0.677 127 773 468 446 364 149 007 370 069 89 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.677 127 773 468 446 364 149 007 370 069 89(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
0.677 127 773 468 446 364 149 007 370 069 89(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: 0.677 127 773 468 446 364 149 007 370 069 89.

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) 0.677 127 773 468 446 364 149 007 370 069 89 × 2 = 1 + 0.354 255 546 936 892 728 298 014 740 139 78;
  • 2) 0.354 255 546 936 892 728 298 014 740 139 78 × 2 = 0 + 0.708 511 093 873 785 456 596 029 480 279 56;
  • 3) 0.708 511 093 873 785 456 596 029 480 279 56 × 2 = 1 + 0.417 022 187 747 570 913 192 058 960 559 12;
  • 4) 0.417 022 187 747 570 913 192 058 960 559 12 × 2 = 0 + 0.834 044 375 495 141 826 384 117 921 118 24;
  • 5) 0.834 044 375 495 141 826 384 117 921 118 24 × 2 = 1 + 0.668 088 750 990 283 652 768 235 842 236 48;
  • 6) 0.668 088 750 990 283 652 768 235 842 236 48 × 2 = 1 + 0.336 177 501 980 567 305 536 471 684 472 96;
  • 7) 0.336 177 501 980 567 305 536 471 684 472 96 × 2 = 0 + 0.672 355 003 961 134 611 072 943 368 945 92;
  • 8) 0.672 355 003 961 134 611 072 943 368 945 92 × 2 = 1 + 0.344 710 007 922 269 222 145 886 737 891 84;
  • 9) 0.344 710 007 922 269 222 145 886 737 891 84 × 2 = 0 + 0.689 420 015 844 538 444 291 773 475 783 68;
  • 10) 0.689 420 015 844 538 444 291 773 475 783 68 × 2 = 1 + 0.378 840 031 689 076 888 583 546 951 567 36;
  • 11) 0.378 840 031 689 076 888 583 546 951 567 36 × 2 = 0 + 0.757 680 063 378 153 777 167 093 903 134 72;
  • 12) 0.757 680 063 378 153 777 167 093 903 134 72 × 2 = 1 + 0.515 360 126 756 307 554 334 187 806 269 44;
  • 13) 0.515 360 126 756 307 554 334 187 806 269 44 × 2 = 1 + 0.030 720 253 512 615 108 668 375 612 538 88;
  • 14) 0.030 720 253 512 615 108 668 375 612 538 88 × 2 = 0 + 0.061 440 507 025 230 217 336 751 225 077 76;
  • 15) 0.061 440 507 025 230 217 336 751 225 077 76 × 2 = 0 + 0.122 881 014 050 460 434 673 502 450 155 52;
  • 16) 0.122 881 014 050 460 434 673 502 450 155 52 × 2 = 0 + 0.245 762 028 100 920 869 347 004 900 311 04;
  • 17) 0.245 762 028 100 920 869 347 004 900 311 04 × 2 = 0 + 0.491 524 056 201 841 738 694 009 800 622 08;
  • 18) 0.491 524 056 201 841 738 694 009 800 622 08 × 2 = 0 + 0.983 048 112 403 683 477 388 019 601 244 16;
  • 19) 0.983 048 112 403 683 477 388 019 601 244 16 × 2 = 1 + 0.966 096 224 807 366 954 776 039 202 488 32;
  • 20) 0.966 096 224 807 366 954 776 039 202 488 32 × 2 = 1 + 0.932 192 449 614 733 909 552 078 404 976 64;
  • 21) 0.932 192 449 614 733 909 552 078 404 976 64 × 2 = 1 + 0.864 384 899 229 467 819 104 156 809 953 28;
  • 22) 0.864 384 899 229 467 819 104 156 809 953 28 × 2 = 1 + 0.728 769 798 458 935 638 208 313 619 906 56;
  • 23) 0.728 769 798 458 935 638 208 313 619 906 56 × 2 = 1 + 0.457 539 596 917 871 276 416 627 239 813 12;
  • 24) 0.457 539 596 917 871 276 416 627 239 813 12 × 2 = 0 + 0.915 079 193 835 742 552 833 254 479 626 24;
  • 25) 0.915 079 193 835 742 552 833 254 479 626 24 × 2 = 1 + 0.830 158 387 671 485 105 666 508 959 252 48;
  • 26) 0.830 158 387 671 485 105 666 508 959 252 48 × 2 = 1 + 0.660 316 775 342 970 211 333 017 918 504 96;
  • 27) 0.660 316 775 342 970 211 333 017 918 504 96 × 2 = 1 + 0.320 633 550 685 940 422 666 035 837 009 92;
  • 28) 0.320 633 550 685 940 422 666 035 837 009 92 × 2 = 0 + 0.641 267 101 371 880 845 332 071 674 019 84;
  • 29) 0.641 267 101 371 880 845 332 071 674 019 84 × 2 = 1 + 0.282 534 202 743 761 690 664 143 348 039 68;
  • 30) 0.282 534 202 743 761 690 664 143 348 039 68 × 2 = 0 + 0.565 068 405 487 523 381 328 286 696 079 36;
  • 31) 0.565 068 405 487 523 381 328 286 696 079 36 × 2 = 1 + 0.130 136 810 975 046 762 656 573 392 158 72;
  • 32) 0.130 136 810 975 046 762 656 573 392 158 72 × 2 = 0 + 0.260 273 621 950 093 525 313 146 784 317 44;
  • 33) 0.260 273 621 950 093 525 313 146 784 317 44 × 2 = 0 + 0.520 547 243 900 187 050 626 293 568 634 88;
  • 34) 0.520 547 243 900 187 050 626 293 568 634 88 × 2 = 1 + 0.041 094 487 800 374 101 252 587 137 269 76;
  • 35) 0.041 094 487 800 374 101 252 587 137 269 76 × 2 = 0 + 0.082 188 975 600 748 202 505 174 274 539 52;
  • 36) 0.082 188 975 600 748 202 505 174 274 539 52 × 2 = 0 + 0.164 377 951 201 496 405 010 348 549 079 04;
  • 37) 0.164 377 951 201 496 405 010 348 549 079 04 × 2 = 0 + 0.328 755 902 402 992 810 020 697 098 158 08;
  • 38) 0.328 755 902 402 992 810 020 697 098 158 08 × 2 = 0 + 0.657 511 804 805 985 620 041 394 196 316 16;
  • 39) 0.657 511 804 805 985 620 041 394 196 316 16 × 2 = 1 + 0.315 023 609 611 971 240 082 788 392 632 32;
  • 40) 0.315 023 609 611 971 240 082 788 392 632 32 × 2 = 0 + 0.630 047 219 223 942 480 165 576 785 264 64;
  • 41) 0.630 047 219 223 942 480 165 576 785 264 64 × 2 = 1 + 0.260 094 438 447 884 960 331 153 570 529 28;
  • 42) 0.260 094 438 447 884 960 331 153 570 529 28 × 2 = 0 + 0.520 188 876 895 769 920 662 307 141 058 56;
  • 43) 0.520 188 876 895 769 920 662 307 141 058 56 × 2 = 1 + 0.040 377 753 791 539 841 324 614 282 117 12;
  • 44) 0.040 377 753 791 539 841 324 614 282 117 12 × 2 = 0 + 0.080 755 507 583 079 682 649 228 564 234 24;
  • 45) 0.080 755 507 583 079 682 649 228 564 234 24 × 2 = 0 + 0.161 511 015 166 159 365 298 457 128 468 48;
  • 46) 0.161 511 015 166 159 365 298 457 128 468 48 × 2 = 0 + 0.323 022 030 332 318 730 596 914 256 936 96;
  • 47) 0.323 022 030 332 318 730 596 914 256 936 96 × 2 = 0 + 0.646 044 060 664 637 461 193 828 513 873 92;
  • 48) 0.646 044 060 664 637 461 193 828 513 873 92 × 2 = 1 + 0.292 088 121 329 274 922 387 657 027 747 84;
  • 49) 0.292 088 121 329 274 922 387 657 027 747 84 × 2 = 0 + 0.584 176 242 658 549 844 775 314 055 495 68;
  • 50) 0.584 176 242 658 549 844 775 314 055 495 68 × 2 = 1 + 0.168 352 485 317 099 689 550 628 110 991 36;
  • 51) 0.168 352 485 317 099 689 550 628 110 991 36 × 2 = 0 + 0.336 704 970 634 199 379 101 256 221 982 72;
  • 52) 0.336 704 970 634 199 379 101 256 221 982 72 × 2 = 0 + 0.673 409 941 268 398 758 202 512 443 965 44;
  • 53) 0.673 409 941 268 398 758 202 512 443 965 44 × 2 = 1 + 0.346 819 882 536 797 516 405 024 887 930 88;

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:


0.677 127 773 468 446 364 149 007 370 069 89(10) =

0.1010 1101 0101 1000 0011 1110 1110 1010 0100 0010 1010 0001 0100 1(2)

5. Positive number before normalization:

0.677 127 773 468 446 364 149 007 370 069 89(10) =

0.1010 1101 0101 1000 0011 1110 1110 1010 0100 0010 1010 0001 0100 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.677 127 773 468 446 364 149 007 370 069 89(10) =

0.1010 1101 0101 1000 0011 1110 1110 1010 0100 0010 1010 0001 0100 1(2) =

0.1010 1101 0101 1000 0011 1110 1110 1010 0100 0010 1010 0001 0100 1(2) × 20 =

1.0101 1010 1011 0000 0111 1101 1101 0100 1000 0101 0100 0010 1001(2) × 2-1


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): -1

Mantissa (not normalized):
1.0101 1010 1011 0000 0111 1101 1101 0100 1000 0101 0100 0010 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 0101 1010 1011 0000 0111 1101 1101 0100 1000 0101 0100 0010 1001 =

0101 1010 1011 0000 0111 1101 1101 0100 1000 0101 0100 0010 1001


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 1110

Mantissa (52 bits) =
0101 1010 1011 0000 0111 1101 1101 0100 1000 0101 0100 0010 1001


Decimal number 0.677 127 773 468 446 364 149 007 370 069 89 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0101 1010 1011 0000 0111 1101 1101 0100 1000 0101 0100 0010 1001

» Convert decimal 0.677 127 773 468 446 364 149 007 370 069 03 to 64 bit double precision IEEE 754 binary floating point representation standard

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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