0.000 000 564 358 690 764 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 690 764(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
0.000 000 564 358 690 764(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: 0.000 000 564 358 690 764.

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) 0.000 000 564 358 690 764 × 2 = 0 + 0.000 001 128 717 381 528;
  • 2) 0.000 001 128 717 381 528 × 2 = 0 + 0.000 002 257 434 763 056;
  • 3) 0.000 002 257 434 763 056 × 2 = 0 + 0.000 004 514 869 526 112;
  • 4) 0.000 004 514 869 526 112 × 2 = 0 + 0.000 009 029 739 052 224;
  • 5) 0.000 009 029 739 052 224 × 2 = 0 + 0.000 018 059 478 104 448;
  • 6) 0.000 018 059 478 104 448 × 2 = 0 + 0.000 036 118 956 208 896;
  • 7) 0.000 036 118 956 208 896 × 2 = 0 + 0.000 072 237 912 417 792;
  • 8) 0.000 072 237 912 417 792 × 2 = 0 + 0.000 144 475 824 835 584;
  • 9) 0.000 144 475 824 835 584 × 2 = 0 + 0.000 288 951 649 671 168;
  • 10) 0.000 288 951 649 671 168 × 2 = 0 + 0.000 577 903 299 342 336;
  • 11) 0.000 577 903 299 342 336 × 2 = 0 + 0.001 155 806 598 684 672;
  • 12) 0.001 155 806 598 684 672 × 2 = 0 + 0.002 311 613 197 369 344;
  • 13) 0.002 311 613 197 369 344 × 2 = 0 + 0.004 623 226 394 738 688;
  • 14) 0.004 623 226 394 738 688 × 2 = 0 + 0.009 246 452 789 477 376;
  • 15) 0.009 246 452 789 477 376 × 2 = 0 + 0.018 492 905 578 954 752;
  • 16) 0.018 492 905 578 954 752 × 2 = 0 + 0.036 985 811 157 909 504;
  • 17) 0.036 985 811 157 909 504 × 2 = 0 + 0.073 971 622 315 819 008;
  • 18) 0.073 971 622 315 819 008 × 2 = 0 + 0.147 943 244 631 638 016;
  • 19) 0.147 943 244 631 638 016 × 2 = 0 + 0.295 886 489 263 276 032;
  • 20) 0.295 886 489 263 276 032 × 2 = 0 + 0.591 772 978 526 552 064;
  • 21) 0.591 772 978 526 552 064 × 2 = 1 + 0.183 545 957 053 104 128;
  • 22) 0.183 545 957 053 104 128 × 2 = 0 + 0.367 091 914 106 208 256;
  • 23) 0.367 091 914 106 208 256 × 2 = 0 + 0.734 183 828 212 416 512;
  • 24) 0.734 183 828 212 416 512 × 2 = 1 + 0.468 367 656 424 833 024;
  • 25) 0.468 367 656 424 833 024 × 2 = 0 + 0.936 735 312 849 666 048;
  • 26) 0.936 735 312 849 666 048 × 2 = 1 + 0.873 470 625 699 332 096;
  • 27) 0.873 470 625 699 332 096 × 2 = 1 + 0.746 941 251 398 664 192;
  • 28) 0.746 941 251 398 664 192 × 2 = 1 + 0.493 882 502 797 328 384;
  • 29) 0.493 882 502 797 328 384 × 2 = 0 + 0.987 765 005 594 656 768;
  • 30) 0.987 765 005 594 656 768 × 2 = 1 + 0.975 530 011 189 313 536;
  • 31) 0.975 530 011 189 313 536 × 2 = 1 + 0.951 060 022 378 627 072;
  • 32) 0.951 060 022 378 627 072 × 2 = 1 + 0.902 120 044 757 254 144;
  • 33) 0.902 120 044 757 254 144 × 2 = 1 + 0.804 240 089 514 508 288;
  • 34) 0.804 240 089 514 508 288 × 2 = 1 + 0.608 480 179 029 016 576;
  • 35) 0.608 480 179 029 016 576 × 2 = 1 + 0.216 960 358 058 033 152;
  • 36) 0.216 960 358 058 033 152 × 2 = 0 + 0.433 920 716 116 066 304;
  • 37) 0.433 920 716 116 066 304 × 2 = 0 + 0.867 841 432 232 132 608;
  • 38) 0.867 841 432 232 132 608 × 2 = 1 + 0.735 682 864 464 265 216;
  • 39) 0.735 682 864 464 265 216 × 2 = 1 + 0.471 365 728 928 530 432;
  • 40) 0.471 365 728 928 530 432 × 2 = 0 + 0.942 731 457 857 060 864;
  • 41) 0.942 731 457 857 060 864 × 2 = 1 + 0.885 462 915 714 121 728;
  • 42) 0.885 462 915 714 121 728 × 2 = 1 + 0.770 925 831 428 243 456;
  • 43) 0.770 925 831 428 243 456 × 2 = 1 + 0.541 851 662 856 486 912;
  • 44) 0.541 851 662 856 486 912 × 2 = 1 + 0.083 703 325 712 973 824;
  • 45) 0.083 703 325 712 973 824 × 2 = 0 + 0.167 406 651 425 947 648;
  • 46) 0.167 406 651 425 947 648 × 2 = 0 + 0.334 813 302 851 895 296;
  • 47) 0.334 813 302 851 895 296 × 2 = 0 + 0.669 626 605 703 790 592;
  • 48) 0.669 626 605 703 790 592 × 2 = 1 + 0.339 253 211 407 581 184;
  • 49) 0.339 253 211 407 581 184 × 2 = 0 + 0.678 506 422 815 162 368;
  • 50) 0.678 506 422 815 162 368 × 2 = 1 + 0.357 012 845 630 324 736;
  • 51) 0.357 012 845 630 324 736 × 2 = 0 + 0.714 025 691 260 649 472;
  • 52) 0.714 025 691 260 649 472 × 2 = 1 + 0.428 051 382 521 298 944;
  • 53) 0.428 051 382 521 298 944 × 2 = 0 + 0.856 102 765 042 597 888;
  • 54) 0.856 102 765 042 597 888 × 2 = 1 + 0.712 205 530 085 195 776;
  • 55) 0.712 205 530 085 195 776 × 2 = 1 + 0.424 411 060 170 391 552;
  • 56) 0.424 411 060 170 391 552 × 2 = 0 + 0.848 822 120 340 783 104;
  • 57) 0.848 822 120 340 783 104 × 2 = 1 + 0.697 644 240 681 566 208;
  • 58) 0.697 644 240 681 566 208 × 2 = 1 + 0.395 288 481 363 132 416;
  • 59) 0.395 288 481 363 132 416 × 2 = 0 + 0.790 576 962 726 264 832;
  • 60) 0.790 576 962 726 264 832 × 2 = 1 + 0.581 153 925 452 529 664;
  • 61) 0.581 153 925 452 529 664 × 2 = 1 + 0.162 307 850 905 059 328;
  • 62) 0.162 307 850 905 059 328 × 2 = 0 + 0.324 615 701 810 118 656;
  • 63) 0.324 615 701 810 118 656 × 2 = 0 + 0.649 231 403 620 237 312;
  • 64) 0.649 231 403 620 237 312 × 2 = 1 + 0.298 462 807 240 474 624;
  • 65) 0.298 462 807 240 474 624 × 2 = 0 + 0.596 925 614 480 949 248;
  • 66) 0.596 925 614 480 949 248 × 2 = 1 + 0.193 851 228 961 898 496;
  • 67) 0.193 851 228 961 898 496 × 2 = 0 + 0.387 702 457 923 796 992;
  • 68) 0.387 702 457 923 796 992 × 2 = 0 + 0.775 404 915 847 593 984;
  • 69) 0.775 404 915 847 593 984 × 2 = 1 + 0.550 809 831 695 187 968;
  • 70) 0.550 809 831 695 187 968 × 2 = 1 + 0.101 619 663 390 375 936;
  • 71) 0.101 619 663 390 375 936 × 2 = 0 + 0.203 239 326 780 751 872;
  • 72) 0.203 239 326 780 751 872 × 2 = 0 + 0.406 478 653 561 503 744;
  • 73) 0.406 478 653 561 503 744 × 2 = 0 + 0.812 957 307 123 007 488;

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:


0.000 000 564 358 690 764(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0110 1101 1001 0100 1100 0(2)

5. Positive number before normalization:

0.000 000 564 358 690 764(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0110 1101 1001 0100 1100 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:


0.000 000 564 358 690 764(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0110 1101 1001 0100 1100 0(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0110 1101 1001 0100 1100 0(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1010 1101 1011 0010 1001 1000(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 1101 1011 0010 1001 1000


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 1101 1011 0010 1001 1000 =

0010 1110 1111 1100 1101 1110 0010 1010 1101 1011 0010 1001 1000


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 1101 1011 0010 1001 1000


Decimal number 0.000 000 564 358 690 764 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1010 1101 1011 0010 1001 1000

Share

» Convert decimal 0.000 000 564 358 690 785 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. Gemini Notebook YouTube Video of 8.15 minutes

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