0.000 000 021 979 552 668 138 406 869 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 552 668 138 406 869(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
0.000 000 021 979 552 668 138 406 869(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: 0.000 000 021 979 552 668 138 406 869.

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) 0.000 000 021 979 552 668 138 406 869 × 2 = 0 + 0.000 000 043 959 105 336 276 813 738;
  • 2) 0.000 000 043 959 105 336 276 813 738 × 2 = 0 + 0.000 000 087 918 210 672 553 627 476;
  • 3) 0.000 000 087 918 210 672 553 627 476 × 2 = 0 + 0.000 000 175 836 421 345 107 254 952;
  • 4) 0.000 000 175 836 421 345 107 254 952 × 2 = 0 + 0.000 000 351 672 842 690 214 509 904;
  • 5) 0.000 000 351 672 842 690 214 509 904 × 2 = 0 + 0.000 000 703 345 685 380 429 019 808;
  • 6) 0.000 000 703 345 685 380 429 019 808 × 2 = 0 + 0.000 001 406 691 370 760 858 039 616;
  • 7) 0.000 001 406 691 370 760 858 039 616 × 2 = 0 + 0.000 002 813 382 741 521 716 079 232;
  • 8) 0.000 002 813 382 741 521 716 079 232 × 2 = 0 + 0.000 005 626 765 483 043 432 158 464;
  • 9) 0.000 005 626 765 483 043 432 158 464 × 2 = 0 + 0.000 011 253 530 966 086 864 316 928;
  • 10) 0.000 011 253 530 966 086 864 316 928 × 2 = 0 + 0.000 022 507 061 932 173 728 633 856;
  • 11) 0.000 022 507 061 932 173 728 633 856 × 2 = 0 + 0.000 045 014 123 864 347 457 267 712;
  • 12) 0.000 045 014 123 864 347 457 267 712 × 2 = 0 + 0.000 090 028 247 728 694 914 535 424;
  • 13) 0.000 090 028 247 728 694 914 535 424 × 2 = 0 + 0.000 180 056 495 457 389 829 070 848;
  • 14) 0.000 180 056 495 457 389 829 070 848 × 2 = 0 + 0.000 360 112 990 914 779 658 141 696;
  • 15) 0.000 360 112 990 914 779 658 141 696 × 2 = 0 + 0.000 720 225 981 829 559 316 283 392;
  • 16) 0.000 720 225 981 829 559 316 283 392 × 2 = 0 + 0.001 440 451 963 659 118 632 566 784;
  • 17) 0.001 440 451 963 659 118 632 566 784 × 2 = 0 + 0.002 880 903 927 318 237 265 133 568;
  • 18) 0.002 880 903 927 318 237 265 133 568 × 2 = 0 + 0.005 761 807 854 636 474 530 267 136;
  • 19) 0.005 761 807 854 636 474 530 267 136 × 2 = 0 + 0.011 523 615 709 272 949 060 534 272;
  • 20) 0.011 523 615 709 272 949 060 534 272 × 2 = 0 + 0.023 047 231 418 545 898 121 068 544;
  • 21) 0.023 047 231 418 545 898 121 068 544 × 2 = 0 + 0.046 094 462 837 091 796 242 137 088;
  • 22) 0.046 094 462 837 091 796 242 137 088 × 2 = 0 + 0.092 188 925 674 183 592 484 274 176;
  • 23) 0.092 188 925 674 183 592 484 274 176 × 2 = 0 + 0.184 377 851 348 367 184 968 548 352;
  • 24) 0.184 377 851 348 367 184 968 548 352 × 2 = 0 + 0.368 755 702 696 734 369 937 096 704;
  • 25) 0.368 755 702 696 734 369 937 096 704 × 2 = 0 + 0.737 511 405 393 468 739 874 193 408;
  • 26) 0.737 511 405 393 468 739 874 193 408 × 2 = 1 + 0.475 022 810 786 937 479 748 386 816;
  • 27) 0.475 022 810 786 937 479 748 386 816 × 2 = 0 + 0.950 045 621 573 874 959 496 773 632;
  • 28) 0.950 045 621 573 874 959 496 773 632 × 2 = 1 + 0.900 091 243 147 749 918 993 547 264;
  • 29) 0.900 091 243 147 749 918 993 547 264 × 2 = 1 + 0.800 182 486 295 499 837 987 094 528;
  • 30) 0.800 182 486 295 499 837 987 094 528 × 2 = 1 + 0.600 364 972 590 999 675 974 189 056;
  • 31) 0.600 364 972 590 999 675 974 189 056 × 2 = 1 + 0.200 729 945 181 999 351 948 378 112;
  • 32) 0.200 729 945 181 999 351 948 378 112 × 2 = 0 + 0.401 459 890 363 998 703 896 756 224;
  • 33) 0.401 459 890 363 998 703 896 756 224 × 2 = 0 + 0.802 919 780 727 997 407 793 512 448;
  • 34) 0.802 919 780 727 997 407 793 512 448 × 2 = 1 + 0.605 839 561 455 994 815 587 024 896;
  • 35) 0.605 839 561 455 994 815 587 024 896 × 2 = 1 + 0.211 679 122 911 989 631 174 049 792;
  • 36) 0.211 679 122 911 989 631 174 049 792 × 2 = 0 + 0.423 358 245 823 979 262 348 099 584;
  • 37) 0.423 358 245 823 979 262 348 099 584 × 2 = 0 + 0.846 716 491 647 958 524 696 199 168;
  • 38) 0.846 716 491 647 958 524 696 199 168 × 2 = 1 + 0.693 432 983 295 917 049 392 398 336;
  • 39) 0.693 432 983 295 917 049 392 398 336 × 2 = 1 + 0.386 865 966 591 834 098 784 796 672;
  • 40) 0.386 865 966 591 834 098 784 796 672 × 2 = 0 + 0.773 731 933 183 668 197 569 593 344;
  • 41) 0.773 731 933 183 668 197 569 593 344 × 2 = 1 + 0.547 463 866 367 336 395 139 186 688;
  • 42) 0.547 463 866 367 336 395 139 186 688 × 2 = 1 + 0.094 927 732 734 672 790 278 373 376;
  • 43) 0.094 927 732 734 672 790 278 373 376 × 2 = 0 + 0.189 855 465 469 345 580 556 746 752;
  • 44) 0.189 855 465 469 345 580 556 746 752 × 2 = 0 + 0.379 710 930 938 691 161 113 493 504;
  • 45) 0.379 710 930 938 691 161 113 493 504 × 2 = 0 + 0.759 421 861 877 382 322 226 987 008;
  • 46) 0.759 421 861 877 382 322 226 987 008 × 2 = 1 + 0.518 843 723 754 764 644 453 974 016;
  • 47) 0.518 843 723 754 764 644 453 974 016 × 2 = 1 + 0.037 687 447 509 529 288 907 948 032;
  • 48) 0.037 687 447 509 529 288 907 948 032 × 2 = 0 + 0.075 374 895 019 058 577 815 896 064;
  • 49) 0.075 374 895 019 058 577 815 896 064 × 2 = 0 + 0.150 749 790 038 117 155 631 792 128;
  • 50) 0.150 749 790 038 117 155 631 792 128 × 2 = 0 + 0.301 499 580 076 234 311 263 584 256;
  • 51) 0.301 499 580 076 234 311 263 584 256 × 2 = 0 + 0.602 999 160 152 468 622 527 168 512;
  • 52) 0.602 999 160 152 468 622 527 168 512 × 2 = 1 + 0.205 998 320 304 937 245 054 337 024;
  • 53) 0.205 998 320 304 937 245 054 337 024 × 2 = 0 + 0.411 996 640 609 874 490 108 674 048;
  • 54) 0.411 996 640 609 874 490 108 674 048 × 2 = 0 + 0.823 993 281 219 748 980 217 348 096;
  • 55) 0.823 993 281 219 748 980 217 348 096 × 2 = 1 + 0.647 986 562 439 497 960 434 696 192;
  • 56) 0.647 986 562 439 497 960 434 696 192 × 2 = 1 + 0.295 973 124 878 995 920 869 392 384;
  • 57) 0.295 973 124 878 995 920 869 392 384 × 2 = 0 + 0.591 946 249 757 991 841 738 784 768;
  • 58) 0.591 946 249 757 991 841 738 784 768 × 2 = 1 + 0.183 892 499 515 983 683 477 569 536;
  • 59) 0.183 892 499 515 983 683 477 569 536 × 2 = 0 + 0.367 784 999 031 967 366 955 139 072;
  • 60) 0.367 784 999 031 967 366 955 139 072 × 2 = 0 + 0.735 569 998 063 934 733 910 278 144;
  • 61) 0.735 569 998 063 934 733 910 278 144 × 2 = 1 + 0.471 139 996 127 869 467 820 556 288;
  • 62) 0.471 139 996 127 869 467 820 556 288 × 2 = 0 + 0.942 279 992 255 738 935 641 112 576;
  • 63) 0.942 279 992 255 738 935 641 112 576 × 2 = 1 + 0.884 559 984 511 477 871 282 225 152;
  • 64) 0.884 559 984 511 477 871 282 225 152 × 2 = 1 + 0.769 119 969 022 955 742 564 450 304;
  • 65) 0.769 119 969 022 955 742 564 450 304 × 2 = 1 + 0.538 239 938 045 911 485 128 900 608;
  • 66) 0.538 239 938 045 911 485 128 900 608 × 2 = 1 + 0.076 479 876 091 822 970 257 801 216;
  • 67) 0.076 479 876 091 822 970 257 801 216 × 2 = 0 + 0.152 959 752 183 645 940 515 602 432;
  • 68) 0.152 959 752 183 645 940 515 602 432 × 2 = 0 + 0.305 919 504 367 291 881 031 204 864;
  • 69) 0.305 919 504 367 291 881 031 204 864 × 2 = 0 + 0.611 839 008 734 583 762 062 409 728;
  • 70) 0.611 839 008 734 583 762 062 409 728 × 2 = 1 + 0.223 678 017 469 167 524 124 819 456;
  • 71) 0.223 678 017 469 167 524 124 819 456 × 2 = 0 + 0.447 356 034 938 335 048 249 638 912;
  • 72) 0.447 356 034 938 335 048 249 638 912 × 2 = 0 + 0.894 712 069 876 670 096 499 277 824;
  • 73) 0.894 712 069 876 670 096 499 277 824 × 2 = 1 + 0.789 424 139 753 340 192 998 555 648;
  • 74) 0.789 424 139 753 340 192 998 555 648 × 2 = 1 + 0.578 848 279 506 680 385 997 111 296;
  • 75) 0.578 848 279 506 680 385 997 111 296 × 2 = 1 + 0.157 696 559 013 360 771 994 222 592;
  • 76) 0.157 696 559 013 360 771 994 222 592 × 2 = 0 + 0.315 393 118 026 721 543 988 445 184;
  • 77) 0.315 393 118 026 721 543 988 445 184 × 2 = 0 + 0.630 786 236 053 443 087 976 890 368;
  • 78) 0.630 786 236 053 443 087 976 890 368 × 2 = 1 + 0.261 572 472 106 886 175 953 780 736;

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:


0.000 000 021 979 552 668 138 406 869(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2)

5. Positive number before normalization:

0.000 000 021 979 552 668 138 406 869(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 021 979 552 668 138 406 869(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2) × 20 =

1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001(2) × 2-26


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): -26

Mantissa (not normalized):
1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

997(10) =

011 1110 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. 0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001 =

0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


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 1110 0101

Mantissa (52 bits) =
0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


Decimal number 0.000 000 021 979 552 668 138 406 869 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001

» Convert decimal 0.000 000 021 979 552 668 138 406 871 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