0.000 046 473 737 954 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 737 954 6(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
0.000 046 473 737 954 6(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: 0.000 046 473 737 954 6.

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) 0.000 046 473 737 954 6 × 2 = 0 + 0.000 092 947 475 909 2;
  • 2) 0.000 092 947 475 909 2 × 2 = 0 + 0.000 185 894 951 818 4;
  • 3) 0.000 185 894 951 818 4 × 2 = 0 + 0.000 371 789 903 636 8;
  • 4) 0.000 371 789 903 636 8 × 2 = 0 + 0.000 743 579 807 273 6;
  • 5) 0.000 743 579 807 273 6 × 2 = 0 + 0.001 487 159 614 547 2;
  • 6) 0.001 487 159 614 547 2 × 2 = 0 + 0.002 974 319 229 094 4;
  • 7) 0.002 974 319 229 094 4 × 2 = 0 + 0.005 948 638 458 188 8;
  • 8) 0.005 948 638 458 188 8 × 2 = 0 + 0.011 897 276 916 377 6;
  • 9) 0.011 897 276 916 377 6 × 2 = 0 + 0.023 794 553 832 755 2;
  • 10) 0.023 794 553 832 755 2 × 2 = 0 + 0.047 589 107 665 510 4;
  • 11) 0.047 589 107 665 510 4 × 2 = 0 + 0.095 178 215 331 020 8;
  • 12) 0.095 178 215 331 020 8 × 2 = 0 + 0.190 356 430 662 041 6;
  • 13) 0.190 356 430 662 041 6 × 2 = 0 + 0.380 712 861 324 083 2;
  • 14) 0.380 712 861 324 083 2 × 2 = 0 + 0.761 425 722 648 166 4;
  • 15) 0.761 425 722 648 166 4 × 2 = 1 + 0.522 851 445 296 332 8;
  • 16) 0.522 851 445 296 332 8 × 2 = 1 + 0.045 702 890 592 665 6;
  • 17) 0.045 702 890 592 665 6 × 2 = 0 + 0.091 405 781 185 331 2;
  • 18) 0.091 405 781 185 331 2 × 2 = 0 + 0.182 811 562 370 662 4;
  • 19) 0.182 811 562 370 662 4 × 2 = 0 + 0.365 623 124 741 324 8;
  • 20) 0.365 623 124 741 324 8 × 2 = 0 + 0.731 246 249 482 649 6;
  • 21) 0.731 246 249 482 649 6 × 2 = 1 + 0.462 492 498 965 299 2;
  • 22) 0.462 492 498 965 299 2 × 2 = 0 + 0.924 984 997 930 598 4;
  • 23) 0.924 984 997 930 598 4 × 2 = 1 + 0.849 969 995 861 196 8;
  • 24) 0.849 969 995 861 196 8 × 2 = 1 + 0.699 939 991 722 393 6;
  • 25) 0.699 939 991 722 393 6 × 2 = 1 + 0.399 879 983 444 787 2;
  • 26) 0.399 879 983 444 787 2 × 2 = 0 + 0.799 759 966 889 574 4;
  • 27) 0.799 759 966 889 574 4 × 2 = 1 + 0.599 519 933 779 148 8;
  • 28) 0.599 519 933 779 148 8 × 2 = 1 + 0.199 039 867 558 297 6;
  • 29) 0.199 039 867 558 297 6 × 2 = 0 + 0.398 079 735 116 595 2;
  • 30) 0.398 079 735 116 595 2 × 2 = 0 + 0.796 159 470 233 190 4;
  • 31) 0.796 159 470 233 190 4 × 2 = 1 + 0.592 318 940 466 380 8;
  • 32) 0.592 318 940 466 380 8 × 2 = 1 + 0.184 637 880 932 761 6;
  • 33) 0.184 637 880 932 761 6 × 2 = 0 + 0.369 275 761 865 523 2;
  • 34) 0.369 275 761 865 523 2 × 2 = 0 + 0.738 551 523 731 046 4;
  • 35) 0.738 551 523 731 046 4 × 2 = 1 + 0.477 103 047 462 092 8;
  • 36) 0.477 103 047 462 092 8 × 2 = 0 + 0.954 206 094 924 185 6;
  • 37) 0.954 206 094 924 185 6 × 2 = 1 + 0.908 412 189 848 371 2;
  • 38) 0.908 412 189 848 371 2 × 2 = 1 + 0.816 824 379 696 742 4;
  • 39) 0.816 824 379 696 742 4 × 2 = 1 + 0.633 648 759 393 484 8;
  • 40) 0.633 648 759 393 484 8 × 2 = 1 + 0.267 297 518 786 969 6;
  • 41) 0.267 297 518 786 969 6 × 2 = 0 + 0.534 595 037 573 939 2;
  • 42) 0.534 595 037 573 939 2 × 2 = 1 + 0.069 190 075 147 878 4;
  • 43) 0.069 190 075 147 878 4 × 2 = 0 + 0.138 380 150 295 756 8;
  • 44) 0.138 380 150 295 756 8 × 2 = 0 + 0.276 760 300 591 513 6;
  • 45) 0.276 760 300 591 513 6 × 2 = 0 + 0.553 520 601 183 027 2;
  • 46) 0.553 520 601 183 027 2 × 2 = 1 + 0.107 041 202 366 054 4;
  • 47) 0.107 041 202 366 054 4 × 2 = 0 + 0.214 082 404 732 108 8;
  • 48) 0.214 082 404 732 108 8 × 2 = 0 + 0.428 164 809 464 217 6;
  • 49) 0.428 164 809 464 217 6 × 2 = 0 + 0.856 329 618 928 435 2;
  • 50) 0.856 329 618 928 435 2 × 2 = 1 + 0.712 659 237 856 870 4;
  • 51) 0.712 659 237 856 870 4 × 2 = 1 + 0.425 318 475 713 740 8;
  • 52) 0.425 318 475 713 740 8 × 2 = 0 + 0.850 636 951 427 481 6;
  • 53) 0.850 636 951 427 481 6 × 2 = 1 + 0.701 273 902 854 963 2;
  • 54) 0.701 273 902 854 963 2 × 2 = 1 + 0.402 547 805 709 926 4;
  • 55) 0.402 547 805 709 926 4 × 2 = 0 + 0.805 095 611 419 852 8;
  • 56) 0.805 095 611 419 852 8 × 2 = 1 + 0.610 191 222 839 705 6;
  • 57) 0.610 191 222 839 705 6 × 2 = 1 + 0.220 382 445 679 411 2;
  • 58) 0.220 382 445 679 411 2 × 2 = 0 + 0.440 764 891 358 822 4;
  • 59) 0.440 764 891 358 822 4 × 2 = 0 + 0.881 529 782 717 644 8;
  • 60) 0.881 529 782 717 644 8 × 2 = 1 + 0.763 059 565 435 289 6;
  • 61) 0.763 059 565 435 289 6 × 2 = 1 + 0.526 119 130 870 579 2;
  • 62) 0.526 119 130 870 579 2 × 2 = 1 + 0.052 238 261 741 158 4;
  • 63) 0.052 238 261 741 158 4 × 2 = 0 + 0.104 476 523 482 316 8;
  • 64) 0.104 476 523 482 316 8 × 2 = 0 + 0.208 953 046 964 633 6;
  • 65) 0.208 953 046 964 633 6 × 2 = 0 + 0.417 906 093 929 267 2;
  • 66) 0.417 906 093 929 267 2 × 2 = 0 + 0.835 812 187 858 534 4;
  • 67) 0.835 812 187 858 534 4 × 2 = 1 + 0.671 624 375 717 068 8;

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:


0.000 046 473 737 954 6(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 0110 1101 1001 1100 001(2)

5. Positive number before normalization:

0.000 046 473 737 954 6(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 0110 1101 1001 1100 001(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:


0.000 046 473 737 954 6(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 0110 1101 1001 1100 001(2) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 0110 1101 1001 1100 001(2) × 20 =

1.1000 0101 1101 1001 1001 0111 1010 0010 0011 0110 1100 1110 0001(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 0010 0011 0110 1100 1110 0001


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 0010 0011 0110 1100 1110 0001 =

1000 0101 1101 1001 1001 0111 1010 0010 0011 0110 1100 1110 0001


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 0010 0011 0110 1100 1110 0001


Decimal number 0.000 046 473 737 954 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 1001 0111 1010 0010 0011 0110 1100 1110 0001

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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