Convert 10.000 000 000 000 005 329 070 518 200 75 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

10.000 000 000 000 005 329 070 518 200 75(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (base 2) the integer part: 10.
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;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

10(10) =


1010(2)


3. Convert to the binary (base 2) the fractional part: 0.000 000 000 000 005 329 070 518 200 75.

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) 0.000 000 000 000 005 329 070 518 200 75 × 2 = 0 + 0.000 000 000 000 010 658 141 036 401 5;
  • 2) 0.000 000 000 000 010 658 141 036 401 5 × 2 = 0 + 0.000 000 000 000 021 316 282 072 803;
  • 3) 0.000 000 000 000 021 316 282 072 803 × 2 = 0 + 0.000 000 000 000 042 632 564 145 606;
  • 4) 0.000 000 000 000 042 632 564 145 606 × 2 = 0 + 0.000 000 000 000 085 265 128 291 212;
  • 5) 0.000 000 000 000 085 265 128 291 212 × 2 = 0 + 0.000 000 000 000 170 530 256 582 424;
  • 6) 0.000 000 000 000 170 530 256 582 424 × 2 = 0 + 0.000 000 000 000 341 060 513 164 848;
  • 7) 0.000 000 000 000 341 060 513 164 848 × 2 = 0 + 0.000 000 000 000 682 121 026 329 696;
  • 8) 0.000 000 000 000 682 121 026 329 696 × 2 = 0 + 0.000 000 000 001 364 242 052 659 392;
  • 9) 0.000 000 000 001 364 242 052 659 392 × 2 = 0 + 0.000 000 000 002 728 484 105 318 784;
  • 10) 0.000 000 000 002 728 484 105 318 784 × 2 = 0 + 0.000 000 000 005 456 968 210 637 568;
  • 11) 0.000 000 000 005 456 968 210 637 568 × 2 = 0 + 0.000 000 000 010 913 936 421 275 136;
  • 12) 0.000 000 000 010 913 936 421 275 136 × 2 = 0 + 0.000 000 000 021 827 872 842 550 272;
  • 13) 0.000 000 000 021 827 872 842 550 272 × 2 = 0 + 0.000 000 000 043 655 745 685 100 544;
  • 14) 0.000 000 000 043 655 745 685 100 544 × 2 = 0 + 0.000 000 000 087 311 491 370 201 088;
  • 15) 0.000 000 000 087 311 491 370 201 088 × 2 = 0 + 0.000 000 000 174 622 982 740 402 176;
  • 16) 0.000 000 000 174 622 982 740 402 176 × 2 = 0 + 0.000 000 000 349 245 965 480 804 352;
  • 17) 0.000 000 000 349 245 965 480 804 352 × 2 = 0 + 0.000 000 000 698 491 930 961 608 704;
  • 18) 0.000 000 000 698 491 930 961 608 704 × 2 = 0 + 0.000 000 001 396 983 861 923 217 408;
  • 19) 0.000 000 001 396 983 861 923 217 408 × 2 = 0 + 0.000 000 002 793 967 723 846 434 816;
  • 20) 0.000 000 002 793 967 723 846 434 816 × 2 = 0 + 0.000 000 005 587 935 447 692 869 632;
  • 21) 0.000 000 005 587 935 447 692 869 632 × 2 = 0 + 0.000 000 011 175 870 895 385 739 264;
  • 22) 0.000 000 011 175 870 895 385 739 264 × 2 = 0 + 0.000 000 022 351 741 790 771 478 528;
  • 23) 0.000 000 022 351 741 790 771 478 528 × 2 = 0 + 0.000 000 044 703 483 581 542 957 056;
  • 24) 0.000 000 044 703 483 581 542 957 056 × 2 = 0 + 0.000 000 089 406 967 163 085 914 112;
  • 25) 0.000 000 089 406 967 163 085 914 112 × 2 = 0 + 0.000 000 178 813 934 326 171 828 224;
  • 26) 0.000 000 178 813 934 326 171 828 224 × 2 = 0 + 0.000 000 357 627 868 652 343 656 448;
  • 27) 0.000 000 357 627 868 652 343 656 448 × 2 = 0 + 0.000 000 715 255 737 304 687 312 896;
  • 28) 0.000 000 715 255 737 304 687 312 896 × 2 = 0 + 0.000 001 430 511 474 609 374 625 792;
  • 29) 0.000 001 430 511 474 609 374 625 792 × 2 = 0 + 0.000 002 861 022 949 218 749 251 584;
  • 30) 0.000 002 861 022 949 218 749 251 584 × 2 = 0 + 0.000 005 722 045 898 437 498 503 168;
  • 31) 0.000 005 722 045 898 437 498 503 168 × 2 = 0 + 0.000 011 444 091 796 874 997 006 336;
  • 32) 0.000 011 444 091 796 874 997 006 336 × 2 = 0 + 0.000 022 888 183 593 749 994 012 672;
  • 33) 0.000 022 888 183 593 749 994 012 672 × 2 = 0 + 0.000 045 776 367 187 499 988 025 344;
  • 34) 0.000 045 776 367 187 499 988 025 344 × 2 = 0 + 0.000 091 552 734 374 999 976 050 688;
  • 35) 0.000 091 552 734 374 999 976 050 688 × 2 = 0 + 0.000 183 105 468 749 999 952 101 376;
  • 36) 0.000 183 105 468 749 999 952 101 376 × 2 = 0 + 0.000 366 210 937 499 999 904 202 752;
  • 37) 0.000 366 210 937 499 999 904 202 752 × 2 = 0 + 0.000 732 421 874 999 999 808 405 504;
  • 38) 0.000 732 421 874 999 999 808 405 504 × 2 = 0 + 0.001 464 843 749 999 999 616 811 008;
  • 39) 0.001 464 843 749 999 999 616 811 008 × 2 = 0 + 0.002 929 687 499 999 999 233 622 016;
  • 40) 0.002 929 687 499 999 999 233 622 016 × 2 = 0 + 0.005 859 374 999 999 998 467 244 032;
  • 41) 0.005 859 374 999 999 998 467 244 032 × 2 = 0 + 0.011 718 749 999 999 996 934 488 064;
  • 42) 0.011 718 749 999 999 996 934 488 064 × 2 = 0 + 0.023 437 499 999 999 993 868 976 128;
  • 43) 0.023 437 499 999 999 993 868 976 128 × 2 = 0 + 0.046 874 999 999 999 987 737 952 256;
  • 44) 0.046 874 999 999 999 987 737 952 256 × 2 = 0 + 0.093 749 999 999 999 975 475 904 512;
  • 45) 0.093 749 999 999 999 975 475 904 512 × 2 = 0 + 0.187 499 999 999 999 950 951 809 024;
  • 46) 0.187 499 999 999 999 950 951 809 024 × 2 = 0 + 0.374 999 999 999 999 901 903 618 048;
  • 47) 0.374 999 999 999 999 901 903 618 048 × 2 = 0 + 0.749 999 999 999 999 803 807 236 096;
  • 48) 0.749 999 999 999 999 803 807 236 096 × 2 = 1 + 0.499 999 999 999 999 607 614 472 192;
  • 49) 0.499 999 999 999 999 607 614 472 192 × 2 = 0 + 0.999 999 999 999 999 215 228 944 384;
  • 50) 0.999 999 999 999 999 215 228 944 384 × 2 = 1 + 0.999 999 999 999 998 430 457 888 768;
  • 51) 0.999 999 999 999 998 430 457 888 768 × 2 = 1 + 0.999 999 999 999 996 860 915 777 536;
  • 52) 0.999 999 999 999 996 860 915 777 536 × 2 = 1 + 0.999 999 999 999 993 721 831 555 072;
  • 53) 0.999 999 999 999 993 721 831 555 072 × 2 = 1 + 0.999 999 999 999 987 443 663 110 144;

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...)


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:

0.000 000 000 000 005 329 070 518 200 75(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0111 1(2)


5. Positive number before normalization:

10.000 000 000 000 005 329 070 518 200 75(10) =


1010.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0111 1(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left so that only one non zero digit remains to the left of it:

10.000 000 000 000 005 329 070 518 200 75(10) =


1010.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0111 1(2) =


1010.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0111 1(2) × 20 =


1.0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1111(2) × 23


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): 3


Mantissa (not normalized):
1.0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


3 + 2(11-1) - 1 =


(3 + 1 023)(10) =


1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1026(10) =


100 0000 0010(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1111 =


0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010


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) =
100 0000 0010


Mantissa (52 bits) =
0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010


Number 10.000 000 000 000 005 329 070 518 200 75 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0010 - 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 1

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 0

      37
    • 0

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 0

      32
    • 0

      31
    • 0

      30
    • 0

      29
    • 0

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 1

      1
    • 0

      0

More operations of this kind:

10.000 000 000 000 005 329 070 518 200 74 = ? ... 10.000 000 000 000 005 329 070 518 200 76 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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