0.119 999 999 894 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.119 999 999 894(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.119 999 999 894(10) to 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: 0.119 999 999 894.

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) 0.119 999 999 894 × 2 = 0 + 0.239 999 999 788;
  • 2) 0.239 999 999 788 × 2 = 0 + 0.479 999 999 576;
  • 3) 0.479 999 999 576 × 2 = 0 + 0.959 999 999 152;
  • 4) 0.959 999 999 152 × 2 = 1 + 0.919 999 998 304;
  • 5) 0.919 999 998 304 × 2 = 1 + 0.839 999 996 608;
  • 6) 0.839 999 996 608 × 2 = 1 + 0.679 999 993 216;
  • 7) 0.679 999 993 216 × 2 = 1 + 0.359 999 986 432;
  • 8) 0.359 999 986 432 × 2 = 0 + 0.719 999 972 864;
  • 9) 0.719 999 972 864 × 2 = 1 + 0.439 999 945 728;
  • 10) 0.439 999 945 728 × 2 = 0 + 0.879 999 891 456;
  • 11) 0.879 999 891 456 × 2 = 1 + 0.759 999 782 912;
  • 12) 0.759 999 782 912 × 2 = 1 + 0.519 999 565 824;
  • 13) 0.519 999 565 824 × 2 = 1 + 0.039 999 131 648;
  • 14) 0.039 999 131 648 × 2 = 0 + 0.079 998 263 296;
  • 15) 0.079 998 263 296 × 2 = 0 + 0.159 996 526 592;
  • 16) 0.159 996 526 592 × 2 = 0 + 0.319 993 053 184;
  • 17) 0.319 993 053 184 × 2 = 0 + 0.639 986 106 368;
  • 18) 0.639 986 106 368 × 2 = 1 + 0.279 972 212 736;
  • 19) 0.279 972 212 736 × 2 = 0 + 0.559 944 425 472;
  • 20) 0.559 944 425 472 × 2 = 1 + 0.119 888 850 944;
  • 21) 0.119 888 850 944 × 2 = 0 + 0.239 777 701 888;
  • 22) 0.239 777 701 888 × 2 = 0 + 0.479 555 403 776;
  • 23) 0.479 555 403 776 × 2 = 0 + 0.959 110 807 552;
  • 24) 0.959 110 807 552 × 2 = 1 + 0.918 221 615 104;
  • 25) 0.918 221 615 104 × 2 = 1 + 0.836 443 230 208;
  • 26) 0.836 443 230 208 × 2 = 1 + 0.672 886 460 416;
  • 27) 0.672 886 460 416 × 2 = 1 + 0.345 772 920 832;
  • 28) 0.345 772 920 832 × 2 = 0 + 0.691 545 841 664;
  • 29) 0.691 545 841 664 × 2 = 1 + 0.383 091 683 328;
  • 30) 0.383 091 683 328 × 2 = 0 + 0.766 183 366 656;
  • 31) 0.766 183 366 656 × 2 = 1 + 0.532 366 733 312;
  • 32) 0.532 366 733 312 × 2 = 1 + 0.064 733 466 624;
  • 33) 0.064 733 466 624 × 2 = 0 + 0.129 466 933 248;
  • 34) 0.129 466 933 248 × 2 = 0 + 0.258 933 866 496;
  • 35) 0.258 933 866 496 × 2 = 0 + 0.517 867 732 992;
  • 36) 0.517 867 732 992 × 2 = 1 + 0.035 735 465 984;
  • 37) 0.035 735 465 984 × 2 = 0 + 0.071 470 931 968;
  • 38) 0.071 470 931 968 × 2 = 0 + 0.142 941 863 936;
  • 39) 0.142 941 863 936 × 2 = 0 + 0.285 883 727 872;
  • 40) 0.285 883 727 872 × 2 = 0 + 0.571 767 455 744;
  • 41) 0.571 767 455 744 × 2 = 1 + 0.143 534 911 488;
  • 42) 0.143 534 911 488 × 2 = 0 + 0.287 069 822 976;
  • 43) 0.287 069 822 976 × 2 = 0 + 0.574 139 645 952;
  • 44) 0.574 139 645 952 × 2 = 1 + 0.148 279 291 904;
  • 45) 0.148 279 291 904 × 2 = 0 + 0.296 558 583 808;
  • 46) 0.296 558 583 808 × 2 = 0 + 0.593 117 167 616;
  • 47) 0.593 117 167 616 × 2 = 1 + 0.186 234 335 232;
  • 48) 0.186 234 335 232 × 2 = 0 + 0.372 468 670 464;
  • 49) 0.372 468 670 464 × 2 = 0 + 0.744 937 340 928;
  • 50) 0.744 937 340 928 × 2 = 1 + 0.489 874 681 856;
  • 51) 0.489 874 681 856 × 2 = 0 + 0.979 749 363 712;
  • 52) 0.979 749 363 712 × 2 = 1 + 0.959 498 727 424;
  • 53) 0.959 498 727 424 × 2 = 1 + 0.918 997 454 848;
  • 54) 0.918 997 454 848 × 2 = 1 + 0.837 994 909 696;
  • 55) 0.837 994 909 696 × 2 = 1 + 0.675 989 819 392;
  • 56) 0.675 989 819 392 × 2 = 1 + 0.351 979 638 784;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.119 999 999 894(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111(2)

5. Positive number before normalization:

0.119 999 999 894(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.119 999 999 894(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111(2) =

0.0001 1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111(2) × 20 =

1.1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111(2) × 2-4


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): -4

Mantissa (not normalized):
1.1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111 =

1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111


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 1011

Mantissa (52 bits) =
1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111


Decimal number 0.119 999 999 894 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1110 1011 1000 0101 0001 1110 1011 0001 0000 1001 0010 0101 1111

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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