0.000 000 564 358 690 722 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 690 722(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
0.000 000 564 358 690 722(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: 0.000 000 564 358 690 722.

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) 0.000 000 564 358 690 722 × 2 = 0 + 0.000 001 128 717 381 444;
  • 2) 0.000 001 128 717 381 444 × 2 = 0 + 0.000 002 257 434 762 888;
  • 3) 0.000 002 257 434 762 888 × 2 = 0 + 0.000 004 514 869 525 776;
  • 4) 0.000 004 514 869 525 776 × 2 = 0 + 0.000 009 029 739 051 552;
  • 5) 0.000 009 029 739 051 552 × 2 = 0 + 0.000 018 059 478 103 104;
  • 6) 0.000 018 059 478 103 104 × 2 = 0 + 0.000 036 118 956 206 208;
  • 7) 0.000 036 118 956 206 208 × 2 = 0 + 0.000 072 237 912 412 416;
  • 8) 0.000 072 237 912 412 416 × 2 = 0 + 0.000 144 475 824 824 832;
  • 9) 0.000 144 475 824 824 832 × 2 = 0 + 0.000 288 951 649 649 664;
  • 10) 0.000 288 951 649 649 664 × 2 = 0 + 0.000 577 903 299 299 328;
  • 11) 0.000 577 903 299 299 328 × 2 = 0 + 0.001 155 806 598 598 656;
  • 12) 0.001 155 806 598 598 656 × 2 = 0 + 0.002 311 613 197 197 312;
  • 13) 0.002 311 613 197 197 312 × 2 = 0 + 0.004 623 226 394 394 624;
  • 14) 0.004 623 226 394 394 624 × 2 = 0 + 0.009 246 452 788 789 248;
  • 15) 0.009 246 452 788 789 248 × 2 = 0 + 0.018 492 905 577 578 496;
  • 16) 0.018 492 905 577 578 496 × 2 = 0 + 0.036 985 811 155 156 992;
  • 17) 0.036 985 811 155 156 992 × 2 = 0 + 0.073 971 622 310 313 984;
  • 18) 0.073 971 622 310 313 984 × 2 = 0 + 0.147 943 244 620 627 968;
  • 19) 0.147 943 244 620 627 968 × 2 = 0 + 0.295 886 489 241 255 936;
  • 20) 0.295 886 489 241 255 936 × 2 = 0 + 0.591 772 978 482 511 872;
  • 21) 0.591 772 978 482 511 872 × 2 = 1 + 0.183 545 956 965 023 744;
  • 22) 0.183 545 956 965 023 744 × 2 = 0 + 0.367 091 913 930 047 488;
  • 23) 0.367 091 913 930 047 488 × 2 = 0 + 0.734 183 827 860 094 976;
  • 24) 0.734 183 827 860 094 976 × 2 = 1 + 0.468 367 655 720 189 952;
  • 25) 0.468 367 655 720 189 952 × 2 = 0 + 0.936 735 311 440 379 904;
  • 26) 0.936 735 311 440 379 904 × 2 = 1 + 0.873 470 622 880 759 808;
  • 27) 0.873 470 622 880 759 808 × 2 = 1 + 0.746 941 245 761 519 616;
  • 28) 0.746 941 245 761 519 616 × 2 = 1 + 0.493 882 491 523 039 232;
  • 29) 0.493 882 491 523 039 232 × 2 = 0 + 0.987 764 983 046 078 464;
  • 30) 0.987 764 983 046 078 464 × 2 = 1 + 0.975 529 966 092 156 928;
  • 31) 0.975 529 966 092 156 928 × 2 = 1 + 0.951 059 932 184 313 856;
  • 32) 0.951 059 932 184 313 856 × 2 = 1 + 0.902 119 864 368 627 712;
  • 33) 0.902 119 864 368 627 712 × 2 = 1 + 0.804 239 728 737 255 424;
  • 34) 0.804 239 728 737 255 424 × 2 = 1 + 0.608 479 457 474 510 848;
  • 35) 0.608 479 457 474 510 848 × 2 = 1 + 0.216 958 914 949 021 696;
  • 36) 0.216 958 914 949 021 696 × 2 = 0 + 0.433 917 829 898 043 392;
  • 37) 0.433 917 829 898 043 392 × 2 = 0 + 0.867 835 659 796 086 784;
  • 38) 0.867 835 659 796 086 784 × 2 = 1 + 0.735 671 319 592 173 568;
  • 39) 0.735 671 319 592 173 568 × 2 = 1 + 0.471 342 639 184 347 136;
  • 40) 0.471 342 639 184 347 136 × 2 = 0 + 0.942 685 278 368 694 272;
  • 41) 0.942 685 278 368 694 272 × 2 = 1 + 0.885 370 556 737 388 544;
  • 42) 0.885 370 556 737 388 544 × 2 = 1 + 0.770 741 113 474 777 088;
  • 43) 0.770 741 113 474 777 088 × 2 = 1 + 0.541 482 226 949 554 176;
  • 44) 0.541 482 226 949 554 176 × 2 = 1 + 0.082 964 453 899 108 352;
  • 45) 0.082 964 453 899 108 352 × 2 = 0 + 0.165 928 907 798 216 704;
  • 46) 0.165 928 907 798 216 704 × 2 = 0 + 0.331 857 815 596 433 408;
  • 47) 0.331 857 815 596 433 408 × 2 = 0 + 0.663 715 631 192 866 816;
  • 48) 0.663 715 631 192 866 816 × 2 = 1 + 0.327 431 262 385 733 632;
  • 49) 0.327 431 262 385 733 632 × 2 = 0 + 0.654 862 524 771 467 264;
  • 50) 0.654 862 524 771 467 264 × 2 = 1 + 0.309 725 049 542 934 528;
  • 51) 0.309 725 049 542 934 528 × 2 = 0 + 0.619 450 099 085 869 056;
  • 52) 0.619 450 099 085 869 056 × 2 = 1 + 0.238 900 198 171 738 112;
  • 53) 0.238 900 198 171 738 112 × 2 = 0 + 0.477 800 396 343 476 224;
  • 54) 0.477 800 396 343 476 224 × 2 = 0 + 0.955 600 792 686 952 448;
  • 55) 0.955 600 792 686 952 448 × 2 = 1 + 0.911 201 585 373 904 896;
  • 56) 0.911 201 585 373 904 896 × 2 = 1 + 0.822 403 170 747 809 792;
  • 57) 0.822 403 170 747 809 792 × 2 = 1 + 0.644 806 341 495 619 584;
  • 58) 0.644 806 341 495 619 584 × 2 = 1 + 0.289 612 682 991 239 168;
  • 59) 0.289 612 682 991 239 168 × 2 = 0 + 0.579 225 365 982 478 336;
  • 60) 0.579 225 365 982 478 336 × 2 = 1 + 0.158 450 731 964 956 672;
  • 61) 0.158 450 731 964 956 672 × 2 = 0 + 0.316 901 463 929 913 344;
  • 62) 0.316 901 463 929 913 344 × 2 = 0 + 0.633 802 927 859 826 688;
  • 63) 0.633 802 927 859 826 688 × 2 = 1 + 0.267 605 855 719 653 376;
  • 64) 0.267 605 855 719 653 376 × 2 = 0 + 0.535 211 711 439 306 752;
  • 65) 0.535 211 711 439 306 752 × 2 = 1 + 0.070 423 422 878 613 504;
  • 66) 0.070 423 422 878 613 504 × 2 = 0 + 0.140 846 845 757 227 008;
  • 67) 0.140 846 845 757 227 008 × 2 = 0 + 0.281 693 691 514 454 016;
  • 68) 0.281 693 691 514 454 016 × 2 = 0 + 0.563 387 383 028 908 032;
  • 69) 0.563 387 383 028 908 032 × 2 = 1 + 0.126 774 766 057 816 064;
  • 70) 0.126 774 766 057 816 064 × 2 = 0 + 0.253 549 532 115 632 128;
  • 71) 0.253 549 532 115 632 128 × 2 = 0 + 0.507 099 064 231 264 256;
  • 72) 0.507 099 064 231 264 256 × 2 = 1 + 0.014 198 128 462 528 512;
  • 73) 0.014 198 128 462 528 512 × 2 = 0 + 0.028 396 256 925 057 024;

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:


0.000 000 564 358 690 722(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0011 1101 0010 1000 1001 0(2)

5. Positive number before normalization:

0.000 000 564 358 690 722(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0011 1101 0010 1000 1001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 564 358 690 722(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0011 1101 0010 1000 1001 0(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0011 1101 0010 1000 1001 0(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1010 0111 1010 0101 0001 0010(2) × 2-21


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): -21

Mantissa (not normalized):
1.0010 1110 1111 1100 1101 1110 0010 1010 0111 1010 0101 0001 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(11-1) - 1 =

(-21 + 1 023)(10) =

1 002(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;
  • 1 002 ÷ 2 = 501 + 0;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1002(10) =

011 1110 1010(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 1110 1111 1100 1101 1110 0010 1010 0111 1010 0101 0001 0010 =

0010 1110 1111 1100 1101 1110 0010 1010 0111 1010 0101 0001 0010


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 1110 1010

Mantissa (52 bits) =
0010 1110 1111 1100 1101 1110 0010 1010 0111 1010 0101 0001 0010


Decimal number 0.000 000 564 358 690 722 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1010 0111 1010 0101 0001 0010

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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