5.919 999 999 999 999 928 945 726 423 989 981 412 887 573 182 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5.919 999 999 999 999 928 945 726 423 989 981 412 887 573 182 5(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
5.919 999 999 999 999 928 945 726 423 989 981 412 887 573 182 5(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. First, convert to binary (in base 2) the integer part: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

5(10) =

101(2)


3. Convert to binary (base 2) the fractional part: 0.919 999 999 999 999 928 945 726 423 989 981 412 887 573 182 5.

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) 0.919 999 999 999 999 928 945 726 423 989 981 412 887 573 182 5 × 2 = 1 + 0.839 999 999 999 999 857 891 452 847 979 962 825 775 146 365;
  • 2) 0.839 999 999 999 999 857 891 452 847 979 962 825 775 146 365 × 2 = 1 + 0.679 999 999 999 999 715 782 905 695 959 925 651 550 292 73;
  • 3) 0.679 999 999 999 999 715 782 905 695 959 925 651 550 292 73 × 2 = 1 + 0.359 999 999 999 999 431 565 811 391 919 851 303 100 585 46;
  • 4) 0.359 999 999 999 999 431 565 811 391 919 851 303 100 585 46 × 2 = 0 + 0.719 999 999 999 998 863 131 622 783 839 702 606 201 170 92;
  • 5) 0.719 999 999 999 998 863 131 622 783 839 702 606 201 170 92 × 2 = 1 + 0.439 999 999 999 997 726 263 245 567 679 405 212 402 341 84;
  • 6) 0.439 999 999 999 997 726 263 245 567 679 405 212 402 341 84 × 2 = 0 + 0.879 999 999 999 995 452 526 491 135 358 810 424 804 683 68;
  • 7) 0.879 999 999 999 995 452 526 491 135 358 810 424 804 683 68 × 2 = 1 + 0.759 999 999 999 990 905 052 982 270 717 620 849 609 367 36;
  • 8) 0.759 999 999 999 990 905 052 982 270 717 620 849 609 367 36 × 2 = 1 + 0.519 999 999 999 981 810 105 964 541 435 241 699 218 734 72;
  • 9) 0.519 999 999 999 981 810 105 964 541 435 241 699 218 734 72 × 2 = 1 + 0.039 999 999 999 963 620 211 929 082 870 483 398 437 469 44;
  • 10) 0.039 999 999 999 963 620 211 929 082 870 483 398 437 469 44 × 2 = 0 + 0.079 999 999 999 927 240 423 858 165 740 966 796 874 938 88;
  • 11) 0.079 999 999 999 927 240 423 858 165 740 966 796 874 938 88 × 2 = 0 + 0.159 999 999 999 854 480 847 716 331 481 933 593 749 877 76;
  • 12) 0.159 999 999 999 854 480 847 716 331 481 933 593 749 877 76 × 2 = 0 + 0.319 999 999 999 708 961 695 432 662 963 867 187 499 755 52;
  • 13) 0.319 999 999 999 708 961 695 432 662 963 867 187 499 755 52 × 2 = 0 + 0.639 999 999 999 417 923 390 865 325 927 734 374 999 511 04;
  • 14) 0.639 999 999 999 417 923 390 865 325 927 734 374 999 511 04 × 2 = 1 + 0.279 999 999 998 835 846 781 730 651 855 468 749 999 022 08;
  • 15) 0.279 999 999 998 835 846 781 730 651 855 468 749 999 022 08 × 2 = 0 + 0.559 999 999 997 671 693 563 461 303 710 937 499 998 044 16;
  • 16) 0.559 999 999 997 671 693 563 461 303 710 937 499 998 044 16 × 2 = 1 + 0.119 999 999 995 343 387 126 922 607 421 874 999 996 088 32;
  • 17) 0.119 999 999 995 343 387 126 922 607 421 874 999 996 088 32 × 2 = 0 + 0.239 999 999 990 686 774 253 845 214 843 749 999 992 176 64;
  • 18) 0.239 999 999 990 686 774 253 845 214 843 749 999 992 176 64 × 2 = 0 + 0.479 999 999 981 373 548 507 690 429 687 499 999 984 353 28;
  • 19) 0.479 999 999 981 373 548 507 690 429 687 499 999 984 353 28 × 2 = 0 + 0.959 999 999 962 747 097 015 380 859 374 999 999 968 706 56;
  • 20) 0.959 999 999 962 747 097 015 380 859 374 999 999 968 706 56 × 2 = 1 + 0.919 999 999 925 494 194 030 761 718 749 999 999 937 413 12;
  • 21) 0.919 999 999 925 494 194 030 761 718 749 999 999 937 413 12 × 2 = 1 + 0.839 999 999 850 988 388 061 523 437 499 999 999 874 826 24;
  • 22) 0.839 999 999 850 988 388 061 523 437 499 999 999 874 826 24 × 2 = 1 + 0.679 999 999 701 976 776 123 046 874 999 999 999 749 652 48;
  • 23) 0.679 999 999 701 976 776 123 046 874 999 999 999 749 652 48 × 2 = 1 + 0.359 999 999 403 953 552 246 093 749 999 999 999 499 304 96;
  • 24) 0.359 999 999 403 953 552 246 093 749 999 999 999 499 304 96 × 2 = 0 + 0.719 999 998 807 907 104 492 187 499 999 999 998 998 609 92;
  • 25) 0.719 999 998 807 907 104 492 187 499 999 999 998 998 609 92 × 2 = 1 + 0.439 999 997 615 814 208 984 374 999 999 999 997 997 219 84;
  • 26) 0.439 999 997 615 814 208 984 374 999 999 999 997 997 219 84 × 2 = 0 + 0.879 999 995 231 628 417 968 749 999 999 999 995 994 439 68;
  • 27) 0.879 999 995 231 628 417 968 749 999 999 999 995 994 439 68 × 2 = 1 + 0.759 999 990 463 256 835 937 499 999 999 999 991 988 879 36;
  • 28) 0.759 999 990 463 256 835 937 499 999 999 999 991 988 879 36 × 2 = 1 + 0.519 999 980 926 513 671 874 999 999 999 999 983 977 758 72;
  • 29) 0.519 999 980 926 513 671 874 999 999 999 999 983 977 758 72 × 2 = 1 + 0.039 999 961 853 027 343 749 999 999 999 999 967 955 517 44;
  • 30) 0.039 999 961 853 027 343 749 999 999 999 999 967 955 517 44 × 2 = 0 + 0.079 999 923 706 054 687 499 999 999 999 999 935 911 034 88;
  • 31) 0.079 999 923 706 054 687 499 999 999 999 999 935 911 034 88 × 2 = 0 + 0.159 999 847 412 109 374 999 999 999 999 999 871 822 069 76;
  • 32) 0.159 999 847 412 109 374 999 999 999 999 999 871 822 069 76 × 2 = 0 + 0.319 999 694 824 218 749 999 999 999 999 999 743 644 139 52;
  • 33) 0.319 999 694 824 218 749 999 999 999 999 999 743 644 139 52 × 2 = 0 + 0.639 999 389 648 437 499 999 999 999 999 999 487 288 279 04;
  • 34) 0.639 999 389 648 437 499 999 999 999 999 999 487 288 279 04 × 2 = 1 + 0.279 998 779 296 874 999 999 999 999 999 998 974 576 558 08;
  • 35) 0.279 998 779 296 874 999 999 999 999 999 998 974 576 558 08 × 2 = 0 + 0.559 997 558 593 749 999 999 999 999 999 997 949 153 116 16;
  • 36) 0.559 997 558 593 749 999 999 999 999 999 997 949 153 116 16 × 2 = 1 + 0.119 995 117 187 499 999 999 999 999 999 995 898 306 232 32;
  • 37) 0.119 995 117 187 499 999 999 999 999 999 995 898 306 232 32 × 2 = 0 + 0.239 990 234 374 999 999 999 999 999 999 991 796 612 464 64;
  • 38) 0.239 990 234 374 999 999 999 999 999 999 991 796 612 464 64 × 2 = 0 + 0.479 980 468 749 999 999 999 999 999 999 983 593 224 929 28;
  • 39) 0.479 980 468 749 999 999 999 999 999 999 983 593 224 929 28 × 2 = 0 + 0.959 960 937 499 999 999 999 999 999 999 967 186 449 858 56;
  • 40) 0.959 960 937 499 999 999 999 999 999 999 967 186 449 858 56 × 2 = 1 + 0.919 921 874 999 999 999 999 999 999 999 934 372 899 717 12;
  • 41) 0.919 921 874 999 999 999 999 999 999 999 934 372 899 717 12 × 2 = 1 + 0.839 843 749 999 999 999 999 999 999 999 868 745 799 434 24;
  • 42) 0.839 843 749 999 999 999 999 999 999 999 868 745 799 434 24 × 2 = 1 + 0.679 687 499 999 999 999 999 999 999 999 737 491 598 868 48;
  • 43) 0.679 687 499 999 999 999 999 999 999 999 737 491 598 868 48 × 2 = 1 + 0.359 374 999 999 999 999 999 999 999 999 474 983 197 736 96;
  • 44) 0.359 374 999 999 999 999 999 999 999 999 474 983 197 736 96 × 2 = 0 + 0.718 749 999 999 999 999 999 999 999 998 949 966 395 473 92;
  • 45) 0.718 749 999 999 999 999 999 999 999 998 949 966 395 473 92 × 2 = 1 + 0.437 499 999 999 999 999 999 999 999 997 899 932 790 947 84;
  • 46) 0.437 499 999 999 999 999 999 999 999 997 899 932 790 947 84 × 2 = 0 + 0.874 999 999 999 999 999 999 999 999 995 799 865 581 895 68;
  • 47) 0.874 999 999 999 999 999 999 999 999 995 799 865 581 895 68 × 2 = 1 + 0.749 999 999 999 999 999 999 999 999 991 599 731 163 791 36;
  • 48) 0.749 999 999 999 999 999 999 999 999 991 599 731 163 791 36 × 2 = 1 + 0.499 999 999 999 999 999 999 999 999 983 199 462 327 582 72;
  • 49) 0.499 999 999 999 999 999 999 999 999 983 199 462 327 582 72 × 2 = 0 + 0.999 999 999 999 999 999 999 999 999 966 398 924 655 165 44;
  • 50) 0.999 999 999 999 999 999 999 999 999 966 398 924 655 165 44 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 932 797 849 310 330 88;
  • 51) 0.999 999 999 999 999 999 999 999 999 932 797 849 310 330 88 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 865 595 698 620 661 76;
  • 52) 0.999 999 999 999 999 999 999 999 999 865 595 698 620 661 76 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 731 191 397 241 323 52;
  • 53) 0.999 999 999 999 999 999 999 999 999 731 191 397 241 323 52 × 2 = 1 + 0.999 999 999 999 999 999 999 999 999 462 382 794 482 647 04;

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).


4. 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:


0.919 999 999 999 999 928 945 726 423 989 981 412 887 573 182 5(10) =

0.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2)

5. Positive number before normalization:

5.919 999 999 999 999 928 945 726 423 989 981 412 887 573 182 5(10) =

101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.919 999 999 999 999 928 945 726 423 989 981 412 887 573 182 5(10) =

101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2) =

101.1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 0111 1(2) × 20 =

1.0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 111(2) × 22


7. 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 0 (a positive number)

Exponent (unadjusted): 2

Mantissa (not normalized):
1.0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(10)


9. 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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1025(10) =

100 0000 0001(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 111 =

0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0001

Mantissa (52 bits) =
0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101


Decimal number 5.919 999 999 999 999 928 945 726 423 989 981 412 887 573 182 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101

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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