-0.000 000 003 27 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 003 27(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
-0.000 000 003 27(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:

|-0.000 000 003 27| = 0.000 000 003 27


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: 0.000 000 003 27.

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) 0.000 000 003 27 × 2 = 0 + 0.000 000 006 54;
  • 2) 0.000 000 006 54 × 2 = 0 + 0.000 000 013 08;
  • 3) 0.000 000 013 08 × 2 = 0 + 0.000 000 026 16;
  • 4) 0.000 000 026 16 × 2 = 0 + 0.000 000 052 32;
  • 5) 0.000 000 052 32 × 2 = 0 + 0.000 000 104 64;
  • 6) 0.000 000 104 64 × 2 = 0 + 0.000 000 209 28;
  • 7) 0.000 000 209 28 × 2 = 0 + 0.000 000 418 56;
  • 8) 0.000 000 418 56 × 2 = 0 + 0.000 000 837 12;
  • 9) 0.000 000 837 12 × 2 = 0 + 0.000 001 674 24;
  • 10) 0.000 001 674 24 × 2 = 0 + 0.000 003 348 48;
  • 11) 0.000 003 348 48 × 2 = 0 + 0.000 006 696 96;
  • 12) 0.000 006 696 96 × 2 = 0 + 0.000 013 393 92;
  • 13) 0.000 013 393 92 × 2 = 0 + 0.000 026 787 84;
  • 14) 0.000 026 787 84 × 2 = 0 + 0.000 053 575 68;
  • 15) 0.000 053 575 68 × 2 = 0 + 0.000 107 151 36;
  • 16) 0.000 107 151 36 × 2 = 0 + 0.000 214 302 72;
  • 17) 0.000 214 302 72 × 2 = 0 + 0.000 428 605 44;
  • 18) 0.000 428 605 44 × 2 = 0 + 0.000 857 210 88;
  • 19) 0.000 857 210 88 × 2 = 0 + 0.001 714 421 76;
  • 20) 0.001 714 421 76 × 2 = 0 + 0.003 428 843 52;
  • 21) 0.003 428 843 52 × 2 = 0 + 0.006 857 687 04;
  • 22) 0.006 857 687 04 × 2 = 0 + 0.013 715 374 08;
  • 23) 0.013 715 374 08 × 2 = 0 + 0.027 430 748 16;
  • 24) 0.027 430 748 16 × 2 = 0 + 0.054 861 496 32;
  • 25) 0.054 861 496 32 × 2 = 0 + 0.109 722 992 64;
  • 26) 0.109 722 992 64 × 2 = 0 + 0.219 445 985 28;
  • 27) 0.219 445 985 28 × 2 = 0 + 0.438 891 970 56;
  • 28) 0.438 891 970 56 × 2 = 0 + 0.877 783 941 12;
  • 29) 0.877 783 941 12 × 2 = 1 + 0.755 567 882 24;
  • 30) 0.755 567 882 24 × 2 = 1 + 0.511 135 764 48;
  • 31) 0.511 135 764 48 × 2 = 1 + 0.022 271 528 96;
  • 32) 0.022 271 528 96 × 2 = 0 + 0.044 543 057 92;
  • 33) 0.044 543 057 92 × 2 = 0 + 0.089 086 115 84;
  • 34) 0.089 086 115 84 × 2 = 0 + 0.178 172 231 68;
  • 35) 0.178 172 231 68 × 2 = 0 + 0.356 344 463 36;
  • 36) 0.356 344 463 36 × 2 = 0 + 0.712 688 926 72;
  • 37) 0.712 688 926 72 × 2 = 1 + 0.425 377 853 44;
  • 38) 0.425 377 853 44 × 2 = 0 + 0.850 755 706 88;
  • 39) 0.850 755 706 88 × 2 = 1 + 0.701 511 413 76;
  • 40) 0.701 511 413 76 × 2 = 1 + 0.403 022 827 52;
  • 41) 0.403 022 827 52 × 2 = 0 + 0.806 045 655 04;
  • 42) 0.806 045 655 04 × 2 = 1 + 0.612 091 310 08;
  • 43) 0.612 091 310 08 × 2 = 1 + 0.224 182 620 16;
  • 44) 0.224 182 620 16 × 2 = 0 + 0.448 365 240 32;
  • 45) 0.448 365 240 32 × 2 = 0 + 0.896 730 480 64;
  • 46) 0.896 730 480 64 × 2 = 1 + 0.793 460 961 28;
  • 47) 0.793 460 961 28 × 2 = 1 + 0.586 921 922 56;
  • 48) 0.586 921 922 56 × 2 = 1 + 0.173 843 845 12;
  • 49) 0.173 843 845 12 × 2 = 0 + 0.347 687 690 24;
  • 50) 0.347 687 690 24 × 2 = 0 + 0.695 375 380 48;
  • 51) 0.695 375 380 48 × 2 = 1 + 0.390 750 760 96;
  • 52) 0.390 750 760 96 × 2 = 0 + 0.781 501 521 92;

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:


0.000 000 003 27(10) =

0.0000 0000 0000 0000 0000 0000 0000 1110 0000 1011 0110 0111 0010(2)

6. Positive number before normalization:

0.000 000 003 27(10) =

0.0000 0000 0000 0000 0000 0000 0000 1110 0000 1011 0110 0111 0010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 003 27(10) =

0.0000 0000 0000 0000 0000 0000 0000 1110 0000 1011 0110 0111 0010(2) =

0.0000 0000 0000 0000 0000 0000 0000 1110 0000 1011 0110 0111 0010(2) × 20 =

1.1100 0001 0110 1100 1110 010(2) × 2-29


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): -29

Mantissa (not normalized):
1.1100 0001 0110 1100 1110 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-29 + 2(8-1) - 1 =

(-29 + 127)(10) =

98(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;
  • 98 ÷ 2 = 49 + 0;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

98(10) =

0110 0010(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. 110 0000 1011 0110 0111 0010 =

110 0000 1011 0110 0111 0010


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 0010

Mantissa (23 bits) =
110 0000 1011 0110 0111 0010


Decimal number -0.000 000 003 27 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0010 - 110 0000 1011 0110 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