0.000 000 000 41 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 41(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 41(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 41.

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 41 × 2 = 0 + 0.000 000 000 82;
  • 2) 0.000 000 000 82 × 2 = 0 + 0.000 000 001 64;
  • 3) 0.000 000 001 64 × 2 = 0 + 0.000 000 003 28;
  • 4) 0.000 000 003 28 × 2 = 0 + 0.000 000 006 56;
  • 5) 0.000 000 006 56 × 2 = 0 + 0.000 000 013 12;
  • 6) 0.000 000 013 12 × 2 = 0 + 0.000 000 026 24;
  • 7) 0.000 000 026 24 × 2 = 0 + 0.000 000 052 48;
  • 8) 0.000 000 052 48 × 2 = 0 + 0.000 000 104 96;
  • 9) 0.000 000 104 96 × 2 = 0 + 0.000 000 209 92;
  • 10) 0.000 000 209 92 × 2 = 0 + 0.000 000 419 84;
  • 11) 0.000 000 419 84 × 2 = 0 + 0.000 000 839 68;
  • 12) 0.000 000 839 68 × 2 = 0 + 0.000 001 679 36;
  • 13) 0.000 001 679 36 × 2 = 0 + 0.000 003 358 72;
  • 14) 0.000 003 358 72 × 2 = 0 + 0.000 006 717 44;
  • 15) 0.000 006 717 44 × 2 = 0 + 0.000 013 434 88;
  • 16) 0.000 013 434 88 × 2 = 0 + 0.000 026 869 76;
  • 17) 0.000 026 869 76 × 2 = 0 + 0.000 053 739 52;
  • 18) 0.000 053 739 52 × 2 = 0 + 0.000 107 479 04;
  • 19) 0.000 107 479 04 × 2 = 0 + 0.000 214 958 08;
  • 20) 0.000 214 958 08 × 2 = 0 + 0.000 429 916 16;
  • 21) 0.000 429 916 16 × 2 = 0 + 0.000 859 832 32;
  • 22) 0.000 859 832 32 × 2 = 0 + 0.001 719 664 64;
  • 23) 0.001 719 664 64 × 2 = 0 + 0.003 439 329 28;
  • 24) 0.003 439 329 28 × 2 = 0 + 0.006 878 658 56;
  • 25) 0.006 878 658 56 × 2 = 0 + 0.013 757 317 12;
  • 26) 0.013 757 317 12 × 2 = 0 + 0.027 514 634 24;
  • 27) 0.027 514 634 24 × 2 = 0 + 0.055 029 268 48;
  • 28) 0.055 029 268 48 × 2 = 0 + 0.110 058 536 96;
  • 29) 0.110 058 536 96 × 2 = 0 + 0.220 117 073 92;
  • 30) 0.220 117 073 92 × 2 = 0 + 0.440 234 147 84;
  • 31) 0.440 234 147 84 × 2 = 0 + 0.880 468 295 68;
  • 32) 0.880 468 295 68 × 2 = 1 + 0.760 936 591 36;
  • 33) 0.760 936 591 36 × 2 = 1 + 0.521 873 182 72;
  • 34) 0.521 873 182 72 × 2 = 1 + 0.043 746 365 44;
  • 35) 0.043 746 365 44 × 2 = 0 + 0.087 492 730 88;
  • 36) 0.087 492 730 88 × 2 = 0 + 0.174 985 461 76;
  • 37) 0.174 985 461 76 × 2 = 0 + 0.349 970 923 52;
  • 38) 0.349 970 923 52 × 2 = 0 + 0.699 941 847 04;
  • 39) 0.699 941 847 04 × 2 = 1 + 0.399 883 694 08;
  • 40) 0.399 883 694 08 × 2 = 0 + 0.799 767 388 16;
  • 41) 0.799 767 388 16 × 2 = 1 + 0.599 534 776 32;
  • 42) 0.599 534 776 32 × 2 = 1 + 0.199 069 552 64;
  • 43) 0.199 069 552 64 × 2 = 0 + 0.398 139 105 28;
  • 44) 0.398 139 105 28 × 2 = 0 + 0.796 278 210 56;
  • 45) 0.796 278 210 56 × 2 = 1 + 0.592 556 421 12;
  • 46) 0.592 556 421 12 × 2 = 1 + 0.185 112 842 24;
  • 47) 0.185 112 842 24 × 2 = 0 + 0.370 225 684 48;
  • 48) 0.370 225 684 48 × 2 = 0 + 0.740 451 368 96;
  • 49) 0.740 451 368 96 × 2 = 1 + 0.480 902 737 92;
  • 50) 0.480 902 737 92 × 2 = 0 + 0.961 805 475 84;
  • 51) 0.961 805 475 84 × 2 = 1 + 0.923 610 951 68;
  • 52) 0.923 610 951 68 × 2 = 1 + 0.847 221 903 36;
  • 53) 0.847 221 903 36 × 2 = 1 + 0.694 443 806 72;
  • 54) 0.694 443 806 72 × 2 = 1 + 0.388 887 613 44;
  • 55) 0.388 887 613 44 × 2 = 0 + 0.777 775 226 88;

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 41(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1100 0010 1100 1100 1011 110(2)

5. Positive number before normalization:

0.000 000 000 41(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1100 0010 1100 1100 1011 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 41(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1100 0010 1100 1100 1011 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1100 0010 1100 1100 1011 110(2) × 20 =

1.1100 0010 1100 1100 1011 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 0010 1100 1100 1011 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 0001 0110 0110 0101 1110 =

110 0001 0110 0110 0101 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 0001 0110 0110 0101 1110


Decimal number 0.000 000 000 41 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0101 1111 - 110 0001 0110 0110 0101 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