-0.000 000 018 616 658 917 718 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 018 616 658 917 718(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
-0.000 000 018 616 658 917 718(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:

|-0.000 000 018 616 658 917 718| = 0.000 000 018 616 658 917 718


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: 0.000 000 018 616 658 917 718.

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) 0.000 000 018 616 658 917 718 × 2 = 0 + 0.000 000 037 233 317 835 436;
  • 2) 0.000 000 037 233 317 835 436 × 2 = 0 + 0.000 000 074 466 635 670 872;
  • 3) 0.000 000 074 466 635 670 872 × 2 = 0 + 0.000 000 148 933 271 341 744;
  • 4) 0.000 000 148 933 271 341 744 × 2 = 0 + 0.000 000 297 866 542 683 488;
  • 5) 0.000 000 297 866 542 683 488 × 2 = 0 + 0.000 000 595 733 085 366 976;
  • 6) 0.000 000 595 733 085 366 976 × 2 = 0 + 0.000 001 191 466 170 733 952;
  • 7) 0.000 001 191 466 170 733 952 × 2 = 0 + 0.000 002 382 932 341 467 904;
  • 8) 0.000 002 382 932 341 467 904 × 2 = 0 + 0.000 004 765 864 682 935 808;
  • 9) 0.000 004 765 864 682 935 808 × 2 = 0 + 0.000 009 531 729 365 871 616;
  • 10) 0.000 009 531 729 365 871 616 × 2 = 0 + 0.000 019 063 458 731 743 232;
  • 11) 0.000 019 063 458 731 743 232 × 2 = 0 + 0.000 038 126 917 463 486 464;
  • 12) 0.000 038 126 917 463 486 464 × 2 = 0 + 0.000 076 253 834 926 972 928;
  • 13) 0.000 076 253 834 926 972 928 × 2 = 0 + 0.000 152 507 669 853 945 856;
  • 14) 0.000 152 507 669 853 945 856 × 2 = 0 + 0.000 305 015 339 707 891 712;
  • 15) 0.000 305 015 339 707 891 712 × 2 = 0 + 0.000 610 030 679 415 783 424;
  • 16) 0.000 610 030 679 415 783 424 × 2 = 0 + 0.001 220 061 358 831 566 848;
  • 17) 0.001 220 061 358 831 566 848 × 2 = 0 + 0.002 440 122 717 663 133 696;
  • 18) 0.002 440 122 717 663 133 696 × 2 = 0 + 0.004 880 245 435 326 267 392;
  • 19) 0.004 880 245 435 326 267 392 × 2 = 0 + 0.009 760 490 870 652 534 784;
  • 20) 0.009 760 490 870 652 534 784 × 2 = 0 + 0.019 520 981 741 305 069 568;
  • 21) 0.019 520 981 741 305 069 568 × 2 = 0 + 0.039 041 963 482 610 139 136;
  • 22) 0.039 041 963 482 610 139 136 × 2 = 0 + 0.078 083 926 965 220 278 272;
  • 23) 0.078 083 926 965 220 278 272 × 2 = 0 + 0.156 167 853 930 440 556 544;
  • 24) 0.156 167 853 930 440 556 544 × 2 = 0 + 0.312 335 707 860 881 113 088;
  • 25) 0.312 335 707 860 881 113 088 × 2 = 0 + 0.624 671 415 721 762 226 176;
  • 26) 0.624 671 415 721 762 226 176 × 2 = 1 + 0.249 342 831 443 524 452 352;
  • 27) 0.249 342 831 443 524 452 352 × 2 = 0 + 0.498 685 662 887 048 904 704;
  • 28) 0.498 685 662 887 048 904 704 × 2 = 0 + 0.997 371 325 774 097 809 408;
  • 29) 0.997 371 325 774 097 809 408 × 2 = 1 + 0.994 742 651 548 195 618 816;
  • 30) 0.994 742 651 548 195 618 816 × 2 = 1 + 0.989 485 303 096 391 237 632;
  • 31) 0.989 485 303 096 391 237 632 × 2 = 1 + 0.978 970 606 192 782 475 264;
  • 32) 0.978 970 606 192 782 475 264 × 2 = 1 + 0.957 941 212 385 564 950 528;
  • 33) 0.957 941 212 385 564 950 528 × 2 = 1 + 0.915 882 424 771 129 901 056;
  • 34) 0.915 882 424 771 129 901 056 × 2 = 1 + 0.831 764 849 542 259 802 112;
  • 35) 0.831 764 849 542 259 802 112 × 2 = 1 + 0.663 529 699 084 519 604 224;
  • 36) 0.663 529 699 084 519 604 224 × 2 = 1 + 0.327 059 398 169 039 208 448;
  • 37) 0.327 059 398 169 039 208 448 × 2 = 0 + 0.654 118 796 338 078 416 896;
  • 38) 0.654 118 796 338 078 416 896 × 2 = 1 + 0.308 237 592 676 156 833 792;
  • 39) 0.308 237 592 676 156 833 792 × 2 = 0 + 0.616 475 185 352 313 667 584;
  • 40) 0.616 475 185 352 313 667 584 × 2 = 1 + 0.232 950 370 704 627 335 168;
  • 41) 0.232 950 370 704 627 335 168 × 2 = 0 + 0.465 900 741 409 254 670 336;
  • 42) 0.465 900 741 409 254 670 336 × 2 = 0 + 0.931 801 482 818 509 340 672;
  • 43) 0.931 801 482 818 509 340 672 × 2 = 1 + 0.863 602 965 637 018 681 344;
  • 44) 0.863 602 965 637 018 681 344 × 2 = 1 + 0.727 205 931 274 037 362 688;
  • 45) 0.727 205 931 274 037 362 688 × 2 = 1 + 0.454 411 862 548 074 725 376;
  • 46) 0.454 411 862 548 074 725 376 × 2 = 0 + 0.908 823 725 096 149 450 752;
  • 47) 0.908 823 725 096 149 450 752 × 2 = 1 + 0.817 647 450 192 298 901 504;
  • 48) 0.817 647 450 192 298 901 504 × 2 = 1 + 0.635 294 900 384 597 803 008;
  • 49) 0.635 294 900 384 597 803 008 × 2 = 1 + 0.270 589 800 769 195 606 016;
  • 50) 0.270 589 800 769 195 606 016 × 2 = 0 + 0.541 179 601 538 391 212 032;
  • 51) 0.541 179 601 538 391 212 032 × 2 = 1 + 0.082 359 203 076 782 424 064;
  • 52) 0.082 359 203 076 782 424 064 × 2 = 0 + 0.164 718 406 153 564 848 128;
  • 53) 0.164 718 406 153 564 848 128 × 2 = 0 + 0.329 436 812 307 129 696 256;
  • 54) 0.329 436 812 307 129 696 256 × 2 = 0 + 0.658 873 624 614 259 392 512;
  • 55) 0.658 873 624 614 259 392 512 × 2 = 1 + 0.317 747 249 228 518 785 024;
  • 56) 0.317 747 249 228 518 785 024 × 2 = 0 + 0.635 494 498 457 037 570 048;
  • 57) 0.635 494 498 457 037 570 048 × 2 = 1 + 0.270 988 996 914 075 140 096;
  • 58) 0.270 988 996 914 075 140 096 × 2 = 0 + 0.541 977 993 828 150 280 192;
  • 59) 0.541 977 993 828 150 280 192 × 2 = 1 + 0.083 955 987 656 300 560 384;
  • 60) 0.083 955 987 656 300 560 384 × 2 = 0 + 0.167 911 975 312 601 120 768;
  • 61) 0.167 911 975 312 601 120 768 × 2 = 0 + 0.335 823 950 625 202 241 536;
  • 62) 0.335 823 950 625 202 241 536 × 2 = 0 + 0.671 647 901 250 404 483 072;
  • 63) 0.671 647 901 250 404 483 072 × 2 = 1 + 0.343 295 802 500 808 966 144;
  • 64) 0.343 295 802 500 808 966 144 × 2 = 0 + 0.686 591 605 001 617 932 288;
  • 65) 0.686 591 605 001 617 932 288 × 2 = 1 + 0.373 183 210 003 235 864 576;
  • 66) 0.373 183 210 003 235 864 576 × 2 = 0 + 0.746 366 420 006 471 729 152;
  • 67) 0.746 366 420 006 471 729 152 × 2 = 1 + 0.492 732 840 012 943 458 304;
  • 68) 0.492 732 840 012 943 458 304 × 2 = 0 + 0.985 465 680 025 886 916 608;
  • 69) 0.985 465 680 025 886 916 608 × 2 = 1 + 0.970 931 360 051 773 833 216;
  • 70) 0.970 931 360 051 773 833 216 × 2 = 1 + 0.941 862 720 103 547 666 432;
  • 71) 0.941 862 720 103 547 666 432 × 2 = 1 + 0.883 725 440 207 095 332 864;
  • 72) 0.883 725 440 207 095 332 864 × 2 = 1 + 0.767 450 880 414 190 665 728;
  • 73) 0.767 450 880 414 190 665 728 × 2 = 1 + 0.534 901 760 828 381 331 456;
  • 74) 0.534 901 760 828 381 331 456 × 2 = 1 + 0.069 803 521 656 762 662 912;
  • 75) 0.069 803 521 656 762 662 912 × 2 = 0 + 0.139 607 043 313 525 325 824;
  • 76) 0.139 607 043 313 525 325 824 × 2 = 0 + 0.279 214 086 627 050 651 648;
  • 77) 0.279 214 086 627 050 651 648 × 2 = 0 + 0.558 428 173 254 101 303 296;
  • 78) 0.558 428 173 254 101 303 296 × 2 = 1 + 0.116 856 346 508 202 606 592;

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:


0.000 000 018 616 658 917 718(10) =

0.0000 0000 0000 0000 0000 0000 0100 1111 1111 0101 0011 1011 1010 0010 1010 0010 1010 1111 1100 01(2)

6. Positive number before normalization:

0.000 000 018 616 658 917 718(10) =

0.0000 0000 0000 0000 0000 0000 0100 1111 1111 0101 0011 1011 1010 0010 1010 0010 1010 1111 1100 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 018 616 658 917 718(10) =

0.0000 0000 0000 0000 0000 0000 0100 1111 1111 0101 0011 1011 1010 0010 1010 0010 1010 1111 1100 01(2) =

0.0000 0000 0000 0000 0000 0000 0100 1111 1111 0101 0011 1011 1010 0010 1010 0010 1010 1111 1100 01(2) × 20 =

1.0011 1111 1101 0100 1110 1110 1000 1010 1000 1010 1011 1111 0001(2) × 2-26


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): -26

Mantissa (not normalized):
1.0011 1111 1101 0100 1110 1110 1000 1010 1000 1010 1011 1111 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 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) =

997(10) =

011 1110 0101(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 1111 1101 0100 1110 1110 1000 1010 1000 1010 1011 1111 0001 =

0011 1111 1101 0100 1110 1110 1000 1010 1000 1010 1011 1111 0001


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 0101

Mantissa (52 bits) =
0011 1111 1101 0100 1110 1110 1000 1010 1000 1010 1011 1111 0001


Decimal number -0.000 000 018 616 658 917 718 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0101 - 0011 1111 1101 0100 1110 1110 1000 1010 1000 1010 1011 1111 0001

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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