0.522 136 891 213 706 920 160 984 39 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.522 136 891 213 706 920 160 984 39(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
0.522 136 891 213 706 920 160 984 39(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: 0.522 136 891 213 706 920 160 984 39.

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) 0.522 136 891 213 706 920 160 984 39 × 2 = 1 + 0.044 273 782 427 413 840 321 968 78;
  • 2) 0.044 273 782 427 413 840 321 968 78 × 2 = 0 + 0.088 547 564 854 827 680 643 937 56;
  • 3) 0.088 547 564 854 827 680 643 937 56 × 2 = 0 + 0.177 095 129 709 655 361 287 875 12;
  • 4) 0.177 095 129 709 655 361 287 875 12 × 2 = 0 + 0.354 190 259 419 310 722 575 750 24;
  • 5) 0.354 190 259 419 310 722 575 750 24 × 2 = 0 + 0.708 380 518 838 621 445 151 500 48;
  • 6) 0.708 380 518 838 621 445 151 500 48 × 2 = 1 + 0.416 761 037 677 242 890 303 000 96;
  • 7) 0.416 761 037 677 242 890 303 000 96 × 2 = 0 + 0.833 522 075 354 485 780 606 001 92;
  • 8) 0.833 522 075 354 485 780 606 001 92 × 2 = 1 + 0.667 044 150 708 971 561 212 003 84;
  • 9) 0.667 044 150 708 971 561 212 003 84 × 2 = 1 + 0.334 088 301 417 943 122 424 007 68;
  • 10) 0.334 088 301 417 943 122 424 007 68 × 2 = 0 + 0.668 176 602 835 886 244 848 015 36;
  • 11) 0.668 176 602 835 886 244 848 015 36 × 2 = 1 + 0.336 353 205 671 772 489 696 030 72;
  • 12) 0.336 353 205 671 772 489 696 030 72 × 2 = 0 + 0.672 706 411 343 544 979 392 061 44;
  • 13) 0.672 706 411 343 544 979 392 061 44 × 2 = 1 + 0.345 412 822 687 089 958 784 122 88;
  • 14) 0.345 412 822 687 089 958 784 122 88 × 2 = 0 + 0.690 825 645 374 179 917 568 245 76;
  • 15) 0.690 825 645 374 179 917 568 245 76 × 2 = 1 + 0.381 651 290 748 359 835 136 491 52;
  • 16) 0.381 651 290 748 359 835 136 491 52 × 2 = 0 + 0.763 302 581 496 719 670 272 983 04;
  • 17) 0.763 302 581 496 719 670 272 983 04 × 2 = 1 + 0.526 605 162 993 439 340 545 966 08;
  • 18) 0.526 605 162 993 439 340 545 966 08 × 2 = 1 + 0.053 210 325 986 878 681 091 932 16;
  • 19) 0.053 210 325 986 878 681 091 932 16 × 2 = 0 + 0.106 420 651 973 757 362 183 864 32;
  • 20) 0.106 420 651 973 757 362 183 864 32 × 2 = 0 + 0.212 841 303 947 514 724 367 728 64;
  • 21) 0.212 841 303 947 514 724 367 728 64 × 2 = 0 + 0.425 682 607 895 029 448 735 457 28;
  • 22) 0.425 682 607 895 029 448 735 457 28 × 2 = 0 + 0.851 365 215 790 058 897 470 914 56;
  • 23) 0.851 365 215 790 058 897 470 914 56 × 2 = 1 + 0.702 730 431 580 117 794 941 829 12;
  • 24) 0.702 730 431 580 117 794 941 829 12 × 2 = 1 + 0.405 460 863 160 235 589 883 658 24;
  • 25) 0.405 460 863 160 235 589 883 658 24 × 2 = 0 + 0.810 921 726 320 471 179 767 316 48;
  • 26) 0.810 921 726 320 471 179 767 316 48 × 2 = 1 + 0.621 843 452 640 942 359 534 632 96;
  • 27) 0.621 843 452 640 942 359 534 632 96 × 2 = 1 + 0.243 686 905 281 884 719 069 265 92;
  • 28) 0.243 686 905 281 884 719 069 265 92 × 2 = 0 + 0.487 373 810 563 769 438 138 531 84;
  • 29) 0.487 373 810 563 769 438 138 531 84 × 2 = 0 + 0.974 747 621 127 538 876 277 063 68;
  • 30) 0.974 747 621 127 538 876 277 063 68 × 2 = 1 + 0.949 495 242 255 077 752 554 127 36;
  • 31) 0.949 495 242 255 077 752 554 127 36 × 2 = 1 + 0.898 990 484 510 155 505 108 254 72;
  • 32) 0.898 990 484 510 155 505 108 254 72 × 2 = 1 + 0.797 980 969 020 311 010 216 509 44;
  • 33) 0.797 980 969 020 311 010 216 509 44 × 2 = 1 + 0.595 961 938 040 622 020 433 018 88;
  • 34) 0.595 961 938 040 622 020 433 018 88 × 2 = 1 + 0.191 923 876 081 244 040 866 037 76;
  • 35) 0.191 923 876 081 244 040 866 037 76 × 2 = 0 + 0.383 847 752 162 488 081 732 075 52;
  • 36) 0.383 847 752 162 488 081 732 075 52 × 2 = 0 + 0.767 695 504 324 976 163 464 151 04;
  • 37) 0.767 695 504 324 976 163 464 151 04 × 2 = 1 + 0.535 391 008 649 952 326 928 302 08;
  • 38) 0.535 391 008 649 952 326 928 302 08 × 2 = 1 + 0.070 782 017 299 904 653 856 604 16;
  • 39) 0.070 782 017 299 904 653 856 604 16 × 2 = 0 + 0.141 564 034 599 809 307 713 208 32;
  • 40) 0.141 564 034 599 809 307 713 208 32 × 2 = 0 + 0.283 128 069 199 618 615 426 416 64;
  • 41) 0.283 128 069 199 618 615 426 416 64 × 2 = 0 + 0.566 256 138 399 237 230 852 833 28;
  • 42) 0.566 256 138 399 237 230 852 833 28 × 2 = 1 + 0.132 512 276 798 474 461 705 666 56;
  • 43) 0.132 512 276 798 474 461 705 666 56 × 2 = 0 + 0.265 024 553 596 948 923 411 333 12;
  • 44) 0.265 024 553 596 948 923 411 333 12 × 2 = 0 + 0.530 049 107 193 897 846 822 666 24;
  • 45) 0.530 049 107 193 897 846 822 666 24 × 2 = 1 + 0.060 098 214 387 795 693 645 332 48;
  • 46) 0.060 098 214 387 795 693 645 332 48 × 2 = 0 + 0.120 196 428 775 591 387 290 664 96;
  • 47) 0.120 196 428 775 591 387 290 664 96 × 2 = 0 + 0.240 392 857 551 182 774 581 329 92;
  • 48) 0.240 392 857 551 182 774 581 329 92 × 2 = 0 + 0.480 785 715 102 365 549 162 659 84;
  • 49) 0.480 785 715 102 365 549 162 659 84 × 2 = 0 + 0.961 571 430 204 731 098 325 319 68;
  • 50) 0.961 571 430 204 731 098 325 319 68 × 2 = 1 + 0.923 142 860 409 462 196 650 639 36;
  • 51) 0.923 142 860 409 462 196 650 639 36 × 2 = 1 + 0.846 285 720 818 924 393 301 278 72;
  • 52) 0.846 285 720 818 924 393 301 278 72 × 2 = 1 + 0.692 571 441 637 848 786 602 557 44;
  • 53) 0.692 571 441 637 848 786 602 557 44 × 2 = 1 + 0.385 142 883 275 697 573 205 114 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:


0.522 136 891 213 706 920 160 984 39(10) =

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

5. Positive number before normalization:

0.522 136 891 213 706 920 160 984 39(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:


0.522 136 891 213 706 920 160 984 39(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 0.522 136 891 213 706 920 160 984 39 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