-0.016 738 891 601 562 496 522 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 522 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
-0.016 738 891 601 562 496 522 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:

|-0.016 738 891 601 562 496 522 3| = 0.016 738 891 601 562 496 522 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: 0.016 738 891 601 562 496 522 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) 0.016 738 891 601 562 496 522 3 × 2 = 0 + 0.033 477 783 203 124 993 044 6;
  • 2) 0.033 477 783 203 124 993 044 6 × 2 = 0 + 0.066 955 566 406 249 986 089 2;
  • 3) 0.066 955 566 406 249 986 089 2 × 2 = 0 + 0.133 911 132 812 499 972 178 4;
  • 4) 0.133 911 132 812 499 972 178 4 × 2 = 0 + 0.267 822 265 624 999 944 356 8;
  • 5) 0.267 822 265 624 999 944 356 8 × 2 = 0 + 0.535 644 531 249 999 888 713 6;
  • 6) 0.535 644 531 249 999 888 713 6 × 2 = 1 + 0.071 289 062 499 999 777 427 2;
  • 7) 0.071 289 062 499 999 777 427 2 × 2 = 0 + 0.142 578 124 999 999 554 854 4;
  • 8) 0.142 578 124 999 999 554 854 4 × 2 = 0 + 0.285 156 249 999 999 109 708 8;
  • 9) 0.285 156 249 999 999 109 708 8 × 2 = 0 + 0.570 312 499 999 998 219 417 6;
  • 10) 0.570 312 499 999 998 219 417 6 × 2 = 1 + 0.140 624 999 999 996 438 835 2;
  • 11) 0.140 624 999 999 996 438 835 2 × 2 = 0 + 0.281 249 999 999 992 877 670 4;
  • 12) 0.281 249 999 999 992 877 670 4 × 2 = 0 + 0.562 499 999 999 985 755 340 8;
  • 13) 0.562 499 999 999 985 755 340 8 × 2 = 1 + 0.124 999 999 999 971 510 681 6;
  • 14) 0.124 999 999 999 971 510 681 6 × 2 = 0 + 0.249 999 999 999 943 021 363 2;
  • 15) 0.249 999 999 999 943 021 363 2 × 2 = 0 + 0.499 999 999 999 886 042 726 4;
  • 16) 0.499 999 999 999 886 042 726 4 × 2 = 0 + 0.999 999 999 999 772 085 452 8;
  • 17) 0.999 999 999 999 772 085 452 8 × 2 = 1 + 0.999 999 999 999 544 170 905 6;
  • 18) 0.999 999 999 999 544 170 905 6 × 2 = 1 + 0.999 999 999 999 088 341 811 2;
  • 19) 0.999 999 999 999 088 341 811 2 × 2 = 1 + 0.999 999 999 998 176 683 622 4;
  • 20) 0.999 999 999 998 176 683 622 4 × 2 = 1 + 0.999 999 999 996 353 367 244 8;
  • 21) 0.999 999 999 996 353 367 244 8 × 2 = 1 + 0.999 999 999 992 706 734 489 6;
  • 22) 0.999 999 999 992 706 734 489 6 × 2 = 1 + 0.999 999 999 985 413 468 979 2;
  • 23) 0.999 999 999 985 413 468 979 2 × 2 = 1 + 0.999 999 999 970 826 937 958 4;
  • 24) 0.999 999 999 970 826 937 958 4 × 2 = 1 + 0.999 999 999 941 653 875 916 8;
  • 25) 0.999 999 999 941 653 875 916 8 × 2 = 1 + 0.999 999 999 883 307 751 833 6;
  • 26) 0.999 999 999 883 307 751 833 6 × 2 = 1 + 0.999 999 999 766 615 503 667 2;
  • 27) 0.999 999 999 766 615 503 667 2 × 2 = 1 + 0.999 999 999 533 231 007 334 4;
  • 28) 0.999 999 999 533 231 007 334 4 × 2 = 1 + 0.999 999 999 066 462 014 668 8;
  • 29) 0.999 999 999 066 462 014 668 8 × 2 = 1 + 0.999 999 998 132 924 029 337 6;
  • 30) 0.999 999 998 132 924 029 337 6 × 2 = 1 + 0.999 999 996 265 848 058 675 2;
  • 31) 0.999 999 996 265 848 058 675 2 × 2 = 1 + 0.999 999 992 531 696 117 350 4;
  • 32) 0.999 999 992 531 696 117 350 4 × 2 = 1 + 0.999 999 985 063 392 234 700 8;
  • 33) 0.999 999 985 063 392 234 700 8 × 2 = 1 + 0.999 999 970 126 784 469 401 6;
  • 34) 0.999 999 970 126 784 469 401 6 × 2 = 1 + 0.999 999 940 253 568 938 803 2;
  • 35) 0.999 999 940 253 568 938 803 2 × 2 = 1 + 0.999 999 880 507 137 877 606 4;
  • 36) 0.999 999 880 507 137 877 606 4 × 2 = 1 + 0.999 999 761 014 275 755 212 8;
  • 37) 0.999 999 761 014 275 755 212 8 × 2 = 1 + 0.999 999 522 028 551 510 425 6;
  • 38) 0.999 999 522 028 551 510 425 6 × 2 = 1 + 0.999 999 044 057 103 020 851 2;
  • 39) 0.999 999 044 057 103 020 851 2 × 2 = 1 + 0.999 998 088 114 206 041 702 4;
  • 40) 0.999 998 088 114 206 041 702 4 × 2 = 1 + 0.999 996 176 228 412 083 404 8;
  • 41) 0.999 996 176 228 412 083 404 8 × 2 = 1 + 0.999 992 352 456 824 166 809 6;
  • 42) 0.999 992 352 456 824 166 809 6 × 2 = 1 + 0.999 984 704 913 648 333 619 2;
  • 43) 0.999 984 704 913 648 333 619 2 × 2 = 1 + 0.999 969 409 827 296 667 238 4;
  • 44) 0.999 969 409 827 296 667 238 4 × 2 = 1 + 0.999 938 819 654 593 334 476 8;
  • 45) 0.999 938 819 654 593 334 476 8 × 2 = 1 + 0.999 877 639 309 186 668 953 6;
  • 46) 0.999 877 639 309 186 668 953 6 × 2 = 1 + 0.999 755 278 618 373 337 907 2;
  • 47) 0.999 755 278 618 373 337 907 2 × 2 = 1 + 0.999 510 557 236 746 675 814 4;
  • 48) 0.999 510 557 236 746 675 814 4 × 2 = 1 + 0.999 021 114 473 493 351 628 8;
  • 49) 0.999 021 114 473 493 351 628 8 × 2 = 1 + 0.998 042 228 946 986 703 257 6;
  • 50) 0.998 042 228 946 986 703 257 6 × 2 = 1 + 0.996 084 457 893 973 406 515 2;
  • 51) 0.996 084 457 893 973 406 515 2 × 2 = 1 + 0.992 168 915 787 946 813 030 4;
  • 52) 0.992 168 915 787 946 813 030 4 × 2 = 1 + 0.984 337 831 575 893 626 060 8;
  • 53) 0.984 337 831 575 893 626 060 8 × 2 = 1 + 0.968 675 663 151 787 252 121 6;
  • 54) 0.968 675 663 151 787 252 121 6 × 2 = 1 + 0.937 351 326 303 574 504 243 2;
  • 55) 0.937 351 326 303 574 504 243 2 × 2 = 1 + 0.874 702 652 607 149 008 486 4;
  • 56) 0.874 702 652 607 149 008 486 4 × 2 = 1 + 0.749 405 305 214 298 016 972 8;
  • 57) 0.749 405 305 214 298 016 972 8 × 2 = 1 + 0.498 810 610 428 596 033 945 6;
  • 58) 0.498 810 610 428 596 033 945 6 × 2 = 0 + 0.997 621 220 857 192 067 891 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:


0.016 738 891 601 562 496 522 3(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 522 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:


0.016 738 891 601 562 496 522 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 -0.016 738 891 601 562 496 522 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