Convert 466 886.886 789 714 7 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

466 886.886 789 714 7(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: 466 886.
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;
  • 466 886 ÷ 2 = 233 443 + 0;
  • 233 443 ÷ 2 = 116 721 + 1;
  • 116 721 ÷ 2 = 58 360 + 1;
  • 58 360 ÷ 2 = 29 180 + 0;
  • 29 180 ÷ 2 = 14 590 + 0;
  • 14 590 ÷ 2 = 7 295 + 0;
  • 7 295 ÷ 2 = 3 647 + 1;
  • 3 647 ÷ 2 = 1 823 + 1;
  • 1 823 ÷ 2 = 911 + 1;
  • 911 ÷ 2 = 455 + 1;
  • 455 ÷ 2 = 227 + 1;
  • 227 ÷ 2 = 113 + 1;
  • 113 ÷ 2 = 56 + 1;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

466 886(10) =


111 0001 1111 1100 0110(2)


3. Convert to the binary (base 2) the fractional part: 0.886 789 714 7.

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) 0.886 789 714 7 × 2 = 1 + 0.773 579 429 4;
  • 2) 0.773 579 429 4 × 2 = 1 + 0.547 158 858 8;
  • 3) 0.547 158 858 8 × 2 = 1 + 0.094 317 717 6;
  • 4) 0.094 317 717 6 × 2 = 0 + 0.188 635 435 2;
  • 5) 0.188 635 435 2 × 2 = 0 + 0.377 270 870 4;
  • 6) 0.377 270 870 4 × 2 = 0 + 0.754 541 740 8;
  • 7) 0.754 541 740 8 × 2 = 1 + 0.509 083 481 6;
  • 8) 0.509 083 481 6 × 2 = 1 + 0.018 166 963 2;
  • 9) 0.018 166 963 2 × 2 = 0 + 0.036 333 926 4;
  • 10) 0.036 333 926 4 × 2 = 0 + 0.072 667 852 8;
  • 11) 0.072 667 852 8 × 2 = 0 + 0.145 335 705 6;
  • 12) 0.145 335 705 6 × 2 = 0 + 0.290 671 411 2;
  • 13) 0.290 671 411 2 × 2 = 0 + 0.581 342 822 4;
  • 14) 0.581 342 822 4 × 2 = 1 + 0.162 685 644 8;
  • 15) 0.162 685 644 8 × 2 = 0 + 0.325 371 289 6;
  • 16) 0.325 371 289 6 × 2 = 0 + 0.650 742 579 2;
  • 17) 0.650 742 579 2 × 2 = 1 + 0.301 485 158 4;
  • 18) 0.301 485 158 4 × 2 = 0 + 0.602 970 316 8;
  • 19) 0.602 970 316 8 × 2 = 1 + 0.205 940 633 6;
  • 20) 0.205 940 633 6 × 2 = 0 + 0.411 881 267 2;
  • 21) 0.411 881 267 2 × 2 = 0 + 0.823 762 534 4;
  • 22) 0.823 762 534 4 × 2 = 1 + 0.647 525 068 8;
  • 23) 0.647 525 068 8 × 2 = 1 + 0.295 050 137 6;
  • 24) 0.295 050 137 6 × 2 = 0 + 0.590 100 275 2;
  • 25) 0.590 100 275 2 × 2 = 1 + 0.180 200 550 4;
  • 26) 0.180 200 550 4 × 2 = 0 + 0.360 401 100 8;
  • 27) 0.360 401 100 8 × 2 = 0 + 0.720 802 201 6;
  • 28) 0.720 802 201 6 × 2 = 1 + 0.441 604 403 2;
  • 29) 0.441 604 403 2 × 2 = 0 + 0.883 208 806 4;
  • 30) 0.883 208 806 4 × 2 = 1 + 0.766 417 612 8;
  • 31) 0.766 417 612 8 × 2 = 1 + 0.532 835 225 6;
  • 32) 0.532 835 225 6 × 2 = 1 + 0.065 670 451 2;
  • 33) 0.065 670 451 2 × 2 = 0 + 0.131 340 902 4;
  • 34) 0.131 340 902 4 × 2 = 0 + 0.262 681 804 8;
  • 35) 0.262 681 804 8 × 2 = 0 + 0.525 363 609 6;
  • 36) 0.525 363 609 6 × 2 = 1 + 0.050 727 219 2;
  • 37) 0.050 727 219 2 × 2 = 0 + 0.101 454 438 4;
  • 38) 0.101 454 438 4 × 2 = 0 + 0.202 908 876 8;
  • 39) 0.202 908 876 8 × 2 = 0 + 0.405 817 753 6;
  • 40) 0.405 817 753 6 × 2 = 0 + 0.811 635 507 2;
  • 41) 0.811 635 507 2 × 2 = 1 + 0.623 271 014 4;
  • 42) 0.623 271 014 4 × 2 = 1 + 0.246 542 028 8;
  • 43) 0.246 542 028 8 × 2 = 0 + 0.493 084 057 6;
  • 44) 0.493 084 057 6 × 2 = 0 + 0.986 168 115 2;
  • 45) 0.986 168 115 2 × 2 = 1 + 0.972 336 230 4;
  • 46) 0.972 336 230 4 × 2 = 1 + 0.944 672 460 8;
  • 47) 0.944 672 460 8 × 2 = 1 + 0.889 344 921 6;
  • 48) 0.889 344 921 6 × 2 = 1 + 0.778 689 843 2;
  • 49) 0.778 689 843 2 × 2 = 1 + 0.557 379 686 4;
  • 50) 0.557 379 686 4 × 2 = 1 + 0.114 759 372 8;
  • 51) 0.114 759 372 8 × 2 = 0 + 0.229 518 745 6;
  • 52) 0.229 518 745 6 × 2 = 0 + 0.459 037 491 2;
  • 53) 0.459 037 491 2 × 2 = 0 + 0.918 074 982 4;

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:

0.886 789 714 7(10) =


0.1110 0011 0000 0100 1010 0110 1001 0111 0001 0000 1100 1111 1100 0(2)


5. Positive number before normalization:

466 886.886 789 714 7(10) =


111 0001 1111 1100 0110.1110 0011 0000 0100 1010 0110 1001 0111 0001 0000 1100 1111 1100 0(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the left so that only one non zero digit remains to the left of it:

466 886.886 789 714 7(10) =


111 0001 1111 1100 0110.1110 0011 0000 0100 1010 0110 1001 0111 0001 0000 1100 1111 1100 0(2) =


111 0001 1111 1100 0110.1110 0011 0000 0100 1010 0110 1001 0111 0001 0000 1100 1111 1100 0(2) × 20 =


1.1100 0111 1111 0001 1011 1000 1100 0001 0010 1001 1010 0101 1100 0100 0011 0011 1111 000(2) × 218


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): 18


Mantissa (not normalized):
1.1100 0111 1111 0001 1011 1000 1100 0001 0010 1001 1010 0101 1100 0100 0011 0011 1111 000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


18 + 2(11-1) - 1 =


(18 + 1 023)(10) =


1 041(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;
  • 1 041 ÷ 2 = 520 + 1;
  • 520 ÷ 2 = 260 + 0;
  • 260 ÷ 2 = 130 + 0;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1041(10) =


100 0001 0001(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 1100 0111 1111 0001 1011 1000 1100 0001 0010 1001 1010 0101 1100 010 0001 1001 1111 1000 =


1100 0111 1111 0001 1011 1000 1100 0001 0010 1001 1010 0101 1100


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) =
100 0001 0001


Mantissa (52 bits) =
1100 0111 1111 0001 1011 1000 1100 0001 0010 1001 1010 0101 1100


Number 466 886.886 789 714 7 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0001 0001 - 1100 0111 1111 0001 1011 1000 1100 0001 0010 1001 1010 0101 1100

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 1

      56
    • 0

      55
    • 0

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 1

      46
    • 1

      45
    • 1

      44
    • 1

      43
    • 1

      42
    • 1

      41
    • 1

      40
    • 0

      39
    • 0

      38
    • 0

      37
    • 1

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 0

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 1

      20
    • 0

      19
    • 0

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 0

      0

More operations of this kind:

466 886.886 789 714 6 = ? ... 466 886.886 789 714 8 = ?


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

466 886.886 789 714 7 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:05 UTC (GMT)
0.000 04 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:05 UTC (GMT)
11 000 000 101 100 999 999 999 999 999 999 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:04 UTC (GMT)
100 000 000 000 000 000 000 008 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:04 UTC (GMT)
10.312 4 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:04 UTC (GMT)
273.77 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:04 UTC (GMT)
-3 518.8 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:02 UTC (GMT)
22.65 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:02 UTC (GMT)
1 000 000 009 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:02 UTC (GMT)
-123 456.78 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:02 UTC (GMT)
500 014 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:01 UTC (GMT)
0.011 718 74 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:01 UTC (GMT)
11 000 345 657 654 343 567 545 456 545 454 345 643 564 301 111.099 945 643 564 356 543 458 82 to 64 bit double precision IEEE 754 binary floating point = ? Mar 03 03:01 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