0.522 136 891 213 706 920 160 983 74 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.522 136 891 213 706 920 160 983 74(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
0.522 136 891 213 706 920 160 983 74(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: 0.522 136 891 213 706 920 160 983 74.

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) 0.522 136 891 213 706 920 160 983 74 × 2 = 1 + 0.044 273 782 427 413 840 321 967 48;
  • 2) 0.044 273 782 427 413 840 321 967 48 × 2 = 0 + 0.088 547 564 854 827 680 643 934 96;
  • 3) 0.088 547 564 854 827 680 643 934 96 × 2 = 0 + 0.177 095 129 709 655 361 287 869 92;
  • 4) 0.177 095 129 709 655 361 287 869 92 × 2 = 0 + 0.354 190 259 419 310 722 575 739 84;
  • 5) 0.354 190 259 419 310 722 575 739 84 × 2 = 0 + 0.708 380 518 838 621 445 151 479 68;
  • 6) 0.708 380 518 838 621 445 151 479 68 × 2 = 1 + 0.416 761 037 677 242 890 302 959 36;
  • 7) 0.416 761 037 677 242 890 302 959 36 × 2 = 0 + 0.833 522 075 354 485 780 605 918 72;
  • 8) 0.833 522 075 354 485 780 605 918 72 × 2 = 1 + 0.667 044 150 708 971 561 211 837 44;
  • 9) 0.667 044 150 708 971 561 211 837 44 × 2 = 1 + 0.334 088 301 417 943 122 423 674 88;
  • 10) 0.334 088 301 417 943 122 423 674 88 × 2 = 0 + 0.668 176 602 835 886 244 847 349 76;
  • 11) 0.668 176 602 835 886 244 847 349 76 × 2 = 1 + 0.336 353 205 671 772 489 694 699 52;
  • 12) 0.336 353 205 671 772 489 694 699 52 × 2 = 0 + 0.672 706 411 343 544 979 389 399 04;
  • 13) 0.672 706 411 343 544 979 389 399 04 × 2 = 1 + 0.345 412 822 687 089 958 778 798 08;
  • 14) 0.345 412 822 687 089 958 778 798 08 × 2 = 0 + 0.690 825 645 374 179 917 557 596 16;
  • 15) 0.690 825 645 374 179 917 557 596 16 × 2 = 1 + 0.381 651 290 748 359 835 115 192 32;
  • 16) 0.381 651 290 748 359 835 115 192 32 × 2 = 0 + 0.763 302 581 496 719 670 230 384 64;
  • 17) 0.763 302 581 496 719 670 230 384 64 × 2 = 1 + 0.526 605 162 993 439 340 460 769 28;
  • 18) 0.526 605 162 993 439 340 460 769 28 × 2 = 1 + 0.053 210 325 986 878 680 921 538 56;
  • 19) 0.053 210 325 986 878 680 921 538 56 × 2 = 0 + 0.106 420 651 973 757 361 843 077 12;
  • 20) 0.106 420 651 973 757 361 843 077 12 × 2 = 0 + 0.212 841 303 947 514 723 686 154 24;
  • 21) 0.212 841 303 947 514 723 686 154 24 × 2 = 0 + 0.425 682 607 895 029 447 372 308 48;
  • 22) 0.425 682 607 895 029 447 372 308 48 × 2 = 0 + 0.851 365 215 790 058 894 744 616 96;
  • 23) 0.851 365 215 790 058 894 744 616 96 × 2 = 1 + 0.702 730 431 580 117 789 489 233 92;
  • 24) 0.702 730 431 580 117 789 489 233 92 × 2 = 1 + 0.405 460 863 160 235 578 978 467 84;
  • 25) 0.405 460 863 160 235 578 978 467 84 × 2 = 0 + 0.810 921 726 320 471 157 956 935 68;
  • 26) 0.810 921 726 320 471 157 956 935 68 × 2 = 1 + 0.621 843 452 640 942 315 913 871 36;
  • 27) 0.621 843 452 640 942 315 913 871 36 × 2 = 1 + 0.243 686 905 281 884 631 827 742 72;
  • 28) 0.243 686 905 281 884 631 827 742 72 × 2 = 0 + 0.487 373 810 563 769 263 655 485 44;
  • 29) 0.487 373 810 563 769 263 655 485 44 × 2 = 0 + 0.974 747 621 127 538 527 310 970 88;
  • 30) 0.974 747 621 127 538 527 310 970 88 × 2 = 1 + 0.949 495 242 255 077 054 621 941 76;
  • 31) 0.949 495 242 255 077 054 621 941 76 × 2 = 1 + 0.898 990 484 510 154 109 243 883 52;
  • 32) 0.898 990 484 510 154 109 243 883 52 × 2 = 1 + 0.797 980 969 020 308 218 487 767 04;
  • 33) 0.797 980 969 020 308 218 487 767 04 × 2 = 1 + 0.595 961 938 040 616 436 975 534 08;
  • 34) 0.595 961 938 040 616 436 975 534 08 × 2 = 1 + 0.191 923 876 081 232 873 951 068 16;
  • 35) 0.191 923 876 081 232 873 951 068 16 × 2 = 0 + 0.383 847 752 162 465 747 902 136 32;
  • 36) 0.383 847 752 162 465 747 902 136 32 × 2 = 0 + 0.767 695 504 324 931 495 804 272 64;
  • 37) 0.767 695 504 324 931 495 804 272 64 × 2 = 1 + 0.535 391 008 649 862 991 608 545 28;
  • 38) 0.535 391 008 649 862 991 608 545 28 × 2 = 1 + 0.070 782 017 299 725 983 217 090 56;
  • 39) 0.070 782 017 299 725 983 217 090 56 × 2 = 0 + 0.141 564 034 599 451 966 434 181 12;
  • 40) 0.141 564 034 599 451 966 434 181 12 × 2 = 0 + 0.283 128 069 198 903 932 868 362 24;
  • 41) 0.283 128 069 198 903 932 868 362 24 × 2 = 0 + 0.566 256 138 397 807 865 736 724 48;
  • 42) 0.566 256 138 397 807 865 736 724 48 × 2 = 1 + 0.132 512 276 795 615 731 473 448 96;
  • 43) 0.132 512 276 795 615 731 473 448 96 × 2 = 0 + 0.265 024 553 591 231 462 946 897 92;
  • 44) 0.265 024 553 591 231 462 946 897 92 × 2 = 0 + 0.530 049 107 182 462 925 893 795 84;
  • 45) 0.530 049 107 182 462 925 893 795 84 × 2 = 1 + 0.060 098 214 364 925 851 787 591 68;
  • 46) 0.060 098 214 364 925 851 787 591 68 × 2 = 0 + 0.120 196 428 729 851 703 575 183 36;
  • 47) 0.120 196 428 729 851 703 575 183 36 × 2 = 0 + 0.240 392 857 459 703 407 150 366 72;
  • 48) 0.240 392 857 459 703 407 150 366 72 × 2 = 0 + 0.480 785 714 919 406 814 300 733 44;
  • 49) 0.480 785 714 919 406 814 300 733 44 × 2 = 0 + 0.961 571 429 838 813 628 601 466 88;
  • 50) 0.961 571 429 838 813 628 601 466 88 × 2 = 1 + 0.923 142 859 677 627 257 202 933 76;
  • 51) 0.923 142 859 677 627 257 202 933 76 × 2 = 1 + 0.846 285 719 355 254 514 405 867 52;
  • 52) 0.846 285 719 355 254 514 405 867 52 × 2 = 1 + 0.692 571 438 710 509 028 811 735 04;
  • 53) 0.692 571 438 710 509 028 811 735 04 × 2 = 1 + 0.385 142 877 421 018 057 623 470 08;

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:


0.522 136 891 213 706 920 160 983 74(10) =

0.1000 0101 1010 1010 1100 0011 0110 0111 1100 1100 0100 1000 0111 1(2)

5. Positive number before normalization:

0.522 136 891 213 706 920 160 983 74(10) =

0.1000 0101 1010 1010 1100 0011 0110 0111 1100 1100 0100 1000 0111 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:


0.522 136 891 213 706 920 160 983 74(10) =

0.1000 0101 1010 1010 1100 0011 0110 0111 1100 1100 0100 1000 0111 1(2) =

0.1000 0101 1010 1010 1100 0011 0110 0111 1100 1100 0100 1000 0111 1(2) × 20 =

1.0000 1011 0101 0101 1000 0110 1100 1111 1001 1000 1001 0000 1111(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):
1.0000 1011 0101 0101 1000 0110 1100 1111 1001 1000 1001 0000 1111


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. 0000 1011 0101 0101 1000 0110 1100 1111 1001 1000 1001 0000 1111 =

0000 1011 0101 0101 1000 0110 1100 1111 1001 1000 1001 0000 1111


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) =
0000 1011 0101 0101 1000 0110 1100 1111 1001 1000 1001 0000 1111


Decimal number 0.522 136 891 213 706 920 160 983 74 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0000 1011 0101 0101 1000 0110 1100 1111 1001 1000 1001 0000 1111

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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