-0.016 738 891 601 562 496 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 06(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
-0.016 738 891 601 562 496 06(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:

|-0.016 738 891 601 562 496 06| = 0.016 738 891 601 562 496 06


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: 0.016 738 891 601 562 496 06.

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) 0.016 738 891 601 562 496 06 × 2 = 0 + 0.033 477 783 203 124 992 12;
  • 2) 0.033 477 783 203 124 992 12 × 2 = 0 + 0.066 955 566 406 249 984 24;
  • 3) 0.066 955 566 406 249 984 24 × 2 = 0 + 0.133 911 132 812 499 968 48;
  • 4) 0.133 911 132 812 499 968 48 × 2 = 0 + 0.267 822 265 624 999 936 96;
  • 5) 0.267 822 265 624 999 936 96 × 2 = 0 + 0.535 644 531 249 999 873 92;
  • 6) 0.535 644 531 249 999 873 92 × 2 = 1 + 0.071 289 062 499 999 747 84;
  • 7) 0.071 289 062 499 999 747 84 × 2 = 0 + 0.142 578 124 999 999 495 68;
  • 8) 0.142 578 124 999 999 495 68 × 2 = 0 + 0.285 156 249 999 998 991 36;
  • 9) 0.285 156 249 999 998 991 36 × 2 = 0 + 0.570 312 499 999 997 982 72;
  • 10) 0.570 312 499 999 997 982 72 × 2 = 1 + 0.140 624 999 999 995 965 44;
  • 11) 0.140 624 999 999 995 965 44 × 2 = 0 + 0.281 249 999 999 991 930 88;
  • 12) 0.281 249 999 999 991 930 88 × 2 = 0 + 0.562 499 999 999 983 861 76;
  • 13) 0.562 499 999 999 983 861 76 × 2 = 1 + 0.124 999 999 999 967 723 52;
  • 14) 0.124 999 999 999 967 723 52 × 2 = 0 + 0.249 999 999 999 935 447 04;
  • 15) 0.249 999 999 999 935 447 04 × 2 = 0 + 0.499 999 999 999 870 894 08;
  • 16) 0.499 999 999 999 870 894 08 × 2 = 0 + 0.999 999 999 999 741 788 16;
  • 17) 0.999 999 999 999 741 788 16 × 2 = 1 + 0.999 999 999 999 483 576 32;
  • 18) 0.999 999 999 999 483 576 32 × 2 = 1 + 0.999 999 999 998 967 152 64;
  • 19) 0.999 999 999 998 967 152 64 × 2 = 1 + 0.999 999 999 997 934 305 28;
  • 20) 0.999 999 999 997 934 305 28 × 2 = 1 + 0.999 999 999 995 868 610 56;
  • 21) 0.999 999 999 995 868 610 56 × 2 = 1 + 0.999 999 999 991 737 221 12;
  • 22) 0.999 999 999 991 737 221 12 × 2 = 1 + 0.999 999 999 983 474 442 24;
  • 23) 0.999 999 999 983 474 442 24 × 2 = 1 + 0.999 999 999 966 948 884 48;
  • 24) 0.999 999 999 966 948 884 48 × 2 = 1 + 0.999 999 999 933 897 768 96;
  • 25) 0.999 999 999 933 897 768 96 × 2 = 1 + 0.999 999 999 867 795 537 92;
  • 26) 0.999 999 999 867 795 537 92 × 2 = 1 + 0.999 999 999 735 591 075 84;
  • 27) 0.999 999 999 735 591 075 84 × 2 = 1 + 0.999 999 999 471 182 151 68;
  • 28) 0.999 999 999 471 182 151 68 × 2 = 1 + 0.999 999 998 942 364 303 36;
  • 29) 0.999 999 998 942 364 303 36 × 2 = 1 + 0.999 999 997 884 728 606 72;
  • 30) 0.999 999 997 884 728 606 72 × 2 = 1 + 0.999 999 995 769 457 213 44;
  • 31) 0.999 999 995 769 457 213 44 × 2 = 1 + 0.999 999 991 538 914 426 88;
  • 32) 0.999 999 991 538 914 426 88 × 2 = 1 + 0.999 999 983 077 828 853 76;
  • 33) 0.999 999 983 077 828 853 76 × 2 = 1 + 0.999 999 966 155 657 707 52;
  • 34) 0.999 999 966 155 657 707 52 × 2 = 1 + 0.999 999 932 311 315 415 04;
  • 35) 0.999 999 932 311 315 415 04 × 2 = 1 + 0.999 999 864 622 630 830 08;
  • 36) 0.999 999 864 622 630 830 08 × 2 = 1 + 0.999 999 729 245 261 660 16;
  • 37) 0.999 999 729 245 261 660 16 × 2 = 1 + 0.999 999 458 490 523 320 32;
  • 38) 0.999 999 458 490 523 320 32 × 2 = 1 + 0.999 998 916 981 046 640 64;
  • 39) 0.999 998 916 981 046 640 64 × 2 = 1 + 0.999 997 833 962 093 281 28;
  • 40) 0.999 997 833 962 093 281 28 × 2 = 1 + 0.999 995 667 924 186 562 56;
  • 41) 0.999 995 667 924 186 562 56 × 2 = 1 + 0.999 991 335 848 373 125 12;
  • 42) 0.999 991 335 848 373 125 12 × 2 = 1 + 0.999 982 671 696 746 250 24;
  • 43) 0.999 982 671 696 746 250 24 × 2 = 1 + 0.999 965 343 393 492 500 48;
  • 44) 0.999 965 343 393 492 500 48 × 2 = 1 + 0.999 930 686 786 985 000 96;
  • 45) 0.999 930 686 786 985 000 96 × 2 = 1 + 0.999 861 373 573 970 001 92;
  • 46) 0.999 861 373 573 970 001 92 × 2 = 1 + 0.999 722 747 147 940 003 84;
  • 47) 0.999 722 747 147 940 003 84 × 2 = 1 + 0.999 445 494 295 880 007 68;
  • 48) 0.999 445 494 295 880 007 68 × 2 = 1 + 0.998 890 988 591 760 015 36;
  • 49) 0.998 890 988 591 760 015 36 × 2 = 1 + 0.997 781 977 183 520 030 72;
  • 50) 0.997 781 977 183 520 030 72 × 2 = 1 + 0.995 563 954 367 040 061 44;
  • 51) 0.995 563 954 367 040 061 44 × 2 = 1 + 0.991 127 908 734 080 122 88;
  • 52) 0.991 127 908 734 080 122 88 × 2 = 1 + 0.982 255 817 468 160 245 76;
  • 53) 0.982 255 817 468 160 245 76 × 2 = 1 + 0.964 511 634 936 320 491 52;
  • 54) 0.964 511 634 936 320 491 52 × 2 = 1 + 0.929 023 269 872 640 983 04;
  • 55) 0.929 023 269 872 640 983 04 × 2 = 1 + 0.858 046 539 745 281 966 08;
  • 56) 0.858 046 539 745 281 966 08 × 2 = 1 + 0.716 093 079 490 563 932 16;
  • 57) 0.716 093 079 490 563 932 16 × 2 = 1 + 0.432 186 158 981 127 864 32;
  • 58) 0.432 186 158 981 127 864 32 × 2 = 0 + 0.864 372 317 962 255 728 64;

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:


0.016 738 891 601 562 496 06(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 06(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 06(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 06 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 1110

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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