0.000 999 999 999 999 999 803 976 247 214 620 798 73 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 999 999 999 999 999 803 976 247 214 620 798 73(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
0.000 999 999 999 999 999 803 976 247 214 620 798 73(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: 0.000 999 999 999 999 999 803 976 247 214 620 798 73.

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) 0.000 999 999 999 999 999 803 976 247 214 620 798 73 × 2 = 0 + 0.001 999 999 999 999 999 607 952 494 429 241 597 46;
  • 2) 0.001 999 999 999 999 999 607 952 494 429 241 597 46 × 2 = 0 + 0.003 999 999 999 999 999 215 904 988 858 483 194 92;
  • 3) 0.003 999 999 999 999 999 215 904 988 858 483 194 92 × 2 = 0 + 0.007 999 999 999 999 998 431 809 977 716 966 389 84;
  • 4) 0.007 999 999 999 999 998 431 809 977 716 966 389 84 × 2 = 0 + 0.015 999 999 999 999 996 863 619 955 433 932 779 68;
  • 5) 0.015 999 999 999 999 996 863 619 955 433 932 779 68 × 2 = 0 + 0.031 999 999 999 999 993 727 239 910 867 865 559 36;
  • 6) 0.031 999 999 999 999 993 727 239 910 867 865 559 36 × 2 = 0 + 0.063 999 999 999 999 987 454 479 821 735 731 118 72;
  • 7) 0.063 999 999 999 999 987 454 479 821 735 731 118 72 × 2 = 0 + 0.127 999 999 999 999 974 908 959 643 471 462 237 44;
  • 8) 0.127 999 999 999 999 974 908 959 643 471 462 237 44 × 2 = 0 + 0.255 999 999 999 999 949 817 919 286 942 924 474 88;
  • 9) 0.255 999 999 999 999 949 817 919 286 942 924 474 88 × 2 = 0 + 0.511 999 999 999 999 899 635 838 573 885 848 949 76;
  • 10) 0.511 999 999 999 999 899 635 838 573 885 848 949 76 × 2 = 1 + 0.023 999 999 999 999 799 271 677 147 771 697 899 52;
  • 11) 0.023 999 999 999 999 799 271 677 147 771 697 899 52 × 2 = 0 + 0.047 999 999 999 999 598 543 354 295 543 395 799 04;
  • 12) 0.047 999 999 999 999 598 543 354 295 543 395 799 04 × 2 = 0 + 0.095 999 999 999 999 197 086 708 591 086 791 598 08;
  • 13) 0.095 999 999 999 999 197 086 708 591 086 791 598 08 × 2 = 0 + 0.191 999 999 999 998 394 173 417 182 173 583 196 16;
  • 14) 0.191 999 999 999 998 394 173 417 182 173 583 196 16 × 2 = 0 + 0.383 999 999 999 996 788 346 834 364 347 166 392 32;
  • 15) 0.383 999 999 999 996 788 346 834 364 347 166 392 32 × 2 = 0 + 0.767 999 999 999 993 576 693 668 728 694 332 784 64;
  • 16) 0.767 999 999 999 993 576 693 668 728 694 332 784 64 × 2 = 1 + 0.535 999 999 999 987 153 387 337 457 388 665 569 28;
  • 17) 0.535 999 999 999 987 153 387 337 457 388 665 569 28 × 2 = 1 + 0.071 999 999 999 974 306 774 674 914 777 331 138 56;
  • 18) 0.071 999 999 999 974 306 774 674 914 777 331 138 56 × 2 = 0 + 0.143 999 999 999 948 613 549 349 829 554 662 277 12;
  • 19) 0.143 999 999 999 948 613 549 349 829 554 662 277 12 × 2 = 0 + 0.287 999 999 999 897 227 098 699 659 109 324 554 24;
  • 20) 0.287 999 999 999 897 227 098 699 659 109 324 554 24 × 2 = 0 + 0.575 999 999 999 794 454 197 399 318 218 649 108 48;
  • 21) 0.575 999 999 999 794 454 197 399 318 218 649 108 48 × 2 = 1 + 0.151 999 999 999 588 908 394 798 636 437 298 216 96;
  • 22) 0.151 999 999 999 588 908 394 798 636 437 298 216 96 × 2 = 0 + 0.303 999 999 999 177 816 789 597 272 874 596 433 92;
  • 23) 0.303 999 999 999 177 816 789 597 272 874 596 433 92 × 2 = 0 + 0.607 999 999 998 355 633 579 194 545 749 192 867 84;
  • 24) 0.607 999 999 998 355 633 579 194 545 749 192 867 84 × 2 = 1 + 0.215 999 999 996 711 267 158 389 091 498 385 735 68;
  • 25) 0.215 999 999 996 711 267 158 389 091 498 385 735 68 × 2 = 0 + 0.431 999 999 993 422 534 316 778 182 996 771 471 36;
  • 26) 0.431 999 999 993 422 534 316 778 182 996 771 471 36 × 2 = 0 + 0.863 999 999 986 845 068 633 556 365 993 542 942 72;
  • 27) 0.863 999 999 986 845 068 633 556 365 993 542 942 72 × 2 = 1 + 0.727 999 999 973 690 137 267 112 731 987 085 885 44;
  • 28) 0.727 999 999 973 690 137 267 112 731 987 085 885 44 × 2 = 1 + 0.455 999 999 947 380 274 534 225 463 974 171 770 88;
  • 29) 0.455 999 999 947 380 274 534 225 463 974 171 770 88 × 2 = 0 + 0.911 999 999 894 760 549 068 450 927 948 343 541 76;
  • 30) 0.911 999 999 894 760 549 068 450 927 948 343 541 76 × 2 = 1 + 0.823 999 999 789 521 098 136 901 855 896 687 083 52;
  • 31) 0.823 999 999 789 521 098 136 901 855 896 687 083 52 × 2 = 1 + 0.647 999 999 579 042 196 273 803 711 793 374 167 04;
  • 32) 0.647 999 999 579 042 196 273 803 711 793 374 167 04 × 2 = 1 + 0.295 999 999 158 084 392 547 607 423 586 748 334 08;
  • 33) 0.295 999 999 158 084 392 547 607 423 586 748 334 08 × 2 = 0 + 0.591 999 998 316 168 785 095 214 847 173 496 668 16;
  • 34) 0.591 999 998 316 168 785 095 214 847 173 496 668 16 × 2 = 1 + 0.183 999 996 632 337 570 190 429 694 346 993 336 32;
  • 35) 0.183 999 996 632 337 570 190 429 694 346 993 336 32 × 2 = 0 + 0.367 999 993 264 675 140 380 859 388 693 986 672 64;
  • 36) 0.367 999 993 264 675 140 380 859 388 693 986 672 64 × 2 = 0 + 0.735 999 986 529 350 280 761 718 777 387 973 345 28;
  • 37) 0.735 999 986 529 350 280 761 718 777 387 973 345 28 × 2 = 1 + 0.471 999 973 058 700 561 523 437 554 775 946 690 56;
  • 38) 0.471 999 973 058 700 561 523 437 554 775 946 690 56 × 2 = 0 + 0.943 999 946 117 401 123 046 875 109 551 893 381 12;
  • 39) 0.943 999 946 117 401 123 046 875 109 551 893 381 12 × 2 = 1 + 0.887 999 892 234 802 246 093 750 219 103 786 762 24;
  • 40) 0.887 999 892 234 802 246 093 750 219 103 786 762 24 × 2 = 1 + 0.775 999 784 469 604 492 187 500 438 207 573 524 48;
  • 41) 0.775 999 784 469 604 492 187 500 438 207 573 524 48 × 2 = 1 + 0.551 999 568 939 208 984 375 000 876 415 147 048 96;
  • 42) 0.551 999 568 939 208 984 375 000 876 415 147 048 96 × 2 = 1 + 0.103 999 137 878 417 968 750 001 752 830 294 097 92;
  • 43) 0.103 999 137 878 417 968 750 001 752 830 294 097 92 × 2 = 0 + 0.207 998 275 756 835 937 500 003 505 660 588 195 84;
  • 44) 0.207 998 275 756 835 937 500 003 505 660 588 195 84 × 2 = 0 + 0.415 996 551 513 671 875 000 007 011 321 176 391 68;
  • 45) 0.415 996 551 513 671 875 000 007 011 321 176 391 68 × 2 = 0 + 0.831 993 103 027 343 750 000 014 022 642 352 783 36;
  • 46) 0.831 993 103 027 343 750 000 014 022 642 352 783 36 × 2 = 1 + 0.663 986 206 054 687 500 000 028 045 284 705 566 72;
  • 47) 0.663 986 206 054 687 500 000 028 045 284 705 566 72 × 2 = 1 + 0.327 972 412 109 375 000 000 056 090 569 411 133 44;
  • 48) 0.327 972 412 109 375 000 000 056 090 569 411 133 44 × 2 = 0 + 0.655 944 824 218 750 000 000 112 181 138 822 266 88;
  • 49) 0.655 944 824 218 750 000 000 112 181 138 822 266 88 × 2 = 1 + 0.311 889 648 437 500 000 000 224 362 277 644 533 76;
  • 50) 0.311 889 648 437 500 000 000 224 362 277 644 533 76 × 2 = 0 + 0.623 779 296 875 000 000 000 448 724 555 289 067 52;
  • 51) 0.623 779 296 875 000 000 000 448 724 555 289 067 52 × 2 = 1 + 0.247 558 593 750 000 000 000 897 449 110 578 135 04;
  • 52) 0.247 558 593 750 000 000 000 897 449 110 578 135 04 × 2 = 0 + 0.495 117 187 500 000 000 001 794 898 221 156 270 08;
  • 53) 0.495 117 187 500 000 000 001 794 898 221 156 270 08 × 2 = 0 + 0.990 234 375 000 000 000 003 589 796 442 312 540 16;
  • 54) 0.990 234 375 000 000 000 003 589 796 442 312 540 16 × 2 = 1 + 0.980 468 750 000 000 000 007 179 592 884 625 080 32;
  • 55) 0.980 468 750 000 000 000 007 179 592 884 625 080 32 × 2 = 1 + 0.960 937 500 000 000 000 014 359 185 769 250 160 64;
  • 56) 0.960 937 500 000 000 000 014 359 185 769 250 160 64 × 2 = 1 + 0.921 875 000 000 000 000 028 718 371 538 500 321 28;
  • 57) 0.921 875 000 000 000 000 028 718 371 538 500 321 28 × 2 = 1 + 0.843 750 000 000 000 000 057 436 743 077 000 642 56;
  • 58) 0.843 750 000 000 000 000 057 436 743 077 000 642 56 × 2 = 1 + 0.687 500 000 000 000 000 114 873 486 154 001 285 12;
  • 59) 0.687 500 000 000 000 000 114 873 486 154 001 285 12 × 2 = 1 + 0.375 000 000 000 000 000 229 746 972 308 002 570 24;
  • 60) 0.375 000 000 000 000 000 229 746 972 308 002 570 24 × 2 = 0 + 0.750 000 000 000 000 000 459 493 944 616 005 140 48;
  • 61) 0.750 000 000 000 000 000 459 493 944 616 005 140 48 × 2 = 1 + 0.500 000 000 000 000 000 918 987 889 232 010 280 96;
  • 62) 0.500 000 000 000 000 000 918 987 889 232 010 280 96 × 2 = 1 + 0.000 000 000 000 000 001 837 975 778 464 020 561 92;

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:


0.000 999 999 999 999 999 803 976 247 214 620 798 73(10) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 11(2)

5. Positive number before normalization:

0.000 999 999 999 999 999 803 976 247 214 620 798 73(10) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 11(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:


0.000 999 999 999 999 999 803 976 247 214 620 798 73(10) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 11(2) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 11(2) × 20 =

1.0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011(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 1011


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 1011 =

0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011


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 1011


Decimal number 0.000 999 999 999 999 999 803 976 247 214 620 798 73 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 1011

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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