0.000 999 999 999 999 999 803 976 247 214 620 793 26 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 999 999 999 999 999 803 976 247 214 620 793 26(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
0.000 999 999 999 999 999 803 976 247 214 620 793 26(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: 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;

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.

0(10) =


0(2)


3. Convert to binary (base 2) the fractional part: 0.000 999 999 999 999 999 803 976 247 214 620 793 26.

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) 0.000 999 999 999 999 999 803 976 247 214 620 793 26 × 2 = 0 + 0.001 999 999 999 999 999 607 952 494 429 241 586 52;
  • 2) 0.001 999 999 999 999 999 607 952 494 429 241 586 52 × 2 = 0 + 0.003 999 999 999 999 999 215 904 988 858 483 173 04;
  • 3) 0.003 999 999 999 999 999 215 904 988 858 483 173 04 × 2 = 0 + 0.007 999 999 999 999 998 431 809 977 716 966 346 08;
  • 4) 0.007 999 999 999 999 998 431 809 977 716 966 346 08 × 2 = 0 + 0.015 999 999 999 999 996 863 619 955 433 932 692 16;
  • 5) 0.015 999 999 999 999 996 863 619 955 433 932 692 16 × 2 = 0 + 0.031 999 999 999 999 993 727 239 910 867 865 384 32;
  • 6) 0.031 999 999 999 999 993 727 239 910 867 865 384 32 × 2 = 0 + 0.063 999 999 999 999 987 454 479 821 735 730 768 64;
  • 7) 0.063 999 999 999 999 987 454 479 821 735 730 768 64 × 2 = 0 + 0.127 999 999 999 999 974 908 959 643 471 461 537 28;
  • 8) 0.127 999 999 999 999 974 908 959 643 471 461 537 28 × 2 = 0 + 0.255 999 999 999 999 949 817 919 286 942 923 074 56;
  • 9) 0.255 999 999 999 999 949 817 919 286 942 923 074 56 × 2 = 0 + 0.511 999 999 999 999 899 635 838 573 885 846 149 12;
  • 10) 0.511 999 999 999 999 899 635 838 573 885 846 149 12 × 2 = 1 + 0.023 999 999 999 999 799 271 677 147 771 692 298 24;
  • 11) 0.023 999 999 999 999 799 271 677 147 771 692 298 24 × 2 = 0 + 0.047 999 999 999 999 598 543 354 295 543 384 596 48;
  • 12) 0.047 999 999 999 999 598 543 354 295 543 384 596 48 × 2 = 0 + 0.095 999 999 999 999 197 086 708 591 086 769 192 96;
  • 13) 0.095 999 999 999 999 197 086 708 591 086 769 192 96 × 2 = 0 + 0.191 999 999 999 998 394 173 417 182 173 538 385 92;
  • 14) 0.191 999 999 999 998 394 173 417 182 173 538 385 92 × 2 = 0 + 0.383 999 999 999 996 788 346 834 364 347 076 771 84;
  • 15) 0.383 999 999 999 996 788 346 834 364 347 076 771 84 × 2 = 0 + 0.767 999 999 999 993 576 693 668 728 694 153 543 68;
  • 16) 0.767 999 999 999 993 576 693 668 728 694 153 543 68 × 2 = 1 + 0.535 999 999 999 987 153 387 337 457 388 307 087 36;
  • 17) 0.535 999 999 999 987 153 387 337 457 388 307 087 36 × 2 = 1 + 0.071 999 999 999 974 306 774 674 914 776 614 174 72;
  • 18) 0.071 999 999 999 974 306 774 674 914 776 614 174 72 × 2 = 0 + 0.143 999 999 999 948 613 549 349 829 553 228 349 44;
  • 19) 0.143 999 999 999 948 613 549 349 829 553 228 349 44 × 2 = 0 + 0.287 999 999 999 897 227 098 699 659 106 456 698 88;
  • 20) 0.287 999 999 999 897 227 098 699 659 106 456 698 88 × 2 = 0 + 0.575 999 999 999 794 454 197 399 318 212 913 397 76;
  • 21) 0.575 999 999 999 794 454 197 399 318 212 913 397 76 × 2 = 1 + 0.151 999 999 999 588 908 394 798 636 425 826 795 52;
  • 22) 0.151 999 999 999 588 908 394 798 636 425 826 795 52 × 2 = 0 + 0.303 999 999 999 177 816 789 597 272 851 653 591 04;
  • 23) 0.303 999 999 999 177 816 789 597 272 851 653 591 04 × 2 = 0 + 0.607 999 999 998 355 633 579 194 545 703 307 182 08;
  • 24) 0.607 999 999 998 355 633 579 194 545 703 307 182 08 × 2 = 1 + 0.215 999 999 996 711 267 158 389 091 406 614 364 16;
  • 25) 0.215 999 999 996 711 267 158 389 091 406 614 364 16 × 2 = 0 + 0.431 999 999 993 422 534 316 778 182 813 228 728 32;
  • 26) 0.431 999 999 993 422 534 316 778 182 813 228 728 32 × 2 = 0 + 0.863 999 999 986 845 068 633 556 365 626 457 456 64;
  • 27) 0.863 999 999 986 845 068 633 556 365 626 457 456 64 × 2 = 1 + 0.727 999 999 973 690 137 267 112 731 252 914 913 28;
  • 28) 0.727 999 999 973 690 137 267 112 731 252 914 913 28 × 2 = 1 + 0.455 999 999 947 380 274 534 225 462 505 829 826 56;
  • 29) 0.455 999 999 947 380 274 534 225 462 505 829 826 56 × 2 = 0 + 0.911 999 999 894 760 549 068 450 925 011 659 653 12;
  • 30) 0.911 999 999 894 760 549 068 450 925 011 659 653 12 × 2 = 1 + 0.823 999 999 789 521 098 136 901 850 023 319 306 24;
  • 31) 0.823 999 999 789 521 098 136 901 850 023 319 306 24 × 2 = 1 + 0.647 999 999 579 042 196 273 803 700 046 638 612 48;
  • 32) 0.647 999 999 579 042 196 273 803 700 046 638 612 48 × 2 = 1 + 0.295 999 999 158 084 392 547 607 400 093 277 224 96;
  • 33) 0.295 999 999 158 084 392 547 607 400 093 277 224 96 × 2 = 0 + 0.591 999 998 316 168 785 095 214 800 186 554 449 92;
  • 34) 0.591 999 998 316 168 785 095 214 800 186 554 449 92 × 2 = 1 + 0.183 999 996 632 337 570 190 429 600 373 108 899 84;
  • 35) 0.183 999 996 632 337 570 190 429 600 373 108 899 84 × 2 = 0 + 0.367 999 993 264 675 140 380 859 200 746 217 799 68;
  • 36) 0.367 999 993 264 675 140 380 859 200 746 217 799 68 × 2 = 0 + 0.735 999 986 529 350 280 761 718 401 492 435 599 36;
  • 37) 0.735 999 986 529 350 280 761 718 401 492 435 599 36 × 2 = 1 + 0.471 999 973 058 700 561 523 436 802 984 871 198 72;
  • 38) 0.471 999 973 058 700 561 523 436 802 984 871 198 72 × 2 = 0 + 0.943 999 946 117 401 123 046 873 605 969 742 397 44;
  • 39) 0.943 999 946 117 401 123 046 873 605 969 742 397 44 × 2 = 1 + 0.887 999 892 234 802 246 093 747 211 939 484 794 88;
  • 40) 0.887 999 892 234 802 246 093 747 211 939 484 794 88 × 2 = 1 + 0.775 999 784 469 604 492 187 494 423 878 969 589 76;
  • 41) 0.775 999 784 469 604 492 187 494 423 878 969 589 76 × 2 = 1 + 0.551 999 568 939 208 984 374 988 847 757 939 179 52;
  • 42) 0.551 999 568 939 208 984 374 988 847 757 939 179 52 × 2 = 1 + 0.103 999 137 878 417 968 749 977 695 515 878 359 04;
  • 43) 0.103 999 137 878 417 968 749 977 695 515 878 359 04 × 2 = 0 + 0.207 998 275 756 835 937 499 955 391 031 756 718 08;
  • 44) 0.207 998 275 756 835 937 499 955 391 031 756 718 08 × 2 = 0 + 0.415 996 551 513 671 874 999 910 782 063 513 436 16;
  • 45) 0.415 996 551 513 671 874 999 910 782 063 513 436 16 × 2 = 0 + 0.831 993 103 027 343 749 999 821 564 127 026 872 32;
  • 46) 0.831 993 103 027 343 749 999 821 564 127 026 872 32 × 2 = 1 + 0.663 986 206 054 687 499 999 643 128 254 053 744 64;
  • 47) 0.663 986 206 054 687 499 999 643 128 254 053 744 64 × 2 = 1 + 0.327 972 412 109 374 999 999 286 256 508 107 489 28;
  • 48) 0.327 972 412 109 374 999 999 286 256 508 107 489 28 × 2 = 0 + 0.655 944 824 218 749 999 998 572 513 016 214 978 56;
  • 49) 0.655 944 824 218 749 999 998 572 513 016 214 978 56 × 2 = 1 + 0.311 889 648 437 499 999 997 145 026 032 429 957 12;
  • 50) 0.311 889 648 437 499 999 997 145 026 032 429 957 12 × 2 = 0 + 0.623 779 296 874 999 999 994 290 052 064 859 914 24;
  • 51) 0.623 779 296 874 999 999 994 290 052 064 859 914 24 × 2 = 1 + 0.247 558 593 749 999 999 988 580 104 129 719 828 48;
  • 52) 0.247 558 593 749 999 999 988 580 104 129 719 828 48 × 2 = 0 + 0.495 117 187 499 999 999 977 160 208 259 439 656 96;
  • 53) 0.495 117 187 499 999 999 977 160 208 259 439 656 96 × 2 = 0 + 0.990 234 374 999 999 999 954 320 416 518 879 313 92;
  • 54) 0.990 234 374 999 999 999 954 320 416 518 879 313 92 × 2 = 1 + 0.980 468 749 999 999 999 908 640 833 037 758 627 84;
  • 55) 0.980 468 749 999 999 999 908 640 833 037 758 627 84 × 2 = 1 + 0.960 937 499 999 999 999 817 281 666 075 517 255 68;
  • 56) 0.960 937 499 999 999 999 817 281 666 075 517 255 68 × 2 = 1 + 0.921 874 999 999 999 999 634 563 332 151 034 511 36;
  • 57) 0.921 874 999 999 999 999 634 563 332 151 034 511 36 × 2 = 1 + 0.843 749 999 999 999 999 269 126 664 302 069 022 72;
  • 58) 0.843 749 999 999 999 999 269 126 664 302 069 022 72 × 2 = 1 + 0.687 499 999 999 999 998 538 253 328 604 138 045 44;
  • 59) 0.687 499 999 999 999 998 538 253 328 604 138 045 44 × 2 = 1 + 0.374 999 999 999 999 997 076 506 657 208 276 090 88;
  • 60) 0.374 999 999 999 999 997 076 506 657 208 276 090 88 × 2 = 0 + 0.749 999 999 999 999 994 153 013 314 416 552 181 76;
  • 61) 0.749 999 999 999 999 994 153 013 314 416 552 181 76 × 2 = 1 + 0.499 999 999 999 999 988 306 026 628 833 104 363 52;
  • 62) 0.499 999 999 999 999 988 306 026 628 833 104 363 52 × 2 = 0 + 0.999 999 999 999 999 976 612 053 257 666 208 727 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:


0.000 999 999 999 999 999 803 976 247 214 620 793 26(10) =


0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 10(2)

5. Positive number before normalization:

0.000 999 999 999 999 999 803 976 247 214 620 793 26(10) =


0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 10(2)

6. Normalize the binary representation of the number.

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


0.000 999 999 999 999 999 803 976 247 214 620 793 26(10) =


0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 10(2) =


0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 10(2) × 20 =


1.0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010(2) × 2-10


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): -10


Mantissa (not normalized):
1.0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010


8. 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) =


1 013(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;
  • 1 013 ÷ 2 = 506 + 1;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =


1013(10) =


011 1111 0101(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, only if necessary (not the case here).


Mantissa (normalized) =


1. 0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010 =


0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010


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) =
011 1111 0101


Mantissa (52 bits) =
0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010


Decimal number 0.000 999 999 999 999 999 803 976 247 214 620 793 26 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0101 - 0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010

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