-0.016 738 891 601 562 496 031 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 031(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
-0.016 738 891 601 562 496 031(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:

|-0.016 738 891 601 562 496 031| = 0.016 738 891 601 562 496 031


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: 0.016 738 891 601 562 496 031.

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) 0.016 738 891 601 562 496 031 × 2 = 0 + 0.033 477 783 203 124 992 062;
  • 2) 0.033 477 783 203 124 992 062 × 2 = 0 + 0.066 955 566 406 249 984 124;
  • 3) 0.066 955 566 406 249 984 124 × 2 = 0 + 0.133 911 132 812 499 968 248;
  • 4) 0.133 911 132 812 499 968 248 × 2 = 0 + 0.267 822 265 624 999 936 496;
  • 5) 0.267 822 265 624 999 936 496 × 2 = 0 + 0.535 644 531 249 999 872 992;
  • 6) 0.535 644 531 249 999 872 992 × 2 = 1 + 0.071 289 062 499 999 745 984;
  • 7) 0.071 289 062 499 999 745 984 × 2 = 0 + 0.142 578 124 999 999 491 968;
  • 8) 0.142 578 124 999 999 491 968 × 2 = 0 + 0.285 156 249 999 998 983 936;
  • 9) 0.285 156 249 999 998 983 936 × 2 = 0 + 0.570 312 499 999 997 967 872;
  • 10) 0.570 312 499 999 997 967 872 × 2 = 1 + 0.140 624 999 999 995 935 744;
  • 11) 0.140 624 999 999 995 935 744 × 2 = 0 + 0.281 249 999 999 991 871 488;
  • 12) 0.281 249 999 999 991 871 488 × 2 = 0 + 0.562 499 999 999 983 742 976;
  • 13) 0.562 499 999 999 983 742 976 × 2 = 1 + 0.124 999 999 999 967 485 952;
  • 14) 0.124 999 999 999 967 485 952 × 2 = 0 + 0.249 999 999 999 934 971 904;
  • 15) 0.249 999 999 999 934 971 904 × 2 = 0 + 0.499 999 999 999 869 943 808;
  • 16) 0.499 999 999 999 869 943 808 × 2 = 0 + 0.999 999 999 999 739 887 616;
  • 17) 0.999 999 999 999 739 887 616 × 2 = 1 + 0.999 999 999 999 479 775 232;
  • 18) 0.999 999 999 999 479 775 232 × 2 = 1 + 0.999 999 999 998 959 550 464;
  • 19) 0.999 999 999 998 959 550 464 × 2 = 1 + 0.999 999 999 997 919 100 928;
  • 20) 0.999 999 999 997 919 100 928 × 2 = 1 + 0.999 999 999 995 838 201 856;
  • 21) 0.999 999 999 995 838 201 856 × 2 = 1 + 0.999 999 999 991 676 403 712;
  • 22) 0.999 999 999 991 676 403 712 × 2 = 1 + 0.999 999 999 983 352 807 424;
  • 23) 0.999 999 999 983 352 807 424 × 2 = 1 + 0.999 999 999 966 705 614 848;
  • 24) 0.999 999 999 966 705 614 848 × 2 = 1 + 0.999 999 999 933 411 229 696;
  • 25) 0.999 999 999 933 411 229 696 × 2 = 1 + 0.999 999 999 866 822 459 392;
  • 26) 0.999 999 999 866 822 459 392 × 2 = 1 + 0.999 999 999 733 644 918 784;
  • 27) 0.999 999 999 733 644 918 784 × 2 = 1 + 0.999 999 999 467 289 837 568;
  • 28) 0.999 999 999 467 289 837 568 × 2 = 1 + 0.999 999 998 934 579 675 136;
  • 29) 0.999 999 998 934 579 675 136 × 2 = 1 + 0.999 999 997 869 159 350 272;
  • 30) 0.999 999 997 869 159 350 272 × 2 = 1 + 0.999 999 995 738 318 700 544;
  • 31) 0.999 999 995 738 318 700 544 × 2 = 1 + 0.999 999 991 476 637 401 088;
  • 32) 0.999 999 991 476 637 401 088 × 2 = 1 + 0.999 999 982 953 274 802 176;
  • 33) 0.999 999 982 953 274 802 176 × 2 = 1 + 0.999 999 965 906 549 604 352;
  • 34) 0.999 999 965 906 549 604 352 × 2 = 1 + 0.999 999 931 813 099 208 704;
  • 35) 0.999 999 931 813 099 208 704 × 2 = 1 + 0.999 999 863 626 198 417 408;
  • 36) 0.999 999 863 626 198 417 408 × 2 = 1 + 0.999 999 727 252 396 834 816;
  • 37) 0.999 999 727 252 396 834 816 × 2 = 1 + 0.999 999 454 504 793 669 632;
  • 38) 0.999 999 454 504 793 669 632 × 2 = 1 + 0.999 998 909 009 587 339 264;
  • 39) 0.999 998 909 009 587 339 264 × 2 = 1 + 0.999 997 818 019 174 678 528;
  • 40) 0.999 997 818 019 174 678 528 × 2 = 1 + 0.999 995 636 038 349 357 056;
  • 41) 0.999 995 636 038 349 357 056 × 2 = 1 + 0.999 991 272 076 698 714 112;
  • 42) 0.999 991 272 076 698 714 112 × 2 = 1 + 0.999 982 544 153 397 428 224;
  • 43) 0.999 982 544 153 397 428 224 × 2 = 1 + 0.999 965 088 306 794 856 448;
  • 44) 0.999 965 088 306 794 856 448 × 2 = 1 + 0.999 930 176 613 589 712 896;
  • 45) 0.999 930 176 613 589 712 896 × 2 = 1 + 0.999 860 353 227 179 425 792;
  • 46) 0.999 860 353 227 179 425 792 × 2 = 1 + 0.999 720 706 454 358 851 584;
  • 47) 0.999 720 706 454 358 851 584 × 2 = 1 + 0.999 441 412 908 717 703 168;
  • 48) 0.999 441 412 908 717 703 168 × 2 = 1 + 0.998 882 825 817 435 406 336;
  • 49) 0.998 882 825 817 435 406 336 × 2 = 1 + 0.997 765 651 634 870 812 672;
  • 50) 0.997 765 651 634 870 812 672 × 2 = 1 + 0.995 531 303 269 741 625 344;
  • 51) 0.995 531 303 269 741 625 344 × 2 = 1 + 0.991 062 606 539 483 250 688;
  • 52) 0.991 062 606 539 483 250 688 × 2 = 1 + 0.982 125 213 078 966 501 376;
  • 53) 0.982 125 213 078 966 501 376 × 2 = 1 + 0.964 250 426 157 933 002 752;
  • 54) 0.964 250 426 157 933 002 752 × 2 = 1 + 0.928 500 852 315 866 005 504;
  • 55) 0.928 500 852 315 866 005 504 × 2 = 1 + 0.857 001 704 631 732 011 008;
  • 56) 0.857 001 704 631 732 011 008 × 2 = 1 + 0.714 003 409 263 464 022 016;
  • 57) 0.714 003 409 263 464 022 016 × 2 = 1 + 0.428 006 818 526 928 044 032;
  • 58) 0.428 006 818 526 928 044 032 × 2 = 0 + 0.856 013 637 053 856 088 064;

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:


0.016 738 891 601 562 496 031(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 031(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:


0.016 738 891 601 562 496 031(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 -0.016 738 891 601 562 496 031 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