0.070 513 669 348 244 859 683 054 324 043 618 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.070 513 669 348 244 859 683 054 324 043 618(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
0.070 513 669 348 244 859 683 054 324 043 618(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: 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.070 513 669 348 244 859 683 054 324 043 618.

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) 0.070 513 669 348 244 859 683 054 324 043 618 × 2 = 0 + 0.141 027 338 696 489 719 366 108 648 087 236;
  • 2) 0.141 027 338 696 489 719 366 108 648 087 236 × 2 = 0 + 0.282 054 677 392 979 438 732 217 296 174 472;
  • 3) 0.282 054 677 392 979 438 732 217 296 174 472 × 2 = 0 + 0.564 109 354 785 958 877 464 434 592 348 944;
  • 4) 0.564 109 354 785 958 877 464 434 592 348 944 × 2 = 1 + 0.128 218 709 571 917 754 928 869 184 697 888;
  • 5) 0.128 218 709 571 917 754 928 869 184 697 888 × 2 = 0 + 0.256 437 419 143 835 509 857 738 369 395 776;
  • 6) 0.256 437 419 143 835 509 857 738 369 395 776 × 2 = 0 + 0.512 874 838 287 671 019 715 476 738 791 552;
  • 7) 0.512 874 838 287 671 019 715 476 738 791 552 × 2 = 1 + 0.025 749 676 575 342 039 430 953 477 583 104;
  • 8) 0.025 749 676 575 342 039 430 953 477 583 104 × 2 = 0 + 0.051 499 353 150 684 078 861 906 955 166 208;
  • 9) 0.051 499 353 150 684 078 861 906 955 166 208 × 2 = 0 + 0.102 998 706 301 368 157 723 813 910 332 416;
  • 10) 0.102 998 706 301 368 157 723 813 910 332 416 × 2 = 0 + 0.205 997 412 602 736 315 447 627 820 664 832;
  • 11) 0.205 997 412 602 736 315 447 627 820 664 832 × 2 = 0 + 0.411 994 825 205 472 630 895 255 641 329 664;
  • 12) 0.411 994 825 205 472 630 895 255 641 329 664 × 2 = 0 + 0.823 989 650 410 945 261 790 511 282 659 328;
  • 13) 0.823 989 650 410 945 261 790 511 282 659 328 × 2 = 1 + 0.647 979 300 821 890 523 581 022 565 318 656;
  • 14) 0.647 979 300 821 890 523 581 022 565 318 656 × 2 = 1 + 0.295 958 601 643 781 047 162 045 130 637 312;
  • 15) 0.295 958 601 643 781 047 162 045 130 637 312 × 2 = 0 + 0.591 917 203 287 562 094 324 090 261 274 624;
  • 16) 0.591 917 203 287 562 094 324 090 261 274 624 × 2 = 1 + 0.183 834 406 575 124 188 648 180 522 549 248;
  • 17) 0.183 834 406 575 124 188 648 180 522 549 248 × 2 = 0 + 0.367 668 813 150 248 377 296 361 045 098 496;
  • 18) 0.367 668 813 150 248 377 296 361 045 098 496 × 2 = 0 + 0.735 337 626 300 496 754 592 722 090 196 992;
  • 19) 0.735 337 626 300 496 754 592 722 090 196 992 × 2 = 1 + 0.470 675 252 600 993 509 185 444 180 393 984;
  • 20) 0.470 675 252 600 993 509 185 444 180 393 984 × 2 = 0 + 0.941 350 505 201 987 018 370 888 360 787 968;
  • 21) 0.941 350 505 201 987 018 370 888 360 787 968 × 2 = 1 + 0.882 701 010 403 974 036 741 776 721 575 936;
  • 22) 0.882 701 010 403 974 036 741 776 721 575 936 × 2 = 1 + 0.765 402 020 807 948 073 483 553 443 151 872;
  • 23) 0.765 402 020 807 948 073 483 553 443 151 872 × 2 = 1 + 0.530 804 041 615 896 146 967 106 886 303 744;
  • 24) 0.530 804 041 615 896 146 967 106 886 303 744 × 2 = 1 + 0.061 608 083 231 792 293 934 213 772 607 488;
  • 25) 0.061 608 083 231 792 293 934 213 772 607 488 × 2 = 0 + 0.123 216 166 463 584 587 868 427 545 214 976;
  • 26) 0.123 216 166 463 584 587 868 427 545 214 976 × 2 = 0 + 0.246 432 332 927 169 175 736 855 090 429 952;
  • 27) 0.246 432 332 927 169 175 736 855 090 429 952 × 2 = 0 + 0.492 864 665 854 338 351 473 710 180 859 904;
  • 28) 0.492 864 665 854 338 351 473 710 180 859 904 × 2 = 0 + 0.985 729 331 708 676 702 947 420 361 719 808;
  • 29) 0.985 729 331 708 676 702 947 420 361 719 808 × 2 = 1 + 0.971 458 663 417 353 405 894 840 723 439 616;
  • 30) 0.971 458 663 417 353 405 894 840 723 439 616 × 2 = 1 + 0.942 917 326 834 706 811 789 681 446 879 232;
  • 31) 0.942 917 326 834 706 811 789 681 446 879 232 × 2 = 1 + 0.885 834 653 669 413 623 579 362 893 758 464;
  • 32) 0.885 834 653 669 413 623 579 362 893 758 464 × 2 = 1 + 0.771 669 307 338 827 247 158 725 787 516 928;
  • 33) 0.771 669 307 338 827 247 158 725 787 516 928 × 2 = 1 + 0.543 338 614 677 654 494 317 451 575 033 856;
  • 34) 0.543 338 614 677 654 494 317 451 575 033 856 × 2 = 1 + 0.086 677 229 355 308 988 634 903 150 067 712;
  • 35) 0.086 677 229 355 308 988 634 903 150 067 712 × 2 = 0 + 0.173 354 458 710 617 977 269 806 300 135 424;
  • 36) 0.173 354 458 710 617 977 269 806 300 135 424 × 2 = 0 + 0.346 708 917 421 235 954 539 612 600 270 848;
  • 37) 0.346 708 917 421 235 954 539 612 600 270 848 × 2 = 0 + 0.693 417 834 842 471 909 079 225 200 541 696;
  • 38) 0.693 417 834 842 471 909 079 225 200 541 696 × 2 = 1 + 0.386 835 669 684 943 818 158 450 401 083 392;
  • 39) 0.386 835 669 684 943 818 158 450 401 083 392 × 2 = 0 + 0.773 671 339 369 887 636 316 900 802 166 784;
  • 40) 0.773 671 339 369 887 636 316 900 802 166 784 × 2 = 1 + 0.547 342 678 739 775 272 633 801 604 333 568;
  • 41) 0.547 342 678 739 775 272 633 801 604 333 568 × 2 = 1 + 0.094 685 357 479 550 545 267 603 208 667 136;
  • 42) 0.094 685 357 479 550 545 267 603 208 667 136 × 2 = 0 + 0.189 370 714 959 101 090 535 206 417 334 272;
  • 43) 0.189 370 714 959 101 090 535 206 417 334 272 × 2 = 0 + 0.378 741 429 918 202 181 070 412 834 668 544;
  • 44) 0.378 741 429 918 202 181 070 412 834 668 544 × 2 = 0 + 0.757 482 859 836 404 362 140 825 669 337 088;
  • 45) 0.757 482 859 836 404 362 140 825 669 337 088 × 2 = 1 + 0.514 965 719 672 808 724 281 651 338 674 176;
  • 46) 0.514 965 719 672 808 724 281 651 338 674 176 × 2 = 1 + 0.029 931 439 345 617 448 563 302 677 348 352;
  • 47) 0.029 931 439 345 617 448 563 302 677 348 352 × 2 = 0 + 0.059 862 878 691 234 897 126 605 354 696 704;
  • 48) 0.059 862 878 691 234 897 126 605 354 696 704 × 2 = 0 + 0.119 725 757 382 469 794 253 210 709 393 408;
  • 49) 0.119 725 757 382 469 794 253 210 709 393 408 × 2 = 0 + 0.239 451 514 764 939 588 506 421 418 786 816;
  • 50) 0.239 451 514 764 939 588 506 421 418 786 816 × 2 = 0 + 0.478 903 029 529 879 177 012 842 837 573 632;
  • 51) 0.478 903 029 529 879 177 012 842 837 573 632 × 2 = 0 + 0.957 806 059 059 758 354 025 685 675 147 264;
  • 52) 0.957 806 059 059 758 354 025 685 675 147 264 × 2 = 1 + 0.915 612 118 119 516 708 051 371 350 294 528;
  • 53) 0.915 612 118 119 516 708 051 371 350 294 528 × 2 = 1 + 0.831 224 236 239 033 416 102 742 700 589 056;
  • 54) 0.831 224 236 239 033 416 102 742 700 589 056 × 2 = 1 + 0.662 448 472 478 066 832 205 485 401 178 112;
  • 55) 0.662 448 472 478 066 832 205 485 401 178 112 × 2 = 1 + 0.324 896 944 956 133 664 410 970 802 356 224;
  • 56) 0.324 896 944 956 133 664 410 970 802 356 224 × 2 = 0 + 0.649 793 889 912 267 328 821 941 604 712 448;

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:


0.070 513 669 348 244 859 683 054 324 043 618(10) =

0.0001 0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110(2)

5. Positive number before normalization:

0.070 513 669 348 244 859 683 054 324 043 618(10) =

0.0001 0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.070 513 669 348 244 859 683 054 324 043 618(10) =

0.0001 0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110(2) =

0.0001 0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110(2) × 20 =

1.0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110(2) × 2-4


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): -4

Mantissa (not normalized):
1.0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1019(10) =

011 1111 1011(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110 =

0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110


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) =
011 1111 1011

Mantissa (52 bits) =
0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110


Decimal number 0.070 513 669 348 244 859 683 054 324 043 618 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 0010 0000 1101 0010 1111 0000 1111 1100 0101 1000 1100 0001 1110

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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