0.000 046 473 737 960 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 737 960 9(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
0.000 046 473 737 960 9(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: 0.000 046 473 737 960 9.

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) 0.000 046 473 737 960 9 × 2 = 0 + 0.000 092 947 475 921 8;
  • 2) 0.000 092 947 475 921 8 × 2 = 0 + 0.000 185 894 951 843 6;
  • 3) 0.000 185 894 951 843 6 × 2 = 0 + 0.000 371 789 903 687 2;
  • 4) 0.000 371 789 903 687 2 × 2 = 0 + 0.000 743 579 807 374 4;
  • 5) 0.000 743 579 807 374 4 × 2 = 0 + 0.001 487 159 614 748 8;
  • 6) 0.001 487 159 614 748 8 × 2 = 0 + 0.002 974 319 229 497 6;
  • 7) 0.002 974 319 229 497 6 × 2 = 0 + 0.005 948 638 458 995 2;
  • 8) 0.005 948 638 458 995 2 × 2 = 0 + 0.011 897 276 917 990 4;
  • 9) 0.011 897 276 917 990 4 × 2 = 0 + 0.023 794 553 835 980 8;
  • 10) 0.023 794 553 835 980 8 × 2 = 0 + 0.047 589 107 671 961 6;
  • 11) 0.047 589 107 671 961 6 × 2 = 0 + 0.095 178 215 343 923 2;
  • 12) 0.095 178 215 343 923 2 × 2 = 0 + 0.190 356 430 687 846 4;
  • 13) 0.190 356 430 687 846 4 × 2 = 0 + 0.380 712 861 375 692 8;
  • 14) 0.380 712 861 375 692 8 × 2 = 0 + 0.761 425 722 751 385 6;
  • 15) 0.761 425 722 751 385 6 × 2 = 1 + 0.522 851 445 502 771 2;
  • 16) 0.522 851 445 502 771 2 × 2 = 1 + 0.045 702 891 005 542 4;
  • 17) 0.045 702 891 005 542 4 × 2 = 0 + 0.091 405 782 011 084 8;
  • 18) 0.091 405 782 011 084 8 × 2 = 0 + 0.182 811 564 022 169 6;
  • 19) 0.182 811 564 022 169 6 × 2 = 0 + 0.365 623 128 044 339 2;
  • 20) 0.365 623 128 044 339 2 × 2 = 0 + 0.731 246 256 088 678 4;
  • 21) 0.731 246 256 088 678 4 × 2 = 1 + 0.462 492 512 177 356 8;
  • 22) 0.462 492 512 177 356 8 × 2 = 0 + 0.924 985 024 354 713 6;
  • 23) 0.924 985 024 354 713 6 × 2 = 1 + 0.849 970 048 709 427 2;
  • 24) 0.849 970 048 709 427 2 × 2 = 1 + 0.699 940 097 418 854 4;
  • 25) 0.699 940 097 418 854 4 × 2 = 1 + 0.399 880 194 837 708 8;
  • 26) 0.399 880 194 837 708 8 × 2 = 0 + 0.799 760 389 675 417 6;
  • 27) 0.799 760 389 675 417 6 × 2 = 1 + 0.599 520 779 350 835 2;
  • 28) 0.599 520 779 350 835 2 × 2 = 1 + 0.199 041 558 701 670 4;
  • 29) 0.199 041 558 701 670 4 × 2 = 0 + 0.398 083 117 403 340 8;
  • 30) 0.398 083 117 403 340 8 × 2 = 0 + 0.796 166 234 806 681 6;
  • 31) 0.796 166 234 806 681 6 × 2 = 1 + 0.592 332 469 613 363 2;
  • 32) 0.592 332 469 613 363 2 × 2 = 1 + 0.184 664 939 226 726 4;
  • 33) 0.184 664 939 226 726 4 × 2 = 0 + 0.369 329 878 453 452 8;
  • 34) 0.369 329 878 453 452 8 × 2 = 0 + 0.738 659 756 906 905 6;
  • 35) 0.738 659 756 906 905 6 × 2 = 1 + 0.477 319 513 813 811 2;
  • 36) 0.477 319 513 813 811 2 × 2 = 0 + 0.954 639 027 627 622 4;
  • 37) 0.954 639 027 627 622 4 × 2 = 1 + 0.909 278 055 255 244 8;
  • 38) 0.909 278 055 255 244 8 × 2 = 1 + 0.818 556 110 510 489 6;
  • 39) 0.818 556 110 510 489 6 × 2 = 1 + 0.637 112 221 020 979 2;
  • 40) 0.637 112 221 020 979 2 × 2 = 1 + 0.274 224 442 041 958 4;
  • 41) 0.274 224 442 041 958 4 × 2 = 0 + 0.548 448 884 083 916 8;
  • 42) 0.548 448 884 083 916 8 × 2 = 1 + 0.096 897 768 167 833 6;
  • 43) 0.096 897 768 167 833 6 × 2 = 0 + 0.193 795 536 335 667 2;
  • 44) 0.193 795 536 335 667 2 × 2 = 0 + 0.387 591 072 671 334 4;
  • 45) 0.387 591 072 671 334 4 × 2 = 0 + 0.775 182 145 342 668 8;
  • 46) 0.775 182 145 342 668 8 × 2 = 1 + 0.550 364 290 685 337 6;
  • 47) 0.550 364 290 685 337 6 × 2 = 1 + 0.100 728 581 370 675 2;
  • 48) 0.100 728 581 370 675 2 × 2 = 0 + 0.201 457 162 741 350 4;
  • 49) 0.201 457 162 741 350 4 × 2 = 0 + 0.402 914 325 482 700 8;
  • 50) 0.402 914 325 482 700 8 × 2 = 0 + 0.805 828 650 965 401 6;
  • 51) 0.805 828 650 965 401 6 × 2 = 1 + 0.611 657 301 930 803 2;
  • 52) 0.611 657 301 930 803 2 × 2 = 1 + 0.223 314 603 861 606 4;
  • 53) 0.223 314 603 861 606 4 × 2 = 0 + 0.446 629 207 723 212 8;
  • 54) 0.446 629 207 723 212 8 × 2 = 0 + 0.893 258 415 446 425 6;
  • 55) 0.893 258 415 446 425 6 × 2 = 1 + 0.786 516 830 892 851 2;
  • 56) 0.786 516 830 892 851 2 × 2 = 1 + 0.573 033 661 785 702 4;
  • 57) 0.573 033 661 785 702 4 × 2 = 1 + 0.146 067 323 571 404 8;
  • 58) 0.146 067 323 571 404 8 × 2 = 0 + 0.292 134 647 142 809 6;
  • 59) 0.292 134 647 142 809 6 × 2 = 0 + 0.584 269 294 285 619 2;
  • 60) 0.584 269 294 285 619 2 × 2 = 1 + 0.168 538 588 571 238 4;
  • 61) 0.168 538 588 571 238 4 × 2 = 0 + 0.337 077 177 142 476 8;
  • 62) 0.337 077 177 142 476 8 × 2 = 0 + 0.674 154 354 284 953 6;
  • 63) 0.674 154 354 284 953 6 × 2 = 1 + 0.348 308 708 569 907 2;
  • 64) 0.348 308 708 569 907 2 × 2 = 0 + 0.696 617 417 139 814 4;
  • 65) 0.696 617 417 139 814 4 × 2 = 1 + 0.393 234 834 279 628 8;
  • 66) 0.393 234 834 279 628 8 × 2 = 0 + 0.786 469 668 559 257 6;
  • 67) 0.786 469 668 559 257 6 × 2 = 1 + 0.572 939 337 118 515 2;

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:


0.000 046 473 737 960 9(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0110 0011 0011 1001 0010 101(2)

5. Positive number before normalization:

0.000 046 473 737 960 9(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0110 0011 0011 1001 0010 101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 046 473 737 960 9(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0110 0011 0011 1001 0010 101(2) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0110 0011 0011 1001 0010 101(2) × 20 =

1.1000 0101 1101 1001 1001 0111 1010 0011 0001 1001 1100 1001 0101(2) × 2-15


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): -15

Mantissa (not normalized):
1.1000 0101 1101 1001 1001 0111 1010 0011 0001 1001 1100 1001 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1008(10) =

011 1111 0000(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. 1000 0101 1101 1001 1001 0111 1010 0011 0001 1001 1100 1001 0101 =

1000 0101 1101 1001 1001 0111 1010 0011 0001 1001 1100 1001 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 1111 0000

Mantissa (52 bits) =
1000 0101 1101 1001 1001 0111 1010 0011 0001 1001 1100 1001 0101


Decimal number 0.000 046 473 737 960 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 1001 0111 1010 0011 0001 1001 1100 1001 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