0.000 999 999 999 999 999 803 976 247 214 620 801 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 999 999 999 999 999 803 976 247 214 620 801 7(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
0.000 999 999 999 999 999 803 976 247 214 620 801 7(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: 0.000 999 999 999 999 999 803 976 247 214 620 801 7.

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) 0.000 999 999 999 999 999 803 976 247 214 620 801 7 × 2 = 0 + 0.001 999 999 999 999 999 607 952 494 429 241 603 4;
  • 2) 0.001 999 999 999 999 999 607 952 494 429 241 603 4 × 2 = 0 + 0.003 999 999 999 999 999 215 904 988 858 483 206 8;
  • 3) 0.003 999 999 999 999 999 215 904 988 858 483 206 8 × 2 = 0 + 0.007 999 999 999 999 998 431 809 977 716 966 413 6;
  • 4) 0.007 999 999 999 999 998 431 809 977 716 966 413 6 × 2 = 0 + 0.015 999 999 999 999 996 863 619 955 433 932 827 2;
  • 5) 0.015 999 999 999 999 996 863 619 955 433 932 827 2 × 2 = 0 + 0.031 999 999 999 999 993 727 239 910 867 865 654 4;
  • 6) 0.031 999 999 999 999 993 727 239 910 867 865 654 4 × 2 = 0 + 0.063 999 999 999 999 987 454 479 821 735 731 308 8;
  • 7) 0.063 999 999 999 999 987 454 479 821 735 731 308 8 × 2 = 0 + 0.127 999 999 999 999 974 908 959 643 471 462 617 6;
  • 8) 0.127 999 999 999 999 974 908 959 643 471 462 617 6 × 2 = 0 + 0.255 999 999 999 999 949 817 919 286 942 925 235 2;
  • 9) 0.255 999 999 999 999 949 817 919 286 942 925 235 2 × 2 = 0 + 0.511 999 999 999 999 899 635 838 573 885 850 470 4;
  • 10) 0.511 999 999 999 999 899 635 838 573 885 850 470 4 × 2 = 1 + 0.023 999 999 999 999 799 271 677 147 771 700 940 8;
  • 11) 0.023 999 999 999 999 799 271 677 147 771 700 940 8 × 2 = 0 + 0.047 999 999 999 999 598 543 354 295 543 401 881 6;
  • 12) 0.047 999 999 999 999 598 543 354 295 543 401 881 6 × 2 = 0 + 0.095 999 999 999 999 197 086 708 591 086 803 763 2;
  • 13) 0.095 999 999 999 999 197 086 708 591 086 803 763 2 × 2 = 0 + 0.191 999 999 999 998 394 173 417 182 173 607 526 4;
  • 14) 0.191 999 999 999 998 394 173 417 182 173 607 526 4 × 2 = 0 + 0.383 999 999 999 996 788 346 834 364 347 215 052 8;
  • 15) 0.383 999 999 999 996 788 346 834 364 347 215 052 8 × 2 = 0 + 0.767 999 999 999 993 576 693 668 728 694 430 105 6;
  • 16) 0.767 999 999 999 993 576 693 668 728 694 430 105 6 × 2 = 1 + 0.535 999 999 999 987 153 387 337 457 388 860 211 2;
  • 17) 0.535 999 999 999 987 153 387 337 457 388 860 211 2 × 2 = 1 + 0.071 999 999 999 974 306 774 674 914 777 720 422 4;
  • 18) 0.071 999 999 999 974 306 774 674 914 777 720 422 4 × 2 = 0 + 0.143 999 999 999 948 613 549 349 829 555 440 844 8;
  • 19) 0.143 999 999 999 948 613 549 349 829 555 440 844 8 × 2 = 0 + 0.287 999 999 999 897 227 098 699 659 110 881 689 6;
  • 20) 0.287 999 999 999 897 227 098 699 659 110 881 689 6 × 2 = 0 + 0.575 999 999 999 794 454 197 399 318 221 763 379 2;
  • 21) 0.575 999 999 999 794 454 197 399 318 221 763 379 2 × 2 = 1 + 0.151 999 999 999 588 908 394 798 636 443 526 758 4;
  • 22) 0.151 999 999 999 588 908 394 798 636 443 526 758 4 × 2 = 0 + 0.303 999 999 999 177 816 789 597 272 887 053 516 8;
  • 23) 0.303 999 999 999 177 816 789 597 272 887 053 516 8 × 2 = 0 + 0.607 999 999 998 355 633 579 194 545 774 107 033 6;
  • 24) 0.607 999 999 998 355 633 579 194 545 774 107 033 6 × 2 = 1 + 0.215 999 999 996 711 267 158 389 091 548 214 067 2;
  • 25) 0.215 999 999 996 711 267 158 389 091 548 214 067 2 × 2 = 0 + 0.431 999 999 993 422 534 316 778 183 096 428 134 4;
  • 26) 0.431 999 999 993 422 534 316 778 183 096 428 134 4 × 2 = 0 + 0.863 999 999 986 845 068 633 556 366 192 856 268 8;
  • 27) 0.863 999 999 986 845 068 633 556 366 192 856 268 8 × 2 = 1 + 0.727 999 999 973 690 137 267 112 732 385 712 537 6;
  • 28) 0.727 999 999 973 690 137 267 112 732 385 712 537 6 × 2 = 1 + 0.455 999 999 947 380 274 534 225 464 771 425 075 2;
  • 29) 0.455 999 999 947 380 274 534 225 464 771 425 075 2 × 2 = 0 + 0.911 999 999 894 760 549 068 450 929 542 850 150 4;
  • 30) 0.911 999 999 894 760 549 068 450 929 542 850 150 4 × 2 = 1 + 0.823 999 999 789 521 098 136 901 859 085 700 300 8;
  • 31) 0.823 999 999 789 521 098 136 901 859 085 700 300 8 × 2 = 1 + 0.647 999 999 579 042 196 273 803 718 171 400 601 6;
  • 32) 0.647 999 999 579 042 196 273 803 718 171 400 601 6 × 2 = 1 + 0.295 999 999 158 084 392 547 607 436 342 801 203 2;
  • 33) 0.295 999 999 158 084 392 547 607 436 342 801 203 2 × 2 = 0 + 0.591 999 998 316 168 785 095 214 872 685 602 406 4;
  • 34) 0.591 999 998 316 168 785 095 214 872 685 602 406 4 × 2 = 1 + 0.183 999 996 632 337 570 190 429 745 371 204 812 8;
  • 35) 0.183 999 996 632 337 570 190 429 745 371 204 812 8 × 2 = 0 + 0.367 999 993 264 675 140 380 859 490 742 409 625 6;
  • 36) 0.367 999 993 264 675 140 380 859 490 742 409 625 6 × 2 = 0 + 0.735 999 986 529 350 280 761 718 981 484 819 251 2;
  • 37) 0.735 999 986 529 350 280 761 718 981 484 819 251 2 × 2 = 1 + 0.471 999 973 058 700 561 523 437 962 969 638 502 4;
  • 38) 0.471 999 973 058 700 561 523 437 962 969 638 502 4 × 2 = 0 + 0.943 999 946 117 401 123 046 875 925 939 277 004 8;
  • 39) 0.943 999 946 117 401 123 046 875 925 939 277 004 8 × 2 = 1 + 0.887 999 892 234 802 246 093 751 851 878 554 009 6;
  • 40) 0.887 999 892 234 802 246 093 751 851 878 554 009 6 × 2 = 1 + 0.775 999 784 469 604 492 187 503 703 757 108 019 2;
  • 41) 0.775 999 784 469 604 492 187 503 703 757 108 019 2 × 2 = 1 + 0.551 999 568 939 208 984 375 007 407 514 216 038 4;
  • 42) 0.551 999 568 939 208 984 375 007 407 514 216 038 4 × 2 = 1 + 0.103 999 137 878 417 968 750 014 815 028 432 076 8;
  • 43) 0.103 999 137 878 417 968 750 014 815 028 432 076 8 × 2 = 0 + 0.207 998 275 756 835 937 500 029 630 056 864 153 6;
  • 44) 0.207 998 275 756 835 937 500 029 630 056 864 153 6 × 2 = 0 + 0.415 996 551 513 671 875 000 059 260 113 728 307 2;
  • 45) 0.415 996 551 513 671 875 000 059 260 113 728 307 2 × 2 = 0 + 0.831 993 103 027 343 750 000 118 520 227 456 614 4;
  • 46) 0.831 993 103 027 343 750 000 118 520 227 456 614 4 × 2 = 1 + 0.663 986 206 054 687 500 000 237 040 454 913 228 8;
  • 47) 0.663 986 206 054 687 500 000 237 040 454 913 228 8 × 2 = 1 + 0.327 972 412 109 375 000 000 474 080 909 826 457 6;
  • 48) 0.327 972 412 109 375 000 000 474 080 909 826 457 6 × 2 = 0 + 0.655 944 824 218 750 000 000 948 161 819 652 915 2;
  • 49) 0.655 944 824 218 750 000 000 948 161 819 652 915 2 × 2 = 1 + 0.311 889 648 437 500 000 001 896 323 639 305 830 4;
  • 50) 0.311 889 648 437 500 000 001 896 323 639 305 830 4 × 2 = 0 + 0.623 779 296 875 000 000 003 792 647 278 611 660 8;
  • 51) 0.623 779 296 875 000 000 003 792 647 278 611 660 8 × 2 = 1 + 0.247 558 593 750 000 000 007 585 294 557 223 321 6;
  • 52) 0.247 558 593 750 000 000 007 585 294 557 223 321 6 × 2 = 0 + 0.495 117 187 500 000 000 015 170 589 114 446 643 2;
  • 53) 0.495 117 187 500 000 000 015 170 589 114 446 643 2 × 2 = 0 + 0.990 234 375 000 000 000 030 341 178 228 893 286 4;
  • 54) 0.990 234 375 000 000 000 030 341 178 228 893 286 4 × 2 = 1 + 0.980 468 750 000 000 000 060 682 356 457 786 572 8;
  • 55) 0.980 468 750 000 000 000 060 682 356 457 786 572 8 × 2 = 1 + 0.960 937 500 000 000 000 121 364 712 915 573 145 6;
  • 56) 0.960 937 500 000 000 000 121 364 712 915 573 145 6 × 2 = 1 + 0.921 875 000 000 000 000 242 729 425 831 146 291 2;
  • 57) 0.921 875 000 000 000 000 242 729 425 831 146 291 2 × 2 = 1 + 0.843 750 000 000 000 000 485 458 851 662 292 582 4;
  • 58) 0.843 750 000 000 000 000 485 458 851 662 292 582 4 × 2 = 1 + 0.687 500 000 000 000 000 970 917 703 324 585 164 8;
  • 59) 0.687 500 000 000 000 000 970 917 703 324 585 164 8 × 2 = 1 + 0.375 000 000 000 000 001 941 835 406 649 170 329 6;
  • 60) 0.375 000 000 000 000 001 941 835 406 649 170 329 6 × 2 = 0 + 0.750 000 000 000 000 003 883 670 813 298 340 659 2;
  • 61) 0.750 000 000 000 000 003 883 670 813 298 340 659 2 × 2 = 1 + 0.500 000 000 000 000 007 767 341 626 596 681 318 4;
  • 62) 0.500 000 000 000 000 007 767 341 626 596 681 318 4 × 2 = 1 + 0.000 000 000 000 000 015 534 683 253 193 362 636 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:


0.000 999 999 999 999 999 803 976 247 214 620 801 7(10) =

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

5. Positive number before normalization:

0.000 999 999 999 999 999 803 976 247 214 620 801 7(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:


0.000 999 999 999 999 999 803 976 247 214 620 801 7(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 0.000 999 999 999 999 999 803 976 247 214 620 801 7 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