0.000 003 814 697 265 604 29 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 814 697 265 604 29(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
0.000 003 814 697 265 604 29(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: 0.000 003 814 697 265 604 29.

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) 0.000 003 814 697 265 604 29 × 2 = 0 + 0.000 007 629 394 531 208 58;
  • 2) 0.000 007 629 394 531 208 58 × 2 = 0 + 0.000 015 258 789 062 417 16;
  • 3) 0.000 015 258 789 062 417 16 × 2 = 0 + 0.000 030 517 578 124 834 32;
  • 4) 0.000 030 517 578 124 834 32 × 2 = 0 + 0.000 061 035 156 249 668 64;
  • 5) 0.000 061 035 156 249 668 64 × 2 = 0 + 0.000 122 070 312 499 337 28;
  • 6) 0.000 122 070 312 499 337 28 × 2 = 0 + 0.000 244 140 624 998 674 56;
  • 7) 0.000 244 140 624 998 674 56 × 2 = 0 + 0.000 488 281 249 997 349 12;
  • 8) 0.000 488 281 249 997 349 12 × 2 = 0 + 0.000 976 562 499 994 698 24;
  • 9) 0.000 976 562 499 994 698 24 × 2 = 0 + 0.001 953 124 999 989 396 48;
  • 10) 0.001 953 124 999 989 396 48 × 2 = 0 + 0.003 906 249 999 978 792 96;
  • 11) 0.003 906 249 999 978 792 96 × 2 = 0 + 0.007 812 499 999 957 585 92;
  • 12) 0.007 812 499 999 957 585 92 × 2 = 0 + 0.015 624 999 999 915 171 84;
  • 13) 0.015 624 999 999 915 171 84 × 2 = 0 + 0.031 249 999 999 830 343 68;
  • 14) 0.031 249 999 999 830 343 68 × 2 = 0 + 0.062 499 999 999 660 687 36;
  • 15) 0.062 499 999 999 660 687 36 × 2 = 0 + 0.124 999 999 999 321 374 72;
  • 16) 0.124 999 999 999 321 374 72 × 2 = 0 + 0.249 999 999 998 642 749 44;
  • 17) 0.249 999 999 998 642 749 44 × 2 = 0 + 0.499 999 999 997 285 498 88;
  • 18) 0.499 999 999 997 285 498 88 × 2 = 0 + 0.999 999 999 994 570 997 76;
  • 19) 0.999 999 999 994 570 997 76 × 2 = 1 + 0.999 999 999 989 141 995 52;
  • 20) 0.999 999 999 989 141 995 52 × 2 = 1 + 0.999 999 999 978 283 991 04;
  • 21) 0.999 999 999 978 283 991 04 × 2 = 1 + 0.999 999 999 956 567 982 08;
  • 22) 0.999 999 999 956 567 982 08 × 2 = 1 + 0.999 999 999 913 135 964 16;
  • 23) 0.999 999 999 913 135 964 16 × 2 = 1 + 0.999 999 999 826 271 928 32;
  • 24) 0.999 999 999 826 271 928 32 × 2 = 1 + 0.999 999 999 652 543 856 64;
  • 25) 0.999 999 999 652 543 856 64 × 2 = 1 + 0.999 999 999 305 087 713 28;
  • 26) 0.999 999 999 305 087 713 28 × 2 = 1 + 0.999 999 998 610 175 426 56;
  • 27) 0.999 999 998 610 175 426 56 × 2 = 1 + 0.999 999 997 220 350 853 12;
  • 28) 0.999 999 997 220 350 853 12 × 2 = 1 + 0.999 999 994 440 701 706 24;
  • 29) 0.999 999 994 440 701 706 24 × 2 = 1 + 0.999 999 988 881 403 412 48;
  • 30) 0.999 999 988 881 403 412 48 × 2 = 1 + 0.999 999 977 762 806 824 96;
  • 31) 0.999 999 977 762 806 824 96 × 2 = 1 + 0.999 999 955 525 613 649 92;
  • 32) 0.999 999 955 525 613 649 92 × 2 = 1 + 0.999 999 911 051 227 299 84;
  • 33) 0.999 999 911 051 227 299 84 × 2 = 1 + 0.999 999 822 102 454 599 68;
  • 34) 0.999 999 822 102 454 599 68 × 2 = 1 + 0.999 999 644 204 909 199 36;
  • 35) 0.999 999 644 204 909 199 36 × 2 = 1 + 0.999 999 288 409 818 398 72;
  • 36) 0.999 999 288 409 818 398 72 × 2 = 1 + 0.999 998 576 819 636 797 44;
  • 37) 0.999 998 576 819 636 797 44 × 2 = 1 + 0.999 997 153 639 273 594 88;
  • 38) 0.999 997 153 639 273 594 88 × 2 = 1 + 0.999 994 307 278 547 189 76;
  • 39) 0.999 994 307 278 547 189 76 × 2 = 1 + 0.999 988 614 557 094 379 52;
  • 40) 0.999 988 614 557 094 379 52 × 2 = 1 + 0.999 977 229 114 188 759 04;
  • 41) 0.999 977 229 114 188 759 04 × 2 = 1 + 0.999 954 458 228 377 518 08;
  • 42) 0.999 954 458 228 377 518 08 × 2 = 1 + 0.999 908 916 456 755 036 16;
  • 43) 0.999 908 916 456 755 036 16 × 2 = 1 + 0.999 817 832 913 510 072 32;
  • 44) 0.999 817 832 913 510 072 32 × 2 = 1 + 0.999 635 665 827 020 144 64;
  • 45) 0.999 635 665 827 020 144 64 × 2 = 1 + 0.999 271 331 654 040 289 28;
  • 46) 0.999 271 331 654 040 289 28 × 2 = 1 + 0.998 542 663 308 080 578 56;
  • 47) 0.998 542 663 308 080 578 56 × 2 = 1 + 0.997 085 326 616 161 157 12;
  • 48) 0.997 085 326 616 161 157 12 × 2 = 1 + 0.994 170 653 232 322 314 24;
  • 49) 0.994 170 653 232 322 314 24 × 2 = 1 + 0.988 341 306 464 644 628 48;
  • 50) 0.988 341 306 464 644 628 48 × 2 = 1 + 0.976 682 612 929 289 256 96;
  • 51) 0.976 682 612 929 289 256 96 × 2 = 1 + 0.953 365 225 858 578 513 92;
  • 52) 0.953 365 225 858 578 513 92 × 2 = 1 + 0.906 730 451 717 157 027 84;
  • 53) 0.906 730 451 717 157 027 84 × 2 = 1 + 0.813 460 903 434 314 055 68;
  • 54) 0.813 460 903 434 314 055 68 × 2 = 1 + 0.626 921 806 868 628 111 36;
  • 55) 0.626 921 806 868 628 111 36 × 2 = 1 + 0.253 843 613 737 256 222 72;
  • 56) 0.253 843 613 737 256 222 72 × 2 = 0 + 0.507 687 227 474 512 445 44;
  • 57) 0.507 687 227 474 512 445 44 × 2 = 1 + 0.015 374 454 949 024 890 88;
  • 58) 0.015 374 454 949 024 890 88 × 2 = 0 + 0.030 748 909 898 049 781 76;
  • 59) 0.030 748 909 898 049 781 76 × 2 = 0 + 0.061 497 819 796 099 563 52;
  • 60) 0.061 497 819 796 099 563 52 × 2 = 0 + 0.122 995 639 592 199 127 04;
  • 61) 0.122 995 639 592 199 127 04 × 2 = 0 + 0.245 991 279 184 398 254 08;
  • 62) 0.245 991 279 184 398 254 08 × 2 = 0 + 0.491 982 558 368 796 508 16;
  • 63) 0.491 982 558 368 796 508 16 × 2 = 0 + 0.983 965 116 737 593 016 32;
  • 64) 0.983 965 116 737 593 016 32 × 2 = 1 + 0.967 930 233 475 186 032 64;
  • 65) 0.967 930 233 475 186 032 64 × 2 = 1 + 0.935 860 466 950 372 065 28;
  • 66) 0.935 860 466 950 372 065 28 × 2 = 1 + 0.871 720 933 900 744 130 56;
  • 67) 0.871 720 933 900 744 130 56 × 2 = 1 + 0.743 441 867 801 488 261 12;
  • 68) 0.743 441 867 801 488 261 12 × 2 = 1 + 0.486 883 735 602 976 522 24;
  • 69) 0.486 883 735 602 976 522 24 × 2 = 0 + 0.973 767 471 205 953 044 48;
  • 70) 0.973 767 471 205 953 044 48 × 2 = 1 + 0.947 534 942 411 906 088 96;
  • 71) 0.947 534 942 411 906 088 96 × 2 = 1 + 0.895 069 884 823 812 177 92;

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:


0.000 003 814 697 265 604 29(10) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 1000 0001 1111 011(2)

5. Positive number before normalization:

0.000 003 814 697 265 604 29(10) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 1000 0001 1111 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 003 814 697 265 604 29(10) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 1000 0001 1111 011(2) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 1000 0001 1111 011(2) × 20 =

1.1111 1111 1111 1111 1111 1111 1111 1111 1111 0100 0000 1111 1011(2) × 2-19


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): -19

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1111 1111 1111 0100 0000 1111 1011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(11-1) - 1 =

(-19 + 1 023)(10) =

1 004(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;
  • 1 004 ÷ 2 = 502 + 0;
  • 502 ÷ 2 = 251 + 0;
  • 251 ÷ 2 = 125 + 1;
  • 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) =

1004(10) =

011 1110 1100(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. 1111 1111 1111 1111 1111 1111 1111 1111 1111 0100 0000 1111 1011 =

1111 1111 1111 1111 1111 1111 1111 1111 1111 0100 0000 1111 1011


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 1100

Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1111 1111 1111 0100 0000 1111 1011


Decimal number 0.000 003 814 697 265 604 29 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1100 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 0100 0000 1111 1011

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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