0.002 204 718 637 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.002 204 718 637 47(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
0.002 204 718 637 47(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: 0.002 204 718 637 47.

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) 0.002 204 718 637 47 × 2 = 0 + 0.004 409 437 274 94;
  • 2) 0.004 409 437 274 94 × 2 = 0 + 0.008 818 874 549 88;
  • 3) 0.008 818 874 549 88 × 2 = 0 + 0.017 637 749 099 76;
  • 4) 0.017 637 749 099 76 × 2 = 0 + 0.035 275 498 199 52;
  • 5) 0.035 275 498 199 52 × 2 = 0 + 0.070 550 996 399 04;
  • 6) 0.070 550 996 399 04 × 2 = 0 + 0.141 101 992 798 08;
  • 7) 0.141 101 992 798 08 × 2 = 0 + 0.282 203 985 596 16;
  • 8) 0.282 203 985 596 16 × 2 = 0 + 0.564 407 971 192 32;
  • 9) 0.564 407 971 192 32 × 2 = 1 + 0.128 815 942 384 64;
  • 10) 0.128 815 942 384 64 × 2 = 0 + 0.257 631 884 769 28;
  • 11) 0.257 631 884 769 28 × 2 = 0 + 0.515 263 769 538 56;
  • 12) 0.515 263 769 538 56 × 2 = 1 + 0.030 527 539 077 12;
  • 13) 0.030 527 539 077 12 × 2 = 0 + 0.061 055 078 154 24;
  • 14) 0.061 055 078 154 24 × 2 = 0 + 0.122 110 156 308 48;
  • 15) 0.122 110 156 308 48 × 2 = 0 + 0.244 220 312 616 96;
  • 16) 0.244 220 312 616 96 × 2 = 0 + 0.488 440 625 233 92;
  • 17) 0.488 440 625 233 92 × 2 = 0 + 0.976 881 250 467 84;
  • 18) 0.976 881 250 467 84 × 2 = 1 + 0.953 762 500 935 68;
  • 19) 0.953 762 500 935 68 × 2 = 1 + 0.907 525 001 871 36;
  • 20) 0.907 525 001 871 36 × 2 = 1 + 0.815 050 003 742 72;
  • 21) 0.815 050 003 742 72 × 2 = 1 + 0.630 100 007 485 44;
  • 22) 0.630 100 007 485 44 × 2 = 1 + 0.260 200 014 970 88;
  • 23) 0.260 200 014 970 88 × 2 = 0 + 0.520 400 029 941 76;
  • 24) 0.520 400 029 941 76 × 2 = 1 + 0.040 800 059 883 52;
  • 25) 0.040 800 059 883 52 × 2 = 0 + 0.081 600 119 767 04;
  • 26) 0.081 600 119 767 04 × 2 = 0 + 0.163 200 239 534 08;
  • 27) 0.163 200 239 534 08 × 2 = 0 + 0.326 400 479 068 16;
  • 28) 0.326 400 479 068 16 × 2 = 0 + 0.652 800 958 136 32;
  • 29) 0.652 800 958 136 32 × 2 = 1 + 0.305 601 916 272 64;
  • 30) 0.305 601 916 272 64 × 2 = 0 + 0.611 203 832 545 28;
  • 31) 0.611 203 832 545 28 × 2 = 1 + 0.222 407 665 090 56;
  • 32) 0.222 407 665 090 56 × 2 = 0 + 0.444 815 330 181 12;
  • 33) 0.444 815 330 181 12 × 2 = 0 + 0.889 630 660 362 24;
  • 34) 0.889 630 660 362 24 × 2 = 1 + 0.779 261 320 724 48;
  • 35) 0.779 261 320 724 48 × 2 = 1 + 0.558 522 641 448 96;
  • 36) 0.558 522 641 448 96 × 2 = 1 + 0.117 045 282 897 92;
  • 37) 0.117 045 282 897 92 × 2 = 0 + 0.234 090 565 795 84;
  • 38) 0.234 090 565 795 84 × 2 = 0 + 0.468 181 131 591 68;
  • 39) 0.468 181 131 591 68 × 2 = 0 + 0.936 362 263 183 36;
  • 40) 0.936 362 263 183 36 × 2 = 1 + 0.872 724 526 366 72;
  • 41) 0.872 724 526 366 72 × 2 = 1 + 0.745 449 052 733 44;
  • 42) 0.745 449 052 733 44 × 2 = 1 + 0.490 898 105 466 88;
  • 43) 0.490 898 105 466 88 × 2 = 0 + 0.981 796 210 933 76;
  • 44) 0.981 796 210 933 76 × 2 = 1 + 0.963 592 421 867 52;
  • 45) 0.963 592 421 867 52 × 2 = 1 + 0.927 184 843 735 04;
  • 46) 0.927 184 843 735 04 × 2 = 1 + 0.854 369 687 470 08;
  • 47) 0.854 369 687 470 08 × 2 = 1 + 0.708 739 374 940 16;
  • 48) 0.708 739 374 940 16 × 2 = 1 + 0.417 478 749 880 32;
  • 49) 0.417 478 749 880 32 × 2 = 0 + 0.834 957 499 760 64;
  • 50) 0.834 957 499 760 64 × 2 = 1 + 0.669 914 999 521 28;
  • 51) 0.669 914 999 521 28 × 2 = 1 + 0.339 829 999 042 56;
  • 52) 0.339 829 999 042 56 × 2 = 0 + 0.679 659 998 085 12;
  • 53) 0.679 659 998 085 12 × 2 = 1 + 0.359 319 996 170 24;
  • 54) 0.359 319 996 170 24 × 2 = 0 + 0.718 639 992 340 48;
  • 55) 0.718 639 992 340 48 × 2 = 1 + 0.437 279 984 680 96;
  • 56) 0.437 279 984 680 96 × 2 = 0 + 0.874 559 969 361 92;
  • 57) 0.874 559 969 361 92 × 2 = 1 + 0.749 119 938 723 84;
  • 58) 0.749 119 938 723 84 × 2 = 1 + 0.498 239 877 447 68;
  • 59) 0.498 239 877 447 68 × 2 = 0 + 0.996 479 754 895 36;
  • 60) 0.996 479 754 895 36 × 2 = 1 + 0.992 959 509 790 72;
  • 61) 0.992 959 509 790 72 × 2 = 1 + 0.985 919 019 581 44;

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:


0.002 204 718 637 47(10) =

0.0000 0000 1001 0000 0111 1101 0000 1010 0111 0001 1101 1111 0110 1010 1101 1(2)

5. Positive number before normalization:

0.002 204 718 637 47(10) =

0.0000 0000 1001 0000 0111 1101 0000 1010 0111 0001 1101 1111 0110 1010 1101 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the right, so that only one non zero digit remains to the left of it:


0.002 204 718 637 47(10) =

0.0000 0000 1001 0000 0111 1101 0000 1010 0111 0001 1101 1111 0110 1010 1101 1(2) =

0.0000 0000 1001 0000 0111 1101 0000 1010 0111 0001 1101 1111 0110 1010 1101 1(2) × 20 =

1.0010 0000 1111 1010 0001 0100 1110 0011 1011 1110 1101 0101 1011(2) × 2-9


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): -9

Mantissa (not normalized):
1.0010 0000 1111 1010 0001 0100 1110 0011 1011 1110 1101 0101 1011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-9 + 2(11-1) - 1 =

(-9 + 1 023)(10) =

1 014(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;
  • 1 014 ÷ 2 = 507 + 0;
  • 507 ÷ 2 = 253 + 1;
  • 253 ÷ 2 = 126 + 1;
  • 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) =

1014(10) =

011 1111 0110(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 0000 1111 1010 0001 0100 1110 0011 1011 1110 1101 0101 1011 =

0010 0000 1111 1010 0001 0100 1110 0011 1011 1110 1101 0101 1011


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 0110

Mantissa (52 bits) =
0010 0000 1111 1010 0001 0100 1110 0011 1011 1110 1101 0101 1011


Decimal number 0.002 204 718 637 47 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0110 - 0010 0000 1111 1010 0001 0100 1110 0011 1011 1110 1101 0101 1011

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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