0.000 000 564 358 690 841 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 690 841(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
0.000 000 564 358 690 841(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: 0.000 000 564 358 690 841.

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) 0.000 000 564 358 690 841 × 2 = 0 + 0.000 001 128 717 381 682;
  • 2) 0.000 001 128 717 381 682 × 2 = 0 + 0.000 002 257 434 763 364;
  • 3) 0.000 002 257 434 763 364 × 2 = 0 + 0.000 004 514 869 526 728;
  • 4) 0.000 004 514 869 526 728 × 2 = 0 + 0.000 009 029 739 053 456;
  • 5) 0.000 009 029 739 053 456 × 2 = 0 + 0.000 018 059 478 106 912;
  • 6) 0.000 018 059 478 106 912 × 2 = 0 + 0.000 036 118 956 213 824;
  • 7) 0.000 036 118 956 213 824 × 2 = 0 + 0.000 072 237 912 427 648;
  • 8) 0.000 072 237 912 427 648 × 2 = 0 + 0.000 144 475 824 855 296;
  • 9) 0.000 144 475 824 855 296 × 2 = 0 + 0.000 288 951 649 710 592;
  • 10) 0.000 288 951 649 710 592 × 2 = 0 + 0.000 577 903 299 421 184;
  • 11) 0.000 577 903 299 421 184 × 2 = 0 + 0.001 155 806 598 842 368;
  • 12) 0.001 155 806 598 842 368 × 2 = 0 + 0.002 311 613 197 684 736;
  • 13) 0.002 311 613 197 684 736 × 2 = 0 + 0.004 623 226 395 369 472;
  • 14) 0.004 623 226 395 369 472 × 2 = 0 + 0.009 246 452 790 738 944;
  • 15) 0.009 246 452 790 738 944 × 2 = 0 + 0.018 492 905 581 477 888;
  • 16) 0.018 492 905 581 477 888 × 2 = 0 + 0.036 985 811 162 955 776;
  • 17) 0.036 985 811 162 955 776 × 2 = 0 + 0.073 971 622 325 911 552;
  • 18) 0.073 971 622 325 911 552 × 2 = 0 + 0.147 943 244 651 823 104;
  • 19) 0.147 943 244 651 823 104 × 2 = 0 + 0.295 886 489 303 646 208;
  • 20) 0.295 886 489 303 646 208 × 2 = 0 + 0.591 772 978 607 292 416;
  • 21) 0.591 772 978 607 292 416 × 2 = 1 + 0.183 545 957 214 584 832;
  • 22) 0.183 545 957 214 584 832 × 2 = 0 + 0.367 091 914 429 169 664;
  • 23) 0.367 091 914 429 169 664 × 2 = 0 + 0.734 183 828 858 339 328;
  • 24) 0.734 183 828 858 339 328 × 2 = 1 + 0.468 367 657 716 678 656;
  • 25) 0.468 367 657 716 678 656 × 2 = 0 + 0.936 735 315 433 357 312;
  • 26) 0.936 735 315 433 357 312 × 2 = 1 + 0.873 470 630 866 714 624;
  • 27) 0.873 470 630 866 714 624 × 2 = 1 + 0.746 941 261 733 429 248;
  • 28) 0.746 941 261 733 429 248 × 2 = 1 + 0.493 882 523 466 858 496;
  • 29) 0.493 882 523 466 858 496 × 2 = 0 + 0.987 765 046 933 716 992;
  • 30) 0.987 765 046 933 716 992 × 2 = 1 + 0.975 530 093 867 433 984;
  • 31) 0.975 530 093 867 433 984 × 2 = 1 + 0.951 060 187 734 867 968;
  • 32) 0.951 060 187 734 867 968 × 2 = 1 + 0.902 120 375 469 735 936;
  • 33) 0.902 120 375 469 735 936 × 2 = 1 + 0.804 240 750 939 471 872;
  • 34) 0.804 240 750 939 471 872 × 2 = 1 + 0.608 481 501 878 943 744;
  • 35) 0.608 481 501 878 943 744 × 2 = 1 + 0.216 963 003 757 887 488;
  • 36) 0.216 963 003 757 887 488 × 2 = 0 + 0.433 926 007 515 774 976;
  • 37) 0.433 926 007 515 774 976 × 2 = 0 + 0.867 852 015 031 549 952;
  • 38) 0.867 852 015 031 549 952 × 2 = 1 + 0.735 704 030 063 099 904;
  • 39) 0.735 704 030 063 099 904 × 2 = 1 + 0.471 408 060 126 199 808;
  • 40) 0.471 408 060 126 199 808 × 2 = 0 + 0.942 816 120 252 399 616;
  • 41) 0.942 816 120 252 399 616 × 2 = 1 + 0.885 632 240 504 799 232;
  • 42) 0.885 632 240 504 799 232 × 2 = 1 + 0.771 264 481 009 598 464;
  • 43) 0.771 264 481 009 598 464 × 2 = 1 + 0.542 528 962 019 196 928;
  • 44) 0.542 528 962 019 196 928 × 2 = 1 + 0.085 057 924 038 393 856;
  • 45) 0.085 057 924 038 393 856 × 2 = 0 + 0.170 115 848 076 787 712;
  • 46) 0.170 115 848 076 787 712 × 2 = 0 + 0.340 231 696 153 575 424;
  • 47) 0.340 231 696 153 575 424 × 2 = 0 + 0.680 463 392 307 150 848;
  • 48) 0.680 463 392 307 150 848 × 2 = 1 + 0.360 926 784 614 301 696;
  • 49) 0.360 926 784 614 301 696 × 2 = 0 + 0.721 853 569 228 603 392;
  • 50) 0.721 853 569 228 603 392 × 2 = 1 + 0.443 707 138 457 206 784;
  • 51) 0.443 707 138 457 206 784 × 2 = 0 + 0.887 414 276 914 413 568;
  • 52) 0.887 414 276 914 413 568 × 2 = 1 + 0.774 828 553 828 827 136;
  • 53) 0.774 828 553 828 827 136 × 2 = 1 + 0.549 657 107 657 654 272;
  • 54) 0.549 657 107 657 654 272 × 2 = 1 + 0.099 314 215 315 308 544;
  • 55) 0.099 314 215 315 308 544 × 2 = 0 + 0.198 628 430 630 617 088;
  • 56) 0.198 628 430 630 617 088 × 2 = 0 + 0.397 256 861 261 234 176;
  • 57) 0.397 256 861 261 234 176 × 2 = 0 + 0.794 513 722 522 468 352;
  • 58) 0.794 513 722 522 468 352 × 2 = 1 + 0.589 027 445 044 936 704;
  • 59) 0.589 027 445 044 936 704 × 2 = 1 + 0.178 054 890 089 873 408;
  • 60) 0.178 054 890 089 873 408 × 2 = 0 + 0.356 109 780 179 746 816;
  • 61) 0.356 109 780 179 746 816 × 2 = 0 + 0.712 219 560 359 493 632;
  • 62) 0.712 219 560 359 493 632 × 2 = 1 + 0.424 439 120 718 987 264;
  • 63) 0.424 439 120 718 987 264 × 2 = 0 + 0.848 878 241 437 974 528;
  • 64) 0.848 878 241 437 974 528 × 2 = 1 + 0.697 756 482 875 949 056;
  • 65) 0.697 756 482 875 949 056 × 2 = 1 + 0.395 512 965 751 898 112;
  • 66) 0.395 512 965 751 898 112 × 2 = 0 + 0.791 025 931 503 796 224;
  • 67) 0.791 025 931 503 796 224 × 2 = 1 + 0.582 051 863 007 592 448;
  • 68) 0.582 051 863 007 592 448 × 2 = 1 + 0.164 103 726 015 184 896;
  • 69) 0.164 103 726 015 184 896 × 2 = 0 + 0.328 207 452 030 369 792;
  • 70) 0.328 207 452 030 369 792 × 2 = 0 + 0.656 414 904 060 739 584;
  • 71) 0.656 414 904 060 739 584 × 2 = 1 + 0.312 829 808 121 479 168;
  • 72) 0.312 829 808 121 479 168 × 2 = 0 + 0.625 659 616 242 958 336;
  • 73) 0.625 659 616 242 958 336 × 2 = 1 + 0.251 319 232 485 916 672;

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:


0.000 000 564 358 690 841(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1100 0110 0101 1011 0010 1(2)

5. Positive number before normalization:

0.000 000 564 358 690 841(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1100 0110 0101 1011 0010 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:


0.000 000 564 358 690 841(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1100 0110 0101 1011 0010 1(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 1100 0110 0101 1011 0010 1(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1011 1000 1100 1011 0110 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 1000 1100 1011 0110 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 1000 1100 1011 0110 0101 =

0010 1110 1111 1100 1101 1110 0010 1011 1000 1100 1011 0110 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 1000 1100 1011 0110 0101


Decimal number 0.000 000 564 358 690 841 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1011 1000 1100 1011 0110 0101

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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