0.119 999 999 995 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.119 999 999 995 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
0.119 999 999 995 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: 0.119 999 999 995 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) 0.119 999 999 995 7 × 2 = 0 + 0.239 999 999 991 4;
  • 2) 0.239 999 999 991 4 × 2 = 0 + 0.479 999 999 982 8;
  • 3) 0.479 999 999 982 8 × 2 = 0 + 0.959 999 999 965 6;
  • 4) 0.959 999 999 965 6 × 2 = 1 + 0.919 999 999 931 2;
  • 5) 0.919 999 999 931 2 × 2 = 1 + 0.839 999 999 862 4;
  • 6) 0.839 999 999 862 4 × 2 = 1 + 0.679 999 999 724 8;
  • 7) 0.679 999 999 724 8 × 2 = 1 + 0.359 999 999 449 6;
  • 8) 0.359 999 999 449 6 × 2 = 0 + 0.719 999 998 899 2;
  • 9) 0.719 999 998 899 2 × 2 = 1 + 0.439 999 997 798 4;
  • 10) 0.439 999 997 798 4 × 2 = 0 + 0.879 999 995 596 8;
  • 11) 0.879 999 995 596 8 × 2 = 1 + 0.759 999 991 193 6;
  • 12) 0.759 999 991 193 6 × 2 = 1 + 0.519 999 982 387 2;
  • 13) 0.519 999 982 387 2 × 2 = 1 + 0.039 999 964 774 4;
  • 14) 0.039 999 964 774 4 × 2 = 0 + 0.079 999 929 548 8;
  • 15) 0.079 999 929 548 8 × 2 = 0 + 0.159 999 859 097 6;
  • 16) 0.159 999 859 097 6 × 2 = 0 + 0.319 999 718 195 2;
  • 17) 0.319 999 718 195 2 × 2 = 0 + 0.639 999 436 390 4;
  • 18) 0.639 999 436 390 4 × 2 = 1 + 0.279 998 872 780 8;
  • 19) 0.279 998 872 780 8 × 2 = 0 + 0.559 997 745 561 6;
  • 20) 0.559 997 745 561 6 × 2 = 1 + 0.119 995 491 123 2;
  • 21) 0.119 995 491 123 2 × 2 = 0 + 0.239 990 982 246 4;
  • 22) 0.239 990 982 246 4 × 2 = 0 + 0.479 981 964 492 8;
  • 23) 0.479 981 964 492 8 × 2 = 0 + 0.959 963 928 985 6;
  • 24) 0.959 963 928 985 6 × 2 = 1 + 0.919 927 857 971 2;
  • 25) 0.919 927 857 971 2 × 2 = 1 + 0.839 855 715 942 4;
  • 26) 0.839 855 715 942 4 × 2 = 1 + 0.679 711 431 884 8;
  • 27) 0.679 711 431 884 8 × 2 = 1 + 0.359 422 863 769 6;
  • 28) 0.359 422 863 769 6 × 2 = 0 + 0.718 845 727 539 2;
  • 29) 0.718 845 727 539 2 × 2 = 1 + 0.437 691 455 078 4;
  • 30) 0.437 691 455 078 4 × 2 = 0 + 0.875 382 910 156 8;
  • 31) 0.875 382 910 156 8 × 2 = 1 + 0.750 765 820 313 6;
  • 32) 0.750 765 820 313 6 × 2 = 1 + 0.501 531 640 627 2;
  • 33) 0.501 531 640 627 2 × 2 = 1 + 0.003 063 281 254 4;
  • 34) 0.003 063 281 254 4 × 2 = 0 + 0.006 126 562 508 8;
  • 35) 0.006 126 562 508 8 × 2 = 0 + 0.012 253 125 017 6;
  • 36) 0.012 253 125 017 6 × 2 = 0 + 0.024 506 250 035 2;
  • 37) 0.024 506 250 035 2 × 2 = 0 + 0.049 012 500 070 4;
  • 38) 0.049 012 500 070 4 × 2 = 0 + 0.098 025 000 140 8;
  • 39) 0.098 025 000 140 8 × 2 = 0 + 0.196 050 000 281 6;
  • 40) 0.196 050 000 281 6 × 2 = 0 + 0.392 100 000 563 2;
  • 41) 0.392 100 000 563 2 × 2 = 0 + 0.784 200 001 126 4;
  • 42) 0.784 200 001 126 4 × 2 = 1 + 0.568 400 002 252 8;
  • 43) 0.568 400 002 252 8 × 2 = 1 + 0.136 800 004 505 6;
  • 44) 0.136 800 004 505 6 × 2 = 0 + 0.273 600 009 011 2;
  • 45) 0.273 600 009 011 2 × 2 = 0 + 0.547 200 018 022 4;
  • 46) 0.547 200 018 022 4 × 2 = 1 + 0.094 400 036 044 8;
  • 47) 0.094 400 036 044 8 × 2 = 0 + 0.188 800 072 089 6;
  • 48) 0.188 800 072 089 6 × 2 = 0 + 0.377 600 144 179 2;
  • 49) 0.377 600 144 179 2 × 2 = 0 + 0.755 200 288 358 4;
  • 50) 0.755 200 288 358 4 × 2 = 1 + 0.510 400 576 716 8;
  • 51) 0.510 400 576 716 8 × 2 = 1 + 0.020 801 153 433 6;
  • 52) 0.020 801 153 433 6 × 2 = 0 + 0.041 602 306 867 2;
  • 53) 0.041 602 306 867 2 × 2 = 0 + 0.083 204 613 734 4;
  • 54) 0.083 204 613 734 4 × 2 = 0 + 0.166 409 227 468 8;
  • 55) 0.166 409 227 468 8 × 2 = 0 + 0.332 818 454 937 6;
  • 56) 0.332 818 454 937 6 × 2 = 0 + 0.665 636 909 875 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:


0.119 999 999 995 7(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0110 0100 0110 0000(2)

5. Positive number before normalization:

0.119 999 999 995 7(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0110 0100 0110 0000(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:


0.119 999 999 995 7(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0110 0100 0110 0000(2) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0110 0100 0110 0000(2) × 20 =

1.1110 1011 1000 0101 0001 1110 1011 1000 0000 0110 0100 0110 0000(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 0000 0110 0100 0110 0000


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 0000 0110 0100 0110 0000 =

1110 1011 1000 0101 0001 1110 1011 1000 0000 0110 0100 0110 0000


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 0000 0110 0100 0110 0000


Decimal number 0.119 999 999 995 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 0000 0110 0100 0110 0000

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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