0.000 003 814 697 265 603 81 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 814 697 265 603 81(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
0.000 003 814 697 265 603 81(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: 0.000 003 814 697 265 603 81.

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) 0.000 003 814 697 265 603 81 × 2 = 0 + 0.000 007 629 394 531 207 62;
  • 2) 0.000 007 629 394 531 207 62 × 2 = 0 + 0.000 015 258 789 062 415 24;
  • 3) 0.000 015 258 789 062 415 24 × 2 = 0 + 0.000 030 517 578 124 830 48;
  • 4) 0.000 030 517 578 124 830 48 × 2 = 0 + 0.000 061 035 156 249 660 96;
  • 5) 0.000 061 035 156 249 660 96 × 2 = 0 + 0.000 122 070 312 499 321 92;
  • 6) 0.000 122 070 312 499 321 92 × 2 = 0 + 0.000 244 140 624 998 643 84;
  • 7) 0.000 244 140 624 998 643 84 × 2 = 0 + 0.000 488 281 249 997 287 68;
  • 8) 0.000 488 281 249 997 287 68 × 2 = 0 + 0.000 976 562 499 994 575 36;
  • 9) 0.000 976 562 499 994 575 36 × 2 = 0 + 0.001 953 124 999 989 150 72;
  • 10) 0.001 953 124 999 989 150 72 × 2 = 0 + 0.003 906 249 999 978 301 44;
  • 11) 0.003 906 249 999 978 301 44 × 2 = 0 + 0.007 812 499 999 956 602 88;
  • 12) 0.007 812 499 999 956 602 88 × 2 = 0 + 0.015 624 999 999 913 205 76;
  • 13) 0.015 624 999 999 913 205 76 × 2 = 0 + 0.031 249 999 999 826 411 52;
  • 14) 0.031 249 999 999 826 411 52 × 2 = 0 + 0.062 499 999 999 652 823 04;
  • 15) 0.062 499 999 999 652 823 04 × 2 = 0 + 0.124 999 999 999 305 646 08;
  • 16) 0.124 999 999 999 305 646 08 × 2 = 0 + 0.249 999 999 998 611 292 16;
  • 17) 0.249 999 999 998 611 292 16 × 2 = 0 + 0.499 999 999 997 222 584 32;
  • 18) 0.499 999 999 997 222 584 32 × 2 = 0 + 0.999 999 999 994 445 168 64;
  • 19) 0.999 999 999 994 445 168 64 × 2 = 1 + 0.999 999 999 988 890 337 28;
  • 20) 0.999 999 999 988 890 337 28 × 2 = 1 + 0.999 999 999 977 780 674 56;
  • 21) 0.999 999 999 977 780 674 56 × 2 = 1 + 0.999 999 999 955 561 349 12;
  • 22) 0.999 999 999 955 561 349 12 × 2 = 1 + 0.999 999 999 911 122 698 24;
  • 23) 0.999 999 999 911 122 698 24 × 2 = 1 + 0.999 999 999 822 245 396 48;
  • 24) 0.999 999 999 822 245 396 48 × 2 = 1 + 0.999 999 999 644 490 792 96;
  • 25) 0.999 999 999 644 490 792 96 × 2 = 1 + 0.999 999 999 288 981 585 92;
  • 26) 0.999 999 999 288 981 585 92 × 2 = 1 + 0.999 999 998 577 963 171 84;
  • 27) 0.999 999 998 577 963 171 84 × 2 = 1 + 0.999 999 997 155 926 343 68;
  • 28) 0.999 999 997 155 926 343 68 × 2 = 1 + 0.999 999 994 311 852 687 36;
  • 29) 0.999 999 994 311 852 687 36 × 2 = 1 + 0.999 999 988 623 705 374 72;
  • 30) 0.999 999 988 623 705 374 72 × 2 = 1 + 0.999 999 977 247 410 749 44;
  • 31) 0.999 999 977 247 410 749 44 × 2 = 1 + 0.999 999 954 494 821 498 88;
  • 32) 0.999 999 954 494 821 498 88 × 2 = 1 + 0.999 999 908 989 642 997 76;
  • 33) 0.999 999 908 989 642 997 76 × 2 = 1 + 0.999 999 817 979 285 995 52;
  • 34) 0.999 999 817 979 285 995 52 × 2 = 1 + 0.999 999 635 958 571 991 04;
  • 35) 0.999 999 635 958 571 991 04 × 2 = 1 + 0.999 999 271 917 143 982 08;
  • 36) 0.999 999 271 917 143 982 08 × 2 = 1 + 0.999 998 543 834 287 964 16;
  • 37) 0.999 998 543 834 287 964 16 × 2 = 1 + 0.999 997 087 668 575 928 32;
  • 38) 0.999 997 087 668 575 928 32 × 2 = 1 + 0.999 994 175 337 151 856 64;
  • 39) 0.999 994 175 337 151 856 64 × 2 = 1 + 0.999 988 350 674 303 713 28;
  • 40) 0.999 988 350 674 303 713 28 × 2 = 1 + 0.999 976 701 348 607 426 56;
  • 41) 0.999 976 701 348 607 426 56 × 2 = 1 + 0.999 953 402 697 214 853 12;
  • 42) 0.999 953 402 697 214 853 12 × 2 = 1 + 0.999 906 805 394 429 706 24;
  • 43) 0.999 906 805 394 429 706 24 × 2 = 1 + 0.999 813 610 788 859 412 48;
  • 44) 0.999 813 610 788 859 412 48 × 2 = 1 + 0.999 627 221 577 718 824 96;
  • 45) 0.999 627 221 577 718 824 96 × 2 = 1 + 0.999 254 443 155 437 649 92;
  • 46) 0.999 254 443 155 437 649 92 × 2 = 1 + 0.998 508 886 310 875 299 84;
  • 47) 0.998 508 886 310 875 299 84 × 2 = 1 + 0.997 017 772 621 750 599 68;
  • 48) 0.997 017 772 621 750 599 68 × 2 = 1 + 0.994 035 545 243 501 199 36;
  • 49) 0.994 035 545 243 501 199 36 × 2 = 1 + 0.988 071 090 487 002 398 72;
  • 50) 0.988 071 090 487 002 398 72 × 2 = 1 + 0.976 142 180 974 004 797 44;
  • 51) 0.976 142 180 974 004 797 44 × 2 = 1 + 0.952 284 361 948 009 594 88;
  • 52) 0.952 284 361 948 009 594 88 × 2 = 1 + 0.904 568 723 896 019 189 76;
  • 53) 0.904 568 723 896 019 189 76 × 2 = 1 + 0.809 137 447 792 038 379 52;
  • 54) 0.809 137 447 792 038 379 52 × 2 = 1 + 0.618 274 895 584 076 759 04;
  • 55) 0.618 274 895 584 076 759 04 × 2 = 1 + 0.236 549 791 168 153 518 08;
  • 56) 0.236 549 791 168 153 518 08 × 2 = 0 + 0.473 099 582 336 307 036 16;
  • 57) 0.473 099 582 336 307 036 16 × 2 = 0 + 0.946 199 164 672 614 072 32;
  • 58) 0.946 199 164 672 614 072 32 × 2 = 1 + 0.892 398 329 345 228 144 64;
  • 59) 0.892 398 329 345 228 144 64 × 2 = 1 + 0.784 796 658 690 456 289 28;
  • 60) 0.784 796 658 690 456 289 28 × 2 = 1 + 0.569 593 317 380 912 578 56;
  • 61) 0.569 593 317 380 912 578 56 × 2 = 1 + 0.139 186 634 761 825 157 12;
  • 62) 0.139 186 634 761 825 157 12 × 2 = 0 + 0.278 373 269 523 650 314 24;
  • 63) 0.278 373 269 523 650 314 24 × 2 = 0 + 0.556 746 539 047 300 628 48;
  • 64) 0.556 746 539 047 300 628 48 × 2 = 1 + 0.113 493 078 094 601 256 96;
  • 65) 0.113 493 078 094 601 256 96 × 2 = 0 + 0.226 986 156 189 202 513 92;
  • 66) 0.226 986 156 189 202 513 92 × 2 = 0 + 0.453 972 312 378 405 027 84;
  • 67) 0.453 972 312 378 405 027 84 × 2 = 0 + 0.907 944 624 756 810 055 68;
  • 68) 0.907 944 624 756 810 055 68 × 2 = 1 + 0.815 889 249 513 620 111 36;
  • 69) 0.815 889 249 513 620 111 36 × 2 = 1 + 0.631 778 499 027 240 222 72;
  • 70) 0.631 778 499 027 240 222 72 × 2 = 1 + 0.263 556 998 054 480 445 44;
  • 71) 0.263 556 998 054 480 445 44 × 2 = 0 + 0.527 113 996 108 960 890 88;

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:


0.000 003 814 697 265 603 81(10) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 0111 1001 0001 110(2)

5. Positive number before normalization:

0.000 003 814 697 265 603 81(10) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 0111 1001 0001 110(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:


0.000 003 814 697 265 603 81(10) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 0111 1001 0001 110(2) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 0111 1001 0001 110(2) × 20 =

1.1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 1100 1000 1110(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 0011 1100 1000 1110


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 0011 1100 1000 1110 =

1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 1100 1000 1110


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 0011 1100 1000 1110


Decimal number 0.000 003 814 697 265 603 81 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 0011 1100 1000 1110

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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