-0.008 788 423 612 913 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 612 913(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
-0.008 788 423 612 913(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:

|-0.008 788 423 612 913| = 0.008 788 423 612 913


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: 0.008 788 423 612 913.

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) 0.008 788 423 612 913 × 2 = 0 + 0.017 576 847 225 826;
  • 2) 0.017 576 847 225 826 × 2 = 0 + 0.035 153 694 451 652;
  • 3) 0.035 153 694 451 652 × 2 = 0 + 0.070 307 388 903 304;
  • 4) 0.070 307 388 903 304 × 2 = 0 + 0.140 614 777 806 608;
  • 5) 0.140 614 777 806 608 × 2 = 0 + 0.281 229 555 613 216;
  • 6) 0.281 229 555 613 216 × 2 = 0 + 0.562 459 111 226 432;
  • 7) 0.562 459 111 226 432 × 2 = 1 + 0.124 918 222 452 864;
  • 8) 0.124 918 222 452 864 × 2 = 0 + 0.249 836 444 905 728;
  • 9) 0.249 836 444 905 728 × 2 = 0 + 0.499 672 889 811 456;
  • 10) 0.499 672 889 811 456 × 2 = 0 + 0.999 345 779 622 912;
  • 11) 0.999 345 779 622 912 × 2 = 1 + 0.998 691 559 245 824;
  • 12) 0.998 691 559 245 824 × 2 = 1 + 0.997 383 118 491 648;
  • 13) 0.997 383 118 491 648 × 2 = 1 + 0.994 766 236 983 296;
  • 14) 0.994 766 236 983 296 × 2 = 1 + 0.989 532 473 966 592;
  • 15) 0.989 532 473 966 592 × 2 = 1 + 0.979 064 947 933 184;
  • 16) 0.979 064 947 933 184 × 2 = 1 + 0.958 129 895 866 368;
  • 17) 0.958 129 895 866 368 × 2 = 1 + 0.916 259 791 732 736;
  • 18) 0.916 259 791 732 736 × 2 = 1 + 0.832 519 583 465 472;
  • 19) 0.832 519 583 465 472 × 2 = 1 + 0.665 039 166 930 944;
  • 20) 0.665 039 166 930 944 × 2 = 1 + 0.330 078 333 861 888;
  • 21) 0.330 078 333 861 888 × 2 = 0 + 0.660 156 667 723 776;
  • 22) 0.660 156 667 723 776 × 2 = 1 + 0.320 313 335 447 552;
  • 23) 0.320 313 335 447 552 × 2 = 0 + 0.640 626 670 895 104;
  • 24) 0.640 626 670 895 104 × 2 = 1 + 0.281 253 341 790 208;
  • 25) 0.281 253 341 790 208 × 2 = 0 + 0.562 506 683 580 416;
  • 26) 0.562 506 683 580 416 × 2 = 1 + 0.125 013 367 160 832;
  • 27) 0.125 013 367 160 832 × 2 = 0 + 0.250 026 734 321 664;
  • 28) 0.250 026 734 321 664 × 2 = 0 + 0.500 053 468 643 328;
  • 29) 0.500 053 468 643 328 × 2 = 1 + 0.000 106 937 286 656;
  • 30) 0.000 106 937 286 656 × 2 = 0 + 0.000 213 874 573 312;
  • 31) 0.000 213 874 573 312 × 2 = 0 + 0.000 427 749 146 624;
  • 32) 0.000 427 749 146 624 × 2 = 0 + 0.000 855 498 293 248;
  • 33) 0.000 855 498 293 248 × 2 = 0 + 0.001 710 996 586 496;
  • 34) 0.001 710 996 586 496 × 2 = 0 + 0.003 421 993 172 992;
  • 35) 0.003 421 993 172 992 × 2 = 0 + 0.006 843 986 345 984;
  • 36) 0.006 843 986 345 984 × 2 = 0 + 0.013 687 972 691 968;
  • 37) 0.013 687 972 691 968 × 2 = 0 + 0.027 375 945 383 936;
  • 38) 0.027 375 945 383 936 × 2 = 0 + 0.054 751 890 767 872;
  • 39) 0.054 751 890 767 872 × 2 = 0 + 0.109 503 781 535 744;
  • 40) 0.109 503 781 535 744 × 2 = 0 + 0.219 007 563 071 488;
  • 41) 0.219 007 563 071 488 × 2 = 0 + 0.438 015 126 142 976;
  • 42) 0.438 015 126 142 976 × 2 = 0 + 0.876 030 252 285 952;
  • 43) 0.876 030 252 285 952 × 2 = 1 + 0.752 060 504 571 904;
  • 44) 0.752 060 504 571 904 × 2 = 1 + 0.504 121 009 143 808;
  • 45) 0.504 121 009 143 808 × 2 = 1 + 0.008 242 018 287 616;
  • 46) 0.008 242 018 287 616 × 2 = 0 + 0.016 484 036 575 232;
  • 47) 0.016 484 036 575 232 × 2 = 0 + 0.032 968 073 150 464;
  • 48) 0.032 968 073 150 464 × 2 = 0 + 0.065 936 146 300 928;
  • 49) 0.065 936 146 300 928 × 2 = 0 + 0.131 872 292 601 856;
  • 50) 0.131 872 292 601 856 × 2 = 0 + 0.263 744 585 203 712;
  • 51) 0.263 744 585 203 712 × 2 = 0 + 0.527 489 170 407 424;
  • 52) 0.527 489 170 407 424 × 2 = 1 + 0.054 978 340 814 848;
  • 53) 0.054 978 340 814 848 × 2 = 0 + 0.109 956 681 629 696;
  • 54) 0.109 956 681 629 696 × 2 = 0 + 0.219 913 363 259 392;
  • 55) 0.219 913 363 259 392 × 2 = 0 + 0.439 826 726 518 784;
  • 56) 0.439 826 726 518 784 × 2 = 0 + 0.879 653 453 037 568;
  • 57) 0.879 653 453 037 568 × 2 = 1 + 0.759 306 906 075 136;
  • 58) 0.759 306 906 075 136 × 2 = 1 + 0.518 613 812 150 272;
  • 59) 0.518 613 812 150 272 × 2 = 1 + 0.037 227 624 300 544;

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:


0.008 788 423 612 913(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0011 1000 0001 0000 111(2)

6. Positive number before normalization:

0.008 788 423 612 913(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0011 1000 0001 0000 111(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:


0.008 788 423 612 913(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0011 1000 0001 0000 111(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0011 1000 0001 0000 111(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0001 1100 0000 1000 0111(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 0001 1100 0000 1000 0111


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 0001 1100 0000 1000 0111 =

0001 1111 1111 1010 1010 0100 0000 0000 0001 1100 0000 1000 0111


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 0001 1100 0000 1000 0111


Decimal number -0.008 788 423 612 913 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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