64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -0.812 499 940 395 355 1 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -0.812 499 940 395 355 1(10) converted and written in 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:

|-0.812 499 940 395 355 1| = 0.812 499 940 395 355 1

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: 0.812 499 940 395 355 1.

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) 0.812 499 940 395 355 1 × 2 = 1 + 0.624 999 880 790 710 2;
  • 2) 0.624 999 880 790 710 2 × 2 = 1 + 0.249 999 761 581 420 4;
  • 3) 0.249 999 761 581 420 4 × 2 = 0 + 0.499 999 523 162 840 8;
  • 4) 0.499 999 523 162 840 8 × 2 = 0 + 0.999 999 046 325 681 6;
  • 5) 0.999 999 046 325 681 6 × 2 = 1 + 0.999 998 092 651 363 2;
  • 6) 0.999 998 092 651 363 2 × 2 = 1 + 0.999 996 185 302 726 4;
  • 7) 0.999 996 185 302 726 4 × 2 = 1 + 0.999 992 370 605 452 8;
  • 8) 0.999 992 370 605 452 8 × 2 = 1 + 0.999 984 741 210 905 6;
  • 9) 0.999 984 741 210 905 6 × 2 = 1 + 0.999 969 482 421 811 2;
  • 10) 0.999 969 482 421 811 2 × 2 = 1 + 0.999 938 964 843 622 4;
  • 11) 0.999 938 964 843 622 4 × 2 = 1 + 0.999 877 929 687 244 8;
  • 12) 0.999 877 929 687 244 8 × 2 = 1 + 0.999 755 859 374 489 6;
  • 13) 0.999 755 859 374 489 6 × 2 = 1 + 0.999 511 718 748 979 2;
  • 14) 0.999 511 718 748 979 2 × 2 = 1 + 0.999 023 437 497 958 4;
  • 15) 0.999 023 437 497 958 4 × 2 = 1 + 0.998 046 874 995 916 8;
  • 16) 0.998 046 874 995 916 8 × 2 = 1 + 0.996 093 749 991 833 6;
  • 17) 0.996 093 749 991 833 6 × 2 = 1 + 0.992 187 499 983 667 2;
  • 18) 0.992 187 499 983 667 2 × 2 = 1 + 0.984 374 999 967 334 4;
  • 19) 0.984 374 999 967 334 4 × 2 = 1 + 0.968 749 999 934 668 8;
  • 20) 0.968 749 999 934 668 8 × 2 = 1 + 0.937 499 999 869 337 6;
  • 21) 0.937 499 999 869 337 6 × 2 = 1 + 0.874 999 999 738 675 2;
  • 22) 0.874 999 999 738 675 2 × 2 = 1 + 0.749 999 999 477 350 4;
  • 23) 0.749 999 999 477 350 4 × 2 = 1 + 0.499 999 998 954 700 8;
  • 24) 0.499 999 998 954 700 8 × 2 = 0 + 0.999 999 997 909 401 6;
  • 25) 0.999 999 997 909 401 6 × 2 = 1 + 0.999 999 995 818 803 2;
  • 26) 0.999 999 995 818 803 2 × 2 = 1 + 0.999 999 991 637 606 4;
  • 27) 0.999 999 991 637 606 4 × 2 = 1 + 0.999 999 983 275 212 8;
  • 28) 0.999 999 983 275 212 8 × 2 = 1 + 0.999 999 966 550 425 6;
  • 29) 0.999 999 966 550 425 6 × 2 = 1 + 0.999 999 933 100 851 2;
  • 30) 0.999 999 933 100 851 2 × 2 = 1 + 0.999 999 866 201 702 4;
  • 31) 0.999 999 866 201 702 4 × 2 = 1 + 0.999 999 732 403 404 8;
  • 32) 0.999 999 732 403 404 8 × 2 = 1 + 0.999 999 464 806 809 6;
  • 33) 0.999 999 464 806 809 6 × 2 = 1 + 0.999 998 929 613 619 2;
  • 34) 0.999 998 929 613 619 2 × 2 = 1 + 0.999 997 859 227 238 4;
  • 35) 0.999 997 859 227 238 4 × 2 = 1 + 0.999 995 718 454 476 8;
  • 36) 0.999 995 718 454 476 8 × 2 = 1 + 0.999 991 436 908 953 6;
  • 37) 0.999 991 436 908 953 6 × 2 = 1 + 0.999 982 873 817 907 2;
  • 38) 0.999 982 873 817 907 2 × 2 = 1 + 0.999 965 747 635 814 4;
  • 39) 0.999 965 747 635 814 4 × 2 = 1 + 0.999 931 495 271 628 8;
  • 40) 0.999 931 495 271 628 8 × 2 = 1 + 0.999 862 990 543 257 6;
  • 41) 0.999 862 990 543 257 6 × 2 = 1 + 0.999 725 981 086 515 2;
  • 42) 0.999 725 981 086 515 2 × 2 = 1 + 0.999 451 962 173 030 4;
  • 43) 0.999 451 962 173 030 4 × 2 = 1 + 0.998 903 924 346 060 8;
  • 44) 0.998 903 924 346 060 8 × 2 = 1 + 0.997 807 848 692 121 6;
  • 45) 0.997 807 848 692 121 6 × 2 = 1 + 0.995 615 697 384 243 2;
  • 46) 0.995 615 697 384 243 2 × 2 = 1 + 0.991 231 394 768 486 4;
  • 47) 0.991 231 394 768 486 4 × 2 = 1 + 0.982 462 789 536 972 8;
  • 48) 0.982 462 789 536 972 8 × 2 = 1 + 0.964 925 579 073 945 6;
  • 49) 0.964 925 579 073 945 6 × 2 = 1 + 0.929 851 158 147 891 2;
  • 50) 0.929 851 158 147 891 2 × 2 = 1 + 0.859 702 316 295 782 4;
  • 51) 0.859 702 316 295 782 4 × 2 = 1 + 0.719 404 632 591 564 8;
  • 52) 0.719 404 632 591 564 8 × 2 = 1 + 0.438 809 265 183 129 6;
  • 53) 0.438 809 265 183 129 6 × 2 = 0 + 0.877 618 530 366 259 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...)


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:


0.812 499 940 395 355 1(10) =


0.1100 1111 1111 1111 1111 1110 1111 1111 1111 1111 1111 1111 1111 0(2)


6. Positive number before normalization:

0.812 499 940 395 355 1(10) =


0.1100 1111 1111 1111 1111 1110 1111 1111 1111 1111 1111 1111 1111 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.812 499 940 395 355 1(10) =


0.1100 1111 1111 1111 1111 1110 1111 1111 1111 1111 1111 1111 1111 0(2) =


0.1100 1111 1111 1111 1111 1110 1111 1111 1111 1111 1111 1111 1111 0(2) × 20 =


1.1001 1111 1111 1111 1111 1101 1111 1111 1111 1111 1111 1111 1110(2) × 2-1


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): -1


Mantissa (not normalized):
1.1001 1111 1111 1111 1111 1101 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 =


-1 + 2(11-1) - 1 =


(-1 + 1 023)(10) =


1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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) =


1022(10) =


011 1111 1110(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. 1001 1111 1111 1111 1111 1101 1111 1111 1111 1111 1111 1111 1110 =


1001 1111 1111 1111 1111 1101 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 1110


Mantissa (52 bits) =
1001 1111 1111 1111 1111 1101 1111 1111 1111 1111 1111 1111 1110


The base ten decimal number -0.812 499 940 395 355 1 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 011 1111 1110 - 1001 1111 1111 1111 1111 1101 1111 1111 1111 1111 1111 1111 1110

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