0.000 000 564 358 690 808 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 690 808(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
0.000 000 564 358 690 808(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: 0.000 000 564 358 690 808.

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) 0.000 000 564 358 690 808 × 2 = 0 + 0.000 001 128 717 381 616;
  • 2) 0.000 001 128 717 381 616 × 2 = 0 + 0.000 002 257 434 763 232;
  • 3) 0.000 002 257 434 763 232 × 2 = 0 + 0.000 004 514 869 526 464;
  • 4) 0.000 004 514 869 526 464 × 2 = 0 + 0.000 009 029 739 052 928;
  • 5) 0.000 009 029 739 052 928 × 2 = 0 + 0.000 018 059 478 105 856;
  • 6) 0.000 018 059 478 105 856 × 2 = 0 + 0.000 036 118 956 211 712;
  • 7) 0.000 036 118 956 211 712 × 2 = 0 + 0.000 072 237 912 423 424;
  • 8) 0.000 072 237 912 423 424 × 2 = 0 + 0.000 144 475 824 846 848;
  • 9) 0.000 144 475 824 846 848 × 2 = 0 + 0.000 288 951 649 693 696;
  • 10) 0.000 288 951 649 693 696 × 2 = 0 + 0.000 577 903 299 387 392;
  • 11) 0.000 577 903 299 387 392 × 2 = 0 + 0.001 155 806 598 774 784;
  • 12) 0.001 155 806 598 774 784 × 2 = 0 + 0.002 311 613 197 549 568;
  • 13) 0.002 311 613 197 549 568 × 2 = 0 + 0.004 623 226 395 099 136;
  • 14) 0.004 623 226 395 099 136 × 2 = 0 + 0.009 246 452 790 198 272;
  • 15) 0.009 246 452 790 198 272 × 2 = 0 + 0.018 492 905 580 396 544;
  • 16) 0.018 492 905 580 396 544 × 2 = 0 + 0.036 985 811 160 793 088;
  • 17) 0.036 985 811 160 793 088 × 2 = 0 + 0.073 971 622 321 586 176;
  • 18) 0.073 971 622 321 586 176 × 2 = 0 + 0.147 943 244 643 172 352;
  • 19) 0.147 943 244 643 172 352 × 2 = 0 + 0.295 886 489 286 344 704;
  • 20) 0.295 886 489 286 344 704 × 2 = 0 + 0.591 772 978 572 689 408;
  • 21) 0.591 772 978 572 689 408 × 2 = 1 + 0.183 545 957 145 378 816;
  • 22) 0.183 545 957 145 378 816 × 2 = 0 + 0.367 091 914 290 757 632;
  • 23) 0.367 091 914 290 757 632 × 2 = 0 + 0.734 183 828 581 515 264;
  • 24) 0.734 183 828 581 515 264 × 2 = 1 + 0.468 367 657 163 030 528;
  • 25) 0.468 367 657 163 030 528 × 2 = 0 + 0.936 735 314 326 061 056;
  • 26) 0.936 735 314 326 061 056 × 2 = 1 + 0.873 470 628 652 122 112;
  • 27) 0.873 470 628 652 122 112 × 2 = 1 + 0.746 941 257 304 244 224;
  • 28) 0.746 941 257 304 244 224 × 2 = 1 + 0.493 882 514 608 488 448;
  • 29) 0.493 882 514 608 488 448 × 2 = 0 + 0.987 765 029 216 976 896;
  • 30) 0.987 765 029 216 976 896 × 2 = 1 + 0.975 530 058 433 953 792;
  • 31) 0.975 530 058 433 953 792 × 2 = 1 + 0.951 060 116 867 907 584;
  • 32) 0.951 060 116 867 907 584 × 2 = 1 + 0.902 120 233 735 815 168;
  • 33) 0.902 120 233 735 815 168 × 2 = 1 + 0.804 240 467 471 630 336;
  • 34) 0.804 240 467 471 630 336 × 2 = 1 + 0.608 480 934 943 260 672;
  • 35) 0.608 480 934 943 260 672 × 2 = 1 + 0.216 961 869 886 521 344;
  • 36) 0.216 961 869 886 521 344 × 2 = 0 + 0.433 923 739 773 042 688;
  • 37) 0.433 923 739 773 042 688 × 2 = 0 + 0.867 847 479 546 085 376;
  • 38) 0.867 847 479 546 085 376 × 2 = 1 + 0.735 694 959 092 170 752;
  • 39) 0.735 694 959 092 170 752 × 2 = 1 + 0.471 389 918 184 341 504;
  • 40) 0.471 389 918 184 341 504 × 2 = 0 + 0.942 779 836 368 683 008;
  • 41) 0.942 779 836 368 683 008 × 2 = 1 + 0.885 559 672 737 366 016;
  • 42) 0.885 559 672 737 366 016 × 2 = 1 + 0.771 119 345 474 732 032;
  • 43) 0.771 119 345 474 732 032 × 2 = 1 + 0.542 238 690 949 464 064;
  • 44) 0.542 238 690 949 464 064 × 2 = 1 + 0.084 477 381 898 928 128;
  • 45) 0.084 477 381 898 928 128 × 2 = 0 + 0.168 954 763 797 856 256;
  • 46) 0.168 954 763 797 856 256 × 2 = 0 + 0.337 909 527 595 712 512;
  • 47) 0.337 909 527 595 712 512 × 2 = 0 + 0.675 819 055 191 425 024;
  • 48) 0.675 819 055 191 425 024 × 2 = 1 + 0.351 638 110 382 850 048;
  • 49) 0.351 638 110 382 850 048 × 2 = 0 + 0.703 276 220 765 700 096;
  • 50) 0.703 276 220 765 700 096 × 2 = 1 + 0.406 552 441 531 400 192;
  • 51) 0.406 552 441 531 400 192 × 2 = 0 + 0.813 104 883 062 800 384;
  • 52) 0.813 104 883 062 800 384 × 2 = 1 + 0.626 209 766 125 600 768;
  • 53) 0.626 209 766 125 600 768 × 2 = 1 + 0.252 419 532 251 201 536;
  • 54) 0.252 419 532 251 201 536 × 2 = 0 + 0.504 839 064 502 403 072;
  • 55) 0.504 839 064 502 403 072 × 2 = 1 + 0.009 678 129 004 806 144;
  • 56) 0.009 678 129 004 806 144 × 2 = 0 + 0.019 356 258 009 612 288;
  • 57) 0.019 356 258 009 612 288 × 2 = 0 + 0.038 712 516 019 224 576;
  • 58) 0.038 712 516 019 224 576 × 2 = 0 + 0.077 425 032 038 449 152;
  • 59) 0.077 425 032 038 449 152 × 2 = 0 + 0.154 850 064 076 898 304;
  • 60) 0.154 850 064 076 898 304 × 2 = 0 + 0.309 700 128 153 796 608;
  • 61) 0.309 700 128 153 796 608 × 2 = 0 + 0.619 400 256 307 593 216;
  • 62) 0.619 400 256 307 593 216 × 2 = 1 + 0.238 800 512 615 186 432;
  • 63) 0.238 800 512 615 186 432 × 2 = 0 + 0.477 601 025 230 372 864;
  • 64) 0.477 601 025 230 372 864 × 2 = 0 + 0.955 202 050 460 745 728;
  • 65) 0.955 202 050 460 745 728 × 2 = 1 + 0.910 404 100 921 491 456;
  • 66) 0.910 404 100 921 491 456 × 2 = 1 + 0.820 808 201 842 982 912;
  • 67) 0.820 808 201 842 982 912 × 2 = 1 + 0.641 616 403 685 965 824;
  • 68) 0.641 616 403 685 965 824 × 2 = 1 + 0.283 232 807 371 931 648;
  • 69) 0.283 232 807 371 931 648 × 2 = 0 + 0.566 465 614 743 863 296;
  • 70) 0.566 465 614 743 863 296 × 2 = 1 + 0.132 931 229 487 726 592;
  • 71) 0.132 931 229 487 726 592 × 2 = 0 + 0.265 862 458 975 453 184;
  • 72) 0.265 862 458 975 453 184 × 2 = 0 + 0.531 724 917 950 906 368;
  • 73) 0.531 724 917 950 906 368 × 2 = 1 + 0.063 449 835 901 812 736;

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:


0.000 000 564 358 690 808(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1010 0000 0100 1111 0100 1(2)

5. Positive number before normalization:

0.000 000 564 358 690 808(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1010 0000 0100 1111 0100 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:


0.000 000 564 358 690 808(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1010 0000 0100 1111 0100 1(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1010 0000 0100 1111 0100 1(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1011 0100 0000 1001 1110 1001(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 0100 0000 1001 1110 1001


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 0100 0000 1001 1110 1001 =

0010 1110 1111 1100 1101 1110 0010 1011 0100 0000 1001 1110 1001


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 0100 0000 1001 1110 1001


Decimal number 0.000 000 564 358 690 808 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1011 0100 0000 1001 1110 1001

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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