-0.016 738 891 601 562 496 542 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 542 9(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
-0.016 738 891 601 562 496 542 9(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:

|-0.016 738 891 601 562 496 542 9| = 0.016 738 891 601 562 496 542 9


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: 0.016 738 891 601 562 496 542 9.

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) 0.016 738 891 601 562 496 542 9 × 2 = 0 + 0.033 477 783 203 124 993 085 8;
  • 2) 0.033 477 783 203 124 993 085 8 × 2 = 0 + 0.066 955 566 406 249 986 171 6;
  • 3) 0.066 955 566 406 249 986 171 6 × 2 = 0 + 0.133 911 132 812 499 972 343 2;
  • 4) 0.133 911 132 812 499 972 343 2 × 2 = 0 + 0.267 822 265 624 999 944 686 4;
  • 5) 0.267 822 265 624 999 944 686 4 × 2 = 0 + 0.535 644 531 249 999 889 372 8;
  • 6) 0.535 644 531 249 999 889 372 8 × 2 = 1 + 0.071 289 062 499 999 778 745 6;
  • 7) 0.071 289 062 499 999 778 745 6 × 2 = 0 + 0.142 578 124 999 999 557 491 2;
  • 8) 0.142 578 124 999 999 557 491 2 × 2 = 0 + 0.285 156 249 999 999 114 982 4;
  • 9) 0.285 156 249 999 999 114 982 4 × 2 = 0 + 0.570 312 499 999 998 229 964 8;
  • 10) 0.570 312 499 999 998 229 964 8 × 2 = 1 + 0.140 624 999 999 996 459 929 6;
  • 11) 0.140 624 999 999 996 459 929 6 × 2 = 0 + 0.281 249 999 999 992 919 859 2;
  • 12) 0.281 249 999 999 992 919 859 2 × 2 = 0 + 0.562 499 999 999 985 839 718 4;
  • 13) 0.562 499 999 999 985 839 718 4 × 2 = 1 + 0.124 999 999 999 971 679 436 8;
  • 14) 0.124 999 999 999 971 679 436 8 × 2 = 0 + 0.249 999 999 999 943 358 873 6;
  • 15) 0.249 999 999 999 943 358 873 6 × 2 = 0 + 0.499 999 999 999 886 717 747 2;
  • 16) 0.499 999 999 999 886 717 747 2 × 2 = 0 + 0.999 999 999 999 773 435 494 4;
  • 17) 0.999 999 999 999 773 435 494 4 × 2 = 1 + 0.999 999 999 999 546 870 988 8;
  • 18) 0.999 999 999 999 546 870 988 8 × 2 = 1 + 0.999 999 999 999 093 741 977 6;
  • 19) 0.999 999 999 999 093 741 977 6 × 2 = 1 + 0.999 999 999 998 187 483 955 2;
  • 20) 0.999 999 999 998 187 483 955 2 × 2 = 1 + 0.999 999 999 996 374 967 910 4;
  • 21) 0.999 999 999 996 374 967 910 4 × 2 = 1 + 0.999 999 999 992 749 935 820 8;
  • 22) 0.999 999 999 992 749 935 820 8 × 2 = 1 + 0.999 999 999 985 499 871 641 6;
  • 23) 0.999 999 999 985 499 871 641 6 × 2 = 1 + 0.999 999 999 970 999 743 283 2;
  • 24) 0.999 999 999 970 999 743 283 2 × 2 = 1 + 0.999 999 999 941 999 486 566 4;
  • 25) 0.999 999 999 941 999 486 566 4 × 2 = 1 + 0.999 999 999 883 998 973 132 8;
  • 26) 0.999 999 999 883 998 973 132 8 × 2 = 1 + 0.999 999 999 767 997 946 265 6;
  • 27) 0.999 999 999 767 997 946 265 6 × 2 = 1 + 0.999 999 999 535 995 892 531 2;
  • 28) 0.999 999 999 535 995 892 531 2 × 2 = 1 + 0.999 999 999 071 991 785 062 4;
  • 29) 0.999 999 999 071 991 785 062 4 × 2 = 1 + 0.999 999 998 143 983 570 124 8;
  • 30) 0.999 999 998 143 983 570 124 8 × 2 = 1 + 0.999 999 996 287 967 140 249 6;
  • 31) 0.999 999 996 287 967 140 249 6 × 2 = 1 + 0.999 999 992 575 934 280 499 2;
  • 32) 0.999 999 992 575 934 280 499 2 × 2 = 1 + 0.999 999 985 151 868 560 998 4;
  • 33) 0.999 999 985 151 868 560 998 4 × 2 = 1 + 0.999 999 970 303 737 121 996 8;
  • 34) 0.999 999 970 303 737 121 996 8 × 2 = 1 + 0.999 999 940 607 474 243 993 6;
  • 35) 0.999 999 940 607 474 243 993 6 × 2 = 1 + 0.999 999 881 214 948 487 987 2;
  • 36) 0.999 999 881 214 948 487 987 2 × 2 = 1 + 0.999 999 762 429 896 975 974 4;
  • 37) 0.999 999 762 429 896 975 974 4 × 2 = 1 + 0.999 999 524 859 793 951 948 8;
  • 38) 0.999 999 524 859 793 951 948 8 × 2 = 1 + 0.999 999 049 719 587 903 897 6;
  • 39) 0.999 999 049 719 587 903 897 6 × 2 = 1 + 0.999 998 099 439 175 807 795 2;
  • 40) 0.999 998 099 439 175 807 795 2 × 2 = 1 + 0.999 996 198 878 351 615 590 4;
  • 41) 0.999 996 198 878 351 615 590 4 × 2 = 1 + 0.999 992 397 756 703 231 180 8;
  • 42) 0.999 992 397 756 703 231 180 8 × 2 = 1 + 0.999 984 795 513 406 462 361 6;
  • 43) 0.999 984 795 513 406 462 361 6 × 2 = 1 + 0.999 969 591 026 812 924 723 2;
  • 44) 0.999 969 591 026 812 924 723 2 × 2 = 1 + 0.999 939 182 053 625 849 446 4;
  • 45) 0.999 939 182 053 625 849 446 4 × 2 = 1 + 0.999 878 364 107 251 698 892 8;
  • 46) 0.999 878 364 107 251 698 892 8 × 2 = 1 + 0.999 756 728 214 503 397 785 6;
  • 47) 0.999 756 728 214 503 397 785 6 × 2 = 1 + 0.999 513 456 429 006 795 571 2;
  • 48) 0.999 513 456 429 006 795 571 2 × 2 = 1 + 0.999 026 912 858 013 591 142 4;
  • 49) 0.999 026 912 858 013 591 142 4 × 2 = 1 + 0.998 053 825 716 027 182 284 8;
  • 50) 0.998 053 825 716 027 182 284 8 × 2 = 1 + 0.996 107 651 432 054 364 569 6;
  • 51) 0.996 107 651 432 054 364 569 6 × 2 = 1 + 0.992 215 302 864 108 729 139 2;
  • 52) 0.992 215 302 864 108 729 139 2 × 2 = 1 + 0.984 430 605 728 217 458 278 4;
  • 53) 0.984 430 605 728 217 458 278 4 × 2 = 1 + 0.968 861 211 456 434 916 556 8;
  • 54) 0.968 861 211 456 434 916 556 8 × 2 = 1 + 0.937 722 422 912 869 833 113 6;
  • 55) 0.937 722 422 912 869 833 113 6 × 2 = 1 + 0.875 444 845 825 739 666 227 2;
  • 56) 0.875 444 845 825 739 666 227 2 × 2 = 1 + 0.750 889 691 651 479 332 454 4;
  • 57) 0.750 889 691 651 479 332 454 4 × 2 = 1 + 0.501 779 383 302 958 664 908 8;
  • 58) 0.501 779 383 302 958 664 908 8 × 2 = 1 + 0.003 558 766 605 917 329 817 6;

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:


0.016 738 891 601 562 496 542 9(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 542 9(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:


0.016 738 891 601 562 496 542 9(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 -0.016 738 891 601 562 496 542 9 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