64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.002 204 718 3 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.002 204 718 3(10) converted and written in 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 3.

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 3 × 2 = 0 + 0.004 409 436 6;
  • 2) 0.004 409 436 6 × 2 = 0 + 0.008 818 873 2;
  • 3) 0.008 818 873 2 × 2 = 0 + 0.017 637 746 4;
  • 4) 0.017 637 746 4 × 2 = 0 + 0.035 275 492 8;
  • 5) 0.035 275 492 8 × 2 = 0 + 0.070 550 985 6;
  • 6) 0.070 550 985 6 × 2 = 0 + 0.141 101 971 2;
  • 7) 0.141 101 971 2 × 2 = 0 + 0.282 203 942 4;
  • 8) 0.282 203 942 4 × 2 = 0 + 0.564 407 884 8;
  • 9) 0.564 407 884 8 × 2 = 1 + 0.128 815 769 6;
  • 10) 0.128 815 769 6 × 2 = 0 + 0.257 631 539 2;
  • 11) 0.257 631 539 2 × 2 = 0 + 0.515 263 078 4;
  • 12) 0.515 263 078 4 × 2 = 1 + 0.030 526 156 8;
  • 13) 0.030 526 156 8 × 2 = 0 + 0.061 052 313 6;
  • 14) 0.061 052 313 6 × 2 = 0 + 0.122 104 627 2;
  • 15) 0.122 104 627 2 × 2 = 0 + 0.244 209 254 4;
  • 16) 0.244 209 254 4 × 2 = 0 + 0.488 418 508 8;
  • 17) 0.488 418 508 8 × 2 = 0 + 0.976 837 017 6;
  • 18) 0.976 837 017 6 × 2 = 1 + 0.953 674 035 2;
  • 19) 0.953 674 035 2 × 2 = 1 + 0.907 348 070 4;
  • 20) 0.907 348 070 4 × 2 = 1 + 0.814 696 140 8;
  • 21) 0.814 696 140 8 × 2 = 1 + 0.629 392 281 6;
  • 22) 0.629 392 281 6 × 2 = 1 + 0.258 784 563 2;
  • 23) 0.258 784 563 2 × 2 = 0 + 0.517 569 126 4;
  • 24) 0.517 569 126 4 × 2 = 1 + 0.035 138 252 8;
  • 25) 0.035 138 252 8 × 2 = 0 + 0.070 276 505 6;
  • 26) 0.070 276 505 6 × 2 = 0 + 0.140 553 011 2;
  • 27) 0.140 553 011 2 × 2 = 0 + 0.281 106 022 4;
  • 28) 0.281 106 022 4 × 2 = 0 + 0.562 212 044 8;
  • 29) 0.562 212 044 8 × 2 = 1 + 0.124 424 089 6;
  • 30) 0.124 424 089 6 × 2 = 0 + 0.248 848 179 2;
  • 31) 0.248 848 179 2 × 2 = 0 + 0.497 696 358 4;
  • 32) 0.497 696 358 4 × 2 = 0 + 0.995 392 716 8;
  • 33) 0.995 392 716 8 × 2 = 1 + 0.990 785 433 6;
  • 34) 0.990 785 433 6 × 2 = 1 + 0.981 570 867 2;
  • 35) 0.981 570 867 2 × 2 = 1 + 0.963 141 734 4;
  • 36) 0.963 141 734 4 × 2 = 1 + 0.926 283 468 8;
  • 37) 0.926 283 468 8 × 2 = 1 + 0.852 566 937 6;
  • 38) 0.852 566 937 6 × 2 = 1 + 0.705 133 875 2;
  • 39) 0.705 133 875 2 × 2 = 1 + 0.410 267 750 4;
  • 40) 0.410 267 750 4 × 2 = 0 + 0.820 535 500 8;
  • 41) 0.820 535 500 8 × 2 = 1 + 0.641 071 001 6;
  • 42) 0.641 071 001 6 × 2 = 1 + 0.282 142 003 2;
  • 43) 0.282 142 003 2 × 2 = 0 + 0.564 284 006 4;
  • 44) 0.564 284 006 4 × 2 = 1 + 0.128 568 012 8;
  • 45) 0.128 568 012 8 × 2 = 0 + 0.257 136 025 6;
  • 46) 0.257 136 025 6 × 2 = 0 + 0.514 272 051 2;
  • 47) 0.514 272 051 2 × 2 = 1 + 0.028 544 102 4;
  • 48) 0.028 544 102 4 × 2 = 0 + 0.057 088 204 8;
  • 49) 0.057 088 204 8 × 2 = 0 + 0.114 176 409 6;
  • 50) 0.114 176 409 6 × 2 = 0 + 0.228 352 819 2;
  • 51) 0.228 352 819 2 × 2 = 0 + 0.456 705 638 4;
  • 52) 0.456 705 638 4 × 2 = 0 + 0.913 411 276 8;
  • 53) 0.913 411 276 8 × 2 = 1 + 0.826 822 553 6;
  • 54) 0.826 822 553 6 × 2 = 1 + 0.653 645 107 2;
  • 55) 0.653 645 107 2 × 2 = 1 + 0.307 290 214 4;
  • 56) 0.307 290 214 4 × 2 = 0 + 0.614 580 428 8;
  • 57) 0.614 580 428 8 × 2 = 1 + 0.229 160 857 6;
  • 58) 0.229 160 857 6 × 2 = 0 + 0.458 321 715 2;
  • 59) 0.458 321 715 2 × 2 = 0 + 0.916 643 430 4;
  • 60) 0.916 643 430 4 × 2 = 1 + 0.833 286 860 8;
  • 61) 0.833 286 860 8 × 2 = 1 + 0.666 573 721 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...)


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 3(10) =

0.0000 0000 1001 0000 0111 1101 0000 1000 1111 1110 1101 0010 0000 1110 1001 1(2)


5. Positive number before normalization:

0.002 204 718 3(10) =

0.0000 0000 1001 0000 0111 1101 0000 1000 1111 1110 1101 0010 0000 1110 1001 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 3(10) =

0.0000 0000 1001 0000 0111 1101 0000 1000 1111 1110 1101 0010 0000 1110 1001 1(2) =

0.0000 0000 1001 0000 0111 1101 0000 1000 1111 1110 1101 0010 0000 1110 1001 1(2) × 20 =

1.0010 0000 1111 1010 0001 0001 1111 1101 1010 0100 0001 1101 0011(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 0001 1111 1101 1010 0100 0001 1101 0011


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 0001 1111 1101 1010 0100 0001 1101 0011 =

0010 0000 1111 1010 0001 0001 1111 1101 1010 0100 0001 1101 0011


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 0001 1111 1101 1010 0100 0001 1101 0011


The base ten decimal number 0.002 204 718 3 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 0110 - 0010 0000 1111 1010 0001 0001 1111 1101 1010 0100 0001 1101 0011

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

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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