-0.016 738 891 601 562 496 534 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 534 8(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
-0.016 738 891 601 562 496 534 8(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:

|-0.016 738 891 601 562 496 534 8| = 0.016 738 891 601 562 496 534 8


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: 0.016 738 891 601 562 496 534 8.

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) 0.016 738 891 601 562 496 534 8 × 2 = 0 + 0.033 477 783 203 124 993 069 6;
  • 2) 0.033 477 783 203 124 993 069 6 × 2 = 0 + 0.066 955 566 406 249 986 139 2;
  • 3) 0.066 955 566 406 249 986 139 2 × 2 = 0 + 0.133 911 132 812 499 972 278 4;
  • 4) 0.133 911 132 812 499 972 278 4 × 2 = 0 + 0.267 822 265 624 999 944 556 8;
  • 5) 0.267 822 265 624 999 944 556 8 × 2 = 0 + 0.535 644 531 249 999 889 113 6;
  • 6) 0.535 644 531 249 999 889 113 6 × 2 = 1 + 0.071 289 062 499 999 778 227 2;
  • 7) 0.071 289 062 499 999 778 227 2 × 2 = 0 + 0.142 578 124 999 999 556 454 4;
  • 8) 0.142 578 124 999 999 556 454 4 × 2 = 0 + 0.285 156 249 999 999 112 908 8;
  • 9) 0.285 156 249 999 999 112 908 8 × 2 = 0 + 0.570 312 499 999 998 225 817 6;
  • 10) 0.570 312 499 999 998 225 817 6 × 2 = 1 + 0.140 624 999 999 996 451 635 2;
  • 11) 0.140 624 999 999 996 451 635 2 × 2 = 0 + 0.281 249 999 999 992 903 270 4;
  • 12) 0.281 249 999 999 992 903 270 4 × 2 = 0 + 0.562 499 999 999 985 806 540 8;
  • 13) 0.562 499 999 999 985 806 540 8 × 2 = 1 + 0.124 999 999 999 971 613 081 6;
  • 14) 0.124 999 999 999 971 613 081 6 × 2 = 0 + 0.249 999 999 999 943 226 163 2;
  • 15) 0.249 999 999 999 943 226 163 2 × 2 = 0 + 0.499 999 999 999 886 452 326 4;
  • 16) 0.499 999 999 999 886 452 326 4 × 2 = 0 + 0.999 999 999 999 772 904 652 8;
  • 17) 0.999 999 999 999 772 904 652 8 × 2 = 1 + 0.999 999 999 999 545 809 305 6;
  • 18) 0.999 999 999 999 545 809 305 6 × 2 = 1 + 0.999 999 999 999 091 618 611 2;
  • 19) 0.999 999 999 999 091 618 611 2 × 2 = 1 + 0.999 999 999 998 183 237 222 4;
  • 20) 0.999 999 999 998 183 237 222 4 × 2 = 1 + 0.999 999 999 996 366 474 444 8;
  • 21) 0.999 999 999 996 366 474 444 8 × 2 = 1 + 0.999 999 999 992 732 948 889 6;
  • 22) 0.999 999 999 992 732 948 889 6 × 2 = 1 + 0.999 999 999 985 465 897 779 2;
  • 23) 0.999 999 999 985 465 897 779 2 × 2 = 1 + 0.999 999 999 970 931 795 558 4;
  • 24) 0.999 999 999 970 931 795 558 4 × 2 = 1 + 0.999 999 999 941 863 591 116 8;
  • 25) 0.999 999 999 941 863 591 116 8 × 2 = 1 + 0.999 999 999 883 727 182 233 6;
  • 26) 0.999 999 999 883 727 182 233 6 × 2 = 1 + 0.999 999 999 767 454 364 467 2;
  • 27) 0.999 999 999 767 454 364 467 2 × 2 = 1 + 0.999 999 999 534 908 728 934 4;
  • 28) 0.999 999 999 534 908 728 934 4 × 2 = 1 + 0.999 999 999 069 817 457 868 8;
  • 29) 0.999 999 999 069 817 457 868 8 × 2 = 1 + 0.999 999 998 139 634 915 737 6;
  • 30) 0.999 999 998 139 634 915 737 6 × 2 = 1 + 0.999 999 996 279 269 831 475 2;
  • 31) 0.999 999 996 279 269 831 475 2 × 2 = 1 + 0.999 999 992 558 539 662 950 4;
  • 32) 0.999 999 992 558 539 662 950 4 × 2 = 1 + 0.999 999 985 117 079 325 900 8;
  • 33) 0.999 999 985 117 079 325 900 8 × 2 = 1 + 0.999 999 970 234 158 651 801 6;
  • 34) 0.999 999 970 234 158 651 801 6 × 2 = 1 + 0.999 999 940 468 317 303 603 2;
  • 35) 0.999 999 940 468 317 303 603 2 × 2 = 1 + 0.999 999 880 936 634 607 206 4;
  • 36) 0.999 999 880 936 634 607 206 4 × 2 = 1 + 0.999 999 761 873 269 214 412 8;
  • 37) 0.999 999 761 873 269 214 412 8 × 2 = 1 + 0.999 999 523 746 538 428 825 6;
  • 38) 0.999 999 523 746 538 428 825 6 × 2 = 1 + 0.999 999 047 493 076 857 651 2;
  • 39) 0.999 999 047 493 076 857 651 2 × 2 = 1 + 0.999 998 094 986 153 715 302 4;
  • 40) 0.999 998 094 986 153 715 302 4 × 2 = 1 + 0.999 996 189 972 307 430 604 8;
  • 41) 0.999 996 189 972 307 430 604 8 × 2 = 1 + 0.999 992 379 944 614 861 209 6;
  • 42) 0.999 992 379 944 614 861 209 6 × 2 = 1 + 0.999 984 759 889 229 722 419 2;
  • 43) 0.999 984 759 889 229 722 419 2 × 2 = 1 + 0.999 969 519 778 459 444 838 4;
  • 44) 0.999 969 519 778 459 444 838 4 × 2 = 1 + 0.999 939 039 556 918 889 676 8;
  • 45) 0.999 939 039 556 918 889 676 8 × 2 = 1 + 0.999 878 079 113 837 779 353 6;
  • 46) 0.999 878 079 113 837 779 353 6 × 2 = 1 + 0.999 756 158 227 675 558 707 2;
  • 47) 0.999 756 158 227 675 558 707 2 × 2 = 1 + 0.999 512 316 455 351 117 414 4;
  • 48) 0.999 512 316 455 351 117 414 4 × 2 = 1 + 0.999 024 632 910 702 234 828 8;
  • 49) 0.999 024 632 910 702 234 828 8 × 2 = 1 + 0.998 049 265 821 404 469 657 6;
  • 50) 0.998 049 265 821 404 469 657 6 × 2 = 1 + 0.996 098 531 642 808 939 315 2;
  • 51) 0.996 098 531 642 808 939 315 2 × 2 = 1 + 0.992 197 063 285 617 878 630 4;
  • 52) 0.992 197 063 285 617 878 630 4 × 2 = 1 + 0.984 394 126 571 235 757 260 8;
  • 53) 0.984 394 126 571 235 757 260 8 × 2 = 1 + 0.968 788 253 142 471 514 521 6;
  • 54) 0.968 788 253 142 471 514 521 6 × 2 = 1 + 0.937 576 506 284 943 029 043 2;
  • 55) 0.937 576 506 284 943 029 043 2 × 2 = 1 + 0.875 153 012 569 886 058 086 4;
  • 56) 0.875 153 012 569 886 058 086 4 × 2 = 1 + 0.750 306 025 139 772 116 172 8;
  • 57) 0.750 306 025 139 772 116 172 8 × 2 = 1 + 0.500 612 050 279 544 232 345 6;
  • 58) 0.500 612 050 279 544 232 345 6 × 2 = 1 + 0.001 224 100 559 088 464 691 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).


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:


0.016 738 891 601 562 496 534 8(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 534 8(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:


0.016 738 891 601 562 496 534 8(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 -0.016 738 891 601 562 496 534 8 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