-0.016 738 891 601 562 498 11 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 498 11(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
-0.016 738 891 601 562 498 11(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. Start with the positive version of the number:

|-0.016 738 891 601 562 498 11| = 0.016 738 891 601 562 498 11


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 498 11.

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) 0.016 738 891 601 562 498 11 × 2 = 0 + 0.033 477 783 203 124 996 22;
  • 2) 0.033 477 783 203 124 996 22 × 2 = 0 + 0.066 955 566 406 249 992 44;
  • 3) 0.066 955 566 406 249 992 44 × 2 = 0 + 0.133 911 132 812 499 984 88;
  • 4) 0.133 911 132 812 499 984 88 × 2 = 0 + 0.267 822 265 624 999 969 76;
  • 5) 0.267 822 265 624 999 969 76 × 2 = 0 + 0.535 644 531 249 999 939 52;
  • 6) 0.535 644 531 249 999 939 52 × 2 = 1 + 0.071 289 062 499 999 879 04;
  • 7) 0.071 289 062 499 999 879 04 × 2 = 0 + 0.142 578 124 999 999 758 08;
  • 8) 0.142 578 124 999 999 758 08 × 2 = 0 + 0.285 156 249 999 999 516 16;
  • 9) 0.285 156 249 999 999 516 16 × 2 = 0 + 0.570 312 499 999 999 032 32;
  • 10) 0.570 312 499 999 999 032 32 × 2 = 1 + 0.140 624 999 999 998 064 64;
  • 11) 0.140 624 999 999 998 064 64 × 2 = 0 + 0.281 249 999 999 996 129 28;
  • 12) 0.281 249 999 999 996 129 28 × 2 = 0 + 0.562 499 999 999 992 258 56;
  • 13) 0.562 499 999 999 992 258 56 × 2 = 1 + 0.124 999 999 999 984 517 12;
  • 14) 0.124 999 999 999 984 517 12 × 2 = 0 + 0.249 999 999 999 969 034 24;
  • 15) 0.249 999 999 999 969 034 24 × 2 = 0 + 0.499 999 999 999 938 068 48;
  • 16) 0.499 999 999 999 938 068 48 × 2 = 0 + 0.999 999 999 999 876 136 96;
  • 17) 0.999 999 999 999 876 136 96 × 2 = 1 + 0.999 999 999 999 752 273 92;
  • 18) 0.999 999 999 999 752 273 92 × 2 = 1 + 0.999 999 999 999 504 547 84;
  • 19) 0.999 999 999 999 504 547 84 × 2 = 1 + 0.999 999 999 999 009 095 68;
  • 20) 0.999 999 999 999 009 095 68 × 2 = 1 + 0.999 999 999 998 018 191 36;
  • 21) 0.999 999 999 998 018 191 36 × 2 = 1 + 0.999 999 999 996 036 382 72;
  • 22) 0.999 999 999 996 036 382 72 × 2 = 1 + 0.999 999 999 992 072 765 44;
  • 23) 0.999 999 999 992 072 765 44 × 2 = 1 + 0.999 999 999 984 145 530 88;
  • 24) 0.999 999 999 984 145 530 88 × 2 = 1 + 0.999 999 999 968 291 061 76;
  • 25) 0.999 999 999 968 291 061 76 × 2 = 1 + 0.999 999 999 936 582 123 52;
  • 26) 0.999 999 999 936 582 123 52 × 2 = 1 + 0.999 999 999 873 164 247 04;
  • 27) 0.999 999 999 873 164 247 04 × 2 = 1 + 0.999 999 999 746 328 494 08;
  • 28) 0.999 999 999 746 328 494 08 × 2 = 1 + 0.999 999 999 492 656 988 16;
  • 29) 0.999 999 999 492 656 988 16 × 2 = 1 + 0.999 999 998 985 313 976 32;
  • 30) 0.999 999 998 985 313 976 32 × 2 = 1 + 0.999 999 997 970 627 952 64;
  • 31) 0.999 999 997 970 627 952 64 × 2 = 1 + 0.999 999 995 941 255 905 28;
  • 32) 0.999 999 995 941 255 905 28 × 2 = 1 + 0.999 999 991 882 511 810 56;
  • 33) 0.999 999 991 882 511 810 56 × 2 = 1 + 0.999 999 983 765 023 621 12;
  • 34) 0.999 999 983 765 023 621 12 × 2 = 1 + 0.999 999 967 530 047 242 24;
  • 35) 0.999 999 967 530 047 242 24 × 2 = 1 + 0.999 999 935 060 094 484 48;
  • 36) 0.999 999 935 060 094 484 48 × 2 = 1 + 0.999 999 870 120 188 968 96;
  • 37) 0.999 999 870 120 188 968 96 × 2 = 1 + 0.999 999 740 240 377 937 92;
  • 38) 0.999 999 740 240 377 937 92 × 2 = 1 + 0.999 999 480 480 755 875 84;
  • 39) 0.999 999 480 480 755 875 84 × 2 = 1 + 0.999 998 960 961 511 751 68;
  • 40) 0.999 998 960 961 511 751 68 × 2 = 1 + 0.999 997 921 923 023 503 36;
  • 41) 0.999 997 921 923 023 503 36 × 2 = 1 + 0.999 995 843 846 047 006 72;
  • 42) 0.999 995 843 846 047 006 72 × 2 = 1 + 0.999 991 687 692 094 013 44;
  • 43) 0.999 991 687 692 094 013 44 × 2 = 1 + 0.999 983 375 384 188 026 88;
  • 44) 0.999 983 375 384 188 026 88 × 2 = 1 + 0.999 966 750 768 376 053 76;
  • 45) 0.999 966 750 768 376 053 76 × 2 = 1 + 0.999 933 501 536 752 107 52;
  • 46) 0.999 933 501 536 752 107 52 × 2 = 1 + 0.999 867 003 073 504 215 04;
  • 47) 0.999 867 003 073 504 215 04 × 2 = 1 + 0.999 734 006 147 008 430 08;
  • 48) 0.999 734 006 147 008 430 08 × 2 = 1 + 0.999 468 012 294 016 860 16;
  • 49) 0.999 468 012 294 016 860 16 × 2 = 1 + 0.998 936 024 588 033 720 32;
  • 50) 0.998 936 024 588 033 720 32 × 2 = 1 + 0.997 872 049 176 067 440 64;
  • 51) 0.997 872 049 176 067 440 64 × 2 = 1 + 0.995 744 098 352 134 881 28;
  • 52) 0.995 744 098 352 134 881 28 × 2 = 1 + 0.991 488 196 704 269 762 56;
  • 53) 0.991 488 196 704 269 762 56 × 2 = 1 + 0.982 976 393 408 539 525 12;
  • 54) 0.982 976 393 408 539 525 12 × 2 = 1 + 0.965 952 786 817 079 050 24;
  • 55) 0.965 952 786 817 079 050 24 × 2 = 1 + 0.931 905 573 634 158 100 48;
  • 56) 0.931 905 573 634 158 100 48 × 2 = 1 + 0.863 811 147 268 316 200 96;
  • 57) 0.863 811 147 268 316 200 96 × 2 = 1 + 0.727 622 294 536 632 401 92;
  • 58) 0.727 622 294 536 632 401 92 × 2 = 1 + 0.455 244 589 073 264 803 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 - the converted number we get in the end will be just a very good approximation of the initial one).


5. 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:


0.016 738 891 601 562 498 11(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 498 11(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 498 11(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(10)


10. 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;
  • 1 017 ÷ 2 = 508 + 1;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1017(10) =

011 1111 1001(2)


12. 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 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number -0.016 738 891 601 562 498 11 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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