0.000 000 021 979 552 668 138 406 941 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 552 668 138 406 941(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
0.000 000 021 979 552 668 138 406 941(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: 0.000 000 021 979 552 668 138 406 941.

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) 0.000 000 021 979 552 668 138 406 941 × 2 = 0 + 0.000 000 043 959 105 336 276 813 882;
  • 2) 0.000 000 043 959 105 336 276 813 882 × 2 = 0 + 0.000 000 087 918 210 672 553 627 764;
  • 3) 0.000 000 087 918 210 672 553 627 764 × 2 = 0 + 0.000 000 175 836 421 345 107 255 528;
  • 4) 0.000 000 175 836 421 345 107 255 528 × 2 = 0 + 0.000 000 351 672 842 690 214 511 056;
  • 5) 0.000 000 351 672 842 690 214 511 056 × 2 = 0 + 0.000 000 703 345 685 380 429 022 112;
  • 6) 0.000 000 703 345 685 380 429 022 112 × 2 = 0 + 0.000 001 406 691 370 760 858 044 224;
  • 7) 0.000 001 406 691 370 760 858 044 224 × 2 = 0 + 0.000 002 813 382 741 521 716 088 448;
  • 8) 0.000 002 813 382 741 521 716 088 448 × 2 = 0 + 0.000 005 626 765 483 043 432 176 896;
  • 9) 0.000 005 626 765 483 043 432 176 896 × 2 = 0 + 0.000 011 253 530 966 086 864 353 792;
  • 10) 0.000 011 253 530 966 086 864 353 792 × 2 = 0 + 0.000 022 507 061 932 173 728 707 584;
  • 11) 0.000 022 507 061 932 173 728 707 584 × 2 = 0 + 0.000 045 014 123 864 347 457 415 168;
  • 12) 0.000 045 014 123 864 347 457 415 168 × 2 = 0 + 0.000 090 028 247 728 694 914 830 336;
  • 13) 0.000 090 028 247 728 694 914 830 336 × 2 = 0 + 0.000 180 056 495 457 389 829 660 672;
  • 14) 0.000 180 056 495 457 389 829 660 672 × 2 = 0 + 0.000 360 112 990 914 779 659 321 344;
  • 15) 0.000 360 112 990 914 779 659 321 344 × 2 = 0 + 0.000 720 225 981 829 559 318 642 688;
  • 16) 0.000 720 225 981 829 559 318 642 688 × 2 = 0 + 0.001 440 451 963 659 118 637 285 376;
  • 17) 0.001 440 451 963 659 118 637 285 376 × 2 = 0 + 0.002 880 903 927 318 237 274 570 752;
  • 18) 0.002 880 903 927 318 237 274 570 752 × 2 = 0 + 0.005 761 807 854 636 474 549 141 504;
  • 19) 0.005 761 807 854 636 474 549 141 504 × 2 = 0 + 0.011 523 615 709 272 949 098 283 008;
  • 20) 0.011 523 615 709 272 949 098 283 008 × 2 = 0 + 0.023 047 231 418 545 898 196 566 016;
  • 21) 0.023 047 231 418 545 898 196 566 016 × 2 = 0 + 0.046 094 462 837 091 796 393 132 032;
  • 22) 0.046 094 462 837 091 796 393 132 032 × 2 = 0 + 0.092 188 925 674 183 592 786 264 064;
  • 23) 0.092 188 925 674 183 592 786 264 064 × 2 = 0 + 0.184 377 851 348 367 185 572 528 128;
  • 24) 0.184 377 851 348 367 185 572 528 128 × 2 = 0 + 0.368 755 702 696 734 371 145 056 256;
  • 25) 0.368 755 702 696 734 371 145 056 256 × 2 = 0 + 0.737 511 405 393 468 742 290 112 512;
  • 26) 0.737 511 405 393 468 742 290 112 512 × 2 = 1 + 0.475 022 810 786 937 484 580 225 024;
  • 27) 0.475 022 810 786 937 484 580 225 024 × 2 = 0 + 0.950 045 621 573 874 969 160 450 048;
  • 28) 0.950 045 621 573 874 969 160 450 048 × 2 = 1 + 0.900 091 243 147 749 938 320 900 096;
  • 29) 0.900 091 243 147 749 938 320 900 096 × 2 = 1 + 0.800 182 486 295 499 876 641 800 192;
  • 30) 0.800 182 486 295 499 876 641 800 192 × 2 = 1 + 0.600 364 972 590 999 753 283 600 384;
  • 31) 0.600 364 972 590 999 753 283 600 384 × 2 = 1 + 0.200 729 945 181 999 506 567 200 768;
  • 32) 0.200 729 945 181 999 506 567 200 768 × 2 = 0 + 0.401 459 890 363 999 013 134 401 536;
  • 33) 0.401 459 890 363 999 013 134 401 536 × 2 = 0 + 0.802 919 780 727 998 026 268 803 072;
  • 34) 0.802 919 780 727 998 026 268 803 072 × 2 = 1 + 0.605 839 561 455 996 052 537 606 144;
  • 35) 0.605 839 561 455 996 052 537 606 144 × 2 = 1 + 0.211 679 122 911 992 105 075 212 288;
  • 36) 0.211 679 122 911 992 105 075 212 288 × 2 = 0 + 0.423 358 245 823 984 210 150 424 576;
  • 37) 0.423 358 245 823 984 210 150 424 576 × 2 = 0 + 0.846 716 491 647 968 420 300 849 152;
  • 38) 0.846 716 491 647 968 420 300 849 152 × 2 = 1 + 0.693 432 983 295 936 840 601 698 304;
  • 39) 0.693 432 983 295 936 840 601 698 304 × 2 = 1 + 0.386 865 966 591 873 681 203 396 608;
  • 40) 0.386 865 966 591 873 681 203 396 608 × 2 = 0 + 0.773 731 933 183 747 362 406 793 216;
  • 41) 0.773 731 933 183 747 362 406 793 216 × 2 = 1 + 0.547 463 866 367 494 724 813 586 432;
  • 42) 0.547 463 866 367 494 724 813 586 432 × 2 = 1 + 0.094 927 732 734 989 449 627 172 864;
  • 43) 0.094 927 732 734 989 449 627 172 864 × 2 = 0 + 0.189 855 465 469 978 899 254 345 728;
  • 44) 0.189 855 465 469 978 899 254 345 728 × 2 = 0 + 0.379 710 930 939 957 798 508 691 456;
  • 45) 0.379 710 930 939 957 798 508 691 456 × 2 = 0 + 0.759 421 861 879 915 597 017 382 912;
  • 46) 0.759 421 861 879 915 597 017 382 912 × 2 = 1 + 0.518 843 723 759 831 194 034 765 824;
  • 47) 0.518 843 723 759 831 194 034 765 824 × 2 = 1 + 0.037 687 447 519 662 388 069 531 648;
  • 48) 0.037 687 447 519 662 388 069 531 648 × 2 = 0 + 0.075 374 895 039 324 776 139 063 296;
  • 49) 0.075 374 895 039 324 776 139 063 296 × 2 = 0 + 0.150 749 790 078 649 552 278 126 592;
  • 50) 0.150 749 790 078 649 552 278 126 592 × 2 = 0 + 0.301 499 580 157 299 104 556 253 184;
  • 51) 0.301 499 580 157 299 104 556 253 184 × 2 = 0 + 0.602 999 160 314 598 209 112 506 368;
  • 52) 0.602 999 160 314 598 209 112 506 368 × 2 = 1 + 0.205 998 320 629 196 418 225 012 736;
  • 53) 0.205 998 320 629 196 418 225 012 736 × 2 = 0 + 0.411 996 641 258 392 836 450 025 472;
  • 54) 0.411 996 641 258 392 836 450 025 472 × 2 = 0 + 0.823 993 282 516 785 672 900 050 944;
  • 55) 0.823 993 282 516 785 672 900 050 944 × 2 = 1 + 0.647 986 565 033 571 345 800 101 888;
  • 56) 0.647 986 565 033 571 345 800 101 888 × 2 = 1 + 0.295 973 130 067 142 691 600 203 776;
  • 57) 0.295 973 130 067 142 691 600 203 776 × 2 = 0 + 0.591 946 260 134 285 383 200 407 552;
  • 58) 0.591 946 260 134 285 383 200 407 552 × 2 = 1 + 0.183 892 520 268 570 766 400 815 104;
  • 59) 0.183 892 520 268 570 766 400 815 104 × 2 = 0 + 0.367 785 040 537 141 532 801 630 208;
  • 60) 0.367 785 040 537 141 532 801 630 208 × 2 = 0 + 0.735 570 081 074 283 065 603 260 416;
  • 61) 0.735 570 081 074 283 065 603 260 416 × 2 = 1 + 0.471 140 162 148 566 131 206 520 832;
  • 62) 0.471 140 162 148 566 131 206 520 832 × 2 = 0 + 0.942 280 324 297 132 262 413 041 664;
  • 63) 0.942 280 324 297 132 262 413 041 664 × 2 = 1 + 0.884 560 648 594 264 524 826 083 328;
  • 64) 0.884 560 648 594 264 524 826 083 328 × 2 = 1 + 0.769 121 297 188 529 049 652 166 656;
  • 65) 0.769 121 297 188 529 049 652 166 656 × 2 = 1 + 0.538 242 594 377 058 099 304 333 312;
  • 66) 0.538 242 594 377 058 099 304 333 312 × 2 = 1 + 0.076 485 188 754 116 198 608 666 624;
  • 67) 0.076 485 188 754 116 198 608 666 624 × 2 = 0 + 0.152 970 377 508 232 397 217 333 248;
  • 68) 0.152 970 377 508 232 397 217 333 248 × 2 = 0 + 0.305 940 755 016 464 794 434 666 496;
  • 69) 0.305 940 755 016 464 794 434 666 496 × 2 = 0 + 0.611 881 510 032 929 588 869 332 992;
  • 70) 0.611 881 510 032 929 588 869 332 992 × 2 = 1 + 0.223 763 020 065 859 177 738 665 984;
  • 71) 0.223 763 020 065 859 177 738 665 984 × 2 = 0 + 0.447 526 040 131 718 355 477 331 968;
  • 72) 0.447 526 040 131 718 355 477 331 968 × 2 = 0 + 0.895 052 080 263 436 710 954 663 936;
  • 73) 0.895 052 080 263 436 710 954 663 936 × 2 = 1 + 0.790 104 160 526 873 421 909 327 872;
  • 74) 0.790 104 160 526 873 421 909 327 872 × 2 = 1 + 0.580 208 321 053 746 843 818 655 744;
  • 75) 0.580 208 321 053 746 843 818 655 744 × 2 = 1 + 0.160 416 642 107 493 687 637 311 488;
  • 76) 0.160 416 642 107 493 687 637 311 488 × 2 = 0 + 0.320 833 284 214 987 375 274 622 976;
  • 77) 0.320 833 284 214 987 375 274 622 976 × 2 = 0 + 0.641 666 568 429 974 750 549 245 952;
  • 78) 0.641 666 568 429 974 750 549 245 952 × 2 = 1 + 0.283 333 136 859 949 501 098 491 904;

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:


0.000 000 021 979 552 668 138 406 941(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:

0.000 000 021 979 552 668 138 406 941(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:


0.000 000 021 979 552 668 138 406 941(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 0.000 000 021 979 552 668 138 406 941 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 0.000 000 021 979 552 668 138 406 933 to 64 bit double precision IEEE 754 binary floating point representation standard

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» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

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