-0.000 072 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 072 93(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 072 93(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 072 93| = 0.000 072 93


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: 0.000 072 93.

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) 0.000 072 93 × 2 = 0 + 0.000 145 86;
  • 2) 0.000 145 86 × 2 = 0 + 0.000 291 72;
  • 3) 0.000 291 72 × 2 = 0 + 0.000 583 44;
  • 4) 0.000 583 44 × 2 = 0 + 0.001 166 88;
  • 5) 0.001 166 88 × 2 = 0 + 0.002 333 76;
  • 6) 0.002 333 76 × 2 = 0 + 0.004 667 52;
  • 7) 0.004 667 52 × 2 = 0 + 0.009 335 04;
  • 8) 0.009 335 04 × 2 = 0 + 0.018 670 08;
  • 9) 0.018 670 08 × 2 = 0 + 0.037 340 16;
  • 10) 0.037 340 16 × 2 = 0 + 0.074 680 32;
  • 11) 0.074 680 32 × 2 = 0 + 0.149 360 64;
  • 12) 0.149 360 64 × 2 = 0 + 0.298 721 28;
  • 13) 0.298 721 28 × 2 = 0 + 0.597 442 56;
  • 14) 0.597 442 56 × 2 = 1 + 0.194 885 12;
  • 15) 0.194 885 12 × 2 = 0 + 0.389 770 24;
  • 16) 0.389 770 24 × 2 = 0 + 0.779 540 48;
  • 17) 0.779 540 48 × 2 = 1 + 0.559 080 96;
  • 18) 0.559 080 96 × 2 = 1 + 0.118 161 92;
  • 19) 0.118 161 92 × 2 = 0 + 0.236 323 84;
  • 20) 0.236 323 84 × 2 = 0 + 0.472 647 68;
  • 21) 0.472 647 68 × 2 = 0 + 0.945 295 36;
  • 22) 0.945 295 36 × 2 = 1 + 0.890 590 72;
  • 23) 0.890 590 72 × 2 = 1 + 0.781 181 44;
  • 24) 0.781 181 44 × 2 = 1 + 0.562 362 88;
  • 25) 0.562 362 88 × 2 = 1 + 0.124 725 76;
  • 26) 0.124 725 76 × 2 = 0 + 0.249 451 52;
  • 27) 0.249 451 52 × 2 = 0 + 0.498 903 04;
  • 28) 0.498 903 04 × 2 = 0 + 0.997 806 08;
  • 29) 0.997 806 08 × 2 = 1 + 0.995 612 16;
  • 30) 0.995 612 16 × 2 = 1 + 0.991 224 32;
  • 31) 0.991 224 32 × 2 = 1 + 0.982 448 64;
  • 32) 0.982 448 64 × 2 = 1 + 0.964 897 28;
  • 33) 0.964 897 28 × 2 = 1 + 0.929 794 56;
  • 34) 0.929 794 56 × 2 = 1 + 0.859 589 12;
  • 35) 0.859 589 12 × 2 = 1 + 0.719 178 24;
  • 36) 0.719 178 24 × 2 = 1 + 0.438 356 48;
  • 37) 0.438 356 48 × 2 = 0 + 0.876 712 96;
  • 38) 0.876 712 96 × 2 = 1 + 0.753 425 92;
  • 39) 0.753 425 92 × 2 = 1 + 0.506 851 84;
  • 40) 0.506 851 84 × 2 = 1 + 0.013 703 68;
  • 41) 0.013 703 68 × 2 = 0 + 0.027 407 36;
  • 42) 0.027 407 36 × 2 = 0 + 0.054 814 72;
  • 43) 0.054 814 72 × 2 = 0 + 0.109 629 44;
  • 44) 0.109 629 44 × 2 = 0 + 0.219 258 88;
  • 45) 0.219 258 88 × 2 = 0 + 0.438 517 76;
  • 46) 0.438 517 76 × 2 = 0 + 0.877 035 52;
  • 47) 0.877 035 52 × 2 = 1 + 0.754 071 04;
  • 48) 0.754 071 04 × 2 = 1 + 0.508 142 08;
  • 49) 0.508 142 08 × 2 = 1 + 0.016 284 16;
  • 50) 0.016 284 16 × 2 = 0 + 0.032 568 32;
  • 51) 0.032 568 32 × 2 = 0 + 0.065 136 64;
  • 52) 0.065 136 64 × 2 = 0 + 0.130 273 28;
  • 53) 0.130 273 28 × 2 = 0 + 0.260 546 56;
  • 54) 0.260 546 56 × 2 = 0 + 0.521 093 12;
  • 55) 0.521 093 12 × 2 = 1 + 0.042 186 24;
  • 56) 0.042 186 24 × 2 = 0 + 0.084 372 48;
  • 57) 0.084 372 48 × 2 = 0 + 0.168 744 96;
  • 58) 0.168 744 96 × 2 = 0 + 0.337 489 92;
  • 59) 0.337 489 92 × 2 = 0 + 0.674 979 84;
  • 60) 0.674 979 84 × 2 = 1 + 0.349 959 68;
  • 61) 0.349 959 68 × 2 = 0 + 0.699 919 36;
  • 62) 0.699 919 36 × 2 = 1 + 0.399 838 72;
  • 63) 0.399 838 72 × 2 = 0 + 0.799 677 44;
  • 64) 0.799 677 44 × 2 = 1 + 0.599 354 88;
  • 65) 0.599 354 88 × 2 = 1 + 0.198 709 76;
  • 66) 0.198 709 76 × 2 = 0 + 0.397 419 52;

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:


0.000 072 93(10) =

0.0000 0000 0000 0100 1100 0111 1000 1111 1111 0111 0000 0011 1000 0010 0001 0101 10(2)

6. Positive number before normalization:

0.000 072 93(10) =

0.0000 0000 0000 0100 1100 0111 1000 1111 1111 0111 0000 0011 1000 0010 0001 0101 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 072 93(10) =

0.0000 0000 0000 0100 1100 0111 1000 1111 1111 0111 0000 0011 1000 0010 0001 0101 10(2) =

0.0000 0000 0000 0100 1100 0111 1000 1111 1111 0111 0000 0011 1000 0010 0001 0101 10(2) × 20 =

1.0011 0001 1110 0011 1111 1101 1100 0000 1110 0000 1000 0101 0110(2) × 2-14


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -14

Mantissa (not normalized):
1.0011 0001 1110 0011 1111 1101 1100 0000 1110 0000 1000 0101 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-14 + 2(11-1) - 1 =

(-14 + 1 023)(10) =

1 009(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 009 ÷ 2 = 504 + 1;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

1009(10) =

011 1111 0001(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 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0011 0001 1110 0011 1111 1101 1100 0000 1110 0000 1000 0101 0110 =

0011 0001 1110 0011 1111 1101 1100 0000 1110 0000 1000 0101 0110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

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

Exponent (11 bits) =
011 1111 0001

Mantissa (52 bits) =
0011 0001 1110 0011 1111 1101 1100 0000 1110 0000 1000 0101 0110


Decimal number -0.000 072 93 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0001 - 0011 0001 1110 0011 1111 1101 1100 0000 1110 0000 1000 0101 0110

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

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

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100