0.000 003 41 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 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
0.000 003 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: 0.000 003 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) 0.000 003 41 × 2 = 0 + 0.000 006 82;
  • 2) 0.000 006 82 × 2 = 0 + 0.000 013 64;
  • 3) 0.000 013 64 × 2 = 0 + 0.000 027 28;
  • 4) 0.000 027 28 × 2 = 0 + 0.000 054 56;
  • 5) 0.000 054 56 × 2 = 0 + 0.000 109 12;
  • 6) 0.000 109 12 × 2 = 0 + 0.000 218 24;
  • 7) 0.000 218 24 × 2 = 0 + 0.000 436 48;
  • 8) 0.000 436 48 × 2 = 0 + 0.000 872 96;
  • 9) 0.000 872 96 × 2 = 0 + 0.001 745 92;
  • 10) 0.001 745 92 × 2 = 0 + 0.003 491 84;
  • 11) 0.003 491 84 × 2 = 0 + 0.006 983 68;
  • 12) 0.006 983 68 × 2 = 0 + 0.013 967 36;
  • 13) 0.013 967 36 × 2 = 0 + 0.027 934 72;
  • 14) 0.027 934 72 × 2 = 0 + 0.055 869 44;
  • 15) 0.055 869 44 × 2 = 0 + 0.111 738 88;
  • 16) 0.111 738 88 × 2 = 0 + 0.223 477 76;
  • 17) 0.223 477 76 × 2 = 0 + 0.446 955 52;
  • 18) 0.446 955 52 × 2 = 0 + 0.893 911 04;
  • 19) 0.893 911 04 × 2 = 1 + 0.787 822 08;
  • 20) 0.787 822 08 × 2 = 1 + 0.575 644 16;
  • 21) 0.575 644 16 × 2 = 1 + 0.151 288 32;
  • 22) 0.151 288 32 × 2 = 0 + 0.302 576 64;
  • 23) 0.302 576 64 × 2 = 0 + 0.605 153 28;
  • 24) 0.605 153 28 × 2 = 1 + 0.210 306 56;
  • 25) 0.210 306 56 × 2 = 0 + 0.420 613 12;
  • 26) 0.420 613 12 × 2 = 0 + 0.841 226 24;
  • 27) 0.841 226 24 × 2 = 1 + 0.682 452 48;
  • 28) 0.682 452 48 × 2 = 1 + 0.364 904 96;
  • 29) 0.364 904 96 × 2 = 0 + 0.729 809 92;
  • 30) 0.729 809 92 × 2 = 1 + 0.459 619 84;
  • 31) 0.459 619 84 × 2 = 0 + 0.919 239 68;
  • 32) 0.919 239 68 × 2 = 1 + 0.838 479 36;
  • 33) 0.838 479 36 × 2 = 1 + 0.676 958 72;
  • 34) 0.676 958 72 × 2 = 1 + 0.353 917 44;
  • 35) 0.353 917 44 × 2 = 0 + 0.707 834 88;
  • 36) 0.707 834 88 × 2 = 1 + 0.415 669 76;
  • 37) 0.415 669 76 × 2 = 0 + 0.831 339 52;
  • 38) 0.831 339 52 × 2 = 1 + 0.662 679 04;
  • 39) 0.662 679 04 × 2 = 1 + 0.325 358 08;
  • 40) 0.325 358 08 × 2 = 0 + 0.650 716 16;
  • 41) 0.650 716 16 × 2 = 1 + 0.301 432 32;
  • 42) 0.301 432 32 × 2 = 0 + 0.602 864 64;

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 41(10) =

0.0000 0000 0000 0000 0011 1001 0011 0101 1101 0110 10(2)

5. Positive number before normalization:

0.000 003 41(10) =

0.0000 0000 0000 0000 0011 1001 0011 0101 1101 0110 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:


0.000 003 41(10) =

0.0000 0000 0000 0000 0011 1001 0011 0101 1101 0110 10(2) =

0.0000 0000 0000 0000 0011 1001 0011 0101 1101 0110 10(2) × 20 =

1.1100 1001 1010 1110 1011 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):
1.1100 1001 1010 1110 1011 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. 110 0100 1101 0111 0101 1010 =

110 0100 1101 0111 0101 1010


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) =
110 0100 1101 0111 0101 1010


Decimal number 0.000 003 41 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1100 - 110 0100 1101 0111 0101 1010

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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