Convert 0.008 388 608 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

0.008 388 608(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: 0.008 388 608.

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) 0.008 388 608 × 2 = 0 + 0.016 777 216;
  • 2) 0.016 777 216 × 2 = 0 + 0.033 554 432;
  • 3) 0.033 554 432 × 2 = 0 + 0.067 108 864;
  • 4) 0.067 108 864 × 2 = 0 + 0.134 217 728;
  • 5) 0.134 217 728 × 2 = 0 + 0.268 435 456;
  • 6) 0.268 435 456 × 2 = 0 + 0.536 870 912;
  • 7) 0.536 870 912 × 2 = 1 + 0.073 741 824;
  • 8) 0.073 741 824 × 2 = 0 + 0.147 483 648;
  • 9) 0.147 483 648 × 2 = 0 + 0.294 967 296;
  • 10) 0.294 967 296 × 2 = 0 + 0.589 934 592;
  • 11) 0.589 934 592 × 2 = 1 + 0.179 869 184;
  • 12) 0.179 869 184 × 2 = 0 + 0.359 738 368;
  • 13) 0.359 738 368 × 2 = 0 + 0.719 476 736;
  • 14) 0.719 476 736 × 2 = 1 + 0.438 953 472;
  • 15) 0.438 953 472 × 2 = 0 + 0.877 906 944;
  • 16) 0.877 906 944 × 2 = 1 + 0.755 813 888;
  • 17) 0.755 813 888 × 2 = 1 + 0.511 627 776;
  • 18) 0.511 627 776 × 2 = 1 + 0.023 255 552;
  • 19) 0.023 255 552 × 2 = 0 + 0.046 511 104;
  • 20) 0.046 511 104 × 2 = 0 + 0.093 022 208;
  • 21) 0.093 022 208 × 2 = 0 + 0.186 044 416;
  • 22) 0.186 044 416 × 2 = 0 + 0.372 088 832;
  • 23) 0.372 088 832 × 2 = 0 + 0.744 177 664;
  • 24) 0.744 177 664 × 2 = 1 + 0.488 355 328;
  • 25) 0.488 355 328 × 2 = 0 + 0.976 710 656;
  • 26) 0.976 710 656 × 2 = 1 + 0.953 421 312;
  • 27) 0.953 421 312 × 2 = 1 + 0.906 842 624;
  • 28) 0.906 842 624 × 2 = 1 + 0.813 685 248;
  • 29) 0.813 685 248 × 2 = 1 + 0.627 370 496;
  • 30) 0.627 370 496 × 2 = 1 + 0.254 740 992;
  • 31) 0.254 740 992 × 2 = 0 + 0.509 481 984;
  • 32) 0.509 481 984 × 2 = 1 + 0.018 963 968;
  • 33) 0.018 963 968 × 2 = 0 + 0.037 927 936;
  • 34) 0.037 927 936 × 2 = 0 + 0.075 855 872;
  • 35) 0.075 855 872 × 2 = 0 + 0.151 711 744;
  • 36) 0.151 711 744 × 2 = 0 + 0.303 423 488;
  • 37) 0.303 423 488 × 2 = 0 + 0.606 846 976;
  • 38) 0.606 846 976 × 2 = 1 + 0.213 693 952;
  • 39) 0.213 693 952 × 2 = 0 + 0.427 387 904;
  • 40) 0.427 387 904 × 2 = 0 + 0.854 775 808;
  • 41) 0.854 775 808 × 2 = 1 + 0.709 551 616;
  • 42) 0.709 551 616 × 2 = 1 + 0.419 103 232;
  • 43) 0.419 103 232 × 2 = 0 + 0.838 206 464;
  • 44) 0.838 206 464 × 2 = 1 + 0.676 412 928;
  • 45) 0.676 412 928 × 2 = 1 + 0.352 825 856;
  • 46) 0.352 825 856 × 2 = 0 + 0.705 651 712;
  • 47) 0.705 651 712 × 2 = 1 + 0.411 303 424;
  • 48) 0.411 303 424 × 2 = 0 + 0.822 606 848;
  • 49) 0.822 606 848 × 2 = 1 + 0.645 213 696;
  • 50) 0.645 213 696 × 2 = 1 + 0.290 427 392;
  • 51) 0.290 427 392 × 2 = 0 + 0.580 854 784;
  • 52) 0.580 854 784 × 2 = 1 + 0.161 709 568;
  • 53) 0.161 709 568 × 2 = 0 + 0.323 419 136;
  • 54) 0.323 419 136 × 2 = 0 + 0.646 838 272;
  • 55) 0.646 838 272 × 2 = 1 + 0.293 676 544;
  • 56) 0.293 676 544 × 2 = 0 + 0.587 353 088;
  • 57) 0.587 353 088 × 2 = 1 + 0.174 706 176;
  • 58) 0.174 706 176 × 2 = 0 + 0.349 412 352;
  • 59) 0.349 412 352 × 2 = 0 + 0.698 824 704;

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:

0.008 388 608(10) =


0.0000 0010 0010 0101 1100 0001 0111 1101 0000 0100 1101 1010 1101 0010 100(2)


5. Positive number before normalization:

0.008 388 608(10) =


0.0000 0010 0010 0101 1100 0001 0111 1101 0000 0100 1101 1010 1101 0010 100(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right so that only one non zero digit remains to the left of it:

0.008 388 608(10) =


0.0000 0010 0010 0101 1100 0001 0111 1101 0000 0100 1101 1010 1101 0010 100(2) =


0.0000 0010 0010 0101 1100 0001 0111 1101 0000 0100 1101 1010 1101 0010 100(2) × 20 =


1.0001 0010 1110 0000 1011 1110 1000 0010 0110 1101 0110 1001 0100(2) × 2-7


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): -7


Mantissa (not normalized):
1.0001 0010 1110 0000 1011 1110 1000 0010 0110 1101 0110 1001 0100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


-7 + 2(11-1) - 1 =


(-7 + 1 023)(10) =


1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =


1016(10) =


011 1111 1000(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. 0001 0010 1110 0000 1011 1110 1000 0010 0110 1101 0110 1001 0100 =


0001 0010 1110 0000 1011 1110 1000 0010 0110 1101 0110 1001 0100


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 1000


Mantissa (52 bits) =
0001 0010 1110 0000 1011 1110 1000 0010 0110 1101 0110 1001 0100


Number 0.008 388 608 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1000 - 0001 0010 1110 0000 1011 1110 1000 0010 0110 1101 0110 1001 0100

(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
    • 1

      55
    • 0

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 0

      46
    • 1

      45
    • 0

      44
    • 1

      43
    • 1

      42
    • 1

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 0

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 1

      31
    • 1

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 0

      0

More operations of this kind:

0.008 388 607 = ? ... 0.008 388 609 = ?


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

0.008 388 608 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:48 UTC (GMT)
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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