Convert 4.072 882 27 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

4.072 882 27(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: 4.
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;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.

4(10) =


100(2)


3. Convert to the binary (base 2) the fractional part: 0.072 882 27.

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) 0.072 882 27 × 2 = 0 + 0.145 764 54;
  • 2) 0.145 764 54 × 2 = 0 + 0.291 529 08;
  • 3) 0.291 529 08 × 2 = 0 + 0.583 058 16;
  • 4) 0.583 058 16 × 2 = 1 + 0.166 116 32;
  • 5) 0.166 116 32 × 2 = 0 + 0.332 232 64;
  • 6) 0.332 232 64 × 2 = 0 + 0.664 465 28;
  • 7) 0.664 465 28 × 2 = 1 + 0.328 930 56;
  • 8) 0.328 930 56 × 2 = 0 + 0.657 861 12;
  • 9) 0.657 861 12 × 2 = 1 + 0.315 722 24;
  • 10) 0.315 722 24 × 2 = 0 + 0.631 444 48;
  • 11) 0.631 444 48 × 2 = 1 + 0.262 888 96;
  • 12) 0.262 888 96 × 2 = 0 + 0.525 777 92;
  • 13) 0.525 777 92 × 2 = 1 + 0.051 555 84;
  • 14) 0.051 555 84 × 2 = 0 + 0.103 111 68;
  • 15) 0.103 111 68 × 2 = 0 + 0.206 223 36;
  • 16) 0.206 223 36 × 2 = 0 + 0.412 446 72;
  • 17) 0.412 446 72 × 2 = 0 + 0.824 893 44;
  • 18) 0.824 893 44 × 2 = 1 + 0.649 786 88;
  • 19) 0.649 786 88 × 2 = 1 + 0.299 573 76;
  • 20) 0.299 573 76 × 2 = 0 + 0.599 147 52;
  • 21) 0.599 147 52 × 2 = 1 + 0.198 295 04;
  • 22) 0.198 295 04 × 2 = 0 + 0.396 590 08;
  • 23) 0.396 590 08 × 2 = 0 + 0.793 180 16;
  • 24) 0.793 180 16 × 2 = 1 + 0.586 360 32;
  • 25) 0.586 360 32 × 2 = 1 + 0.172 720 64;
  • 26) 0.172 720 64 × 2 = 0 + 0.345 441 28;
  • 27) 0.345 441 28 × 2 = 0 + 0.690 882 56;
  • 28) 0.690 882 56 × 2 = 1 + 0.381 765 12;
  • 29) 0.381 765 12 × 2 = 0 + 0.763 530 24;
  • 30) 0.763 530 24 × 2 = 1 + 0.527 060 48;
  • 31) 0.527 060 48 × 2 = 1 + 0.054 120 96;
  • 32) 0.054 120 96 × 2 = 0 + 0.108 241 92;
  • 33) 0.108 241 92 × 2 = 0 + 0.216 483 84;
  • 34) 0.216 483 84 × 2 = 0 + 0.432 967 68;
  • 35) 0.432 967 68 × 2 = 0 + 0.865 935 36;
  • 36) 0.865 935 36 × 2 = 1 + 0.731 870 72;
  • 37) 0.731 870 72 × 2 = 1 + 0.463 741 44;
  • 38) 0.463 741 44 × 2 = 0 + 0.927 482 88;
  • 39) 0.927 482 88 × 2 = 1 + 0.854 965 76;
  • 40) 0.854 965 76 × 2 = 1 + 0.709 931 52;
  • 41) 0.709 931 52 × 2 = 1 + 0.419 863 04;
  • 42) 0.419 863 04 × 2 = 0 + 0.839 726 08;
  • 43) 0.839 726 08 × 2 = 1 + 0.679 452 16;
  • 44) 0.679 452 16 × 2 = 1 + 0.358 904 32;
  • 45) 0.358 904 32 × 2 = 0 + 0.717 808 64;
  • 46) 0.717 808 64 × 2 = 1 + 0.435 617 28;
  • 47) 0.435 617 28 × 2 = 0 + 0.871 234 56;
  • 48) 0.871 234 56 × 2 = 1 + 0.742 469 12;
  • 49) 0.742 469 12 × 2 = 1 + 0.484 938 24;
  • 50) 0.484 938 24 × 2 = 0 + 0.969 876 48;
  • 51) 0.969 876 48 × 2 = 1 + 0.939 752 96;
  • 52) 0.939 752 96 × 2 = 1 + 0.879 505 92;
  • 53) 0.879 505 92 × 2 = 1 + 0.759 011 84;

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:

0.072 882 27(10) =


0.0001 0010 1010 1000 0110 1001 1001 0110 0001 1011 1011 0101 1011 1(2)


5. Positive number before normalization:

4.072 882 27(10) =


100.0001 0010 1010 1000 0110 1001 1001 0110 0001 1011 1011 0101 1011 1(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left so that only one non zero digit remains to the left of it:

4.072 882 27(10) =


100.0001 0010 1010 1000 0110 1001 1001 0110 0001 1011 1011 0101 1011 1(2) =


100.0001 0010 1010 1000 0110 1001 1001 0110 0001 1011 1011 0101 1011 1(2) × 20 =


1.0000 0100 1010 1010 0001 1010 0110 0101 1000 0110 1110 1101 0110 111(2) × 22


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): 2


Mantissa (not normalized):
1.0000 0100 1010 1010 0001 1010 0110 0101 1000 0110 1110 1101 0110 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


2 + 2(11-1) - 1 =


(2 + 1 023)(10) =


1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 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) =


1025(10) =


100 0000 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. 0000 0100 1010 1010 0001 1010 0110 0101 1000 0110 1110 1101 0110 111 =


0000 0100 1010 1010 0001 1010 0110 0101 1000 0110 1110 1101 0110


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 0000 0001


Mantissa (52 bits) =
0000 0100 1010 1010 0001 1010 0110 0101 1000 0110 1110 1101 0110


Number 4.072 882 27 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0001 - 0000 0100 1010 1010 0001 1010 0110 0101 1000 0110 1110 1101 0110

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 1

      46
    • 0

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 0

      40
    • 1

      39
    • 0

      38
    • 1

      37
    • 0

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 1

      31
    • 0

      30
    • 1

      29
    • 0

      28
    • 0

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 0

      23
    • 1

      22
    • 0

      21
    • 1

      20
    • 1

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 1

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 1

      1
    • 0

      0

More operations of this kind:

4.072 882 26 = ? ... 4.072 882 28 = ?


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

4.072 882 27 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:56 UTC (GMT)
0.000 000 000 000 000 000 000 013 806 448 5 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:56 UTC (GMT)
2.536 98 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:56 UTC (GMT)
-99.45 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:56 UTC (GMT)
0.324 218 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:55 UTC (GMT)
179 769 313 486 231 550 856 124 328 384 506 240 234 343 437 157 459 335 924 404 872 448 581 845 754 556 114 388 470 639 943 126 220 321 960 804 027 157 371 570 809 852 884 964 511 743 044 087 662 767 600 909 594 331 927 728 237 078 876 188 760 579 532 563 768 698 654 064 825 262 115 771 015 791 463 983 014 857 704 008 123 419 459 386 245 141 723 703 148 097 529 108 423 358 883 457 665 451 722 744 025 579 516 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:55 UTC (GMT)
-90.69 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:55 UTC (GMT)
-17.77 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:55 UTC (GMT)
32.008 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:55 UTC (GMT)
90 071 992 547 409 929 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:55 UTC (GMT)
0.140 624 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:55 UTC (GMT)
1 505 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:54 UTC (GMT)
64.457 565 667 99 to 64 bit double precision IEEE 754 binary floating point = ? Apr 14 10:54 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