-0.000 000 000 31 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 31(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 31(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 31| = 0.000 000 000 31


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 31.

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 31 × 2 = 0 + 0.000 000 000 62;
  • 2) 0.000 000 000 62 × 2 = 0 + 0.000 000 001 24;
  • 3) 0.000 000 001 24 × 2 = 0 + 0.000 000 002 48;
  • 4) 0.000 000 002 48 × 2 = 0 + 0.000 000 004 96;
  • 5) 0.000 000 004 96 × 2 = 0 + 0.000 000 009 92;
  • 6) 0.000 000 009 92 × 2 = 0 + 0.000 000 019 84;
  • 7) 0.000 000 019 84 × 2 = 0 + 0.000 000 039 68;
  • 8) 0.000 000 039 68 × 2 = 0 + 0.000 000 079 36;
  • 9) 0.000 000 079 36 × 2 = 0 + 0.000 000 158 72;
  • 10) 0.000 000 158 72 × 2 = 0 + 0.000 000 317 44;
  • 11) 0.000 000 317 44 × 2 = 0 + 0.000 000 634 88;
  • 12) 0.000 000 634 88 × 2 = 0 + 0.000 001 269 76;
  • 13) 0.000 001 269 76 × 2 = 0 + 0.000 002 539 52;
  • 14) 0.000 002 539 52 × 2 = 0 + 0.000 005 079 04;
  • 15) 0.000 005 079 04 × 2 = 0 + 0.000 010 158 08;
  • 16) 0.000 010 158 08 × 2 = 0 + 0.000 020 316 16;
  • 17) 0.000 020 316 16 × 2 = 0 + 0.000 040 632 32;
  • 18) 0.000 040 632 32 × 2 = 0 + 0.000 081 264 64;
  • 19) 0.000 081 264 64 × 2 = 0 + 0.000 162 529 28;
  • 20) 0.000 162 529 28 × 2 = 0 + 0.000 325 058 56;
  • 21) 0.000 325 058 56 × 2 = 0 + 0.000 650 117 12;
  • 22) 0.000 650 117 12 × 2 = 0 + 0.001 300 234 24;
  • 23) 0.001 300 234 24 × 2 = 0 + 0.002 600 468 48;
  • 24) 0.002 600 468 48 × 2 = 0 + 0.005 200 936 96;
  • 25) 0.005 200 936 96 × 2 = 0 + 0.010 401 873 92;
  • 26) 0.010 401 873 92 × 2 = 0 + 0.020 803 747 84;
  • 27) 0.020 803 747 84 × 2 = 0 + 0.041 607 495 68;
  • 28) 0.041 607 495 68 × 2 = 0 + 0.083 214 991 36;
  • 29) 0.083 214 991 36 × 2 = 0 + 0.166 429 982 72;
  • 30) 0.166 429 982 72 × 2 = 0 + 0.332 859 965 44;
  • 31) 0.332 859 965 44 × 2 = 0 + 0.665 719 930 88;
  • 32) 0.665 719 930 88 × 2 = 1 + 0.331 439 861 76;
  • 33) 0.331 439 861 76 × 2 = 0 + 0.662 879 723 52;
  • 34) 0.662 879 723 52 × 2 = 1 + 0.325 759 447 04;
  • 35) 0.325 759 447 04 × 2 = 0 + 0.651 518 894 08;
  • 36) 0.651 518 894 08 × 2 = 1 + 0.303 037 788 16;
  • 37) 0.303 037 788 16 × 2 = 0 + 0.606 075 576 32;
  • 38) 0.606 075 576 32 × 2 = 1 + 0.212 151 152 64;
  • 39) 0.212 151 152 64 × 2 = 0 + 0.424 302 305 28;
  • 40) 0.424 302 305 28 × 2 = 0 + 0.848 604 610 56;
  • 41) 0.848 604 610 56 × 2 = 1 + 0.697 209 221 12;
  • 42) 0.697 209 221 12 × 2 = 1 + 0.394 418 442 24;
  • 43) 0.394 418 442 24 × 2 = 0 + 0.788 836 884 48;
  • 44) 0.788 836 884 48 × 2 = 1 + 0.577 673 768 96;
  • 45) 0.577 673 768 96 × 2 = 1 + 0.155 347 537 92;
  • 46) 0.155 347 537 92 × 2 = 0 + 0.310 695 075 84;
  • 47) 0.310 695 075 84 × 2 = 0 + 0.621 390 151 68;
  • 48) 0.621 390 151 68 × 2 = 1 + 0.242 780 303 36;
  • 49) 0.242 780 303 36 × 2 = 0 + 0.485 560 606 72;
  • 50) 0.485 560 606 72 × 2 = 0 + 0.971 121 213 44;
  • 51) 0.971 121 213 44 × 2 = 1 + 0.942 242 426 88;
  • 52) 0.942 242 426 88 × 2 = 1 + 0.884 484 853 76;
  • 53) 0.884 484 853 76 × 2 = 1 + 0.768 969 707 52;
  • 54) 0.768 969 707 52 × 2 = 1 + 0.537 939 415 04;
  • 55) 0.537 939 415 04 × 2 = 1 + 0.075 878 830 08;

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 31(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0101 0100 1101 1001 0011 111(2)

6. Positive number before normalization:

0.000 000 000 31(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0101 0100 1101 1001 0011 111(2)

7. 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 31(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0101 0100 1101 1001 0011 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0101 0100 1101 1001 0011 111(2) × 20 =

1.0101 0100 1101 1001 0011 111(2) × 2-32


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): -32

Mantissa (not normalized):
1.0101 0100 1101 1001 0011 111


9. 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)


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;
  • 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;

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) =

95(10) =

0101 1111(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. 010 1010 0110 1100 1001 1111 =

010 1010 0110 1100 1001 1111


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) =
0101 1111

Mantissa (23 bits) =
010 1010 0110 1100 1001 1111


Decimal number -0.000 000 000 31 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1111 - 010 1010 0110 1100 1001 1111

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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