0.119 999 999 996 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.119 999 999 996 7(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
0.119 999 999 996 7(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: 0.119 999 999 996 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) 0.119 999 999 996 7 × 2 = 0 + 0.239 999 999 993 4;
  • 2) 0.239 999 999 993 4 × 2 = 0 + 0.479 999 999 986 8;
  • 3) 0.479 999 999 986 8 × 2 = 0 + 0.959 999 999 973 6;
  • 4) 0.959 999 999 973 6 × 2 = 1 + 0.919 999 999 947 2;
  • 5) 0.919 999 999 947 2 × 2 = 1 + 0.839 999 999 894 4;
  • 6) 0.839 999 999 894 4 × 2 = 1 + 0.679 999 999 788 8;
  • 7) 0.679 999 999 788 8 × 2 = 1 + 0.359 999 999 577 6;
  • 8) 0.359 999 999 577 6 × 2 = 0 + 0.719 999 999 155 2;
  • 9) 0.719 999 999 155 2 × 2 = 1 + 0.439 999 998 310 4;
  • 10) 0.439 999 998 310 4 × 2 = 0 + 0.879 999 996 620 8;
  • 11) 0.879 999 996 620 8 × 2 = 1 + 0.759 999 993 241 6;
  • 12) 0.759 999 993 241 6 × 2 = 1 + 0.519 999 986 483 2;
  • 13) 0.519 999 986 483 2 × 2 = 1 + 0.039 999 972 966 4;
  • 14) 0.039 999 972 966 4 × 2 = 0 + 0.079 999 945 932 8;
  • 15) 0.079 999 945 932 8 × 2 = 0 + 0.159 999 891 865 6;
  • 16) 0.159 999 891 865 6 × 2 = 0 + 0.319 999 783 731 2;
  • 17) 0.319 999 783 731 2 × 2 = 0 + 0.639 999 567 462 4;
  • 18) 0.639 999 567 462 4 × 2 = 1 + 0.279 999 134 924 8;
  • 19) 0.279 999 134 924 8 × 2 = 0 + 0.559 998 269 849 6;
  • 20) 0.559 998 269 849 6 × 2 = 1 + 0.119 996 539 699 2;
  • 21) 0.119 996 539 699 2 × 2 = 0 + 0.239 993 079 398 4;
  • 22) 0.239 993 079 398 4 × 2 = 0 + 0.479 986 158 796 8;
  • 23) 0.479 986 158 796 8 × 2 = 0 + 0.959 972 317 593 6;
  • 24) 0.959 972 317 593 6 × 2 = 1 + 0.919 944 635 187 2;
  • 25) 0.919 944 635 187 2 × 2 = 1 + 0.839 889 270 374 4;
  • 26) 0.839 889 270 374 4 × 2 = 1 + 0.679 778 540 748 8;
  • 27) 0.679 778 540 748 8 × 2 = 1 + 0.359 557 081 497 6;
  • 28) 0.359 557 081 497 6 × 2 = 0 + 0.719 114 162 995 2;
  • 29) 0.719 114 162 995 2 × 2 = 1 + 0.438 228 325 990 4;
  • 30) 0.438 228 325 990 4 × 2 = 0 + 0.876 456 651 980 8;
  • 31) 0.876 456 651 980 8 × 2 = 1 + 0.752 913 303 961 6;
  • 32) 0.752 913 303 961 6 × 2 = 1 + 0.505 826 607 923 2;
  • 33) 0.505 826 607 923 2 × 2 = 1 + 0.011 653 215 846 4;
  • 34) 0.011 653 215 846 4 × 2 = 0 + 0.023 306 431 692 8;
  • 35) 0.023 306 431 692 8 × 2 = 0 + 0.046 612 863 385 6;
  • 36) 0.046 612 863 385 6 × 2 = 0 + 0.093 225 726 771 2;
  • 37) 0.093 225 726 771 2 × 2 = 0 + 0.186 451 453 542 4;
  • 38) 0.186 451 453 542 4 × 2 = 0 + 0.372 902 907 084 8;
  • 39) 0.372 902 907 084 8 × 2 = 0 + 0.745 805 814 169 6;
  • 40) 0.745 805 814 169 6 × 2 = 1 + 0.491 611 628 339 2;
  • 41) 0.491 611 628 339 2 × 2 = 0 + 0.983 223 256 678 4;
  • 42) 0.983 223 256 678 4 × 2 = 1 + 0.966 446 513 356 8;
  • 43) 0.966 446 513 356 8 × 2 = 1 + 0.932 893 026 713 6;
  • 44) 0.932 893 026 713 6 × 2 = 1 + 0.865 786 053 427 2;
  • 45) 0.865 786 053 427 2 × 2 = 1 + 0.731 572 106 854 4;
  • 46) 0.731 572 106 854 4 × 2 = 1 + 0.463 144 213 708 8;
  • 47) 0.463 144 213 708 8 × 2 = 0 + 0.926 288 427 417 6;
  • 48) 0.926 288 427 417 6 × 2 = 1 + 0.852 576 854 835 2;
  • 49) 0.852 576 854 835 2 × 2 = 1 + 0.705 153 709 670 4;
  • 50) 0.705 153 709 670 4 × 2 = 1 + 0.410 307 419 340 8;
  • 51) 0.410 307 419 340 8 × 2 = 0 + 0.820 614 838 681 6;
  • 52) 0.820 614 838 681 6 × 2 = 1 + 0.641 229 677 363 2;
  • 53) 0.641 229 677 363 2 × 2 = 1 + 0.282 459 354 726 4;
  • 54) 0.282 459 354 726 4 × 2 = 0 + 0.564 918 709 452 8;
  • 55) 0.564 918 709 452 8 × 2 = 1 + 0.129 837 418 905 6;
  • 56) 0.129 837 418 905 6 × 2 = 0 + 0.259 674 837 811 2;

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:


0.119 999 999 996 7(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0001 0111 1101 1101 1010(2)

5. Positive number before normalization:

0.119 999 999 996 7(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0001 0111 1101 1101 1010(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:


0.119 999 999 996 7(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0001 0111 1101 1101 1010(2) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0001 0111 1101 1101 1010(2) × 20 =

1.1110 1011 1000 0101 0001 1110 1011 1000 0001 0111 1101 1101 1010(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 1000 0001 0111 1101 1101 1010


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 1000 0001 0111 1101 1101 1010 =

1110 1011 1000 0101 0001 1110 1011 1000 0001 0111 1101 1101 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 1011

Mantissa (52 bits) =
1110 1011 1000 0101 0001 1110 1011 1000 0001 0111 1101 1101 1010


Decimal number 0.119 999 999 996 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1110 1011 1000 0101 0001 1110 1011 1000 0001 0111 1101 1101 1010

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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