-0.016 738 891 601 562 497 96 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 497 96(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
-0.016 738 891 601 562 497 96(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:

|-0.016 738 891 601 562 497 96| = 0.016 738 891 601 562 497 96


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: 0.016 738 891 601 562 497 96.

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) 0.016 738 891 601 562 497 96 × 2 = 0 + 0.033 477 783 203 124 995 92;
  • 2) 0.033 477 783 203 124 995 92 × 2 = 0 + 0.066 955 566 406 249 991 84;
  • 3) 0.066 955 566 406 249 991 84 × 2 = 0 + 0.133 911 132 812 499 983 68;
  • 4) 0.133 911 132 812 499 983 68 × 2 = 0 + 0.267 822 265 624 999 967 36;
  • 5) 0.267 822 265 624 999 967 36 × 2 = 0 + 0.535 644 531 249 999 934 72;
  • 6) 0.535 644 531 249 999 934 72 × 2 = 1 + 0.071 289 062 499 999 869 44;
  • 7) 0.071 289 062 499 999 869 44 × 2 = 0 + 0.142 578 124 999 999 738 88;
  • 8) 0.142 578 124 999 999 738 88 × 2 = 0 + 0.285 156 249 999 999 477 76;
  • 9) 0.285 156 249 999 999 477 76 × 2 = 0 + 0.570 312 499 999 998 955 52;
  • 10) 0.570 312 499 999 998 955 52 × 2 = 1 + 0.140 624 999 999 997 911 04;
  • 11) 0.140 624 999 999 997 911 04 × 2 = 0 + 0.281 249 999 999 995 822 08;
  • 12) 0.281 249 999 999 995 822 08 × 2 = 0 + 0.562 499 999 999 991 644 16;
  • 13) 0.562 499 999 999 991 644 16 × 2 = 1 + 0.124 999 999 999 983 288 32;
  • 14) 0.124 999 999 999 983 288 32 × 2 = 0 + 0.249 999 999 999 966 576 64;
  • 15) 0.249 999 999 999 966 576 64 × 2 = 0 + 0.499 999 999 999 933 153 28;
  • 16) 0.499 999 999 999 933 153 28 × 2 = 0 + 0.999 999 999 999 866 306 56;
  • 17) 0.999 999 999 999 866 306 56 × 2 = 1 + 0.999 999 999 999 732 613 12;
  • 18) 0.999 999 999 999 732 613 12 × 2 = 1 + 0.999 999 999 999 465 226 24;
  • 19) 0.999 999 999 999 465 226 24 × 2 = 1 + 0.999 999 999 998 930 452 48;
  • 20) 0.999 999 999 998 930 452 48 × 2 = 1 + 0.999 999 999 997 860 904 96;
  • 21) 0.999 999 999 997 860 904 96 × 2 = 1 + 0.999 999 999 995 721 809 92;
  • 22) 0.999 999 999 995 721 809 92 × 2 = 1 + 0.999 999 999 991 443 619 84;
  • 23) 0.999 999 999 991 443 619 84 × 2 = 1 + 0.999 999 999 982 887 239 68;
  • 24) 0.999 999 999 982 887 239 68 × 2 = 1 + 0.999 999 999 965 774 479 36;
  • 25) 0.999 999 999 965 774 479 36 × 2 = 1 + 0.999 999 999 931 548 958 72;
  • 26) 0.999 999 999 931 548 958 72 × 2 = 1 + 0.999 999 999 863 097 917 44;
  • 27) 0.999 999 999 863 097 917 44 × 2 = 1 + 0.999 999 999 726 195 834 88;
  • 28) 0.999 999 999 726 195 834 88 × 2 = 1 + 0.999 999 999 452 391 669 76;
  • 29) 0.999 999 999 452 391 669 76 × 2 = 1 + 0.999 999 998 904 783 339 52;
  • 30) 0.999 999 998 904 783 339 52 × 2 = 1 + 0.999 999 997 809 566 679 04;
  • 31) 0.999 999 997 809 566 679 04 × 2 = 1 + 0.999 999 995 619 133 358 08;
  • 32) 0.999 999 995 619 133 358 08 × 2 = 1 + 0.999 999 991 238 266 716 16;
  • 33) 0.999 999 991 238 266 716 16 × 2 = 1 + 0.999 999 982 476 533 432 32;
  • 34) 0.999 999 982 476 533 432 32 × 2 = 1 + 0.999 999 964 953 066 864 64;
  • 35) 0.999 999 964 953 066 864 64 × 2 = 1 + 0.999 999 929 906 133 729 28;
  • 36) 0.999 999 929 906 133 729 28 × 2 = 1 + 0.999 999 859 812 267 458 56;
  • 37) 0.999 999 859 812 267 458 56 × 2 = 1 + 0.999 999 719 624 534 917 12;
  • 38) 0.999 999 719 624 534 917 12 × 2 = 1 + 0.999 999 439 249 069 834 24;
  • 39) 0.999 999 439 249 069 834 24 × 2 = 1 + 0.999 998 878 498 139 668 48;
  • 40) 0.999 998 878 498 139 668 48 × 2 = 1 + 0.999 997 756 996 279 336 96;
  • 41) 0.999 997 756 996 279 336 96 × 2 = 1 + 0.999 995 513 992 558 673 92;
  • 42) 0.999 995 513 992 558 673 92 × 2 = 1 + 0.999 991 027 985 117 347 84;
  • 43) 0.999 991 027 985 117 347 84 × 2 = 1 + 0.999 982 055 970 234 695 68;
  • 44) 0.999 982 055 970 234 695 68 × 2 = 1 + 0.999 964 111 940 469 391 36;
  • 45) 0.999 964 111 940 469 391 36 × 2 = 1 + 0.999 928 223 880 938 782 72;
  • 46) 0.999 928 223 880 938 782 72 × 2 = 1 + 0.999 856 447 761 877 565 44;
  • 47) 0.999 856 447 761 877 565 44 × 2 = 1 + 0.999 712 895 523 755 130 88;
  • 48) 0.999 712 895 523 755 130 88 × 2 = 1 + 0.999 425 791 047 510 261 76;
  • 49) 0.999 425 791 047 510 261 76 × 2 = 1 + 0.998 851 582 095 020 523 52;
  • 50) 0.998 851 582 095 020 523 52 × 2 = 1 + 0.997 703 164 190 041 047 04;
  • 51) 0.997 703 164 190 041 047 04 × 2 = 1 + 0.995 406 328 380 082 094 08;
  • 52) 0.995 406 328 380 082 094 08 × 2 = 1 + 0.990 812 656 760 164 188 16;
  • 53) 0.990 812 656 760 164 188 16 × 2 = 1 + 0.981 625 313 520 328 376 32;
  • 54) 0.981 625 313 520 328 376 32 × 2 = 1 + 0.963 250 627 040 656 752 64;
  • 55) 0.963 250 627 040 656 752 64 × 2 = 1 + 0.926 501 254 081 313 505 28;
  • 56) 0.926 501 254 081 313 505 28 × 2 = 1 + 0.853 002 508 162 627 010 56;
  • 57) 0.853 002 508 162 627 010 56 × 2 = 1 + 0.706 005 016 325 254 021 12;
  • 58) 0.706 005 016 325 254 021 12 × 2 = 1 + 0.412 010 032 650 508 042 24;

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:


0.016 738 891 601 562 497 96(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 497 96(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:


0.016 738 891 601 562 497 96(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 -0.016 738 891 601 562 497 96 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