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

Convert decimal -0.000 000 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 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. Start with the positive version of the number:

|-0.000 000 41| = 0.000 000 41


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 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 41 × 2 = 0 + 0.000 000 82;
  • 2) 0.000 000 82 × 2 = 0 + 0.000 001 64;
  • 3) 0.000 001 64 × 2 = 0 + 0.000 003 28;
  • 4) 0.000 003 28 × 2 = 0 + 0.000 006 56;
  • 5) 0.000 006 56 × 2 = 0 + 0.000 013 12;
  • 6) 0.000 013 12 × 2 = 0 + 0.000 026 24;
  • 7) 0.000 026 24 × 2 = 0 + 0.000 052 48;
  • 8) 0.000 052 48 × 2 = 0 + 0.000 104 96;
  • 9) 0.000 104 96 × 2 = 0 + 0.000 209 92;
  • 10) 0.000 209 92 × 2 = 0 + 0.000 419 84;
  • 11) 0.000 419 84 × 2 = 0 + 0.000 839 68;
  • 12) 0.000 839 68 × 2 = 0 + 0.001 679 36;
  • 13) 0.001 679 36 × 2 = 0 + 0.003 358 72;
  • 14) 0.003 358 72 × 2 = 0 + 0.006 717 44;
  • 15) 0.006 717 44 × 2 = 0 + 0.013 434 88;
  • 16) 0.013 434 88 × 2 = 0 + 0.026 869 76;
  • 17) 0.026 869 76 × 2 = 0 + 0.053 739 52;
  • 18) 0.053 739 52 × 2 = 0 + 0.107 479 04;
  • 19) 0.107 479 04 × 2 = 0 + 0.214 958 08;
  • 20) 0.214 958 08 × 2 = 0 + 0.429 916 16;
  • 21) 0.429 916 16 × 2 = 0 + 0.859 832 32;
  • 22) 0.859 832 32 × 2 = 1 + 0.719 664 64;
  • 23) 0.719 664 64 × 2 = 1 + 0.439 329 28;
  • 24) 0.439 329 28 × 2 = 0 + 0.878 658 56;
  • 25) 0.878 658 56 × 2 = 1 + 0.757 317 12;
  • 26) 0.757 317 12 × 2 = 1 + 0.514 634 24;
  • 27) 0.514 634 24 × 2 = 1 + 0.029 268 48;
  • 28) 0.029 268 48 × 2 = 0 + 0.058 536 96;
  • 29) 0.058 536 96 × 2 = 0 + 0.117 073 92;
  • 30) 0.117 073 92 × 2 = 0 + 0.234 147 84;
  • 31) 0.234 147 84 × 2 = 0 + 0.468 295 68;
  • 32) 0.468 295 68 × 2 = 0 + 0.936 591 36;
  • 33) 0.936 591 36 × 2 = 1 + 0.873 182 72;
  • 34) 0.873 182 72 × 2 = 1 + 0.746 365 44;
  • 35) 0.746 365 44 × 2 = 1 + 0.492 730 88;
  • 36) 0.492 730 88 × 2 = 0 + 0.985 461 76;
  • 37) 0.985 461 76 × 2 = 1 + 0.970 923 52;
  • 38) 0.970 923 52 × 2 = 1 + 0.941 847 04;
  • 39) 0.941 847 04 × 2 = 1 + 0.883 694 08;
  • 40) 0.883 694 08 × 2 = 1 + 0.767 388 16;
  • 41) 0.767 388 16 × 2 = 1 + 0.534 776 32;
  • 42) 0.534 776 32 × 2 = 1 + 0.069 552 64;
  • 43) 0.069 552 64 × 2 = 0 + 0.139 105 28;
  • 44) 0.139 105 28 × 2 = 0 + 0.278 210 56;
  • 45) 0.278 210 56 × 2 = 0 + 0.556 421 12;

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).


5. 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 41(10) =

0.0000 0000 0000 0000 0000 0110 1110 0000 1110 1111 1100 0(2)

6. Positive number before normalization:

0.000 000 41(10) =

0.0000 0000 0000 0000 0000 0110 1110 0000 1110 1111 1100 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 41(10) =

0.0000 0000 0000 0000 0000 0110 1110 0000 1110 1111 1100 0(2) =

0.0000 0000 0000 0000 0000 0110 1110 0000 1110 1111 1100 0(2) × 20 =

1.1011 1000 0011 1011 1111 000(2) × 2-22


8. 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 1 (a negative number)

Exponent (unadjusted): -22

Mantissa (not normalized):
1.1011 1000 0011 1011 1111 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(10)


10. 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;
  • 105 ÷ 2 = 52 + 1;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

105(10) =

0110 1001(2)


12. 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 1100 0001 1101 1111 1000 =

101 1100 0001 1101 1111 1000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 1001

Mantissa (23 bits) =
101 1100 0001 1101 1111 1000


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

1 - 0110 1001 - 101 1100 0001 1101 1111 1000

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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