0.009 234 567 810 987 345 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.009 234 567 810 987 345(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
0.009 234 567 810 987 345(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: 0.009 234 567 810 987 345.

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) 0.009 234 567 810 987 345 × 2 = 0 + 0.018 469 135 621 974 69;
  • 2) 0.018 469 135 621 974 69 × 2 = 0 + 0.036 938 271 243 949 38;
  • 3) 0.036 938 271 243 949 38 × 2 = 0 + 0.073 876 542 487 898 76;
  • 4) 0.073 876 542 487 898 76 × 2 = 0 + 0.147 753 084 975 797 52;
  • 5) 0.147 753 084 975 797 52 × 2 = 0 + 0.295 506 169 951 595 04;
  • 6) 0.295 506 169 951 595 04 × 2 = 0 + 0.591 012 339 903 190 08;
  • 7) 0.591 012 339 903 190 08 × 2 = 1 + 0.182 024 679 806 380 16;
  • 8) 0.182 024 679 806 380 16 × 2 = 0 + 0.364 049 359 612 760 32;
  • 9) 0.364 049 359 612 760 32 × 2 = 0 + 0.728 098 719 225 520 64;
  • 10) 0.728 098 719 225 520 64 × 2 = 1 + 0.456 197 438 451 041 28;
  • 11) 0.456 197 438 451 041 28 × 2 = 0 + 0.912 394 876 902 082 56;
  • 12) 0.912 394 876 902 082 56 × 2 = 1 + 0.824 789 753 804 165 12;
  • 13) 0.824 789 753 804 165 12 × 2 = 1 + 0.649 579 507 608 330 24;
  • 14) 0.649 579 507 608 330 24 × 2 = 1 + 0.299 159 015 216 660 48;
  • 15) 0.299 159 015 216 660 48 × 2 = 0 + 0.598 318 030 433 320 96;
  • 16) 0.598 318 030 433 320 96 × 2 = 1 + 0.196 636 060 866 641 92;
  • 17) 0.196 636 060 866 641 92 × 2 = 0 + 0.393 272 121 733 283 84;
  • 18) 0.393 272 121 733 283 84 × 2 = 0 + 0.786 544 243 466 567 68;
  • 19) 0.786 544 243 466 567 68 × 2 = 1 + 0.573 088 486 933 135 36;
  • 20) 0.573 088 486 933 135 36 × 2 = 1 + 0.146 176 973 866 270 72;
  • 21) 0.146 176 973 866 270 72 × 2 = 0 + 0.292 353 947 732 541 44;
  • 22) 0.292 353 947 732 541 44 × 2 = 0 + 0.584 707 895 465 082 88;
  • 23) 0.584 707 895 465 082 88 × 2 = 1 + 0.169 415 790 930 165 76;
  • 24) 0.169 415 790 930 165 76 × 2 = 0 + 0.338 831 581 860 331 52;
  • 25) 0.338 831 581 860 331 52 × 2 = 0 + 0.677 663 163 720 663 04;
  • 26) 0.677 663 163 720 663 04 × 2 = 1 + 0.355 326 327 441 326 08;
  • 27) 0.355 326 327 441 326 08 × 2 = 0 + 0.710 652 654 882 652 16;
  • 28) 0.710 652 654 882 652 16 × 2 = 1 + 0.421 305 309 765 304 32;
  • 29) 0.421 305 309 765 304 32 × 2 = 0 + 0.842 610 619 530 608 64;
  • 30) 0.842 610 619 530 608 64 × 2 = 1 + 0.685 221 239 061 217 28;
  • 31) 0.685 221 239 061 217 28 × 2 = 1 + 0.370 442 478 122 434 56;
  • 32) 0.370 442 478 122 434 56 × 2 = 0 + 0.740 884 956 244 869 12;
  • 33) 0.740 884 956 244 869 12 × 2 = 1 + 0.481 769 912 489 738 24;
  • 34) 0.481 769 912 489 738 24 × 2 = 0 + 0.963 539 824 979 476 48;
  • 35) 0.963 539 824 979 476 48 × 2 = 1 + 0.927 079 649 958 952 96;
  • 36) 0.927 079 649 958 952 96 × 2 = 1 + 0.854 159 299 917 905 92;
  • 37) 0.854 159 299 917 905 92 × 2 = 1 + 0.708 318 599 835 811 84;
  • 38) 0.708 318 599 835 811 84 × 2 = 1 + 0.416 637 199 671 623 68;
  • 39) 0.416 637 199 671 623 68 × 2 = 0 + 0.833 274 399 343 247 36;
  • 40) 0.833 274 399 343 247 36 × 2 = 1 + 0.666 548 798 686 494 72;
  • 41) 0.666 548 798 686 494 72 × 2 = 1 + 0.333 097 597 372 989 44;
  • 42) 0.333 097 597 372 989 44 × 2 = 0 + 0.666 195 194 745 978 88;
  • 43) 0.666 195 194 745 978 88 × 2 = 1 + 0.332 390 389 491 957 76;
  • 44) 0.332 390 389 491 957 76 × 2 = 0 + 0.664 780 778 983 915 52;
  • 45) 0.664 780 778 983 915 52 × 2 = 1 + 0.329 561 557 967 831 04;
  • 46) 0.329 561 557 967 831 04 × 2 = 0 + 0.659 123 115 935 662 08;
  • 47) 0.659 123 115 935 662 08 × 2 = 1 + 0.318 246 231 871 324 16;
  • 48) 0.318 246 231 871 324 16 × 2 = 0 + 0.636 492 463 742 648 32;
  • 49) 0.636 492 463 742 648 32 × 2 = 1 + 0.272 984 927 485 296 64;
  • 50) 0.272 984 927 485 296 64 × 2 = 0 + 0.545 969 854 970 593 28;
  • 51) 0.545 969 854 970 593 28 × 2 = 1 + 0.091 939 709 941 186 56;
  • 52) 0.091 939 709 941 186 56 × 2 = 0 + 0.183 879 419 882 373 12;
  • 53) 0.183 879 419 882 373 12 × 2 = 0 + 0.367 758 839 764 746 24;
  • 54) 0.367 758 839 764 746 24 × 2 = 0 + 0.735 517 679 529 492 48;
  • 55) 0.735 517 679 529 492 48 × 2 = 1 + 0.471 035 359 058 984 96;
  • 56) 0.471 035 359 058 984 96 × 2 = 0 + 0.942 070 718 117 969 92;
  • 57) 0.942 070 718 117 969 92 × 2 = 1 + 0.884 141 436 235 939 84;
  • 58) 0.884 141 436 235 939 84 × 2 = 1 + 0.768 282 872 471 879 68;
  • 59) 0.768 282 872 471 879 68 × 2 = 1 + 0.536 565 744 943 759 36;

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:


0.009 234 567 810 987 345(10) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1101 1010 1010 1010 0010 111(2)

5. Positive number before normalization:

0.009 234 567 810 987 345(10) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1101 1010 1010 1010 0010 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.009 234 567 810 987 345(10) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1101 1010 1010 1010 0010 111(2) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1101 1010 1010 1010 0010 111(2) × 20 =

1.0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0101 0001 0111(2) × 2-7


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): -7

Mantissa (not normalized):
1.0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0101 0001 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 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) =

1016(10) =

011 1111 1000(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 1001 1001 0010 1011 0101 1110 1101 0101 0101 0001 0111 =

0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0101 0001 0111


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 1111 1000

Mantissa (52 bits) =
0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0101 0001 0111


Decimal number 0.009 234 567 810 987 345 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1000 - 0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0101 0001 0111

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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