-0.008 788 423 613 55 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 613 55(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
-0.008 788 423 613 55(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:

|-0.008 788 423 613 55| = 0.008 788 423 613 55


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: 0.008 788 423 613 55.

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) 0.008 788 423 613 55 × 2 = 0 + 0.017 576 847 227 1;
  • 2) 0.017 576 847 227 1 × 2 = 0 + 0.035 153 694 454 2;
  • 3) 0.035 153 694 454 2 × 2 = 0 + 0.070 307 388 908 4;
  • 4) 0.070 307 388 908 4 × 2 = 0 + 0.140 614 777 816 8;
  • 5) 0.140 614 777 816 8 × 2 = 0 + 0.281 229 555 633 6;
  • 6) 0.281 229 555 633 6 × 2 = 0 + 0.562 459 111 267 2;
  • 7) 0.562 459 111 267 2 × 2 = 1 + 0.124 918 222 534 4;
  • 8) 0.124 918 222 534 4 × 2 = 0 + 0.249 836 445 068 8;
  • 9) 0.249 836 445 068 8 × 2 = 0 + 0.499 672 890 137 6;
  • 10) 0.499 672 890 137 6 × 2 = 0 + 0.999 345 780 275 2;
  • 11) 0.999 345 780 275 2 × 2 = 1 + 0.998 691 560 550 4;
  • 12) 0.998 691 560 550 4 × 2 = 1 + 0.997 383 121 100 8;
  • 13) 0.997 383 121 100 8 × 2 = 1 + 0.994 766 242 201 6;
  • 14) 0.994 766 242 201 6 × 2 = 1 + 0.989 532 484 403 2;
  • 15) 0.989 532 484 403 2 × 2 = 1 + 0.979 064 968 806 4;
  • 16) 0.979 064 968 806 4 × 2 = 1 + 0.958 129 937 612 8;
  • 17) 0.958 129 937 612 8 × 2 = 1 + 0.916 259 875 225 6;
  • 18) 0.916 259 875 225 6 × 2 = 1 + 0.832 519 750 451 2;
  • 19) 0.832 519 750 451 2 × 2 = 1 + 0.665 039 500 902 4;
  • 20) 0.665 039 500 902 4 × 2 = 1 + 0.330 079 001 804 8;
  • 21) 0.330 079 001 804 8 × 2 = 0 + 0.660 158 003 609 6;
  • 22) 0.660 158 003 609 6 × 2 = 1 + 0.320 316 007 219 2;
  • 23) 0.320 316 007 219 2 × 2 = 0 + 0.640 632 014 438 4;
  • 24) 0.640 632 014 438 4 × 2 = 1 + 0.281 264 028 876 8;
  • 25) 0.281 264 028 876 8 × 2 = 0 + 0.562 528 057 753 6;
  • 26) 0.562 528 057 753 6 × 2 = 1 + 0.125 056 115 507 2;
  • 27) 0.125 056 115 507 2 × 2 = 0 + 0.250 112 231 014 4;
  • 28) 0.250 112 231 014 4 × 2 = 0 + 0.500 224 462 028 8;
  • 29) 0.500 224 462 028 8 × 2 = 1 + 0.000 448 924 057 6;
  • 30) 0.000 448 924 057 6 × 2 = 0 + 0.000 897 848 115 2;
  • 31) 0.000 897 848 115 2 × 2 = 0 + 0.001 795 696 230 4;
  • 32) 0.001 795 696 230 4 × 2 = 0 + 0.003 591 392 460 8;
  • 33) 0.003 591 392 460 8 × 2 = 0 + 0.007 182 784 921 6;
  • 34) 0.007 182 784 921 6 × 2 = 0 + 0.014 365 569 843 2;
  • 35) 0.014 365 569 843 2 × 2 = 0 + 0.028 731 139 686 4;
  • 36) 0.028 731 139 686 4 × 2 = 0 + 0.057 462 279 372 8;
  • 37) 0.057 462 279 372 8 × 2 = 0 + 0.114 924 558 745 6;
  • 38) 0.114 924 558 745 6 × 2 = 0 + 0.229 849 117 491 2;
  • 39) 0.229 849 117 491 2 × 2 = 0 + 0.459 698 234 982 4;
  • 40) 0.459 698 234 982 4 × 2 = 0 + 0.919 396 469 964 8;
  • 41) 0.919 396 469 964 8 × 2 = 1 + 0.838 792 939 929 6;
  • 42) 0.838 792 939 929 6 × 2 = 1 + 0.677 585 879 859 2;
  • 43) 0.677 585 879 859 2 × 2 = 1 + 0.355 171 759 718 4;
  • 44) 0.355 171 759 718 4 × 2 = 0 + 0.710 343 519 436 8;
  • 45) 0.710 343 519 436 8 × 2 = 1 + 0.420 687 038 873 6;
  • 46) 0.420 687 038 873 6 × 2 = 0 + 0.841 374 077 747 2;
  • 47) 0.841 374 077 747 2 × 2 = 1 + 0.682 748 155 494 4;
  • 48) 0.682 748 155 494 4 × 2 = 1 + 0.365 496 310 988 8;
  • 49) 0.365 496 310 988 8 × 2 = 0 + 0.730 992 621 977 6;
  • 50) 0.730 992 621 977 6 × 2 = 1 + 0.461 985 243 955 2;
  • 51) 0.461 985 243 955 2 × 2 = 0 + 0.923 970 487 910 4;
  • 52) 0.923 970 487 910 4 × 2 = 1 + 0.847 940 975 820 8;
  • 53) 0.847 940 975 820 8 × 2 = 1 + 0.695 881 951 641 6;
  • 54) 0.695 881 951 641 6 × 2 = 1 + 0.391 763 903 283 2;
  • 55) 0.391 763 903 283 2 × 2 = 0 + 0.783 527 806 566 4;
  • 56) 0.783 527 806 566 4 × 2 = 1 + 0.567 055 613 132 8;
  • 57) 0.567 055 613 132 8 × 2 = 1 + 0.134 111 226 265 6;
  • 58) 0.134 111 226 265 6 × 2 = 0 + 0.268 222 452 531 2;
  • 59) 0.268 222 452 531 2 × 2 = 0 + 0.536 444 905 062 4;

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:


0.008 788 423 613 55(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1110 1011 0101 1101 100(2)

6. Positive number before normalization:

0.008 788 423 613 55(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1110 1011 0101 1101 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.008 788 423 613 55(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1110 1011 0101 1101 100(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1110 1011 0101 1101 100(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0111 0101 1010 1110 1100(2) × 2-7


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): -7

Mantissa (not normalized):
1.0001 1111 1111 1010 1010 0100 0000 0000 0111 0101 1010 1110 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1016(10) =

011 1111 1000(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. 0001 1111 1111 1010 1010 0100 0000 0000 0111 0101 1010 1110 1100 =

0001 1111 1111 1010 1010 0100 0000 0000 0111 0101 1010 1110 1100


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 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0100 0000 0000 0111 0101 1010 1110 1100


Decimal number -0.008 788 423 613 55 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 0111 0101 1010 1110 1100

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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