0.000 569 761 701 604 429 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 569 761 701 604 429(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
0.000 569 761 701 604 429(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: 0.000 569 761 701 604 429.

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) 0.000 569 761 701 604 429 × 2 = 0 + 0.001 139 523 403 208 858;
  • 2) 0.001 139 523 403 208 858 × 2 = 0 + 0.002 279 046 806 417 716;
  • 3) 0.002 279 046 806 417 716 × 2 = 0 + 0.004 558 093 612 835 432;
  • 4) 0.004 558 093 612 835 432 × 2 = 0 + 0.009 116 187 225 670 864;
  • 5) 0.009 116 187 225 670 864 × 2 = 0 + 0.018 232 374 451 341 728;
  • 6) 0.018 232 374 451 341 728 × 2 = 0 + 0.036 464 748 902 683 456;
  • 7) 0.036 464 748 902 683 456 × 2 = 0 + 0.072 929 497 805 366 912;
  • 8) 0.072 929 497 805 366 912 × 2 = 0 + 0.145 858 995 610 733 824;
  • 9) 0.145 858 995 610 733 824 × 2 = 0 + 0.291 717 991 221 467 648;
  • 10) 0.291 717 991 221 467 648 × 2 = 0 + 0.583 435 982 442 935 296;
  • 11) 0.583 435 982 442 935 296 × 2 = 1 + 0.166 871 964 885 870 592;
  • 12) 0.166 871 964 885 870 592 × 2 = 0 + 0.333 743 929 771 741 184;
  • 13) 0.333 743 929 771 741 184 × 2 = 0 + 0.667 487 859 543 482 368;
  • 14) 0.667 487 859 543 482 368 × 2 = 1 + 0.334 975 719 086 964 736;
  • 15) 0.334 975 719 086 964 736 × 2 = 0 + 0.669 951 438 173 929 472;
  • 16) 0.669 951 438 173 929 472 × 2 = 1 + 0.339 902 876 347 858 944;
  • 17) 0.339 902 876 347 858 944 × 2 = 0 + 0.679 805 752 695 717 888;
  • 18) 0.679 805 752 695 717 888 × 2 = 1 + 0.359 611 505 391 435 776;
  • 19) 0.359 611 505 391 435 776 × 2 = 0 + 0.719 223 010 782 871 552;
  • 20) 0.719 223 010 782 871 552 × 2 = 1 + 0.438 446 021 565 743 104;
  • 21) 0.438 446 021 565 743 104 × 2 = 0 + 0.876 892 043 131 486 208;
  • 22) 0.876 892 043 131 486 208 × 2 = 1 + 0.753 784 086 262 972 416;
  • 23) 0.753 784 086 262 972 416 × 2 = 1 + 0.507 568 172 525 944 832;
  • 24) 0.507 568 172 525 944 832 × 2 = 1 + 0.015 136 345 051 889 664;
  • 25) 0.015 136 345 051 889 664 × 2 = 0 + 0.030 272 690 103 779 328;
  • 26) 0.030 272 690 103 779 328 × 2 = 0 + 0.060 545 380 207 558 656;
  • 27) 0.060 545 380 207 558 656 × 2 = 0 + 0.121 090 760 415 117 312;
  • 28) 0.121 090 760 415 117 312 × 2 = 0 + 0.242 181 520 830 234 624;
  • 29) 0.242 181 520 830 234 624 × 2 = 0 + 0.484 363 041 660 469 248;
  • 30) 0.484 363 041 660 469 248 × 2 = 0 + 0.968 726 083 320 938 496;
  • 31) 0.968 726 083 320 938 496 × 2 = 1 + 0.937 452 166 641 876 992;
  • 32) 0.937 452 166 641 876 992 × 2 = 1 + 0.874 904 333 283 753 984;
  • 33) 0.874 904 333 283 753 984 × 2 = 1 + 0.749 808 666 567 507 968;
  • 34) 0.749 808 666 567 507 968 × 2 = 1 + 0.499 617 333 135 015 936;
  • 35) 0.499 617 333 135 015 936 × 2 = 0 + 0.999 234 666 270 031 872;
  • 36) 0.999 234 666 270 031 872 × 2 = 1 + 0.998 469 332 540 063 744;
  • 37) 0.998 469 332 540 063 744 × 2 = 1 + 0.996 938 665 080 127 488;
  • 38) 0.996 938 665 080 127 488 × 2 = 1 + 0.993 877 330 160 254 976;
  • 39) 0.993 877 330 160 254 976 × 2 = 1 + 0.987 754 660 320 509 952;
  • 40) 0.987 754 660 320 509 952 × 2 = 1 + 0.975 509 320 641 019 904;
  • 41) 0.975 509 320 641 019 904 × 2 = 1 + 0.951 018 641 282 039 808;
  • 42) 0.951 018 641 282 039 808 × 2 = 1 + 0.902 037 282 564 079 616;
  • 43) 0.902 037 282 564 079 616 × 2 = 1 + 0.804 074 565 128 159 232;
  • 44) 0.804 074 565 128 159 232 × 2 = 1 + 0.608 149 130 256 318 464;
  • 45) 0.608 149 130 256 318 464 × 2 = 1 + 0.216 298 260 512 636 928;
  • 46) 0.216 298 260 512 636 928 × 2 = 0 + 0.432 596 521 025 273 856;
  • 47) 0.432 596 521 025 273 856 × 2 = 0 + 0.865 193 042 050 547 712;
  • 48) 0.865 193 042 050 547 712 × 2 = 1 + 0.730 386 084 101 095 424;
  • 49) 0.730 386 084 101 095 424 × 2 = 1 + 0.460 772 168 202 190 848;
  • 50) 0.460 772 168 202 190 848 × 2 = 0 + 0.921 544 336 404 381 696;
  • 51) 0.921 544 336 404 381 696 × 2 = 1 + 0.843 088 672 808 763 392;
  • 52) 0.843 088 672 808 763 392 × 2 = 1 + 0.686 177 345 617 526 784;
  • 53) 0.686 177 345 617 526 784 × 2 = 1 + 0.372 354 691 235 053 568;
  • 54) 0.372 354 691 235 053 568 × 2 = 0 + 0.744 709 382 470 107 136;
  • 55) 0.744 709 382 470 107 136 × 2 = 1 + 0.489 418 764 940 214 272;
  • 56) 0.489 418 764 940 214 272 × 2 = 0 + 0.978 837 529 880 428 544;
  • 57) 0.978 837 529 880 428 544 × 2 = 1 + 0.957 675 059 760 857 088;
  • 58) 0.957 675 059 760 857 088 × 2 = 1 + 0.915 350 119 521 714 176;
  • 59) 0.915 350 119 521 714 176 × 2 = 1 + 0.830 700 239 043 428 352;
  • 60) 0.830 700 239 043 428 352 × 2 = 1 + 0.661 400 478 086 856 704;
  • 61) 0.661 400 478 086 856 704 × 2 = 1 + 0.322 800 956 173 713 408;
  • 62) 0.322 800 956 173 713 408 × 2 = 0 + 0.645 601 912 347 426 816;
  • 63) 0.645 601 912 347 426 816 × 2 = 1 + 0.291 203 824 694 853 632;

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:


0.000 569 761 701 604 429(10) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1010 1111 101(2)

5. Positive number before normalization:

0.000 569 761 701 604 429(10) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1010 1111 101(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:


0.000 569 761 701 604 429(10) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1010 1111 101(2) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1010 1111 101(2) × 20 =

1.0010 1010 1011 1000 0001 1110 1111 1111 1100 1101 1101 0111 1101(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 1101 0111 1101


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 1101 0111 1101 =

0010 1010 1011 1000 0001 1110 1111 1111 1100 1101 1101 0111 1101


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 1101 0111 1101


Decimal number 0.000 569 761 701 604 429 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 1101 0111 1101

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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