0.000 046 473 737 956 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 737 956 2(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
0.000 046 473 737 956 2(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: 0.000 046 473 737 956 2.

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) 0.000 046 473 737 956 2 × 2 = 0 + 0.000 092 947 475 912 4;
  • 2) 0.000 092 947 475 912 4 × 2 = 0 + 0.000 185 894 951 824 8;
  • 3) 0.000 185 894 951 824 8 × 2 = 0 + 0.000 371 789 903 649 6;
  • 4) 0.000 371 789 903 649 6 × 2 = 0 + 0.000 743 579 807 299 2;
  • 5) 0.000 743 579 807 299 2 × 2 = 0 + 0.001 487 159 614 598 4;
  • 6) 0.001 487 159 614 598 4 × 2 = 0 + 0.002 974 319 229 196 8;
  • 7) 0.002 974 319 229 196 8 × 2 = 0 + 0.005 948 638 458 393 6;
  • 8) 0.005 948 638 458 393 6 × 2 = 0 + 0.011 897 276 916 787 2;
  • 9) 0.011 897 276 916 787 2 × 2 = 0 + 0.023 794 553 833 574 4;
  • 10) 0.023 794 553 833 574 4 × 2 = 0 + 0.047 589 107 667 148 8;
  • 11) 0.047 589 107 667 148 8 × 2 = 0 + 0.095 178 215 334 297 6;
  • 12) 0.095 178 215 334 297 6 × 2 = 0 + 0.190 356 430 668 595 2;
  • 13) 0.190 356 430 668 595 2 × 2 = 0 + 0.380 712 861 337 190 4;
  • 14) 0.380 712 861 337 190 4 × 2 = 0 + 0.761 425 722 674 380 8;
  • 15) 0.761 425 722 674 380 8 × 2 = 1 + 0.522 851 445 348 761 6;
  • 16) 0.522 851 445 348 761 6 × 2 = 1 + 0.045 702 890 697 523 2;
  • 17) 0.045 702 890 697 523 2 × 2 = 0 + 0.091 405 781 395 046 4;
  • 18) 0.091 405 781 395 046 4 × 2 = 0 + 0.182 811 562 790 092 8;
  • 19) 0.182 811 562 790 092 8 × 2 = 0 + 0.365 623 125 580 185 6;
  • 20) 0.365 623 125 580 185 6 × 2 = 0 + 0.731 246 251 160 371 2;
  • 21) 0.731 246 251 160 371 2 × 2 = 1 + 0.462 492 502 320 742 4;
  • 22) 0.462 492 502 320 742 4 × 2 = 0 + 0.924 985 004 641 484 8;
  • 23) 0.924 985 004 641 484 8 × 2 = 1 + 0.849 970 009 282 969 6;
  • 24) 0.849 970 009 282 969 6 × 2 = 1 + 0.699 940 018 565 939 2;
  • 25) 0.699 940 018 565 939 2 × 2 = 1 + 0.399 880 037 131 878 4;
  • 26) 0.399 880 037 131 878 4 × 2 = 0 + 0.799 760 074 263 756 8;
  • 27) 0.799 760 074 263 756 8 × 2 = 1 + 0.599 520 148 527 513 6;
  • 28) 0.599 520 148 527 513 6 × 2 = 1 + 0.199 040 297 055 027 2;
  • 29) 0.199 040 297 055 027 2 × 2 = 0 + 0.398 080 594 110 054 4;
  • 30) 0.398 080 594 110 054 4 × 2 = 0 + 0.796 161 188 220 108 8;
  • 31) 0.796 161 188 220 108 8 × 2 = 1 + 0.592 322 376 440 217 6;
  • 32) 0.592 322 376 440 217 6 × 2 = 1 + 0.184 644 752 880 435 2;
  • 33) 0.184 644 752 880 435 2 × 2 = 0 + 0.369 289 505 760 870 4;
  • 34) 0.369 289 505 760 870 4 × 2 = 0 + 0.738 579 011 521 740 8;
  • 35) 0.738 579 011 521 740 8 × 2 = 1 + 0.477 158 023 043 481 6;
  • 36) 0.477 158 023 043 481 6 × 2 = 0 + 0.954 316 046 086 963 2;
  • 37) 0.954 316 046 086 963 2 × 2 = 1 + 0.908 632 092 173 926 4;
  • 38) 0.908 632 092 173 926 4 × 2 = 1 + 0.817 264 184 347 852 8;
  • 39) 0.817 264 184 347 852 8 × 2 = 1 + 0.634 528 368 695 705 6;
  • 40) 0.634 528 368 695 705 6 × 2 = 1 + 0.269 056 737 391 411 2;
  • 41) 0.269 056 737 391 411 2 × 2 = 0 + 0.538 113 474 782 822 4;
  • 42) 0.538 113 474 782 822 4 × 2 = 1 + 0.076 226 949 565 644 8;
  • 43) 0.076 226 949 565 644 8 × 2 = 0 + 0.152 453 899 131 289 6;
  • 44) 0.152 453 899 131 289 6 × 2 = 0 + 0.304 907 798 262 579 2;
  • 45) 0.304 907 798 262 579 2 × 2 = 0 + 0.609 815 596 525 158 4;
  • 46) 0.609 815 596 525 158 4 × 2 = 1 + 0.219 631 193 050 316 8;
  • 47) 0.219 631 193 050 316 8 × 2 = 0 + 0.439 262 386 100 633 6;
  • 48) 0.439 262 386 100 633 6 × 2 = 0 + 0.878 524 772 201 267 2;
  • 49) 0.878 524 772 201 267 2 × 2 = 1 + 0.757 049 544 402 534 4;
  • 50) 0.757 049 544 402 534 4 × 2 = 1 + 0.514 099 088 805 068 8;
  • 51) 0.514 099 088 805 068 8 × 2 = 1 + 0.028 198 177 610 137 6;
  • 52) 0.028 198 177 610 137 6 × 2 = 0 + 0.056 396 355 220 275 2;
  • 53) 0.056 396 355 220 275 2 × 2 = 0 + 0.112 792 710 440 550 4;
  • 54) 0.112 792 710 440 550 4 × 2 = 0 + 0.225 585 420 881 100 8;
  • 55) 0.225 585 420 881 100 8 × 2 = 0 + 0.451 170 841 762 201 6;
  • 56) 0.451 170 841 762 201 6 × 2 = 0 + 0.902 341 683 524 403 2;
  • 57) 0.902 341 683 524 403 2 × 2 = 1 + 0.804 683 367 048 806 4;
  • 58) 0.804 683 367 048 806 4 × 2 = 1 + 0.609 366 734 097 612 8;
  • 59) 0.609 366 734 097 612 8 × 2 = 1 + 0.218 733 468 195 225 6;
  • 60) 0.218 733 468 195 225 6 × 2 = 0 + 0.437 466 936 390 451 2;
  • 61) 0.437 466 936 390 451 2 × 2 = 0 + 0.874 933 872 780 902 4;
  • 62) 0.874 933 872 780 902 4 × 2 = 1 + 0.749 867 745 561 804 8;
  • 63) 0.749 867 745 561 804 8 × 2 = 1 + 0.499 735 491 123 609 6;
  • 64) 0.499 735 491 123 609 6 × 2 = 0 + 0.999 470 982 247 219 2;
  • 65) 0.999 470 982 247 219 2 × 2 = 1 + 0.998 941 964 494 438 4;
  • 66) 0.998 941 964 494 438 4 × 2 = 1 + 0.997 883 928 988 876 8;
  • 67) 0.997 883 928 988 876 8 × 2 = 1 + 0.995 767 857 977 753 6;

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:


0.000 046 473 737 956 2(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 1110 0000 1110 0110 111(2)

5. Positive number before normalization:

0.000 046 473 737 956 2(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 1110 0000 1110 0110 111(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:


0.000 046 473 737 956 2(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 1110 0000 1110 0110 111(2) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 1110 0000 1110 0110 111(2) × 20 =

1.1000 0101 1101 1001 1001 0111 1010 0010 0111 0000 0111 0011 0111(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 0111 0000 0111 0011 0111


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 0111 0000 0111 0011 0111 =

1000 0101 1101 1001 1001 0111 1010 0010 0111 0000 0111 0011 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 0000

Mantissa (52 bits) =
1000 0101 1101 1001 1001 0111 1010 0010 0111 0000 0111 0011 0111


Decimal number 0.000 046 473 737 956 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 1001 0111 1010 0010 0111 0000 0111 0011 0111

» Convert decimal 0.000 046 473 737 949 8 to 64 bit double precision IEEE 754 binary floating point representation standard

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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