0.000 569 761 701 604 388 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 569 761 701 604 388(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
0.000 569 761 701 604 388(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: 0.000 569 761 701 604 388.

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) 0.000 569 761 701 604 388 × 2 = 0 + 0.001 139 523 403 208 776;
  • 2) 0.001 139 523 403 208 776 × 2 = 0 + 0.002 279 046 806 417 552;
  • 3) 0.002 279 046 806 417 552 × 2 = 0 + 0.004 558 093 612 835 104;
  • 4) 0.004 558 093 612 835 104 × 2 = 0 + 0.009 116 187 225 670 208;
  • 5) 0.009 116 187 225 670 208 × 2 = 0 + 0.018 232 374 451 340 416;
  • 6) 0.018 232 374 451 340 416 × 2 = 0 + 0.036 464 748 902 680 832;
  • 7) 0.036 464 748 902 680 832 × 2 = 0 + 0.072 929 497 805 361 664;
  • 8) 0.072 929 497 805 361 664 × 2 = 0 + 0.145 858 995 610 723 328;
  • 9) 0.145 858 995 610 723 328 × 2 = 0 + 0.291 717 991 221 446 656;
  • 10) 0.291 717 991 221 446 656 × 2 = 0 + 0.583 435 982 442 893 312;
  • 11) 0.583 435 982 442 893 312 × 2 = 1 + 0.166 871 964 885 786 624;
  • 12) 0.166 871 964 885 786 624 × 2 = 0 + 0.333 743 929 771 573 248;
  • 13) 0.333 743 929 771 573 248 × 2 = 0 + 0.667 487 859 543 146 496;
  • 14) 0.667 487 859 543 146 496 × 2 = 1 + 0.334 975 719 086 292 992;
  • 15) 0.334 975 719 086 292 992 × 2 = 0 + 0.669 951 438 172 585 984;
  • 16) 0.669 951 438 172 585 984 × 2 = 1 + 0.339 902 876 345 171 968;
  • 17) 0.339 902 876 345 171 968 × 2 = 0 + 0.679 805 752 690 343 936;
  • 18) 0.679 805 752 690 343 936 × 2 = 1 + 0.359 611 505 380 687 872;
  • 19) 0.359 611 505 380 687 872 × 2 = 0 + 0.719 223 010 761 375 744;
  • 20) 0.719 223 010 761 375 744 × 2 = 1 + 0.438 446 021 522 751 488;
  • 21) 0.438 446 021 522 751 488 × 2 = 0 + 0.876 892 043 045 502 976;
  • 22) 0.876 892 043 045 502 976 × 2 = 1 + 0.753 784 086 091 005 952;
  • 23) 0.753 784 086 091 005 952 × 2 = 1 + 0.507 568 172 182 011 904;
  • 24) 0.507 568 172 182 011 904 × 2 = 1 + 0.015 136 344 364 023 808;
  • 25) 0.015 136 344 364 023 808 × 2 = 0 + 0.030 272 688 728 047 616;
  • 26) 0.030 272 688 728 047 616 × 2 = 0 + 0.060 545 377 456 095 232;
  • 27) 0.060 545 377 456 095 232 × 2 = 0 + 0.121 090 754 912 190 464;
  • 28) 0.121 090 754 912 190 464 × 2 = 0 + 0.242 181 509 824 380 928;
  • 29) 0.242 181 509 824 380 928 × 2 = 0 + 0.484 363 019 648 761 856;
  • 30) 0.484 363 019 648 761 856 × 2 = 0 + 0.968 726 039 297 523 712;
  • 31) 0.968 726 039 297 523 712 × 2 = 1 + 0.937 452 078 595 047 424;
  • 32) 0.937 452 078 595 047 424 × 2 = 1 + 0.874 904 157 190 094 848;
  • 33) 0.874 904 157 190 094 848 × 2 = 1 + 0.749 808 314 380 189 696;
  • 34) 0.749 808 314 380 189 696 × 2 = 1 + 0.499 616 628 760 379 392;
  • 35) 0.499 616 628 760 379 392 × 2 = 0 + 0.999 233 257 520 758 784;
  • 36) 0.999 233 257 520 758 784 × 2 = 1 + 0.998 466 515 041 517 568;
  • 37) 0.998 466 515 041 517 568 × 2 = 1 + 0.996 933 030 083 035 136;
  • 38) 0.996 933 030 083 035 136 × 2 = 1 + 0.993 866 060 166 070 272;
  • 39) 0.993 866 060 166 070 272 × 2 = 1 + 0.987 732 120 332 140 544;
  • 40) 0.987 732 120 332 140 544 × 2 = 1 + 0.975 464 240 664 281 088;
  • 41) 0.975 464 240 664 281 088 × 2 = 1 + 0.950 928 481 328 562 176;
  • 42) 0.950 928 481 328 562 176 × 2 = 1 + 0.901 856 962 657 124 352;
  • 43) 0.901 856 962 657 124 352 × 2 = 1 + 0.803 713 925 314 248 704;
  • 44) 0.803 713 925 314 248 704 × 2 = 1 + 0.607 427 850 628 497 408;
  • 45) 0.607 427 850 628 497 408 × 2 = 1 + 0.214 855 701 256 994 816;
  • 46) 0.214 855 701 256 994 816 × 2 = 0 + 0.429 711 402 513 989 632;
  • 47) 0.429 711 402 513 989 632 × 2 = 0 + 0.859 422 805 027 979 264;
  • 48) 0.859 422 805 027 979 264 × 2 = 1 + 0.718 845 610 055 958 528;
  • 49) 0.718 845 610 055 958 528 × 2 = 1 + 0.437 691 220 111 917 056;
  • 50) 0.437 691 220 111 917 056 × 2 = 0 + 0.875 382 440 223 834 112;
  • 51) 0.875 382 440 223 834 112 × 2 = 1 + 0.750 764 880 447 668 224;
  • 52) 0.750 764 880 447 668 224 × 2 = 1 + 0.501 529 760 895 336 448;
  • 53) 0.501 529 760 895 336 448 × 2 = 1 + 0.003 059 521 790 672 896;
  • 54) 0.003 059 521 790 672 896 × 2 = 0 + 0.006 119 043 581 345 792;
  • 55) 0.006 119 043 581 345 792 × 2 = 0 + 0.012 238 087 162 691 584;
  • 56) 0.012 238 087 162 691 584 × 2 = 0 + 0.024 476 174 325 383 168;
  • 57) 0.024 476 174 325 383 168 × 2 = 0 + 0.048 952 348 650 766 336;
  • 58) 0.048 952 348 650 766 336 × 2 = 0 + 0.097 904 697 301 532 672;
  • 59) 0.097 904 697 301 532 672 × 2 = 0 + 0.195 809 394 603 065 344;
  • 60) 0.195 809 394 603 065 344 × 2 = 0 + 0.391 618 789 206 130 688;
  • 61) 0.391 618 789 206 130 688 × 2 = 0 + 0.783 237 578 412 261 376;
  • 62) 0.783 237 578 412 261 376 × 2 = 1 + 0.566 475 156 824 522 752;
  • 63) 0.566 475 156 824 522 752 × 2 = 1 + 0.132 950 313 649 045 504;

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:


0.000 569 761 701 604 388(10) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1000 0000 011(2)

5. Positive number before normalization:

0.000 569 761 701 604 388(10) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1000 0000 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:


0.000 569 761 701 604 388(10) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1000 0000 011(2) =

0.0000 0000 0010 0101 0101 0111 0000 0011 1101 1111 1111 1001 1011 1000 0000 011(2) × 20 =

1.0010 1010 1011 1000 0001 1110 1111 1111 1100 1101 1100 0000 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 1100 0000 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 1100 0000 0011 =

0010 1010 1011 1000 0001 1110 1111 1111 1100 1101 1100 0000 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 1100 0000 0011


Decimal number 0.000 569 761 701 604 388 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 1100 0000 0011

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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