0.000 000 013 44 × 2 = 0 + 0.000 000 026 88;
  • 7) 0.000 000 026 88 × 2 = 0 + 0.000 000 053 76;
  • 8) 0.000 000 053 76 × 2 = 0 + 0.000 000 107 52;
  • 9) 0.000 000 107 52 × 2 = 0 + 0.000 000 215 04;
  • 10) 0.000 000 215 04 × 2 = 0 + 0.000 000 430 08;
  • 11) 0.000 000 430 08 × 2 = 0 + 0.000 000 860 16;
  • 12) 0.000 000 860 16 × 2 = 0 + 0.000 001 720 32;
  • 13) 0.000 001 720 32 × 2 = 0 + 0.000 003 440 64;
  • 14) 0.000 003 440 64 × 2 = 0 + 0.000 006 881 28;
  • 15) 0.000 006 881 28 × 2 = 0 + 0.000 013 762 56;
  • 16) 0.000 013 762 56 × 2 = 0 + 0.000 027 525 12;
  • 17) 0.000 027 525 12 × 2 = 0 + 0.000 055 050 24;
  • 18) 0.000 055 050 24 × 2 = 0 + 0.000 110 100 48;
  • 19) 0.000 110 100 48 × 2 = 0 + 0.000 220 200 96;
  • 20) 0.000 220 200 96 × 2 = 0 + 0.000 440 401 92;
  • 21) 0.000 440 401 92 × 2 = 0 + 0.000 880 803 84;
  • 22) 0.000 880 803 84 × 2 = 0 + 0.001 761 607 68;
  • 23) 0.001 761 607 68 × 2 = 0 + 0.003 523 215 36;
  • 24) 0.003 523 215 36 × 2 = 0 + 0.007 046 430 72;
  • 25) 0.007 046 430 72 × 2 = 0 + 0.014 092 861 44;
  • 26) 0.014 092 861 44 × 2 = 0 + 0.028 185 722 88;
  • 27) 0.028 185 722 88 × 2 = 0 + 0.056 371 445 76;
  • 28) 0.056 371 445 76 × 2 = 0 + 0.112 742 891 52;
  • 29) 0.112 742 891 52 × 2 = 0 + 0.225 485 783 04;
  • 30) 0.225 485 783 04 × 2 = 0 + 0.450 971 566 08;
  • 31) 0.450 971 566 08 × 2 = 0 + 0.901 943 132 16;
  • 32) 0.901 943 132 16 × 2 = 1 + 0.803 886 264 32;
  • 33) 0.803 886 264 32 × 2 = 1 + 0.607 772 528 64;
  • 34) 0.607 772 528 64 × 2 = 1 + 0.215 545 057 28;
  • 35) 0.215 545 057 28 × 2 = 0 + 0.431 090 114 56;
  • 36) 0.431 090 114 56 × 2 = 0 + 0.862 180 229 12;
  • 37) 0.862 180 229 12 × 2 = 1 + 0.724 360 458 24;
  • 38) 0.724 360 458 24 × 2 = 1 + 0.448 720 916 48;
  • 39) 0.448 720 916 48 × 2 = 0 + 0.897 441 832 96;
  • 40) 0.897 441 832 96 × 2 = 1 + 0.794 883 665 92;
  • 41) 0.794 883 665 92 × 2 = 1 + 0.589 767 331 84;
  • 42) 0.589 767 331 84 × 2 = 1 + 0.179 534 663 68;
  • 43) 0.179 534 663 68 × 2 = 0 + 0.359 069 327 36;
  • 44) 0.359 069 327 36 × 2 = 0 + 0.718 138 654 72;
  • 45) 0.718 138 654 72 × 2 = 1 + 0.436 277 309 44;
  • 46) 0.436 277 309 44 × 2 = 0 + 0.872 554 618 88;
  • 47) 0.872 554 618 88 × 2 = 1 + 0.745 109 237 76;
  • 48) 0.745 109 237 76 × 2 = 1 + 0.490 218 475 52;
  • 49) 0.490 218 475 52 × 2 = 0 + 0.980 436 951 04;
  • 50) 0.980 436 951 04 × 2 = 1 + 0.960 873 902 08;
  • 51) 0.960 873 902 08 × 2 = 1 + 0.921 747 804 16;
  • 52) 0.921 747 804 16 × 2 = 1 + 0.843 495 608 32;
  • 53) 0.843 495 608 32 × 2 = 1 + 0.686 991 216 64;
  • 54) 0.686 991 216 64 × 2 = 1 + 0.373 982 433 28;
  • 55) 0.373 982 433 28 × 2 = 0 + 0.747 964 866 56;

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:


0.000 000 000 42(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1100 1101 1100 1011 0111 110(2)

5. Positive number before normalization:

0.000 000 000 42(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1100 1101 1100 1011 0111 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 42(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1100 1101 1100 1011 0111 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1100 1101 1100 1011 0111 110(2) × 20 =

1.1100 1101 1100 1011 0111 110(2) × 2-32


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): -32

Mantissa (not normalized):
1.1100 1101 1100 1011 0111 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(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;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

95(10) =

0101 1111(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. 110 0110 1110 0101 1011 1110 =

110 0110 1110 0101 1011 1110


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) =
0101 1111

Mantissa (23 bits) =
110 0110 1110 0101 1011 1110


Decimal number 0.000 000 000 42 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0101 1111 - 110 0110 1110 0101 1011 1110

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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