Decimal to 64 Bit IEEE 754 Binary: Convert Number 1 111 100 110 111 100 001 100 111 099 894 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 1 111 100 110 111 100 001 100 111 099 894(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. 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;
  • 1 111 100 110 111 100 001 100 111 099 894 ÷ 2 = 555 550 055 055 550 000 550 055 549 947 + 0;
  • 555 550 055 055 550 000 550 055 549 947 ÷ 2 = 277 775 027 527 775 000 275 027 774 973 + 1;
  • 277 775 027 527 775 000 275 027 774 973 ÷ 2 = 138 887 513 763 887 500 137 513 887 486 + 1;
  • 138 887 513 763 887 500 137 513 887 486 ÷ 2 = 69 443 756 881 943 750 068 756 943 743 + 0;
  • 69 443 756 881 943 750 068 756 943 743 ÷ 2 = 34 721 878 440 971 875 034 378 471 871 + 1;
  • 34 721 878 440 971 875 034 378 471 871 ÷ 2 = 17 360 939 220 485 937 517 189 235 935 + 1;
  • 17 360 939 220 485 937 517 189 235 935 ÷ 2 = 8 680 469 610 242 968 758 594 617 967 + 1;
  • 8 680 469 610 242 968 758 594 617 967 ÷ 2 = 4 340 234 805 121 484 379 297 308 983 + 1;
  • 4 340 234 805 121 484 379 297 308 983 ÷ 2 = 2 170 117 402 560 742 189 648 654 491 + 1;
  • 2 170 117 402 560 742 189 648 654 491 ÷ 2 = 1 085 058 701 280 371 094 824 327 245 + 1;
  • 1 085 058 701 280 371 094 824 327 245 ÷ 2 = 542 529 350 640 185 547 412 163 622 + 1;
  • 542 529 350 640 185 547 412 163 622 ÷ 2 = 271 264 675 320 092 773 706 081 811 + 0;
  • 271 264 675 320 092 773 706 081 811 ÷ 2 = 135 632 337 660 046 386 853 040 905 + 1;
  • 135 632 337 660 046 386 853 040 905 ÷ 2 = 67 816 168 830 023 193 426 520 452 + 1;
  • 67 816 168 830 023 193 426 520 452 ÷ 2 = 33 908 084 415 011 596 713 260 226 + 0;
  • 33 908 084 415 011 596 713 260 226 ÷ 2 = 16 954 042 207 505 798 356 630 113 + 0;
  • 16 954 042 207 505 798 356 630 113 ÷ 2 = 8 477 021 103 752 899 178 315 056 + 1;
  • 8 477 021 103 752 899 178 315 056 ÷ 2 = 4 238 510 551 876 449 589 157 528 + 0;
  • 4 238 510 551 876 449 589 157 528 ÷ 2 = 2 119 255 275 938 224 794 578 764 + 0;
  • 2 119 255 275 938 224 794 578 764 ÷ 2 = 1 059 627 637 969 112 397 289 382 + 0;
  • 1 059 627 637 969 112 397 289 382 ÷ 2 = 529 813 818 984 556 198 644 691 + 0;
  • 529 813 818 984 556 198 644 691 ÷ 2 = 264 906 909 492 278 099 322 345 + 1;
  • 264 906 909 492 278 099 322 345 ÷ 2 = 132 453 454 746 139 049 661 172 + 1;
  • 132 453 454 746 139 049 661 172 ÷ 2 = 66 226 727 373 069 524 830 586 + 0;
  • 66 226 727 373 069 524 830 586 ÷ 2 = 33 113 363 686 534 762 415 293 + 0;
  • 33 113 363 686 534 762 415 293 ÷ 2 = 16 556 681 843 267 381 207 646 + 1;
  • 16 556 681 843 267 381 207 646 ÷ 2 = 8 278 340 921 633 690 603 823 + 0;
  • 8 278 340 921 633 690 603 823 ÷ 2 = 4 139 170 460 816 845 301 911 + 1;
  • 4 139 170 460 816 845 301 911 ÷ 2 = 2 069 585 230 408 422 650 955 + 1;
  • 2 069 585 230 408 422 650 955 ÷ 2 = 1 034 792 615 204 211 325 477 + 1;
  • 1 034 792 615 204 211 325 477 ÷ 2 = 517 396 307 602 105 662 738 + 1;
  • 517 396 307 602 105 662 738 ÷ 2 = 258 698 153 801 052 831 369 + 0;
  • 258 698 153 801 052 831 369 ÷ 2 = 129 349 076 900 526 415 684 + 1;
  • 129 349 076 900 526 415 684 ÷ 2 = 64 674 538 450 263 207 842 + 0;
  • 64 674 538 450 263 207 842 ÷ 2 = 32 337 269 225 131 603 921 + 0;
  • 32 337 269 225 131 603 921 ÷ 2 = 16 168 634 612 565 801 960 + 1;
  • 16 168 634 612 565 801 960 ÷ 2 = 8 084 317 306 282 900 980 + 0;
  • 8 084 317 306 282 900 980 ÷ 2 = 4 042 158 653 141 450 490 + 0;
  • 4 042 158 653 141 450 490 ÷ 2 = 2 021 079 326 570 725 245 + 0;
  • 2 021 079 326 570 725 245 ÷ 2 = 1 010 539 663 285 362 622 + 1;
  • 1 010 539 663 285 362 622 ÷ 2 = 505 269 831 642 681 311 + 0;
  • 505 269 831 642 681 311 ÷ 2 = 252 634 915 821 340 655 + 1;
  • 252 634 915 821 340 655 ÷ 2 = 126 317 457 910 670 327 + 1;
  • 126 317 457 910 670 327 ÷ 2 = 63 158 728 955 335 163 + 1;
  • 63 158 728 955 335 163 ÷ 2 = 31 579 364 477 667 581 + 1;
  • 31 579 364 477 667 581 ÷ 2 = 15 789 682 238 833 790 + 1;
  • 15 789 682 238 833 790 ÷ 2 = 7 894 841 119 416 895 + 0;
  • 7 894 841 119 416 895 ÷ 2 = 3 947 420 559 708 447 + 1;
  • 3 947 420 559 708 447 ÷ 2 = 1 973 710 279 854 223 + 1;
  • 1 973 710 279 854 223 ÷ 2 = 986 855 139 927 111 + 1;
  • 986 855 139 927 111 ÷ 2 = 493 427 569 963 555 + 1;
  • 493 427 569 963 555 ÷ 2 = 246 713 784 981 777 + 1;
  • 246 713 784 981 777 ÷ 2 = 123 356 892 490 888 + 1;
  • 123 356 892 490 888 ÷ 2 = 61 678 446 245 444 + 0;
  • 61 678 446 245 444 ÷ 2 = 30 839 223 122 722 + 0;
  • 30 839 223 122 722 ÷ 2 = 15 419 611 561 361 + 0;
  • 15 419 611 561 361 ÷ 2 = 7 709 805 780 680 + 1;
  • 7 709 805 780 680 ÷ 2 = 3 854 902 890 340 + 0;
  • 3 854 902 890 340 ÷ 2 = 1 927 451 445 170 + 0;
  • 1 927 451 445 170 ÷ 2 = 963 725 722 585 + 0;
  • 963 725 722 585 ÷ 2 = 481 862 861 292 + 1;
  • 481 862 861 292 ÷ 2 = 240 931 430 646 + 0;
  • 240 931 430 646 ÷ 2 = 120 465 715 323 + 0;
  • 120 465 715 323 ÷ 2 = 60 232 857 661 + 1;
  • 60 232 857 661 ÷ 2 = 30 116 428 830 + 1;
  • 30 116 428 830 ÷ 2 = 15 058 214 415 + 0;
  • 15 058 214 415 ÷ 2 = 7 529 107 207 + 1;
  • 7 529 107 207 ÷ 2 = 3 764 553 603 + 1;
  • 3 764 553 603 ÷ 2 = 1 882 276 801 + 1;
  • 1 882 276 801 ÷ 2 = 941 138 400 + 1;
  • 941 138 400 ÷ 2 = 470 569 200 + 0;
  • 470 569 200 ÷ 2 = 235 284 600 + 0;
  • 235 284 600 ÷ 2 = 117 642 300 + 0;
  • 117 642 300 ÷ 2 = 58 821 150 + 0;
  • 58 821 150 ÷ 2 = 29 410 575 + 0;
  • 29 410 575 ÷ 2 = 14 705 287 + 1;
  • 14 705 287 ÷ 2 = 7 352 643 + 1;
  • 7 352 643 ÷ 2 = 3 676 321 + 1;
  • 3 676 321 ÷ 2 = 1 838 160 + 1;
  • 1 838 160 ÷ 2 = 919 080 + 0;
  • 919 080 ÷ 2 = 459 540 + 0;
  • 459 540 ÷ 2 = 229 770 + 0;
  • 229 770 ÷ 2 = 114 885 + 0;
  • 114 885 ÷ 2 = 57 442 + 1;
  • 57 442 ÷ 2 = 28 721 + 0;
  • 28 721 ÷ 2 = 14 360 + 1;
  • 14 360 ÷ 2 = 7 180 + 0;
  • 7 180 ÷ 2 = 3 590 + 0;
  • 3 590 ÷ 2 = 1 795 + 0;
  • 1 795 ÷ 2 = 897 + 1;
  • 897 ÷ 2 = 448 + 1;
  • 448 ÷ 2 = 224 + 0;
  • 224 ÷ 2 = 112 + 0;
  • 112 ÷ 2 = 56 + 0;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

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

1 111 100 110 111 100 001 100 111 099 894(10) =


1110 0000 0110 0010 1000 0111 1000 0011 1101 1001 0001 0001 1111 1011 1110 1000 1001 0111 1010 0110 0001 0011 0111 1111 0110(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 99 positions to the left, so that only one non zero digit remains to the left of it:


1 111 100 110 111 100 001 100 111 099 894(10) =


1110 0000 0110 0010 1000 0111 1000 0011 1101 1001 0001 0001 1111 1011 1110 1000 1001 0111 1010 0110 0001 0011 0111 1111 0110(2) =


1110 0000 0110 0010 1000 0111 1000 0011 1101 1001 0001 0001 1111 1011 1110 1000 1001 0111 1010 0110 0001 0011 0111 1111 0110(2) × 20 =


1.1100 0000 1100 0101 0000 1111 0000 0111 1011 0010 0010 0011 1111 0111 1101 0001 0010 1111 0100 1100 0010 0110 1111 1110 110(2) × 299


4. 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): 99


Mantissa (not normalized):
1.1100 0000 1100 0101 0000 1111 0000 0111 1011 0010 0010 0011 1111 0111 1101 0001 0010 1111 0100 1100 0010 0110 1111 1110 110


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


99 + 2(11-1) - 1 =


(99 + 1 023)(10) =


1 122(10)


6. 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;
  • 1 122 ÷ 2 = 561 + 0;
  • 561 ÷ 2 = 280 + 1;
  • 280 ÷ 2 = 140 + 0;
  • 140 ÷ 2 = 70 + 0;
  • 70 ÷ 2 = 35 + 0;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


1122(10) =


100 0110 0010(2)


8. 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. 1100 0000 1100 0101 0000 1111 0000 0111 1011 0010 0010 0011 1111 011 1110 1000 1001 0111 1010 0110 0001 0011 0111 1111 0110 =


1100 0000 1100 0101 0000 1111 0000 0111 1011 0010 0010 0011 1111


9. 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 0110 0010


Mantissa (52 bits) =
1100 0000 1100 0101 0000 1111 0000 0111 1011 0010 0010 0011 1111


The base ten decimal number 1 111 100 110 111 100 001 100 111 099 894 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0110 0010 - 1100 0000 1100 0101 0000 1111 0000 0111 1011 0010 0010 0011 1111

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