Convert 0.002 204 718 632 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

0.002 204 718 632(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.002 204 718 632.

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) 0.002 204 718 632 × 2 = 0 + 0.004 409 437 264;
  • 2) 0.004 409 437 264 × 2 = 0 + 0.008 818 874 528;
  • 3) 0.008 818 874 528 × 2 = 0 + 0.017 637 749 056;
  • 4) 0.017 637 749 056 × 2 = 0 + 0.035 275 498 112;
  • 5) 0.035 275 498 112 × 2 = 0 + 0.070 550 996 224;
  • 6) 0.070 550 996 224 × 2 = 0 + 0.141 101 992 448;
  • 7) 0.141 101 992 448 × 2 = 0 + 0.282 203 984 896;
  • 8) 0.282 203 984 896 × 2 = 0 + 0.564 407 969 792;
  • 9) 0.564 407 969 792 × 2 = 1 + 0.128 815 939 584;
  • 10) 0.128 815 939 584 × 2 = 0 + 0.257 631 879 168;
  • 11) 0.257 631 879 168 × 2 = 0 + 0.515 263 758 336;
  • 12) 0.515 263 758 336 × 2 = 1 + 0.030 527 516 672;
  • 13) 0.030 527 516 672 × 2 = 0 + 0.061 055 033 344;
  • 14) 0.061 055 033 344 × 2 = 0 + 0.122 110 066 688;
  • 15) 0.122 110 066 688 × 2 = 0 + 0.244 220 133 376;
  • 16) 0.244 220 133 376 × 2 = 0 + 0.488 440 266 752;
  • 17) 0.488 440 266 752 × 2 = 0 + 0.976 880 533 504;
  • 18) 0.976 880 533 504 × 2 = 1 + 0.953 761 067 008;
  • 19) 0.953 761 067 008 × 2 = 1 + 0.907 522 134 016;
  • 20) 0.907 522 134 016 × 2 = 1 + 0.815 044 268 032;
  • 21) 0.815 044 268 032 × 2 = 1 + 0.630 088 536 064;
  • 22) 0.630 088 536 064 × 2 = 1 + 0.260 177 072 128;
  • 23) 0.260 177 072 128 × 2 = 0 + 0.520 354 144 256;
  • 24) 0.520 354 144 256 × 2 = 1 + 0.040 708 288 512;
  • 25) 0.040 708 288 512 × 2 = 0 + 0.081 416 577 024;
  • 26) 0.081 416 577 024 × 2 = 0 + 0.162 833 154 048;
  • 27) 0.162 833 154 048 × 2 = 0 + 0.325 666 308 096;
  • 28) 0.325 666 308 096 × 2 = 0 + 0.651 332 616 192;
  • 29) 0.651 332 616 192 × 2 = 1 + 0.302 665 232 384;
  • 30) 0.302 665 232 384 × 2 = 0 + 0.605 330 464 768;
  • 31) 0.605 330 464 768 × 2 = 1 + 0.210 660 929 536;
  • 32) 0.210 660 929 536 × 2 = 0 + 0.421 321 859 072;
  • 33) 0.421 321 859 072 × 2 = 0 + 0.842 643 718 144;
  • 34) 0.842 643 718 144 × 2 = 1 + 0.685 287 436 288;
  • 35) 0.685 287 436 288 × 2 = 1 + 0.370 574 872 576;
  • 36) 0.370 574 872 576 × 2 = 0 + 0.741 149 745 152;
  • 37) 0.741 149 745 152 × 2 = 1 + 0.482 299 490 304;
  • 38) 0.482 299 490 304 × 2 = 0 + 0.964 598 980 608;
  • 39) 0.964 598 980 608 × 2 = 1 + 0.929 197 961 216;
  • 40) 0.929 197 961 216 × 2 = 1 + 0.858 395 922 432;
  • 41) 0.858 395 922 432 × 2 = 1 + 0.716 791 844 864;
  • 42) 0.716 791 844 864 × 2 = 1 + 0.433 583 689 728;
  • 43) 0.433 583 689 728 × 2 = 0 + 0.867 167 379 456;
  • 44) 0.867 167 379 456 × 2 = 1 + 0.734 334 758 912;
  • 45) 0.734 334 758 912 × 2 = 1 + 0.468 669 517 824;
  • 46) 0.468 669 517 824 × 2 = 0 + 0.937 339 035 648;
  • 47) 0.937 339 035 648 × 2 = 1 + 0.874 678 071 296;
  • 48) 0.874 678 071 296 × 2 = 1 + 0.749 356 142 592;
  • 49) 0.749 356 142 592 × 2 = 1 + 0.498 712 285 184;
  • 50) 0.498 712 285 184 × 2 = 0 + 0.997 424 570 368;
  • 51) 0.997 424 570 368 × 2 = 1 + 0.994 849 140 736;
  • 52) 0.994 849 140 736 × 2 = 1 + 0.989 698 281 472;
  • 53) 0.989 698 281 472 × 2 = 1 + 0.979 396 562 944;
  • 54) 0.979 396 562 944 × 2 = 1 + 0.958 793 125 888;
  • 55) 0.958 793 125 888 × 2 = 1 + 0.917 586 251 776;
  • 56) 0.917 586 251 776 × 2 = 1 + 0.835 172 503 552;
  • 57) 0.835 172 503 552 × 2 = 1 + 0.670 345 007 104;
  • 58) 0.670 345 007 104 × 2 = 1 + 0.340 690 014 208;
  • 59) 0.340 690 014 208 × 2 = 0 + 0.681 380 028 416;
  • 60) 0.681 380 028 416 × 2 = 1 + 0.362 760 056 832;
  • 61) 0.362 760 056 832 × 2 = 0 + 0.725 520 113 664;

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:

0.002 204 718 632(10) =


0.0000 0000 1001 0000 0111 1101 0000 1010 0110 1011 1101 1011 1011 1111 1101 0(2)


5. Positive number before normalization:

0.002 204 718 632(10) =


0.0000 0000 1001 0000 0111 1101 0000 1010 0110 1011 1101 1011 1011 1111 1101 0(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:

0.002 204 718 632(10) =


0.0000 0000 1001 0000 0111 1101 0000 1010 0110 1011 1101 1011 1011 1111 1101 0(2) =


0.0000 0000 1001 0000 0111 1101 0000 1010 0110 1011 1101 1011 1011 1111 1101 0(2) × 20 =


1.0010 0000 1111 1010 0001 0100 1101 0111 1011 0111 0111 1111 1010(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 1101 0111 1011 0111 0111 1111 1010


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 1101 0111 1011 0111 0111 1111 1010 =


0010 0000 1111 1010 0001 0100 1101 0111 1011 0111 0111 1111 1010


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 1101 0111 1011 0111 0111 1111 1010


Number 0.002 204 718 632 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 0110 - 0010 0000 1111 1010 0001 0100 1101 0111 1011 0111 0111 1111 1010

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 0

      55
    • 1

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 1

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 1

      43
    • 1

      42
    • 1

      41
    • 1

      40
    • 1

      39
    • 0

      38
    • 1

      37
    • 0

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 0

      31
    • 1

      30
    • 0

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 0

      23
    • 1

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 0

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 1

      7
    • 1

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 0

      0

More operations of this kind:

0.002 204 718 631 = ? ... 0.002 204 718 633 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

0.002 204 718 632 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
11 110 110 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
0.696 350 260 7 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
111 011 010 000 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
1 111 010 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
1.999 999 940 1 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
1 111.111 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
2.333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 333 332 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
111 011 010 001.001 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
1 110 107 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:20 UTC (GMT)
11 101 009 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:19 UTC (GMT)
11 101 007 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:19 UTC (GMT)
-18 446 744 073 709 551 606 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:19 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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