-0.000 000 005 917 031 261 54 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 005 917 031 261 54(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
-0.000 000 005 917 031 261 54(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:

|-0.000 000 005 917 031 261 54| = 0.000 000 005 917 031 261 54


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: 0.000 000 005 917 031 261 54.

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) 0.000 000 005 917 031 261 54 × 2 = 0 + 0.000 000 011 834 062 523 08;
  • 2) 0.000 000 011 834 062 523 08 × 2 = 0 + 0.000 000 023 668 125 046 16;
  • 3) 0.000 000 023 668 125 046 16 × 2 = 0 + 0.000 000 047 336 250 092 32;
  • 4) 0.000 000 047 336 250 092 32 × 2 = 0 + 0.000 000 094 672 500 184 64;
  • 5) 0.000 000 094 672 500 184 64 × 2 = 0 + 0.000 000 189 345 000 369 28;
  • 6) 0.000 000 189 345 000 369 28 × 2 = 0 + 0.000 000 378 690 000 738 56;
  • 7) 0.000 000 378 690 000 738 56 × 2 = 0 + 0.000 000 757 380 001 477 12;
  • 8) 0.000 000 757 380 001 477 12 × 2 = 0 + 0.000 001 514 760 002 954 24;
  • 9) 0.000 001 514 760 002 954 24 × 2 = 0 + 0.000 003 029 520 005 908 48;
  • 10) 0.000 003 029 520 005 908 48 × 2 = 0 + 0.000 006 059 040 011 816 96;
  • 11) 0.000 006 059 040 011 816 96 × 2 = 0 + 0.000 012 118 080 023 633 92;
  • 12) 0.000 012 118 080 023 633 92 × 2 = 0 + 0.000 024 236 160 047 267 84;
  • 13) 0.000 024 236 160 047 267 84 × 2 = 0 + 0.000 048 472 320 094 535 68;
  • 14) 0.000 048 472 320 094 535 68 × 2 = 0 + 0.000 096 944 640 189 071 36;
  • 15) 0.000 096 944 640 189 071 36 × 2 = 0 + 0.000 193 889 280 378 142 72;
  • 16) 0.000 193 889 280 378 142 72 × 2 = 0 + 0.000 387 778 560 756 285 44;
  • 17) 0.000 387 778 560 756 285 44 × 2 = 0 + 0.000 775 557 121 512 570 88;
  • 18) 0.000 775 557 121 512 570 88 × 2 = 0 + 0.001 551 114 243 025 141 76;
  • 19) 0.001 551 114 243 025 141 76 × 2 = 0 + 0.003 102 228 486 050 283 52;
  • 20) 0.003 102 228 486 050 283 52 × 2 = 0 + 0.006 204 456 972 100 567 04;
  • 21) 0.006 204 456 972 100 567 04 × 2 = 0 + 0.012 408 913 944 201 134 08;
  • 22) 0.012 408 913 944 201 134 08 × 2 = 0 + 0.024 817 827 888 402 268 16;
  • 23) 0.024 817 827 888 402 268 16 × 2 = 0 + 0.049 635 655 776 804 536 32;
  • 24) 0.049 635 655 776 804 536 32 × 2 = 0 + 0.099 271 311 553 609 072 64;
  • 25) 0.099 271 311 553 609 072 64 × 2 = 0 + 0.198 542 623 107 218 145 28;
  • 26) 0.198 542 623 107 218 145 28 × 2 = 0 + 0.397 085 246 214 436 290 56;
  • 27) 0.397 085 246 214 436 290 56 × 2 = 0 + 0.794 170 492 428 872 581 12;
  • 28) 0.794 170 492 428 872 581 12 × 2 = 1 + 0.588 340 984 857 745 162 24;
  • 29) 0.588 340 984 857 745 162 24 × 2 = 1 + 0.176 681 969 715 490 324 48;
  • 30) 0.176 681 969 715 490 324 48 × 2 = 0 + 0.353 363 939 430 980 648 96;
  • 31) 0.353 363 939 430 980 648 96 × 2 = 0 + 0.706 727 878 861 961 297 92;
  • 32) 0.706 727 878 861 961 297 92 × 2 = 1 + 0.413 455 757 723 922 595 84;
  • 33) 0.413 455 757 723 922 595 84 × 2 = 0 + 0.826 911 515 447 845 191 68;
  • 34) 0.826 911 515 447 845 191 68 × 2 = 1 + 0.653 823 030 895 690 383 36;
  • 35) 0.653 823 030 895 690 383 36 × 2 = 1 + 0.307 646 061 791 380 766 72;
  • 36) 0.307 646 061 791 380 766 72 × 2 = 0 + 0.615 292 123 582 761 533 44;
  • 37) 0.615 292 123 582 761 533 44 × 2 = 1 + 0.230 584 247 165 523 066 88;
  • 38) 0.230 584 247 165 523 066 88 × 2 = 0 + 0.461 168 494 331 046 133 76;
  • 39) 0.461 168 494 331 046 133 76 × 2 = 0 + 0.922 336 988 662 092 267 52;
  • 40) 0.922 336 988 662 092 267 52 × 2 = 1 + 0.844 673 977 324 184 535 04;
  • 41) 0.844 673 977 324 184 535 04 × 2 = 1 + 0.689 347 954 648 369 070 08;
  • 42) 0.689 347 954 648 369 070 08 × 2 = 1 + 0.378 695 909 296 738 140 16;
  • 43) 0.378 695 909 296 738 140 16 × 2 = 0 + 0.757 391 818 593 476 280 32;
  • 44) 0.757 391 818 593 476 280 32 × 2 = 1 + 0.514 783 637 186 952 560 64;
  • 45) 0.514 783 637 186 952 560 64 × 2 = 1 + 0.029 567 274 373 905 121 28;
  • 46) 0.029 567 274 373 905 121 28 × 2 = 0 + 0.059 134 548 747 810 242 56;
  • 47) 0.059 134 548 747 810 242 56 × 2 = 0 + 0.118 269 097 495 620 485 12;
  • 48) 0.118 269 097 495 620 485 12 × 2 = 0 + 0.236 538 194 991 240 970 24;
  • 49) 0.236 538 194 991 240 970 24 × 2 = 0 + 0.473 076 389 982 481 940 48;
  • 50) 0.473 076 389 982 481 940 48 × 2 = 0 + 0.946 152 779 964 963 880 96;
  • 51) 0.946 152 779 964 963 880 96 × 2 = 1 + 0.892 305 559 929 927 761 92;
  • 52) 0.892 305 559 929 927 761 92 × 2 = 1 + 0.784 611 119 859 855 523 84;
  • 53) 0.784 611 119 859 855 523 84 × 2 = 1 + 0.569 222 239 719 711 047 68;
  • 54) 0.569 222 239 719 711 047 68 × 2 = 1 + 0.138 444 479 439 422 095 36;
  • 55) 0.138 444 479 439 422 095 36 × 2 = 0 + 0.276 888 958 878 844 190 72;
  • 56) 0.276 888 958 878 844 190 72 × 2 = 0 + 0.553 777 917 757 688 381 44;
  • 57) 0.553 777 917 757 688 381 44 × 2 = 1 + 0.107 555 835 515 376 762 88;
  • 58) 0.107 555 835 515 376 762 88 × 2 = 0 + 0.215 111 671 030 753 525 76;
  • 59) 0.215 111 671 030 753 525 76 × 2 = 0 + 0.430 223 342 061 507 051 52;
  • 60) 0.430 223 342 061 507 051 52 × 2 = 0 + 0.860 446 684 123 014 103 04;
  • 61) 0.860 446 684 123 014 103 04 × 2 = 1 + 0.720 893 368 246 028 206 08;
  • 62) 0.720 893 368 246 028 206 08 × 2 = 1 + 0.441 786 736 492 056 412 16;
  • 63) 0.441 786 736 492 056 412 16 × 2 = 0 + 0.883 573 472 984 112 824 32;
  • 64) 0.883 573 472 984 112 824 32 × 2 = 1 + 0.767 146 945 968 225 648 64;
  • 65) 0.767 146 945 968 225 648 64 × 2 = 1 + 0.534 293 891 936 451 297 28;
  • 66) 0.534 293 891 936 451 297 28 × 2 = 1 + 0.068 587 783 872 902 594 56;
  • 67) 0.068 587 783 872 902 594 56 × 2 = 0 + 0.137 175 567 745 805 189 12;
  • 68) 0.137 175 567 745 805 189 12 × 2 = 0 + 0.274 351 135 491 610 378 24;
  • 69) 0.274 351 135 491 610 378 24 × 2 = 0 + 0.548 702 270 983 220 756 48;
  • 70) 0.548 702 270 983 220 756 48 × 2 = 1 + 0.097 404 541 966 441 512 96;
  • 71) 0.097 404 541 966 441 512 96 × 2 = 0 + 0.194 809 083 932 883 025 92;
  • 72) 0.194 809 083 932 883 025 92 × 2 = 0 + 0.389 618 167 865 766 051 84;
  • 73) 0.389 618 167 865 766 051 84 × 2 = 0 + 0.779 236 335 731 532 103 68;
  • 74) 0.779 236 335 731 532 103 68 × 2 = 1 + 0.558 472 671 463 064 207 36;
  • 75) 0.558 472 671 463 064 207 36 × 2 = 1 + 0.116 945 342 926 128 414 72;
  • 76) 0.116 945 342 926 128 414 72 × 2 = 0 + 0.233 890 685 852 256 829 44;
  • 77) 0.233 890 685 852 256 829 44 × 2 = 0 + 0.467 781 371 704 513 658 88;
  • 78) 0.467 781 371 704 513 658 88 × 2 = 0 + 0.935 562 743 409 027 317 76;
  • 79) 0.935 562 743 409 027 317 76 × 2 = 1 + 0.871 125 486 818 054 635 52;
  • 80) 0.871 125 486 818 054 635 52 × 2 = 1 + 0.742 250 973 636 109 271 04;

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:


0.000 000 005 917 031 261 54(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011(2)

6. Positive number before normalization:

0.000 000 005 917 031 261 54(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 005 917 031 261 54(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011(2) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011(2) × 20 =

1.1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011(2) × 2-28


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): -28

Mantissa (not normalized):
1.1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(11-1) - 1 =

(-28 + 1 023)(10) =

995(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;
  • 995 ÷ 2 = 497 + 1;
  • 497 ÷ 2 = 248 + 1;
  • 248 ÷ 2 = 124 + 0;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

995(10) =

011 1110 0011(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. 1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011 =

1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011


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 1110 0011

Mantissa (52 bits) =
1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011


Decimal number -0.000 000 005 917 031 261 54 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0011 - 1001 0110 1001 1101 1000 0011 1100 1000 1101 1100 0100 0110 0011

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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