-0.000 000 005 917 031 260 533 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 005 917 031 260 533(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
-0.000 000 005 917 031 260 533(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:

|-0.000 000 005 917 031 260 533| = 0.000 000 005 917 031 260 533


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: 0.000 000 005 917 031 260 533.

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) 0.000 000 005 917 031 260 533 × 2 = 0 + 0.000 000 011 834 062 521 066;
  • 2) 0.000 000 011 834 062 521 066 × 2 = 0 + 0.000 000 023 668 125 042 132;
  • 3) 0.000 000 023 668 125 042 132 × 2 = 0 + 0.000 000 047 336 250 084 264;
  • 4) 0.000 000 047 336 250 084 264 × 2 = 0 + 0.000 000 094 672 500 168 528;
  • 5) 0.000 000 094 672 500 168 528 × 2 = 0 + 0.000 000 189 345 000 337 056;
  • 6) 0.000 000 189 345 000 337 056 × 2 = 0 + 0.000 000 378 690 000 674 112;
  • 7) 0.000 000 378 690 000 674 112 × 2 = 0 + 0.000 000 757 380 001 348 224;
  • 8) 0.000 000 757 380 001 348 224 × 2 = 0 + 0.000 001 514 760 002 696 448;
  • 9) 0.000 001 514 760 002 696 448 × 2 = 0 + 0.000 003 029 520 005 392 896;
  • 10) 0.000 003 029 520 005 392 896 × 2 = 0 + 0.000 006 059 040 010 785 792;
  • 11) 0.000 006 059 040 010 785 792 × 2 = 0 + 0.000 012 118 080 021 571 584;
  • 12) 0.000 012 118 080 021 571 584 × 2 = 0 + 0.000 024 236 160 043 143 168;
  • 13) 0.000 024 236 160 043 143 168 × 2 = 0 + 0.000 048 472 320 086 286 336;
  • 14) 0.000 048 472 320 086 286 336 × 2 = 0 + 0.000 096 944 640 172 572 672;
  • 15) 0.000 096 944 640 172 572 672 × 2 = 0 + 0.000 193 889 280 345 145 344;
  • 16) 0.000 193 889 280 345 145 344 × 2 = 0 + 0.000 387 778 560 690 290 688;
  • 17) 0.000 387 778 560 690 290 688 × 2 = 0 + 0.000 775 557 121 380 581 376;
  • 18) 0.000 775 557 121 380 581 376 × 2 = 0 + 0.001 551 114 242 761 162 752;
  • 19) 0.001 551 114 242 761 162 752 × 2 = 0 + 0.003 102 228 485 522 325 504;
  • 20) 0.003 102 228 485 522 325 504 × 2 = 0 + 0.006 204 456 971 044 651 008;
  • 21) 0.006 204 456 971 044 651 008 × 2 = 0 + 0.012 408 913 942 089 302 016;
  • 22) 0.012 408 913 942 089 302 016 × 2 = 0 + 0.024 817 827 884 178 604 032;
  • 23) 0.024 817 827 884 178 604 032 × 2 = 0 + 0.049 635 655 768 357 208 064;
  • 24) 0.049 635 655 768 357 208 064 × 2 = 0 + 0.099 271 311 536 714 416 128;
  • 25) 0.099 271 311 536 714 416 128 × 2 = 0 + 0.198 542 623 073 428 832 256;
  • 26) 0.198 542 623 073 428 832 256 × 2 = 0 + 0.397 085 246 146 857 664 512;
  • 27) 0.397 085 246 146 857 664 512 × 2 = 0 + 0.794 170 492 293 715 329 024;
  • 28) 0.794 170 492 293 715 329 024 × 2 = 1 + 0.588 340 984 587 430 658 048;
  • 29) 0.588 340 984 587 430 658 048 × 2 = 1 + 0.176 681 969 174 861 316 096;
  • 30) 0.176 681 969 174 861 316 096 × 2 = 0 + 0.353 363 938 349 722 632 192;
  • 31) 0.353 363 938 349 722 632 192 × 2 = 0 + 0.706 727 876 699 445 264 384;
  • 32) 0.706 727 876 699 445 264 384 × 2 = 1 + 0.413 455 753 398 890 528 768;
  • 33) 0.413 455 753 398 890 528 768 × 2 = 0 + 0.826 911 506 797 781 057 536;
  • 34) 0.826 911 506 797 781 057 536 × 2 = 1 + 0.653 823 013 595 562 115 072;
  • 35) 0.653 823 013 595 562 115 072 × 2 = 1 + 0.307 646 027 191 124 230 144;
  • 36) 0.307 646 027 191 124 230 144 × 2 = 0 + 0.615 292 054 382 248 460 288;
  • 37) 0.615 292 054 382 248 460 288 × 2 = 1 + 0.230 584 108 764 496 920 576;
  • 38) 0.230 584 108 764 496 920 576 × 2 = 0 + 0.461 168 217 528 993 841 152;
  • 39) 0.461 168 217 528 993 841 152 × 2 = 0 + 0.922 336 435 057 987 682 304;
  • 40) 0.922 336 435 057 987 682 304 × 2 = 1 + 0.844 672 870 115 975 364 608;
  • 41) 0.844 672 870 115 975 364 608 × 2 = 1 + 0.689 345 740 231 950 729 216;
  • 42) 0.689 345 740 231 950 729 216 × 2 = 1 + 0.378 691 480 463 901 458 432;
  • 43) 0.378 691 480 463 901 458 432 × 2 = 0 + 0.757 382 960 927 802 916 864;
  • 44) 0.757 382 960 927 802 916 864 × 2 = 1 + 0.514 765 921 855 605 833 728;
  • 45) 0.514 765 921 855 605 833 728 × 2 = 1 + 0.029 531 843 711 211 667 456;
  • 46) 0.029 531 843 711 211 667 456 × 2 = 0 + 0.059 063 687 422 423 334 912;
  • 47) 0.059 063 687 422 423 334 912 × 2 = 0 + 0.118 127 374 844 846 669 824;
  • 48) 0.118 127 374 844 846 669 824 × 2 = 0 + 0.236 254 749 689 693 339 648;
  • 49) 0.236 254 749 689 693 339 648 × 2 = 0 + 0.472 509 499 379 386 679 296;
  • 50) 0.472 509 499 379 386 679 296 × 2 = 0 + 0.945 018 998 758 773 358 592;
  • 51) 0.945 018 998 758 773 358 592 × 2 = 1 + 0.890 037 997 517 546 717 184;
  • 52) 0.890 037 997 517 546 717 184 × 2 = 1 + 0.780 075 995 035 093 434 368;
  • 53) 0.780 075 995 035 093 434 368 × 2 = 1 + 0.560 151 990 070 186 868 736;
  • 54) 0.560 151 990 070 186 868 736 × 2 = 1 + 0.120 303 980 140 373 737 472;
  • 55) 0.120 303 980 140 373 737 472 × 2 = 0 + 0.240 607 960 280 747 474 944;
  • 56) 0.240 607 960 280 747 474 944 × 2 = 0 + 0.481 215 920 561 494 949 888;
  • 57) 0.481 215 920 561 494 949 888 × 2 = 0 + 0.962 431 841 122 989 899 776;
  • 58) 0.962 431 841 122 989 899 776 × 2 = 1 + 0.924 863 682 245 979 799 552;
  • 59) 0.924 863 682 245 979 799 552 × 2 = 1 + 0.849 727 364 491 959 599 104;
  • 60) 0.849 727 364 491 959 599 104 × 2 = 1 + 0.699 454 728 983 919 198 208;
  • 61) 0.699 454 728 983 919 198 208 × 2 = 1 + 0.398 909 457 967 838 396 416;
  • 62) 0.398 909 457 967 838 396 416 × 2 = 0 + 0.797 818 915 935 676 792 832;
  • 63) 0.797 818 915 935 676 792 832 × 2 = 1 + 0.595 637 831 871 353 585 664;
  • 64) 0.595 637 831 871 353 585 664 × 2 = 1 + 0.191 275 663 742 707 171 328;
  • 65) 0.191 275 663 742 707 171 328 × 2 = 0 + 0.382 551 327 485 414 342 656;
  • 66) 0.382 551 327 485 414 342 656 × 2 = 0 + 0.765 102 654 970 828 685 312;
  • 67) 0.765 102 654 970 828 685 312 × 2 = 1 + 0.530 205 309 941 657 370 624;
  • 68) 0.530 205 309 941 657 370 624 × 2 = 1 + 0.060 410 619 883 314 741 248;
  • 69) 0.060 410 619 883 314 741 248 × 2 = 0 + 0.120 821 239 766 629 482 496;
  • 70) 0.120 821 239 766 629 482 496 × 2 = 0 + 0.241 642 479 533 258 964 992;
  • 71) 0.241 642 479 533 258 964 992 × 2 = 0 + 0.483 284 959 066 517 929 984;
  • 72) 0.483 284 959 066 517 929 984 × 2 = 0 + 0.966 569 918 133 035 859 968;
  • 73) 0.966 569 918 133 035 859 968 × 2 = 1 + 0.933 139 836 266 071 719 936;
  • 74) 0.933 139 836 266 071 719 936 × 2 = 1 + 0.866 279 672 532 143 439 872;
  • 75) 0.866 279 672 532 143 439 872 × 2 = 1 + 0.732 559 345 064 286 879 744;
  • 76) 0.732 559 345 064 286 879 744 × 2 = 1 + 0.465 118 690 128 573 759 488;
  • 77) 0.465 118 690 128 573 759 488 × 2 = 0 + 0.930 237 380 257 147 518 976;
  • 78) 0.930 237 380 257 147 518 976 × 2 = 1 + 0.860 474 760 514 295 037 952;
  • 79) 0.860 474 760 514 295 037 952 × 2 = 1 + 0.720 949 521 028 590 075 904;
  • 80) 0.720 949 521 028 590 075 904 × 2 = 1 + 0.441 899 042 057 180 151 808;

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:


0.000 000 005 917 031 260 533(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1011 0011 0000 1111 0111(2)

6. Positive number before normalization:

0.000 000 005 917 031 260 533(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1011 0011 0000 1111 0111(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:


0.000 000 005 917 031 260 533(10) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1011 0011 0000 1111 0111(2) =

0.0000 0000 0000 0000 0000 0000 0001 1001 0110 1001 1101 1000 0011 1100 0111 1011 0011 0000 1111 0111(2) × 20 =

1.1001 0110 1001 1101 1000 0011 1100 0111 1011 0011 0000 1111 0111(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 0111 1011 0011 0000 1111 0111


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 0111 1011 0011 0000 1111 0111 =

1001 0110 1001 1101 1000 0011 1100 0111 1011 0011 0000 1111 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 1110 0011

Mantissa (52 bits) =
1001 0110 1001 1101 1000 0011 1100 0111 1011 0011 0000 1111 0111


Decimal number -0.000 000 005 917 031 260 533 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0011 - 1001 0110 1001 1101 1000 0011 1100 0111 1011 0011 0000 1111 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