5.919 999 999 999 999 928 945 726 423 989 981 412 887 554 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5.919 999 999 999 999 928 945 726 423 989 981 412 887 554 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
5.919 999 999 999 999 928 945 726 423 989 981 412 887 554 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. First, convert to binary (in base 2) the integer part: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

5(10) =


101(2)


3. Convert to binary (base 2) the fractional part: 0.919 999 999 999 999 928 945 726 423 989 981 412 887 554 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) 0.919 999 999 999 999 928 945 726 423 989 981 412 887 554 5 × 2 = 1 + 0.839 999 999 999 999 857 891 452 847 979 962 825 775 109;
  • 2) 0.839 999 999 999 999 857 891 452 847 979 962 825 775 109 × 2 = 1 + 0.679 999 999 999 999 715 782 905 695 959 925 651 550 218;
  • 3) 0.679 999 999 999 999 715 782 905 695 959 925 651 550 218 × 2 = 1 + 0.359 999 999 999 999 431 565 811 391 919 851 303 100 436;
  • 4) 0.359 999 999 999 999 431 565 811 391 919 851 303 100 436 × 2 = 0 + 0.719 999 999 999 998 863 131 622 783 839 702 606 200 872;
  • 5) 0.719 999 999 999 998 863 131 622 783 839 702 606 200 872 × 2 = 1 + 0.439 999 999 999 997 726 263 245 567 679 405 212 401 744;
  • 6) 0.439 999 999 999 997 726 263 245 567 679 405 212 401 744 × 2 = 0 + 0.879 999 999 999 995 452 526 491 135 358 810 424 803 488;
  • 7) 0.879 999 999 999 995 452 526 491 135 358 810 424 803 488 × 2 = 1 + 0.759 999 999 999 990 905 052 982 270 717 620 849 606 976;
  • 8) 0.759 999 999 999 990 905 052 982 270 717 620 849 606 976 × 2 = 1 + 0.519 999 999 999 981 810 105 964 541 435 241 699 213 952;
  • 9) 0.519 999 999 999 981 810 105 964 541 435 241 699 213 952 × 2 = 1 + 0.039 999 999 999 963 620 211 929 082 870 483 398 427 904;
  • 10) 0.039 999 999 999 963 620 211 929 082 870 483 398 427 904 × 2 = 0 + 0.079 999 999 999 927 240 423 858 165 740 966 796 855 808;
  • 11) 0.079 999 999 999 927 240 423 858 165 740 966 796 855 808 × 2 = 0 + 0.159 999 999 999 854 480 847 716 331 481 933 593 711 616;
  • 12) 0.159 999 999 999 854 480 847 716 331 481 933 593 711 616 × 2 = 0 + 0.319 999 999 999 708 961 695 432 662 963 867 187 423 232;
  • 13) 0.319 999 999 999 708 961 695 432 662 963 867 187 423 232 × 2 = 0 + 0.639 999 999 999 417 923 390 865 325 927 734 374 846 464;
  • 14) 0.639 999 999 999 417 923 390 865 325 927 734 374 846 464 × 2 = 1 + 0.279 999 999 998 835 846 781 730 651 855 468 749 692 928;
  • 15) 0.279 999 999 998 835 846 781 730 651 855 468 749 692 928 × 2 = 0 + 0.559 999 999 997 671 693 563 461 303 710 937 499 385 856;
  • 16) 0.559 999 999 997 671 693 563 461 303 710 937 499 385 856 × 2 = 1 + 0.119 999 999 995 343 387 126 922 607 421 874 998 771 712;
  • 17) 0.119 999 999 995 343 387 126 922 607 421 874 998 771 712 × 2 = 0 + 0.239 999 999 990 686 774 253 845 214 843 749 997 543 424;
  • 18) 0.239 999 999 990 686 774 253 845 214 843 749 997 543 424 × 2 = 0 + 0.479 999 999 981 373 548 507 690 429 687 499 995 086 848;
  • 19) 0.479 999 999 981 373 548 507 690 429 687 499 995 086 848 × 2 = 0 + 0.959 999 999 962 747 097 015 380 859 374 999 990 173 696;
  • 20) 0.959 999 999 962 747 097 015 380 859 374 999 990 173 696 × 2 = 1 + 0.919 999 999 925 494 194 030 761 718 749 999 980 347 392;
  • 21) 0.919 999 999 925 494 194 030 761 718 749 999 980 347 392 × 2 = 1 + 0.839 999 999 850 988 388 061 523 437 499 999 960 694 784;
  • 22) 0.839 999 999 850 988 388 061 523 437 499 999 960 694 784 × 2 = 1 + 0.679 999 999 701 976 776 123 046 874 999 999 921 389 568;
  • 23) 0.679 999 999 701 976 776 123 046 874 999 999 921 389 568 × 2 = 1 + 0.359 999 999 403 953 552 246 093 749 999 999 842 779 136;
  • 24) 0.359 999 999 403 953 552 246 093 749 999 999 842 779 136 × 2 = 0 + 0.719 999 998 807 907 104 492 187 499 999 999 685 558 272;
  • 25) 0.719 999 998 807 907 104 492 187 499 999 999 685 558 272 × 2 = 1 + 0.439 999 997 615 814 208 984 374 999 999 999 371 116 544;
  • 26) 0.439 999 997 615 814 208 984 374 999 999 999 371 116 544 × 2 = 0 + 0.879 999 995 231 628 417 968 749 999 999 998 742 233 088;
  • 27) 0.879 999 995 231 628 417 968 749 999 999 998 742 233 088 × 2 = 1 + 0.759 999 990 463 256 835 937 499 999 999 997 484 466 176;
  • 28) 0.759 999 990 463 256 835 937 499 999 999 997 484 466 176 × 2 = 1 + 0.519 999 980 926 513 671 874 999 999 999 994 968 932 352;
  • 29) 0.519 999 980 926 513 671 874 999 999 999 994 968 932 352 × 2 = 1 + 0.039 999 961 853 027 343 749 999 999 999 989 937 864 704;
  • 30) 0.039 999 961 853 027 343 749 999 999 999 989 937 864 704 × 2 = 0 + 0.079 999 923 706 054 687 499 999 999 999 979 875 729 408;
  • 31) 0.079 999 923 706 054 687 499 999 999 999 979 875 729 408 × 2 = 0 + 0.159 999 847 412 109 374 999 999 999 999 959 751 458 816;
  • 32) 0.159 999 847 412 109 374 999 999 999 999 959 751 458 816 × 2 = 0 + 0.319 999 694 824 218 749 999 999 999 999 919 502 917 632;
  • 33) 0.319 999 694 824 218 749 999 999 999 999 919 502 917 632 × 2 = 0 + 0.639 999 389 648 437 499 999 999 999 999 839 005 835 264;
  • 34) 0.639 999 389 648 437 499 999 999 999 999 839 005 835 264 × 2 = 1 + 0.279 998 779 296 874 999 999 999 999 999 678 011 670 528;
  • 35) 0.279 998 779 296 874 999 999 999 999 999 678 011 670 528 × 2 = 0 + 0.559 997 558 593 749 999 999 999 999 999 356 023 341 056;
  • 36) 0.559 997 558 593 749 999 999 999 999 999 356 023 341 056 × 2 = 1 + 0.119 995 117 187 499 999 999 999 999 998 712 046 682 112;
  • 37) 0.119 995 117 187 499 999 999 999 999 998 712 046 682 112 × 2 = 0 + 0.239 990 234 374 999 999 999 999 999 997 424 093 364 224;
  • 38) 0.239 990 234 374 999 999 999 999 999 997 424 093 364 224 × 2 = 0 + 0.479 980 468 749 999 999 999 999 999 994 848 186 728 448;
  • 39) 0.479 980 468 749 999 999 999 999 999 994 848 186 728 448 × 2 = 0 + 0.959 960 937 499 999 999 999 999 999 989 696 373 456 896;
  • 40) 0.959 960 937 499 999 999 999 999 999 989 696 373 456 896 × 2 = 1 + 0.919 921 874 999 999 999 999 999 999 979 392 746 913 792;
  • 41) 0.919 921 874 999 999 999 999 999 999 979 392 746 913 792 × 2 = 1 + 0.839 843 749 999 999 999 999 999 999 958 785 493 827 584;
  • 42) 0.839 843 749 999 999 999 999 999 999 958 785 493 827 584 × 2 = 1 + 0.679 687 499 999 999 999 999 999 999 917 570 987 655 168;
  • 43) 0.679 687 499 999 999 999 999 999 999 917 570 987 655 168 × 2 = 1 + 0.359 374 999 999 999 999 999 999 999 835 141 975 310 336;
  • 44) 0.359 374 999 999 999 999 999 999 999 835 141 975 310 336 × 2 = 0 + 0.718 749 999 999 999 999 999 999 999 670 283 950 620 672;
  • 45) 0.718 749 999 999 999 999 999 999 999 670 283 950 620 672 × 2 = 1 + 0.437 499 999 999 999 999 999 999 999 340 567 901 241 344;
  • 46) 0.437 499 999 999 999 999 999 999 999 340 567 901 241 344 × 2 = 0 + 0.874 999 999 999 999 999 999 999 998 681 135 802 482 688;
  • 47) 0.874 999 999 999 999 999 999 999 998 681 135 802 482 688 × 2 = 1 + 0.749 999 999 999 999 999 999 999 997 362 271 604 965 376;
  • 48) 0.749 999 999 999 999 999 999 999 997 362 271 604 965 376 × 2 = 1 + 0.499 999 999 999 999 999 999 999 994 724 543 209 930 752;
  • 49) 0.499 999 999 999 999 999 999 999 994 724 543 209 930 752 × 2 = 0 + 0.999 999 999 999 999 999 999 999 989 449 086 419 861 504;
  • 50) 0.999 999 999 999 999 999 999 999 989 449 086 419 861 504 × 2 = 1 + 0.999 999 999 999 999 999 999 999 978 898 172 839 723 008;
  • 51) 0.999 999 999 999 999 999 999 999 978 898 172 839 723 008 × 2 = 1 + 0.999 999 999 999 999 999 999 999 957 796 345 679 446 016;
  • 52) 0.999 999 999 999 999 999 999 999 957 796 345 679 446 016 × 2 = 1 + 0.999 999 999 999 999 999 999 999 915 592 691 358 892 032;
  • 53) 0.999 999 999 999 999 999 999 999 915 592 691 358 892 032 × 2 = 1 + 0.999 999 999 999 999 999 999 999 831 185 382 717 784 064;

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).


4. 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:


0.919 999 999 999 999 928 945 726 423 989 981 412 887 554 5(10) =


0.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2)

5. Positive number before normalization:

5.919 999 999 999 999 928 945 726 423 989 981 412 887 554 5(10) =


101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.919 999 999 999 999 928 945 726 423 989 981 412 887 554 5(10) =


101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2) =


101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2) × 20 =


1.0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 111(2) × 22


7. 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 0 (a positive number)


Exponent (unadjusted): 2


Mantissa (not normalized):
1.0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


2 + 2(11-1) - 1 =


(2 + 1 023)(10) =


1 025(10)


9. 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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1025(10) =


100 0000 0001(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 111 =


0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (11 bits) =
100 0000 0001


Mantissa (52 bits) =
0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101


Decimal number 5.919 999 999 999 999 928 945 726 423 989 981 412 887 554 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101


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