0.000 000 564 358 690 797 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 690 797(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
0.000 000 564 358 690 797(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: 0.000 000 564 358 690 797.

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) 0.000 000 564 358 690 797 × 2 = 0 + 0.000 001 128 717 381 594;
  • 2) 0.000 001 128 717 381 594 × 2 = 0 + 0.000 002 257 434 763 188;
  • 3) 0.000 002 257 434 763 188 × 2 = 0 + 0.000 004 514 869 526 376;
  • 4) 0.000 004 514 869 526 376 × 2 = 0 + 0.000 009 029 739 052 752;
  • 5) 0.000 009 029 739 052 752 × 2 = 0 + 0.000 018 059 478 105 504;
  • 6) 0.000 018 059 478 105 504 × 2 = 0 + 0.000 036 118 956 211 008;
  • 7) 0.000 036 118 956 211 008 × 2 = 0 + 0.000 072 237 912 422 016;
  • 8) 0.000 072 237 912 422 016 × 2 = 0 + 0.000 144 475 824 844 032;
  • 9) 0.000 144 475 824 844 032 × 2 = 0 + 0.000 288 951 649 688 064;
  • 10) 0.000 288 951 649 688 064 × 2 = 0 + 0.000 577 903 299 376 128;
  • 11) 0.000 577 903 299 376 128 × 2 = 0 + 0.001 155 806 598 752 256;
  • 12) 0.001 155 806 598 752 256 × 2 = 0 + 0.002 311 613 197 504 512;
  • 13) 0.002 311 613 197 504 512 × 2 = 0 + 0.004 623 226 395 009 024;
  • 14) 0.004 623 226 395 009 024 × 2 = 0 + 0.009 246 452 790 018 048;
  • 15) 0.009 246 452 790 018 048 × 2 = 0 + 0.018 492 905 580 036 096;
  • 16) 0.018 492 905 580 036 096 × 2 = 0 + 0.036 985 811 160 072 192;
  • 17) 0.036 985 811 160 072 192 × 2 = 0 + 0.073 971 622 320 144 384;
  • 18) 0.073 971 622 320 144 384 × 2 = 0 + 0.147 943 244 640 288 768;
  • 19) 0.147 943 244 640 288 768 × 2 = 0 + 0.295 886 489 280 577 536;
  • 20) 0.295 886 489 280 577 536 × 2 = 0 + 0.591 772 978 561 155 072;
  • 21) 0.591 772 978 561 155 072 × 2 = 1 + 0.183 545 957 122 310 144;
  • 22) 0.183 545 957 122 310 144 × 2 = 0 + 0.367 091 914 244 620 288;
  • 23) 0.367 091 914 244 620 288 × 2 = 0 + 0.734 183 828 489 240 576;
  • 24) 0.734 183 828 489 240 576 × 2 = 1 + 0.468 367 656 978 481 152;
  • 25) 0.468 367 656 978 481 152 × 2 = 0 + 0.936 735 313 956 962 304;
  • 26) 0.936 735 313 956 962 304 × 2 = 1 + 0.873 470 627 913 924 608;
  • 27) 0.873 470 627 913 924 608 × 2 = 1 + 0.746 941 255 827 849 216;
  • 28) 0.746 941 255 827 849 216 × 2 = 1 + 0.493 882 511 655 698 432;
  • 29) 0.493 882 511 655 698 432 × 2 = 0 + 0.987 765 023 311 396 864;
  • 30) 0.987 765 023 311 396 864 × 2 = 1 + 0.975 530 046 622 793 728;
  • 31) 0.975 530 046 622 793 728 × 2 = 1 + 0.951 060 093 245 587 456;
  • 32) 0.951 060 093 245 587 456 × 2 = 1 + 0.902 120 186 491 174 912;
  • 33) 0.902 120 186 491 174 912 × 2 = 1 + 0.804 240 372 982 349 824;
  • 34) 0.804 240 372 982 349 824 × 2 = 1 + 0.608 480 745 964 699 648;
  • 35) 0.608 480 745 964 699 648 × 2 = 1 + 0.216 961 491 929 399 296;
  • 36) 0.216 961 491 929 399 296 × 2 = 0 + 0.433 922 983 858 798 592;
  • 37) 0.433 922 983 858 798 592 × 2 = 0 + 0.867 845 967 717 597 184;
  • 38) 0.867 845 967 717 597 184 × 2 = 1 + 0.735 691 935 435 194 368;
  • 39) 0.735 691 935 435 194 368 × 2 = 1 + 0.471 383 870 870 388 736;
  • 40) 0.471 383 870 870 388 736 × 2 = 0 + 0.942 767 741 740 777 472;
  • 41) 0.942 767 741 740 777 472 × 2 = 1 + 0.885 535 483 481 554 944;
  • 42) 0.885 535 483 481 554 944 × 2 = 1 + 0.771 070 966 963 109 888;
  • 43) 0.771 070 966 963 109 888 × 2 = 1 + 0.542 141 933 926 219 776;
  • 44) 0.542 141 933 926 219 776 × 2 = 1 + 0.084 283 867 852 439 552;
  • 45) 0.084 283 867 852 439 552 × 2 = 0 + 0.168 567 735 704 879 104;
  • 46) 0.168 567 735 704 879 104 × 2 = 0 + 0.337 135 471 409 758 208;
  • 47) 0.337 135 471 409 758 208 × 2 = 0 + 0.674 270 942 819 516 416;
  • 48) 0.674 270 942 819 516 416 × 2 = 1 + 0.348 541 885 639 032 832;
  • 49) 0.348 541 885 639 032 832 × 2 = 0 + 0.697 083 771 278 065 664;
  • 50) 0.697 083 771 278 065 664 × 2 = 1 + 0.394 167 542 556 131 328;
  • 51) 0.394 167 542 556 131 328 × 2 = 0 + 0.788 335 085 112 262 656;
  • 52) 0.788 335 085 112 262 656 × 2 = 1 + 0.576 670 170 224 525 312;
  • 53) 0.576 670 170 224 525 312 × 2 = 1 + 0.153 340 340 449 050 624;
  • 54) 0.153 340 340 449 050 624 × 2 = 0 + 0.306 680 680 898 101 248;
  • 55) 0.306 680 680 898 101 248 × 2 = 0 + 0.613 361 361 796 202 496;
  • 56) 0.613 361 361 796 202 496 × 2 = 1 + 0.226 722 723 592 404 992;
  • 57) 0.226 722 723 592 404 992 × 2 = 0 + 0.453 445 447 184 809 984;
  • 58) 0.453 445 447 184 809 984 × 2 = 0 + 0.906 890 894 369 619 968;
  • 59) 0.906 890 894 369 619 968 × 2 = 1 + 0.813 781 788 739 239 936;
  • 60) 0.813 781 788 739 239 936 × 2 = 1 + 0.627 563 577 478 479 872;
  • 61) 0.627 563 577 478 479 872 × 2 = 1 + 0.255 127 154 956 959 744;
  • 62) 0.255 127 154 956 959 744 × 2 = 0 + 0.510 254 309 913 919 488;
  • 63) 0.510 254 309 913 919 488 × 2 = 1 + 0.020 508 619 827 838 976;
  • 64) 0.020 508 619 827 838 976 × 2 = 0 + 0.041 017 239 655 677 952;
  • 65) 0.041 017 239 655 677 952 × 2 = 0 + 0.082 034 479 311 355 904;
  • 66) 0.082 034 479 311 355 904 × 2 = 0 + 0.164 068 958 622 711 808;
  • 67) 0.164 068 958 622 711 808 × 2 = 0 + 0.328 137 917 245 423 616;
  • 68) 0.328 137 917 245 423 616 × 2 = 0 + 0.656 275 834 490 847 232;
  • 69) 0.656 275 834 490 847 232 × 2 = 1 + 0.312 551 668 981 694 464;
  • 70) 0.312 551 668 981 694 464 × 2 = 0 + 0.625 103 337 963 388 928;
  • 71) 0.625 103 337 963 388 928 × 2 = 1 + 0.250 206 675 926 777 856;
  • 72) 0.250 206 675 926 777 856 × 2 = 0 + 0.500 413 351 853 555 712;
  • 73) 0.500 413 351 853 555 712 × 2 = 1 + 0.000 826 703 707 111 424;

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:


0.000 000 564 358 690 797(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1001 0011 1010 0000 1010 1(2)

5. Positive number before normalization:

0.000 000 564 358 690 797(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1001 0011 1010 0000 1010 1(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:


0.000 000 564 358 690 797(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1001 0011 1010 0000 1010 1(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1001 0011 1010 0000 1010 1(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1011 0010 0111 0100 0001 0101(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 1011 0010 0111 0100 0001 0101


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 1011 0010 0111 0100 0001 0101 =

0010 1110 1111 1100 1101 1110 0010 1011 0010 0111 0100 0001 0101


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 1011 0010 0111 0100 0001 0101


Decimal number 0.000 000 564 358 690 797 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1011 0010 0111 0100 0001 0101

Share

» Convert decimal 0.000 000 564 358 690 715 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: Reputation: Bridge Between Company's Actions & Market Value. Gemini Notebook YouTube Video of 6.55 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