-0.016 738 891 601 562 496 527 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 527 3(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
-0.016 738 891 601 562 496 527 3(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:

|-0.016 738 891 601 562 496 527 3| = 0.016 738 891 601 562 496 527 3


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: 0.016 738 891 601 562 496 527 3.

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) 0.016 738 891 601 562 496 527 3 × 2 = 0 + 0.033 477 783 203 124 993 054 6;
  • 2) 0.033 477 783 203 124 993 054 6 × 2 = 0 + 0.066 955 566 406 249 986 109 2;
  • 3) 0.066 955 566 406 249 986 109 2 × 2 = 0 + 0.133 911 132 812 499 972 218 4;
  • 4) 0.133 911 132 812 499 972 218 4 × 2 = 0 + 0.267 822 265 624 999 944 436 8;
  • 5) 0.267 822 265 624 999 944 436 8 × 2 = 0 + 0.535 644 531 249 999 888 873 6;
  • 6) 0.535 644 531 249 999 888 873 6 × 2 = 1 + 0.071 289 062 499 999 777 747 2;
  • 7) 0.071 289 062 499 999 777 747 2 × 2 = 0 + 0.142 578 124 999 999 555 494 4;
  • 8) 0.142 578 124 999 999 555 494 4 × 2 = 0 + 0.285 156 249 999 999 110 988 8;
  • 9) 0.285 156 249 999 999 110 988 8 × 2 = 0 + 0.570 312 499 999 998 221 977 6;
  • 10) 0.570 312 499 999 998 221 977 6 × 2 = 1 + 0.140 624 999 999 996 443 955 2;
  • 11) 0.140 624 999 999 996 443 955 2 × 2 = 0 + 0.281 249 999 999 992 887 910 4;
  • 12) 0.281 249 999 999 992 887 910 4 × 2 = 0 + 0.562 499 999 999 985 775 820 8;
  • 13) 0.562 499 999 999 985 775 820 8 × 2 = 1 + 0.124 999 999 999 971 551 641 6;
  • 14) 0.124 999 999 999 971 551 641 6 × 2 = 0 + 0.249 999 999 999 943 103 283 2;
  • 15) 0.249 999 999 999 943 103 283 2 × 2 = 0 + 0.499 999 999 999 886 206 566 4;
  • 16) 0.499 999 999 999 886 206 566 4 × 2 = 0 + 0.999 999 999 999 772 413 132 8;
  • 17) 0.999 999 999 999 772 413 132 8 × 2 = 1 + 0.999 999 999 999 544 826 265 6;
  • 18) 0.999 999 999 999 544 826 265 6 × 2 = 1 + 0.999 999 999 999 089 652 531 2;
  • 19) 0.999 999 999 999 089 652 531 2 × 2 = 1 + 0.999 999 999 998 179 305 062 4;
  • 20) 0.999 999 999 998 179 305 062 4 × 2 = 1 + 0.999 999 999 996 358 610 124 8;
  • 21) 0.999 999 999 996 358 610 124 8 × 2 = 1 + 0.999 999 999 992 717 220 249 6;
  • 22) 0.999 999 999 992 717 220 249 6 × 2 = 1 + 0.999 999 999 985 434 440 499 2;
  • 23) 0.999 999 999 985 434 440 499 2 × 2 = 1 + 0.999 999 999 970 868 880 998 4;
  • 24) 0.999 999 999 970 868 880 998 4 × 2 = 1 + 0.999 999 999 941 737 761 996 8;
  • 25) 0.999 999 999 941 737 761 996 8 × 2 = 1 + 0.999 999 999 883 475 523 993 6;
  • 26) 0.999 999 999 883 475 523 993 6 × 2 = 1 + 0.999 999 999 766 951 047 987 2;
  • 27) 0.999 999 999 766 951 047 987 2 × 2 = 1 + 0.999 999 999 533 902 095 974 4;
  • 28) 0.999 999 999 533 902 095 974 4 × 2 = 1 + 0.999 999 999 067 804 191 948 8;
  • 29) 0.999 999 999 067 804 191 948 8 × 2 = 1 + 0.999 999 998 135 608 383 897 6;
  • 30) 0.999 999 998 135 608 383 897 6 × 2 = 1 + 0.999 999 996 271 216 767 795 2;
  • 31) 0.999 999 996 271 216 767 795 2 × 2 = 1 + 0.999 999 992 542 433 535 590 4;
  • 32) 0.999 999 992 542 433 535 590 4 × 2 = 1 + 0.999 999 985 084 867 071 180 8;
  • 33) 0.999 999 985 084 867 071 180 8 × 2 = 1 + 0.999 999 970 169 734 142 361 6;
  • 34) 0.999 999 970 169 734 142 361 6 × 2 = 1 + 0.999 999 940 339 468 284 723 2;
  • 35) 0.999 999 940 339 468 284 723 2 × 2 = 1 + 0.999 999 880 678 936 569 446 4;
  • 36) 0.999 999 880 678 936 569 446 4 × 2 = 1 + 0.999 999 761 357 873 138 892 8;
  • 37) 0.999 999 761 357 873 138 892 8 × 2 = 1 + 0.999 999 522 715 746 277 785 6;
  • 38) 0.999 999 522 715 746 277 785 6 × 2 = 1 + 0.999 999 045 431 492 555 571 2;
  • 39) 0.999 999 045 431 492 555 571 2 × 2 = 1 + 0.999 998 090 862 985 111 142 4;
  • 40) 0.999 998 090 862 985 111 142 4 × 2 = 1 + 0.999 996 181 725 970 222 284 8;
  • 41) 0.999 996 181 725 970 222 284 8 × 2 = 1 + 0.999 992 363 451 940 444 569 6;
  • 42) 0.999 992 363 451 940 444 569 6 × 2 = 1 + 0.999 984 726 903 880 889 139 2;
  • 43) 0.999 984 726 903 880 889 139 2 × 2 = 1 + 0.999 969 453 807 761 778 278 4;
  • 44) 0.999 969 453 807 761 778 278 4 × 2 = 1 + 0.999 938 907 615 523 556 556 8;
  • 45) 0.999 938 907 615 523 556 556 8 × 2 = 1 + 0.999 877 815 231 047 113 113 6;
  • 46) 0.999 877 815 231 047 113 113 6 × 2 = 1 + 0.999 755 630 462 094 226 227 2;
  • 47) 0.999 755 630 462 094 226 227 2 × 2 = 1 + 0.999 511 260 924 188 452 454 4;
  • 48) 0.999 511 260 924 188 452 454 4 × 2 = 1 + 0.999 022 521 848 376 904 908 8;
  • 49) 0.999 022 521 848 376 904 908 8 × 2 = 1 + 0.998 045 043 696 753 809 817 6;
  • 50) 0.998 045 043 696 753 809 817 6 × 2 = 1 + 0.996 090 087 393 507 619 635 2;
  • 51) 0.996 090 087 393 507 619 635 2 × 2 = 1 + 0.992 180 174 787 015 239 270 4;
  • 52) 0.992 180 174 787 015 239 270 4 × 2 = 1 + 0.984 360 349 574 030 478 540 8;
  • 53) 0.984 360 349 574 030 478 540 8 × 2 = 1 + 0.968 720 699 148 060 957 081 6;
  • 54) 0.968 720 699 148 060 957 081 6 × 2 = 1 + 0.937 441 398 296 121 914 163 2;
  • 55) 0.937 441 398 296 121 914 163 2 × 2 = 1 + 0.874 882 796 592 243 828 326 4;
  • 56) 0.874 882 796 592 243 828 326 4 × 2 = 1 + 0.749 765 593 184 487 656 652 8;
  • 57) 0.749 765 593 184 487 656 652 8 × 2 = 1 + 0.499 531 186 368 975 313 305 6;
  • 58) 0.499 531 186 368 975 313 305 6 × 2 = 0 + 0.999 062 372 737 950 626 611 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:


0.016 738 891 601 562 496 527 3(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 527 3(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:


0.016 738 891 601 562 496 527 3(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 -0.016 738 891 601 562 496 527 3 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