0.000 569 761 701 604 447 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 569 761 701 604 447(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
0.000 569 761 701 604 447(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: 0.000 569 761 701 604 447.

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) 0.000 569 761 701 604 447 × 2 = 0 + 0.001 139 523 403 208 894;
  • 2) 0.001 139 523 403 208 894 × 2 = 0 + 0.002 279 046 806 417 788;
  • 3) 0.002 279 046 806 417 788 × 2 = 0 + 0.004 558 093 612 835 576;
  • 4) 0.004 558 093 612 835 576 × 2 = 0 + 0.009 116 187 225 671 152;
  • 5) 0.009 116 187 225 671 152 × 2 = 0 + 0.018 232 374 451 342 304;
  • 6) 0.018 232 374 451 342 304 × 2 = 0 + 0.036 464 748 902 684 608;
  • 7) 0.036 464 748 902 684 608 × 2 = 0 + 0.072 929 497 805 369 216;
  • 8) 0.072 929 497 805 369 216 × 2 = 0 + 0.145 858 995 610 738 432;
  • 9) 0.145 858 995 610 738 432 × 2 = 0 + 0.291 717 991 221 476 864;
  • 10) 0.291 717 991 221 476 864 × 2 = 0 + 0.583 435 982 442 953 728;
  • 11) 0.583 435 982 442 953 728 × 2 = 1 + 0.166 871 964 885 907 456;
  • 12) 0.166 871 964 885 907 456 × 2 = 0 + 0.333 743 929 771 814 912;
  • 13) 0.333 743 929 771 814 912 × 2 = 0 + 0.667 487 859 543 629 824;
  • 14) 0.667 487 859 543 629 824 × 2 = 1 + 0.334 975 719 087 259 648;
  • 15) 0.334 975 719 087 259 648 × 2 = 0 + 0.669 951 438 174 519 296;
  • 16) 0.669 951 438 174 519 296 × 2 = 1 + 0.339 902 876 349 038 592;
  • 17) 0.339 902 876 349 038 592 × 2 = 0 + 0.679 805 752 698 077 184;
  • 18) 0.679 805 752 698 077 184 × 2 = 1 + 0.359 611 505 396 154 368;
  • 19) 0.359 611 505 396 154 368 × 2 = 0 + 0.719 223 010 792 308 736;
  • 20) 0.719 223 010 792 308 736 × 2 = 1 + 0.438 446 021 584 617 472;
  • 21) 0.438 446 021 584 617 472 × 2 = 0 + 0.876 892 043 169 234 944;
  • 22) 0.876 892 043 169 234 944 × 2 = 1 + 0.753 784 086 338 469 888;
  • 23) 0.753 784 086 338 469 888 × 2 = 1 + 0.507 568 172 676 939 776;
  • 24) 0.507 568 172 676 939 776 × 2 = 1 + 0.015 136 345 353 879 552;
  • 25) 0.015 136 345 353 879 552 × 2 = 0 + 0.030 272 690 707 759 104;
  • 26) 0.030 272 690 707 759 104 × 2 = 0 + 0.060 545 381 415 518 208;
  • 27) 0.060 545 381 415 518 208 × 2 = 0 + 0.121 090 762 831 036 416;
  • 28) 0.121 090 762 831 036 416 × 2 = 0 + 0.242 181 525 662 072 832;
  • 29) 0.242 181 525 662 072 832 × 2 = 0 + 0.484 363 051 324 145 664;
  • 30) 0.484 363 051 324 145 664 × 2 = 0 + 0.968 726 102 648 291 328;
  • 31) 0.968 726 102 648 291 328 × 2 = 1 + 0.937 452 205 296 582 656;
  • 32) 0.937 452 205 296 582 656 × 2 = 1 + 0.874 904 410 593 165 312;
  • 33) 0.874 904 410 593 165 312 × 2 = 1 + 0.749 808 821 186 330 624;
  • 34) 0.749 808 821 186 330 624 × 2 = 1 + 0.499 617 642 372 661 248;
  • 35) 0.499 617 642 372 661 248 × 2 = 0 + 0.999 235 284 745 322 496;
  • 36) 0.999 235 284 745 322 496 × 2 = 1 + 0.998 470 569 490 644 992;
  • 37) 0.998 470 569 490 644 992 × 2 = 1 + 0.996 941 138 981 289 984;
  • 38) 0.996 941 138 981 289 984 × 2 = 1 + 0.993 882 277 962 579 968;
  • 39) 0.993 882 277 962 579 968 × 2 = 1 + 0.987 764 555 925 159 936;
  • 40) 0.987 764 555 925 159 936 × 2 = 1 + 0.975 529 111 850 319 872;
  • 41) 0.975 529 111 850 319 872 × 2 = 1 + 0.951 058 223 700 639 744;
  • 42) 0.951 058 223 700 639 744 × 2 = 1 + 0.902 116 447 401 279 488;
  • 43) 0.902 116 447 401 279 488 × 2 = 1 + 0.804 232 894 802 558 976;
  • 44) 0.804 232 894 802 558 976 × 2 = 1 + 0.608 465 789 605 117 952;
  • 45) 0.608 465 789 605 117 952 × 2 = 1 + 0.216 931 579 210 235 904;
  • 46) 0.216 931 579 210 235 904 × 2 = 0 + 0.433 863 158 420 471 808;
  • 47) 0.433 863 158 420 471 808 × 2 = 0 + 0.867 726 316 840 943 616;
  • 48) 0.867 726 316 840 943 616 × 2 = 1 + 0.735 452 633 681 887 232;
  • 49) 0.735 452 633 681 887 232 × 2 = 1 + 0.470 905 267 363 774 464;
  • 50) 0.470 905 267 363 774 464 × 2 = 0 + 0.941 810 534 727 548 928;
  • 51) 0.941 810 534 727 548 928 × 2 = 1 + 0.883 621 069 455 097 856;
  • 52) 0.883 621 069 455 097 856 × 2 = 1 + 0.767 242 138 910 195 712;
  • 53) 0.767 242 138 910 195 712 × 2 = 1 + 0.534 484 277 820 391 424;
  • 54) 0.534 484 277 820 391 424 × 2 = 1 + 0.068 968 555 640 782 848;
  • 55) 0.068 968 555 640 782 848 × 2 = 0 + 0.137 937 111 281 565 696;
  • 56) 0.137 937 111 281 565 696 × 2 = 0 + 0.275 874 222 563 131 392;
  • 57) 0.275 874 222 563 131 392 × 2 = 0 + 0.551 748 445 126 262 784;
  • 58) 0.551 748 445 126 262 784 × 2 = 1 + 0.103 496 890 252 525 568;
  • 59) 0.103 496 890 252 525 568 × 2 = 0 + 0.206 993 780 505 051 136;
  • 60) 0.206 993 780 505 051 136 × 2 = 0 + 0.413 987 561 010 102 272;
  • 61) 0.413 987 561 010 102 272 × 2 = 0 + 0.827 975 122 020 204 544;
  • 62) 0.827 975 122 020 204 544 × 2 = 1 + 0.655 950 244 040 409 088;
  • 63) 0.655 950 244 040 409 088 × 2 = 1 + 0.311 900 488 080 818 176;

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:


0.000 569 761 701 604 447(10) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1100 0100 011(2)

5. Positive number before normalization:

0.000 569 761 701 604 447(10) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1100 0100 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 569 761 701 604 447(10) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1100 0100 011(2) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1100 0100 011(2) × 20 =

1.0010 1010 1011 1000 0001 1110 1111 1111 1100 1101 1110 0010 0011(2) × 2-11


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): -11

Mantissa (not normalized):
1.0010 1010 1011 1000 0001 1110 1111 1111 1100 1101 1110 0010 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-11 + 1 023)(10) =

1 012(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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1012(10) =

011 1111 0100(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 1010 1011 1000 0001 1110 1111 1111 1100 1101 1110 0010 0011 =

0010 1010 1011 1000 0001 1110 1111 1111 1100 1101 1110 0010 0011


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 0100

Mantissa (52 bits) =
0010 1010 1011 1000 0001 1110 1111 1111 1100 1101 1110 0010 0011


Decimal number 0.000 569 761 701 604 447 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0100 - 0010 1010 1011 1000 0001 1110 1111 1111 1100 1101 1110 0010 0011

Share

» Convert decimal 0.000 569 761 701 604 429 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 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - 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