0.000 999 999 999 999 999 803 976 247 214 620 802 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 999 999 999 999 999 803 976 247 214 620 802 2(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
0.000 999 999 999 999 999 803 976 247 214 620 802 2(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: 0.000 999 999 999 999 999 803 976 247 214 620 802 2.

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) 0.000 999 999 999 999 999 803 976 247 214 620 802 2 × 2 = 0 + 0.001 999 999 999 999 999 607 952 494 429 241 604 4;
  • 2) 0.001 999 999 999 999 999 607 952 494 429 241 604 4 × 2 = 0 + 0.003 999 999 999 999 999 215 904 988 858 483 208 8;
  • 3) 0.003 999 999 999 999 999 215 904 988 858 483 208 8 × 2 = 0 + 0.007 999 999 999 999 998 431 809 977 716 966 417 6;
  • 4) 0.007 999 999 999 999 998 431 809 977 716 966 417 6 × 2 = 0 + 0.015 999 999 999 999 996 863 619 955 433 932 835 2;
  • 5) 0.015 999 999 999 999 996 863 619 955 433 932 835 2 × 2 = 0 + 0.031 999 999 999 999 993 727 239 910 867 865 670 4;
  • 6) 0.031 999 999 999 999 993 727 239 910 867 865 670 4 × 2 = 0 + 0.063 999 999 999 999 987 454 479 821 735 731 340 8;
  • 7) 0.063 999 999 999 999 987 454 479 821 735 731 340 8 × 2 = 0 + 0.127 999 999 999 999 974 908 959 643 471 462 681 6;
  • 8) 0.127 999 999 999 999 974 908 959 643 471 462 681 6 × 2 = 0 + 0.255 999 999 999 999 949 817 919 286 942 925 363 2;
  • 9) 0.255 999 999 999 999 949 817 919 286 942 925 363 2 × 2 = 0 + 0.511 999 999 999 999 899 635 838 573 885 850 726 4;
  • 10) 0.511 999 999 999 999 899 635 838 573 885 850 726 4 × 2 = 1 + 0.023 999 999 999 999 799 271 677 147 771 701 452 8;
  • 11) 0.023 999 999 999 999 799 271 677 147 771 701 452 8 × 2 = 0 + 0.047 999 999 999 999 598 543 354 295 543 402 905 6;
  • 12) 0.047 999 999 999 999 598 543 354 295 543 402 905 6 × 2 = 0 + 0.095 999 999 999 999 197 086 708 591 086 805 811 2;
  • 13) 0.095 999 999 999 999 197 086 708 591 086 805 811 2 × 2 = 0 + 0.191 999 999 999 998 394 173 417 182 173 611 622 4;
  • 14) 0.191 999 999 999 998 394 173 417 182 173 611 622 4 × 2 = 0 + 0.383 999 999 999 996 788 346 834 364 347 223 244 8;
  • 15) 0.383 999 999 999 996 788 346 834 364 347 223 244 8 × 2 = 0 + 0.767 999 999 999 993 576 693 668 728 694 446 489 6;
  • 16) 0.767 999 999 999 993 576 693 668 728 694 446 489 6 × 2 = 1 + 0.535 999 999 999 987 153 387 337 457 388 892 979 2;
  • 17) 0.535 999 999 999 987 153 387 337 457 388 892 979 2 × 2 = 1 + 0.071 999 999 999 974 306 774 674 914 777 785 958 4;
  • 18) 0.071 999 999 999 974 306 774 674 914 777 785 958 4 × 2 = 0 + 0.143 999 999 999 948 613 549 349 829 555 571 916 8;
  • 19) 0.143 999 999 999 948 613 549 349 829 555 571 916 8 × 2 = 0 + 0.287 999 999 999 897 227 098 699 659 111 143 833 6;
  • 20) 0.287 999 999 999 897 227 098 699 659 111 143 833 6 × 2 = 0 + 0.575 999 999 999 794 454 197 399 318 222 287 667 2;
  • 21) 0.575 999 999 999 794 454 197 399 318 222 287 667 2 × 2 = 1 + 0.151 999 999 999 588 908 394 798 636 444 575 334 4;
  • 22) 0.151 999 999 999 588 908 394 798 636 444 575 334 4 × 2 = 0 + 0.303 999 999 999 177 816 789 597 272 889 150 668 8;
  • 23) 0.303 999 999 999 177 816 789 597 272 889 150 668 8 × 2 = 0 + 0.607 999 999 998 355 633 579 194 545 778 301 337 6;
  • 24) 0.607 999 999 998 355 633 579 194 545 778 301 337 6 × 2 = 1 + 0.215 999 999 996 711 267 158 389 091 556 602 675 2;
  • 25) 0.215 999 999 996 711 267 158 389 091 556 602 675 2 × 2 = 0 + 0.431 999 999 993 422 534 316 778 183 113 205 350 4;
  • 26) 0.431 999 999 993 422 534 316 778 183 113 205 350 4 × 2 = 0 + 0.863 999 999 986 845 068 633 556 366 226 410 700 8;
  • 27) 0.863 999 999 986 845 068 633 556 366 226 410 700 8 × 2 = 1 + 0.727 999 999 973 690 137 267 112 732 452 821 401 6;
  • 28) 0.727 999 999 973 690 137 267 112 732 452 821 401 6 × 2 = 1 + 0.455 999 999 947 380 274 534 225 464 905 642 803 2;
  • 29) 0.455 999 999 947 380 274 534 225 464 905 642 803 2 × 2 = 0 + 0.911 999 999 894 760 549 068 450 929 811 285 606 4;
  • 30) 0.911 999 999 894 760 549 068 450 929 811 285 606 4 × 2 = 1 + 0.823 999 999 789 521 098 136 901 859 622 571 212 8;
  • 31) 0.823 999 999 789 521 098 136 901 859 622 571 212 8 × 2 = 1 + 0.647 999 999 579 042 196 273 803 719 245 142 425 6;
  • 32) 0.647 999 999 579 042 196 273 803 719 245 142 425 6 × 2 = 1 + 0.295 999 999 158 084 392 547 607 438 490 284 851 2;
  • 33) 0.295 999 999 158 084 392 547 607 438 490 284 851 2 × 2 = 0 + 0.591 999 998 316 168 785 095 214 876 980 569 702 4;
  • 34) 0.591 999 998 316 168 785 095 214 876 980 569 702 4 × 2 = 1 + 0.183 999 996 632 337 570 190 429 753 961 139 404 8;
  • 35) 0.183 999 996 632 337 570 190 429 753 961 139 404 8 × 2 = 0 + 0.367 999 993 264 675 140 380 859 507 922 278 809 6;
  • 36) 0.367 999 993 264 675 140 380 859 507 922 278 809 6 × 2 = 0 + 0.735 999 986 529 350 280 761 719 015 844 557 619 2;
  • 37) 0.735 999 986 529 350 280 761 719 015 844 557 619 2 × 2 = 1 + 0.471 999 973 058 700 561 523 438 031 689 115 238 4;
  • 38) 0.471 999 973 058 700 561 523 438 031 689 115 238 4 × 2 = 0 + 0.943 999 946 117 401 123 046 876 063 378 230 476 8;
  • 39) 0.943 999 946 117 401 123 046 876 063 378 230 476 8 × 2 = 1 + 0.887 999 892 234 802 246 093 752 126 756 460 953 6;
  • 40) 0.887 999 892 234 802 246 093 752 126 756 460 953 6 × 2 = 1 + 0.775 999 784 469 604 492 187 504 253 512 921 907 2;
  • 41) 0.775 999 784 469 604 492 187 504 253 512 921 907 2 × 2 = 1 + 0.551 999 568 939 208 984 375 008 507 025 843 814 4;
  • 42) 0.551 999 568 939 208 984 375 008 507 025 843 814 4 × 2 = 1 + 0.103 999 137 878 417 968 750 017 014 051 687 628 8;
  • 43) 0.103 999 137 878 417 968 750 017 014 051 687 628 8 × 2 = 0 + 0.207 998 275 756 835 937 500 034 028 103 375 257 6;
  • 44) 0.207 998 275 756 835 937 500 034 028 103 375 257 6 × 2 = 0 + 0.415 996 551 513 671 875 000 068 056 206 750 515 2;
  • 45) 0.415 996 551 513 671 875 000 068 056 206 750 515 2 × 2 = 0 + 0.831 993 103 027 343 750 000 136 112 413 501 030 4;
  • 46) 0.831 993 103 027 343 750 000 136 112 413 501 030 4 × 2 = 1 + 0.663 986 206 054 687 500 000 272 224 827 002 060 8;
  • 47) 0.663 986 206 054 687 500 000 272 224 827 002 060 8 × 2 = 1 + 0.327 972 412 109 375 000 000 544 449 654 004 121 6;
  • 48) 0.327 972 412 109 375 000 000 544 449 654 004 121 6 × 2 = 0 + 0.655 944 824 218 750 000 001 088 899 308 008 243 2;
  • 49) 0.655 944 824 218 750 000 001 088 899 308 008 243 2 × 2 = 1 + 0.311 889 648 437 500 000 002 177 798 616 016 486 4;
  • 50) 0.311 889 648 437 500 000 002 177 798 616 016 486 4 × 2 = 0 + 0.623 779 296 875 000 000 004 355 597 232 032 972 8;
  • 51) 0.623 779 296 875 000 000 004 355 597 232 032 972 8 × 2 = 1 + 0.247 558 593 750 000 000 008 711 194 464 065 945 6;
  • 52) 0.247 558 593 750 000 000 008 711 194 464 065 945 6 × 2 = 0 + 0.495 117 187 500 000 000 017 422 388 928 131 891 2;
  • 53) 0.495 117 187 500 000 000 017 422 388 928 131 891 2 × 2 = 0 + 0.990 234 375 000 000 000 034 844 777 856 263 782 4;
  • 54) 0.990 234 375 000 000 000 034 844 777 856 263 782 4 × 2 = 1 + 0.980 468 750 000 000 000 069 689 555 712 527 564 8;
  • 55) 0.980 468 750 000 000 000 069 689 555 712 527 564 8 × 2 = 1 + 0.960 937 500 000 000 000 139 379 111 425 055 129 6;
  • 56) 0.960 937 500 000 000 000 139 379 111 425 055 129 6 × 2 = 1 + 0.921 875 000 000 000 000 278 758 222 850 110 259 2;
  • 57) 0.921 875 000 000 000 000 278 758 222 850 110 259 2 × 2 = 1 + 0.843 750 000 000 000 000 557 516 445 700 220 518 4;
  • 58) 0.843 750 000 000 000 000 557 516 445 700 220 518 4 × 2 = 1 + 0.687 500 000 000 000 001 115 032 891 400 441 036 8;
  • 59) 0.687 500 000 000 000 001 115 032 891 400 441 036 8 × 2 = 1 + 0.375 000 000 000 000 002 230 065 782 800 882 073 6;
  • 60) 0.375 000 000 000 000 002 230 065 782 800 882 073 6 × 2 = 0 + 0.750 000 000 000 000 004 460 131 565 601 764 147 2;
  • 61) 0.750 000 000 000 000 004 460 131 565 601 764 147 2 × 2 = 1 + 0.500 000 000 000 000 008 920 263 131 203 528 294 4;
  • 62) 0.500 000 000 000 000 008 920 263 131 203 528 294 4 × 2 = 1 + 0.000 000 000 000 000 017 840 526 262 407 056 588 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).


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:


0.000 999 999 999 999 999 803 976 247 214 620 802 2(10) =

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

5. Positive number before normalization:

0.000 999 999 999 999 999 803 976 247 214 620 802 2(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:


0.000 999 999 999 999 999 803 976 247 214 620 802 2(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 0.000 999 999 999 999 999 803 976 247 214 620 802 2 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