-0.000 000 000 71 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 71(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 71(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:

|-0.000 000 000 71| = 0.000 000 000 71


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: 0.000 000 000 71.

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 71 × 2 = 0 + 0.000 000 001 42;
  • 2) 0.000 000 001 42 × 2 = 0 + 0.000 000 002 84;
  • 3) 0.000 000 002 84 × 2 = 0 + 0.000 000 005 68;
  • 4) 0.000 000 005 68 × 2 = 0 + 0.000 000 011 36;
  • 5) 0.000 000 011 36 × 2 = 0 + 0.000 000 022 72;
  • 6) 0.000 000 022 72 × 2 = 0 + 0.000 000 045 44;
  • 7) 0.000 000 045 44 × 2 = 0 + 0.000 000 090 88;
  • 8) 0.000 000 090 88 × 2 = 0 + 0.000 000 181 76;
  • 9) 0.000 000 181 76 × 2 = 0 + 0.000 000 363 52;
  • 10) 0.000 000 363 52 × 2 = 0 + 0.000 000 727 04;
  • 11) 0.000 000 727 04 × 2 = 0 + 0.000 001 454 08;
  • 12) 0.000 001 454 08 × 2 = 0 + 0.000 002 908 16;
  • 13) 0.000 002 908 16 × 2 = 0 + 0.000 005 816 32;
  • 14) 0.000 005 816 32 × 2 = 0 + 0.000 011 632 64;
  • 15) 0.000 011 632 64 × 2 = 0 + 0.000 023 265 28;
  • 16) 0.000 023 265 28 × 2 = 0 + 0.000 046 530 56;
  • 17) 0.000 046 530 56 × 2 = 0 + 0.000 093 061 12;
  • 18) 0.000 093 061 12 × 2 = 0 + 0.000 186 122 24;
  • 19) 0.000 186 122 24 × 2 = 0 + 0.000 372 244 48;
  • 20) 0.000 372 244 48 × 2 = 0 + 0.000 744 488 96;
  • 21) 0.000 744 488 96 × 2 = 0 + 0.001 488 977 92;
  • 22) 0.001 488 977 92 × 2 = 0 + 0.002 977 955 84;
  • 23) 0.002 977 955 84 × 2 = 0 + 0.005 955 911 68;
  • 24) 0.005 955 911 68 × 2 = 0 + 0.011 911 823 36;
  • 25) 0.011 911 823 36 × 2 = 0 + 0.023 823 646 72;
  • 26) 0.023 823 646 72 × 2 = 0 + 0.047 647 293 44;
  • 27) 0.047 647 293 44 × 2 = 0 + 0.095 294 586 88;
  • 28) 0.095 294 586 88 × 2 = 0 + 0.190 589 173 76;
  • 29) 0.190 589 173 76 × 2 = 0 + 0.381 178 347 52;
  • 30) 0.381 178 347 52 × 2 = 0 + 0.762 356 695 04;
  • 31) 0.762 356 695 04 × 2 = 1 + 0.524 713 390 08;
  • 32) 0.524 713 390 08 × 2 = 1 + 0.049 426 780 16;
  • 33) 0.049 426 780 16 × 2 = 0 + 0.098 853 560 32;
  • 34) 0.098 853 560 32 × 2 = 0 + 0.197 707 120 64;
  • 35) 0.197 707 120 64 × 2 = 0 + 0.395 414 241 28;
  • 36) 0.395 414 241 28 × 2 = 0 + 0.790 828 482 56;
  • 37) 0.790 828 482 56 × 2 = 1 + 0.581 656 965 12;
  • 38) 0.581 656 965 12 × 2 = 1 + 0.163 313 930 24;
  • 39) 0.163 313 930 24 × 2 = 0 + 0.326 627 860 48;
  • 40) 0.326 627 860 48 × 2 = 0 + 0.653 255 720 96;
  • 41) 0.653 255 720 96 × 2 = 1 + 0.306 511 441 92;
  • 42) 0.306 511 441 92 × 2 = 0 + 0.613 022 883 84;
  • 43) 0.613 022 883 84 × 2 = 1 + 0.226 045 767 68;
  • 44) 0.226 045 767 68 × 2 = 0 + 0.452 091 535 36;
  • 45) 0.452 091 535 36 × 2 = 0 + 0.904 183 070 72;
  • 46) 0.904 183 070 72 × 2 = 1 + 0.808 366 141 44;
  • 47) 0.808 366 141 44 × 2 = 1 + 0.616 732 282 88;
  • 48) 0.616 732 282 88 × 2 = 1 + 0.233 464 565 76;
  • 49) 0.233 464 565 76 × 2 = 0 + 0.466 929 131 52;
  • 50) 0.466 929 131 52 × 2 = 0 + 0.933 858 263 04;
  • 51) 0.933 858 263 04 × 2 = 1 + 0.867 716 526 08;
  • 52) 0.867 716 526 08 × 2 = 1 + 0.735 433 052 16;
  • 53) 0.735 433 052 16 × 2 = 1 + 0.470 866 104 32;
  • 54) 0.470 866 104 32 × 2 = 0 + 0.941 732 208 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).


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:


0.000 000 000 71(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 1010 0111 0011 10(2)

6. Positive number before normalization:

0.000 000 000 71(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 1010 0111 0011 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 71(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 1010 0111 0011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 1010 0111 0011 10(2) × 20 =

1.1000 0110 0101 0011 1001 110(2) × 2-31


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): -31

Mantissa (not normalized):
1.1000 0110 0101 0011 1001 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 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) =

96(10) =

0110 0000(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. 100 0011 0010 1001 1100 1110 =

100 0011 0010 1001 1100 1110


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 0000

Mantissa (23 bits) =
100 0011 0010 1001 1100 1110


Decimal number -0.000 000 000 71 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 0011 0010 1001 1100 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