-1 036.699 999 999 999 818 101 059 645 414 352 416 992 154 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1 036.699 999 999 999 818 101 059 645 414 352 416 992 154 4(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
-1 036.699 999 999 999 818 101 059 645 414 352 416 992 154 4(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:

|-1 036.699 999 999 999 818 101 059 645 414 352 416 992 154 4| = 1 036.699 999 999 999 818 101 059 645 414 352 416 992 154 4


2. First, convert to binary (in base 2) the integer part: 1 036.
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;
  • 1 036 ÷ 2 = 518 + 0;
  • 518 ÷ 2 = 259 + 0;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 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;

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.

1 036(10) =

100 0000 1100(2)


4. Convert to binary (base 2) the fractional part: 0.699 999 999 999 818 101 059 645 414 352 416 992 154 4.

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) 0.699 999 999 999 818 101 059 645 414 352 416 992 154 4 × 2 = 1 + 0.399 999 999 999 636 202 119 290 828 704 833 984 308 8;
  • 2) 0.399 999 999 999 636 202 119 290 828 704 833 984 308 8 × 2 = 0 + 0.799 999 999 999 272 404 238 581 657 409 667 968 617 6;
  • 3) 0.799 999 999 999 272 404 238 581 657 409 667 968 617 6 × 2 = 1 + 0.599 999 999 998 544 808 477 163 314 819 335 937 235 2;
  • 4) 0.599 999 999 998 544 808 477 163 314 819 335 937 235 2 × 2 = 1 + 0.199 999 999 997 089 616 954 326 629 638 671 874 470 4;
  • 5) 0.199 999 999 997 089 616 954 326 629 638 671 874 470 4 × 2 = 0 + 0.399 999 999 994 179 233 908 653 259 277 343 748 940 8;
  • 6) 0.399 999 999 994 179 233 908 653 259 277 343 748 940 8 × 2 = 0 + 0.799 999 999 988 358 467 817 306 518 554 687 497 881 6;
  • 7) 0.799 999 999 988 358 467 817 306 518 554 687 497 881 6 × 2 = 1 + 0.599 999 999 976 716 935 634 613 037 109 374 995 763 2;
  • 8) 0.599 999 999 976 716 935 634 613 037 109 374 995 763 2 × 2 = 1 + 0.199 999 999 953 433 871 269 226 074 218 749 991 526 4;
  • 9) 0.199 999 999 953 433 871 269 226 074 218 749 991 526 4 × 2 = 0 + 0.399 999 999 906 867 742 538 452 148 437 499 983 052 8;
  • 10) 0.399 999 999 906 867 742 538 452 148 437 499 983 052 8 × 2 = 0 + 0.799 999 999 813 735 485 076 904 296 874 999 966 105 6;
  • 11) 0.799 999 999 813 735 485 076 904 296 874 999 966 105 6 × 2 = 1 + 0.599 999 999 627 470 970 153 808 593 749 999 932 211 2;
  • 12) 0.599 999 999 627 470 970 153 808 593 749 999 932 211 2 × 2 = 1 + 0.199 999 999 254 941 940 307 617 187 499 999 864 422 4;
  • 13) 0.199 999 999 254 941 940 307 617 187 499 999 864 422 4 × 2 = 0 + 0.399 999 998 509 883 880 615 234 374 999 999 728 844 8;
  • 14) 0.399 999 998 509 883 880 615 234 374 999 999 728 844 8 × 2 = 0 + 0.799 999 997 019 767 761 230 468 749 999 999 457 689 6;
  • 15) 0.799 999 997 019 767 761 230 468 749 999 999 457 689 6 × 2 = 1 + 0.599 999 994 039 535 522 460 937 499 999 998 915 379 2;
  • 16) 0.599 999 994 039 535 522 460 937 499 999 998 915 379 2 × 2 = 1 + 0.199 999 988 079 071 044 921 874 999 999 997 830 758 4;
  • 17) 0.199 999 988 079 071 044 921 874 999 999 997 830 758 4 × 2 = 0 + 0.399 999 976 158 142 089 843 749 999 999 995 661 516 8;
  • 18) 0.399 999 976 158 142 089 843 749 999 999 995 661 516 8 × 2 = 0 + 0.799 999 952 316 284 179 687 499 999 999 991 323 033 6;
  • 19) 0.799 999 952 316 284 179 687 499 999 999 991 323 033 6 × 2 = 1 + 0.599 999 904 632 568 359 374 999 999 999 982 646 067 2;
  • 20) 0.599 999 904 632 568 359 374 999 999 999 982 646 067 2 × 2 = 1 + 0.199 999 809 265 136 718 749 999 999 999 965 292 134 4;
  • 21) 0.199 999 809 265 136 718 749 999 999 999 965 292 134 4 × 2 = 0 + 0.399 999 618 530 273 437 499 999 999 999 930 584 268 8;
  • 22) 0.399 999 618 530 273 437 499 999 999 999 930 584 268 8 × 2 = 0 + 0.799 999 237 060 546 874 999 999 999 999 861 168 537 6;
  • 23) 0.799 999 237 060 546 874 999 999 999 999 861 168 537 6 × 2 = 1 + 0.599 998 474 121 093 749 999 999 999 999 722 337 075 2;
  • 24) 0.599 998 474 121 093 749 999 999 999 999 722 337 075 2 × 2 = 1 + 0.199 996 948 242 187 499 999 999 999 999 444 674 150 4;
  • 25) 0.199 996 948 242 187 499 999 999 999 999 444 674 150 4 × 2 = 0 + 0.399 993 896 484 374 999 999 999 999 998 889 348 300 8;
  • 26) 0.399 993 896 484 374 999 999 999 999 998 889 348 300 8 × 2 = 0 + 0.799 987 792 968 749 999 999 999 999 997 778 696 601 6;
  • 27) 0.799 987 792 968 749 999 999 999 999 997 778 696 601 6 × 2 = 1 + 0.599 975 585 937 499 999 999 999 999 995 557 393 203 2;
  • 28) 0.599 975 585 937 499 999 999 999 999 995 557 393 203 2 × 2 = 1 + 0.199 951 171 874 999 999 999 999 999 991 114 786 406 4;
  • 29) 0.199 951 171 874 999 999 999 999 999 991 114 786 406 4 × 2 = 0 + 0.399 902 343 749 999 999 999 999 999 982 229 572 812 8;
  • 30) 0.399 902 343 749 999 999 999 999 999 982 229 572 812 8 × 2 = 0 + 0.799 804 687 499 999 999 999 999 999 964 459 145 625 6;
  • 31) 0.799 804 687 499 999 999 999 999 999 964 459 145 625 6 × 2 = 1 + 0.599 609 374 999 999 999 999 999 999 928 918 291 251 2;
  • 32) 0.599 609 374 999 999 999 999 999 999 928 918 291 251 2 × 2 = 1 + 0.199 218 749 999 999 999 999 999 999 857 836 582 502 4;
  • 33) 0.199 218 749 999 999 999 999 999 999 857 836 582 502 4 × 2 = 0 + 0.398 437 499 999 999 999 999 999 999 715 673 165 004 8;
  • 34) 0.398 437 499 999 999 999 999 999 999 715 673 165 004 8 × 2 = 0 + 0.796 874 999 999 999 999 999 999 999 431 346 330 009 6;
  • 35) 0.796 874 999 999 999 999 999 999 999 431 346 330 009 6 × 2 = 1 + 0.593 749 999 999 999 999 999 999 998 862 692 660 019 2;
  • 36) 0.593 749 999 999 999 999 999 999 998 862 692 660 019 2 × 2 = 1 + 0.187 499 999 999 999 999 999 999 997 725 385 320 038 4;
  • 37) 0.187 499 999 999 999 999 999 999 997 725 385 320 038 4 × 2 = 0 + 0.374 999 999 999 999 999 999 999 995 450 770 640 076 8;
  • 38) 0.374 999 999 999 999 999 999 999 995 450 770 640 076 8 × 2 = 0 + 0.749 999 999 999 999 999 999 999 990 901 541 280 153 6;
  • 39) 0.749 999 999 999 999 999 999 999 990 901 541 280 153 6 × 2 = 1 + 0.499 999 999 999 999 999 999 999 981 803 082 560 307 2;
  • 40) 0.499 999 999 999 999 999 999 999 981 803 082 560 307 2 × 2 = 0 + 0.999 999 999 999 999 999 999 999 963 606 165 120 614 4;
  • 41) 0.999 999 999 999 999 999 999 999 963 606 165 120 614 4 × 2 = 1 + 0.999 999 999 999 999 999 999 999 927 212 330 241 228 8;
  • 42) 0.999 999 999 999 999 999 999 999 927 212 330 241 228 8 × 2 = 1 + 0.999 999 999 999 999 999 999 999 854 424 660 482 457 6;
  • 43) 0.999 999 999 999 999 999 999 999 854 424 660 482 457 6 × 2 = 1 + 0.999 999 999 999 999 999 999 999 708 849 320 964 915 2;
  • 44) 0.999 999 999 999 999 999 999 999 708 849 320 964 915 2 × 2 = 1 + 0.999 999 999 999 999 999 999 999 417 698 641 929 830 4;
  • 45) 0.999 999 999 999 999 999 999 999 417 698 641 929 830 4 × 2 = 1 + 0.999 999 999 999 999 999 999 998 835 397 283 859 660 8;
  • 46) 0.999 999 999 999 999 999 999 998 835 397 283 859 660 8 × 2 = 1 + 0.999 999 999 999 999 999 999 997 670 794 567 719 321 6;
  • 47) 0.999 999 999 999 999 999 999 997 670 794 567 719 321 6 × 2 = 1 + 0.999 999 999 999 999 999 999 995 341 589 135 438 643 2;
  • 48) 0.999 999 999 999 999 999 999 995 341 589 135 438 643 2 × 2 = 1 + 0.999 999 999 999 999 999 999 990 683 178 270 877 286 4;
  • 49) 0.999 999 999 999 999 999 999 990 683 178 270 877 286 4 × 2 = 1 + 0.999 999 999 999 999 999 999 981 366 356 541 754 572 8;
  • 50) 0.999 999 999 999 999 999 999 981 366 356 541 754 572 8 × 2 = 1 + 0.999 999 999 999 999 999 999 962 732 713 083 509 145 6;
  • 51) 0.999 999 999 999 999 999 999 962 732 713 083 509 145 6 × 2 = 1 + 0.999 999 999 999 999 999 999 925 465 426 167 018 291 2;
  • 52) 0.999 999 999 999 999 999 999 925 465 426 167 018 291 2 × 2 = 1 + 0.999 999 999 999 999 999 999 850 930 852 334 036 582 4;
  • 53) 0.999 999 999 999 999 999 999 850 930 852 334 036 582 4 × 2 = 1 + 0.999 999 999 999 999 999 999 701 861 704 668 073 164 8;

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:


0.699 999 999 999 818 101 059 645 414 352 416 992 154 4(10) =

0.1011 0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 1111 1111 1(2)

6. Positive number before normalization:

1 036.699 999 999 999 818 101 059 645 414 352 416 992 154 4(10) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 1111 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the left, so that only one non zero digit remains to the left of it:


1 036.699 999 999 999 818 101 059 645 414 352 416 992 154 4(10) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 1111 1111 1(2) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 1111 1111 1(2) × 20 =

1.0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1111 111(2) × 210


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): 10

Mantissa (not normalized):
1.0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1111 111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

10 + 2(11-1) - 1 =

(10 + 1 023)(10) =

1 033(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;
  • 1 033 ÷ 2 = 516 + 1;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 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;

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) =

1033(10) =

100 0000 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, 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. 0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 111 1111 1111 =

0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011


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) =
100 0000 1001

Mantissa (52 bits) =
0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011


Decimal number -1 036.699 999 999 999 818 101 059 645 414 352 416 992 154 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 1001 - 0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011

» Convert decimal -1 036.699 999 999 999 818 101 059 645 414 352 416 992 164 2 to 64 bit double precision IEEE 754 binary floating point representation standard

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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