-0.000 806 264 623 585 362 514 063 654 156 850 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 806 264 623 585 362 514 063 654 156 850 5(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
-0.000 806 264 623 585 362 514 063 654 156 850 5(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:

|-0.000 806 264 623 585 362 514 063 654 156 850 5| = 0.000 806 264 623 585 362 514 063 654 156 850 5


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: 0.000 806 264 623 585 362 514 063 654 156 850 5.

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) 0.000 806 264 623 585 362 514 063 654 156 850 5 × 2 = 0 + 0.001 612 529 247 170 725 028 127 308 313 701;
  • 2) 0.001 612 529 247 170 725 028 127 308 313 701 × 2 = 0 + 0.003 225 058 494 341 450 056 254 616 627 402;
  • 3) 0.003 225 058 494 341 450 056 254 616 627 402 × 2 = 0 + 0.006 450 116 988 682 900 112 509 233 254 804;
  • 4) 0.006 450 116 988 682 900 112 509 233 254 804 × 2 = 0 + 0.012 900 233 977 365 800 225 018 466 509 608;
  • 5) 0.012 900 233 977 365 800 225 018 466 509 608 × 2 = 0 + 0.025 800 467 954 731 600 450 036 933 019 216;
  • 6) 0.025 800 467 954 731 600 450 036 933 019 216 × 2 = 0 + 0.051 600 935 909 463 200 900 073 866 038 432;
  • 7) 0.051 600 935 909 463 200 900 073 866 038 432 × 2 = 0 + 0.103 201 871 818 926 401 800 147 732 076 864;
  • 8) 0.103 201 871 818 926 401 800 147 732 076 864 × 2 = 0 + 0.206 403 743 637 852 803 600 295 464 153 728;
  • 9) 0.206 403 743 637 852 803 600 295 464 153 728 × 2 = 0 + 0.412 807 487 275 705 607 200 590 928 307 456;
  • 10) 0.412 807 487 275 705 607 200 590 928 307 456 × 2 = 0 + 0.825 614 974 551 411 214 401 181 856 614 912;
  • 11) 0.825 614 974 551 411 214 401 181 856 614 912 × 2 = 1 + 0.651 229 949 102 822 428 802 363 713 229 824;
  • 12) 0.651 229 949 102 822 428 802 363 713 229 824 × 2 = 1 + 0.302 459 898 205 644 857 604 727 426 459 648;
  • 13) 0.302 459 898 205 644 857 604 727 426 459 648 × 2 = 0 + 0.604 919 796 411 289 715 209 454 852 919 296;
  • 14) 0.604 919 796 411 289 715 209 454 852 919 296 × 2 = 1 + 0.209 839 592 822 579 430 418 909 705 838 592;
  • 15) 0.209 839 592 822 579 430 418 909 705 838 592 × 2 = 0 + 0.419 679 185 645 158 860 837 819 411 677 184;
  • 16) 0.419 679 185 645 158 860 837 819 411 677 184 × 2 = 0 + 0.839 358 371 290 317 721 675 638 823 354 368;
  • 17) 0.839 358 371 290 317 721 675 638 823 354 368 × 2 = 1 + 0.678 716 742 580 635 443 351 277 646 708 736;
  • 18) 0.678 716 742 580 635 443 351 277 646 708 736 × 2 = 1 + 0.357 433 485 161 270 886 702 555 293 417 472;
  • 19) 0.357 433 485 161 270 886 702 555 293 417 472 × 2 = 0 + 0.714 866 970 322 541 773 405 110 586 834 944;
  • 20) 0.714 866 970 322 541 773 405 110 586 834 944 × 2 = 1 + 0.429 733 940 645 083 546 810 221 173 669 888;
  • 21) 0.429 733 940 645 083 546 810 221 173 669 888 × 2 = 0 + 0.859 467 881 290 167 093 620 442 347 339 776;
  • 22) 0.859 467 881 290 167 093 620 442 347 339 776 × 2 = 1 + 0.718 935 762 580 334 187 240 884 694 679 552;
  • 23) 0.718 935 762 580 334 187 240 884 694 679 552 × 2 = 1 + 0.437 871 525 160 668 374 481 769 389 359 104;
  • 24) 0.437 871 525 160 668 374 481 769 389 359 104 × 2 = 0 + 0.875 743 050 321 336 748 963 538 778 718 208;
  • 25) 0.875 743 050 321 336 748 963 538 778 718 208 × 2 = 1 + 0.751 486 100 642 673 497 927 077 557 436 416;
  • 26) 0.751 486 100 642 673 497 927 077 557 436 416 × 2 = 1 + 0.502 972 201 285 346 995 854 155 114 872 832;
  • 27) 0.502 972 201 285 346 995 854 155 114 872 832 × 2 = 1 + 0.005 944 402 570 693 991 708 310 229 745 664;
  • 28) 0.005 944 402 570 693 991 708 310 229 745 664 × 2 = 0 + 0.011 888 805 141 387 983 416 620 459 491 328;
  • 29) 0.011 888 805 141 387 983 416 620 459 491 328 × 2 = 0 + 0.023 777 610 282 775 966 833 240 918 982 656;
  • 30) 0.023 777 610 282 775 966 833 240 918 982 656 × 2 = 0 + 0.047 555 220 565 551 933 666 481 837 965 312;
  • 31) 0.047 555 220 565 551 933 666 481 837 965 312 × 2 = 0 + 0.095 110 441 131 103 867 332 963 675 930 624;
  • 32) 0.095 110 441 131 103 867 332 963 675 930 624 × 2 = 0 + 0.190 220 882 262 207 734 665 927 351 861 248;
  • 33) 0.190 220 882 262 207 734 665 927 351 861 248 × 2 = 0 + 0.380 441 764 524 415 469 331 854 703 722 496;
  • 34) 0.380 441 764 524 415 469 331 854 703 722 496 × 2 = 0 + 0.760 883 529 048 830 938 663 709 407 444 992;
  • 35) 0.760 883 529 048 830 938 663 709 407 444 992 × 2 = 1 + 0.521 767 058 097 661 877 327 418 814 889 984;
  • 36) 0.521 767 058 097 661 877 327 418 814 889 984 × 2 = 1 + 0.043 534 116 195 323 754 654 837 629 779 968;
  • 37) 0.043 534 116 195 323 754 654 837 629 779 968 × 2 = 0 + 0.087 068 232 390 647 509 309 675 259 559 936;
  • 38) 0.087 068 232 390 647 509 309 675 259 559 936 × 2 = 0 + 0.174 136 464 781 295 018 619 350 519 119 872;
  • 39) 0.174 136 464 781 295 018 619 350 519 119 872 × 2 = 0 + 0.348 272 929 562 590 037 238 701 038 239 744;
  • 40) 0.348 272 929 562 590 037 238 701 038 239 744 × 2 = 0 + 0.696 545 859 125 180 074 477 402 076 479 488;
  • 41) 0.696 545 859 125 180 074 477 402 076 479 488 × 2 = 1 + 0.393 091 718 250 360 148 954 804 152 958 976;
  • 42) 0.393 091 718 250 360 148 954 804 152 958 976 × 2 = 0 + 0.786 183 436 500 720 297 909 608 305 917 952;
  • 43) 0.786 183 436 500 720 297 909 608 305 917 952 × 2 = 1 + 0.572 366 873 001 440 595 819 216 611 835 904;
  • 44) 0.572 366 873 001 440 595 819 216 611 835 904 × 2 = 1 + 0.144 733 746 002 881 191 638 433 223 671 808;
  • 45) 0.144 733 746 002 881 191 638 433 223 671 808 × 2 = 0 + 0.289 467 492 005 762 383 276 866 447 343 616;
  • 46) 0.289 467 492 005 762 383 276 866 447 343 616 × 2 = 0 + 0.578 934 984 011 524 766 553 732 894 687 232;
  • 47) 0.578 934 984 011 524 766 553 732 894 687 232 × 2 = 1 + 0.157 869 968 023 049 533 107 465 789 374 464;
  • 48) 0.157 869 968 023 049 533 107 465 789 374 464 × 2 = 0 + 0.315 739 936 046 099 066 214 931 578 748 928;
  • 49) 0.315 739 936 046 099 066 214 931 578 748 928 × 2 = 0 + 0.631 479 872 092 198 132 429 863 157 497 856;
  • 50) 0.631 479 872 092 198 132 429 863 157 497 856 × 2 = 1 + 0.262 959 744 184 396 264 859 726 314 995 712;
  • 51) 0.262 959 744 184 396 264 859 726 314 995 712 × 2 = 0 + 0.525 919 488 368 792 529 719 452 629 991 424;
  • 52) 0.525 919 488 368 792 529 719 452 629 991 424 × 2 = 1 + 0.051 838 976 737 585 059 438 905 259 982 848;
  • 53) 0.051 838 976 737 585 059 438 905 259 982 848 × 2 = 0 + 0.103 677 953 475 170 118 877 810 519 965 696;
  • 54) 0.103 677 953 475 170 118 877 810 519 965 696 × 2 = 0 + 0.207 355 906 950 340 237 755 621 039 931 392;
  • 55) 0.207 355 906 950 340 237 755 621 039 931 392 × 2 = 0 + 0.414 711 813 900 680 475 511 242 079 862 784;
  • 56) 0.414 711 813 900 680 475 511 242 079 862 784 × 2 = 0 + 0.829 423 627 801 360 951 022 484 159 725 568;
  • 57) 0.829 423 627 801 360 951 022 484 159 725 568 × 2 = 1 + 0.658 847 255 602 721 902 044 968 319 451 136;
  • 58) 0.658 847 255 602 721 902 044 968 319 451 136 × 2 = 1 + 0.317 694 511 205 443 804 089 936 638 902 272;
  • 59) 0.317 694 511 205 443 804 089 936 638 902 272 × 2 = 0 + 0.635 389 022 410 887 608 179 873 277 804 544;
  • 60) 0.635 389 022 410 887 608 179 873 277 804 544 × 2 = 1 + 0.270 778 044 821 775 216 359 746 555 609 088;
  • 61) 0.270 778 044 821 775 216 359 746 555 609 088 × 2 = 0 + 0.541 556 089 643 550 432 719 493 111 218 176;
  • 62) 0.541 556 089 643 550 432 719 493 111 218 176 × 2 = 1 + 0.083 112 179 287 100 865 438 986 222 436 352;
  • 63) 0.083 112 179 287 100 865 438 986 222 436 352 × 2 = 0 + 0.166 224 358 574 201 730 877 972 444 872 704;

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:


0.000 806 264 623 585 362 514 063 654 156 850 5(10) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2)

6. Positive number before normalization:

0.000 806 264 623 585 362 514 063 654 156 850 5(10) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 806 264 623 585 362 514 063 654 156 850 5(10) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2) × 20 =

1.1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010(2) × 2-11


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): -11

Mantissa (not normalized):
1.1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-11 + 1 023)(10) =

1 012(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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1012(10) =

011 1111 0100(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. 1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010 =

1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010


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 0100

Mantissa (52 bits) =
1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010


Decimal number -0.000 806 264 623 585 362 514 063 654 156 850 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0100 - 1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010

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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