-0.016 738 891 601 562 496 531 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 531 47(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
-0.016 738 891 601 562 496 531 47(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:

|-0.016 738 891 601 562 496 531 47| = 0.016 738 891 601 562 496 531 47


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: 0.016 738 891 601 562 496 531 47.

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) 0.016 738 891 601 562 496 531 47 × 2 = 0 + 0.033 477 783 203 124 993 062 94;
  • 2) 0.033 477 783 203 124 993 062 94 × 2 = 0 + 0.066 955 566 406 249 986 125 88;
  • 3) 0.066 955 566 406 249 986 125 88 × 2 = 0 + 0.133 911 132 812 499 972 251 76;
  • 4) 0.133 911 132 812 499 972 251 76 × 2 = 0 + 0.267 822 265 624 999 944 503 52;
  • 5) 0.267 822 265 624 999 944 503 52 × 2 = 0 + 0.535 644 531 249 999 889 007 04;
  • 6) 0.535 644 531 249 999 889 007 04 × 2 = 1 + 0.071 289 062 499 999 778 014 08;
  • 7) 0.071 289 062 499 999 778 014 08 × 2 = 0 + 0.142 578 124 999 999 556 028 16;
  • 8) 0.142 578 124 999 999 556 028 16 × 2 = 0 + 0.285 156 249 999 999 112 056 32;
  • 9) 0.285 156 249 999 999 112 056 32 × 2 = 0 + 0.570 312 499 999 998 224 112 64;
  • 10) 0.570 312 499 999 998 224 112 64 × 2 = 1 + 0.140 624 999 999 996 448 225 28;
  • 11) 0.140 624 999 999 996 448 225 28 × 2 = 0 + 0.281 249 999 999 992 896 450 56;
  • 12) 0.281 249 999 999 992 896 450 56 × 2 = 0 + 0.562 499 999 999 985 792 901 12;
  • 13) 0.562 499 999 999 985 792 901 12 × 2 = 1 + 0.124 999 999 999 971 585 802 24;
  • 14) 0.124 999 999 999 971 585 802 24 × 2 = 0 + 0.249 999 999 999 943 171 604 48;
  • 15) 0.249 999 999 999 943 171 604 48 × 2 = 0 + 0.499 999 999 999 886 343 208 96;
  • 16) 0.499 999 999 999 886 343 208 96 × 2 = 0 + 0.999 999 999 999 772 686 417 92;
  • 17) 0.999 999 999 999 772 686 417 92 × 2 = 1 + 0.999 999 999 999 545 372 835 84;
  • 18) 0.999 999 999 999 545 372 835 84 × 2 = 1 + 0.999 999 999 999 090 745 671 68;
  • 19) 0.999 999 999 999 090 745 671 68 × 2 = 1 + 0.999 999 999 998 181 491 343 36;
  • 20) 0.999 999 999 998 181 491 343 36 × 2 = 1 + 0.999 999 999 996 362 982 686 72;
  • 21) 0.999 999 999 996 362 982 686 72 × 2 = 1 + 0.999 999 999 992 725 965 373 44;
  • 22) 0.999 999 999 992 725 965 373 44 × 2 = 1 + 0.999 999 999 985 451 930 746 88;
  • 23) 0.999 999 999 985 451 930 746 88 × 2 = 1 + 0.999 999 999 970 903 861 493 76;
  • 24) 0.999 999 999 970 903 861 493 76 × 2 = 1 + 0.999 999 999 941 807 722 987 52;
  • 25) 0.999 999 999 941 807 722 987 52 × 2 = 1 + 0.999 999 999 883 615 445 975 04;
  • 26) 0.999 999 999 883 615 445 975 04 × 2 = 1 + 0.999 999 999 767 230 891 950 08;
  • 27) 0.999 999 999 767 230 891 950 08 × 2 = 1 + 0.999 999 999 534 461 783 900 16;
  • 28) 0.999 999 999 534 461 783 900 16 × 2 = 1 + 0.999 999 999 068 923 567 800 32;
  • 29) 0.999 999 999 068 923 567 800 32 × 2 = 1 + 0.999 999 998 137 847 135 600 64;
  • 30) 0.999 999 998 137 847 135 600 64 × 2 = 1 + 0.999 999 996 275 694 271 201 28;
  • 31) 0.999 999 996 275 694 271 201 28 × 2 = 1 + 0.999 999 992 551 388 542 402 56;
  • 32) 0.999 999 992 551 388 542 402 56 × 2 = 1 + 0.999 999 985 102 777 084 805 12;
  • 33) 0.999 999 985 102 777 084 805 12 × 2 = 1 + 0.999 999 970 205 554 169 610 24;
  • 34) 0.999 999 970 205 554 169 610 24 × 2 = 1 + 0.999 999 940 411 108 339 220 48;
  • 35) 0.999 999 940 411 108 339 220 48 × 2 = 1 + 0.999 999 880 822 216 678 440 96;
  • 36) 0.999 999 880 822 216 678 440 96 × 2 = 1 + 0.999 999 761 644 433 356 881 92;
  • 37) 0.999 999 761 644 433 356 881 92 × 2 = 1 + 0.999 999 523 288 866 713 763 84;
  • 38) 0.999 999 523 288 866 713 763 84 × 2 = 1 + 0.999 999 046 577 733 427 527 68;
  • 39) 0.999 999 046 577 733 427 527 68 × 2 = 1 + 0.999 998 093 155 466 855 055 36;
  • 40) 0.999 998 093 155 466 855 055 36 × 2 = 1 + 0.999 996 186 310 933 710 110 72;
  • 41) 0.999 996 186 310 933 710 110 72 × 2 = 1 + 0.999 992 372 621 867 420 221 44;
  • 42) 0.999 992 372 621 867 420 221 44 × 2 = 1 + 0.999 984 745 243 734 840 442 88;
  • 43) 0.999 984 745 243 734 840 442 88 × 2 = 1 + 0.999 969 490 487 469 680 885 76;
  • 44) 0.999 969 490 487 469 680 885 76 × 2 = 1 + 0.999 938 980 974 939 361 771 52;
  • 45) 0.999 938 980 974 939 361 771 52 × 2 = 1 + 0.999 877 961 949 878 723 543 04;
  • 46) 0.999 877 961 949 878 723 543 04 × 2 = 1 + 0.999 755 923 899 757 447 086 08;
  • 47) 0.999 755 923 899 757 447 086 08 × 2 = 1 + 0.999 511 847 799 514 894 172 16;
  • 48) 0.999 511 847 799 514 894 172 16 × 2 = 1 + 0.999 023 695 599 029 788 344 32;
  • 49) 0.999 023 695 599 029 788 344 32 × 2 = 1 + 0.998 047 391 198 059 576 688 64;
  • 50) 0.998 047 391 198 059 576 688 64 × 2 = 1 + 0.996 094 782 396 119 153 377 28;
  • 51) 0.996 094 782 396 119 153 377 28 × 2 = 1 + 0.992 189 564 792 238 306 754 56;
  • 52) 0.992 189 564 792 238 306 754 56 × 2 = 1 + 0.984 379 129 584 476 613 509 12;
  • 53) 0.984 379 129 584 476 613 509 12 × 2 = 1 + 0.968 758 259 168 953 227 018 24;
  • 54) 0.968 758 259 168 953 227 018 24 × 2 = 1 + 0.937 516 518 337 906 454 036 48;
  • 55) 0.937 516 518 337 906 454 036 48 × 2 = 1 + 0.875 033 036 675 812 908 072 96;
  • 56) 0.875 033 036 675 812 908 072 96 × 2 = 1 + 0.750 066 073 351 625 816 145 92;
  • 57) 0.750 066 073 351 625 816 145 92 × 2 = 1 + 0.500 132 146 703 251 632 291 84;
  • 58) 0.500 132 146 703 251 632 291 84 × 2 = 1 + 0.000 264 293 406 503 264 583 68;

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:


0.016 738 891 601 562 496 531 47(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 531 47(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:


0.016 738 891 601 562 496 531 47(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 -0.016 738 891 601 562 496 531 47 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

Share

» Convert decimal -0.016 738 891 601 562 496 530 57 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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