-0.000 806 264 623 585 362 514 063 654 156 856 085 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 806 264 623 585 362 514 063 654 156 856 085 3(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
-0.000 806 264 623 585 362 514 063 654 156 856 085 3(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. Start with the positive version of the number:

|-0.000 806 264 623 585 362 514 063 654 156 856 085 3| = 0.000 806 264 623 585 362 514 063 654 156 856 085 3


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 806 264 623 585 362 514 063 654 156 856 085 3.

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) 0.000 806 264 623 585 362 514 063 654 156 856 085 3 × 2 = 0 + 0.001 612 529 247 170 725 028 127 308 313 712 170 6;
  • 2) 0.001 612 529 247 170 725 028 127 308 313 712 170 6 × 2 = 0 + 0.003 225 058 494 341 450 056 254 616 627 424 341 2;
  • 3) 0.003 225 058 494 341 450 056 254 616 627 424 341 2 × 2 = 0 + 0.006 450 116 988 682 900 112 509 233 254 848 682 4;
  • 4) 0.006 450 116 988 682 900 112 509 233 254 848 682 4 × 2 = 0 + 0.012 900 233 977 365 800 225 018 466 509 697 364 8;
  • 5) 0.012 900 233 977 365 800 225 018 466 509 697 364 8 × 2 = 0 + 0.025 800 467 954 731 600 450 036 933 019 394 729 6;
  • 6) 0.025 800 467 954 731 600 450 036 933 019 394 729 6 × 2 = 0 + 0.051 600 935 909 463 200 900 073 866 038 789 459 2;
  • 7) 0.051 600 935 909 463 200 900 073 866 038 789 459 2 × 2 = 0 + 0.103 201 871 818 926 401 800 147 732 077 578 918 4;
  • 8) 0.103 201 871 818 926 401 800 147 732 077 578 918 4 × 2 = 0 + 0.206 403 743 637 852 803 600 295 464 155 157 836 8;
  • 9) 0.206 403 743 637 852 803 600 295 464 155 157 836 8 × 2 = 0 + 0.412 807 487 275 705 607 200 590 928 310 315 673 6;
  • 10) 0.412 807 487 275 705 607 200 590 928 310 315 673 6 × 2 = 0 + 0.825 614 974 551 411 214 401 181 856 620 631 347 2;
  • 11) 0.825 614 974 551 411 214 401 181 856 620 631 347 2 × 2 = 1 + 0.651 229 949 102 822 428 802 363 713 241 262 694 4;
  • 12) 0.651 229 949 102 822 428 802 363 713 241 262 694 4 × 2 = 1 + 0.302 459 898 205 644 857 604 727 426 482 525 388 8;
  • 13) 0.302 459 898 205 644 857 604 727 426 482 525 388 8 × 2 = 0 + 0.604 919 796 411 289 715 209 454 852 965 050 777 6;
  • 14) 0.604 919 796 411 289 715 209 454 852 965 050 777 6 × 2 = 1 + 0.209 839 592 822 579 430 418 909 705 930 101 555 2;
  • 15) 0.209 839 592 822 579 430 418 909 705 930 101 555 2 × 2 = 0 + 0.419 679 185 645 158 860 837 819 411 860 203 110 4;
  • 16) 0.419 679 185 645 158 860 837 819 411 860 203 110 4 × 2 = 0 + 0.839 358 371 290 317 721 675 638 823 720 406 220 8;
  • 17) 0.839 358 371 290 317 721 675 638 823 720 406 220 8 × 2 = 1 + 0.678 716 742 580 635 443 351 277 647 440 812 441 6;
  • 18) 0.678 716 742 580 635 443 351 277 647 440 812 441 6 × 2 = 1 + 0.357 433 485 161 270 886 702 555 294 881 624 883 2;
  • 19) 0.357 433 485 161 270 886 702 555 294 881 624 883 2 × 2 = 0 + 0.714 866 970 322 541 773 405 110 589 763 249 766 4;
  • 20) 0.714 866 970 322 541 773 405 110 589 763 249 766 4 × 2 = 1 + 0.429 733 940 645 083 546 810 221 179 526 499 532 8;
  • 21) 0.429 733 940 645 083 546 810 221 179 526 499 532 8 × 2 = 0 + 0.859 467 881 290 167 093 620 442 359 052 999 065 6;
  • 22) 0.859 467 881 290 167 093 620 442 359 052 999 065 6 × 2 = 1 + 0.718 935 762 580 334 187 240 884 718 105 998 131 2;
  • 23) 0.718 935 762 580 334 187 240 884 718 105 998 131 2 × 2 = 1 + 0.437 871 525 160 668 374 481 769 436 211 996 262 4;
  • 24) 0.437 871 525 160 668 374 481 769 436 211 996 262 4 × 2 = 0 + 0.875 743 050 321 336 748 963 538 872 423 992 524 8;
  • 25) 0.875 743 050 321 336 748 963 538 872 423 992 524 8 × 2 = 1 + 0.751 486 100 642 673 497 927 077 744 847 985 049 6;
  • 26) 0.751 486 100 642 673 497 927 077 744 847 985 049 6 × 2 = 1 + 0.502 972 201 285 346 995 854 155 489 695 970 099 2;
  • 27) 0.502 972 201 285 346 995 854 155 489 695 970 099 2 × 2 = 1 + 0.005 944 402 570 693 991 708 310 979 391 940 198 4;
  • 28) 0.005 944 402 570 693 991 708 310 979 391 940 198 4 × 2 = 0 + 0.011 888 805 141 387 983 416 621 958 783 880 396 8;
  • 29) 0.011 888 805 141 387 983 416 621 958 783 880 396 8 × 2 = 0 + 0.023 777 610 282 775 966 833 243 917 567 760 793 6;
  • 30) 0.023 777 610 282 775 966 833 243 917 567 760 793 6 × 2 = 0 + 0.047 555 220 565 551 933 666 487 835 135 521 587 2;
  • 31) 0.047 555 220 565 551 933 666 487 835 135 521 587 2 × 2 = 0 + 0.095 110 441 131 103 867 332 975 670 271 043 174 4;
  • 32) 0.095 110 441 131 103 867 332 975 670 271 043 174 4 × 2 = 0 + 0.190 220 882 262 207 734 665 951 340 542 086 348 8;
  • 33) 0.190 220 882 262 207 734 665 951 340 542 086 348 8 × 2 = 0 + 0.380 441 764 524 415 469 331 902 681 084 172 697 6;
  • 34) 0.380 441 764 524 415 469 331 902 681 084 172 697 6 × 2 = 0 + 0.760 883 529 048 830 938 663 805 362 168 345 395 2;
  • 35) 0.760 883 529 048 830 938 663 805 362 168 345 395 2 × 2 = 1 + 0.521 767 058 097 661 877 327 610 724 336 690 790 4;
  • 36) 0.521 767 058 097 661 877 327 610 724 336 690 790 4 × 2 = 1 + 0.043 534 116 195 323 754 655 221 448 673 381 580 8;
  • 37) 0.043 534 116 195 323 754 655 221 448 673 381 580 8 × 2 = 0 + 0.087 068 232 390 647 509 310 442 897 346 763 161 6;
  • 38) 0.087 068 232 390 647 509 310 442 897 346 763 161 6 × 2 = 0 + 0.174 136 464 781 295 018 620 885 794 693 526 323 2;
  • 39) 0.174 136 464 781 295 018 620 885 794 693 526 323 2 × 2 = 0 + 0.348 272 929 562 590 037 241 771 589 387 052 646 4;
  • 40) 0.348 272 929 562 590 037 241 771 589 387 052 646 4 × 2 = 0 + 0.696 545 859 125 180 074 483 543 178 774 105 292 8;
  • 41) 0.696 545 859 125 180 074 483 543 178 774 105 292 8 × 2 = 1 + 0.393 091 718 250 360 148 967 086 357 548 210 585 6;
  • 42) 0.393 091 718 250 360 148 967 086 357 548 210 585 6 × 2 = 0 + 0.786 183 436 500 720 297 934 172 715 096 421 171 2;
  • 43) 0.786 183 436 500 720 297 934 172 715 096 421 171 2 × 2 = 1 + 0.572 366 873 001 440 595 868 345 430 192 842 342 4;
  • 44) 0.572 366 873 001 440 595 868 345 430 192 842 342 4 × 2 = 1 + 0.144 733 746 002 881 191 736 690 860 385 684 684 8;
  • 45) 0.144 733 746 002 881 191 736 690 860 385 684 684 8 × 2 = 0 + 0.289 467 492 005 762 383 473 381 720 771 369 369 6;
  • 46) 0.289 467 492 005 762 383 473 381 720 771 369 369 6 × 2 = 0 + 0.578 934 984 011 524 766 946 763 441 542 738 739 2;
  • 47) 0.578 934 984 011 524 766 946 763 441 542 738 739 2 × 2 = 1 + 0.157 869 968 023 049 533 893 526 883 085 477 478 4;
  • 48) 0.157 869 968 023 049 533 893 526 883 085 477 478 4 × 2 = 0 + 0.315 739 936 046 099 067 787 053 766 170 954 956 8;
  • 49) 0.315 739 936 046 099 067 787 053 766 170 954 956 8 × 2 = 0 + 0.631 479 872 092 198 135 574 107 532 341 909 913 6;
  • 50) 0.631 479 872 092 198 135 574 107 532 341 909 913 6 × 2 = 1 + 0.262 959 744 184 396 271 148 215 064 683 819 827 2;
  • 51) 0.262 959 744 184 396 271 148 215 064 683 819 827 2 × 2 = 0 + 0.525 919 488 368 792 542 296 430 129 367 639 654 4;
  • 52) 0.525 919 488 368 792 542 296 430 129 367 639 654 4 × 2 = 1 + 0.051 838 976 737 585 084 592 860 258 735 279 308 8;
  • 53) 0.051 838 976 737 585 084 592 860 258 735 279 308 8 × 2 = 0 + 0.103 677 953 475 170 169 185 720 517 470 558 617 6;
  • 54) 0.103 677 953 475 170 169 185 720 517 470 558 617 6 × 2 = 0 + 0.207 355 906 950 340 338 371 441 034 941 117 235 2;
  • 55) 0.207 355 906 950 340 338 371 441 034 941 117 235 2 × 2 = 0 + 0.414 711 813 900 680 676 742 882 069 882 234 470 4;
  • 56) 0.414 711 813 900 680 676 742 882 069 882 234 470 4 × 2 = 0 + 0.829 423 627 801 361 353 485 764 139 764 468 940 8;
  • 57) 0.829 423 627 801 361 353 485 764 139 764 468 940 8 × 2 = 1 + 0.658 847 255 602 722 706 971 528 279 528 937 881 6;
  • 58) 0.658 847 255 602 722 706 971 528 279 528 937 881 6 × 2 = 1 + 0.317 694 511 205 445 413 943 056 559 057 875 763 2;
  • 59) 0.317 694 511 205 445 413 943 056 559 057 875 763 2 × 2 = 0 + 0.635 389 022 410 890 827 886 113 118 115 751 526 4;
  • 60) 0.635 389 022 410 890 827 886 113 118 115 751 526 4 × 2 = 1 + 0.270 778 044 821 781 655 772 226 236 231 503 052 8;
  • 61) 0.270 778 044 821 781 655 772 226 236 231 503 052 8 × 2 = 0 + 0.541 556 089 643 563 311 544 452 472 463 006 105 6;
  • 62) 0.541 556 089 643 563 311 544 452 472 463 006 105 6 × 2 = 1 + 0.083 112 179 287 126 623 088 904 944 926 012 211 2;
  • 63) 0.083 112 179 287 126 623 088 904 944 926 012 211 2 × 2 = 0 + 0.166 224 358 574 253 246 177 809 889 852 024 422 4;

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).


5. 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:


0.000 806 264 623 585 362 514 063 654 156 856 085 3(10) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2)

6. Positive number before normalization:

0.000 806 264 623 585 362 514 063 654 156 856 085 3(10) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 806 264 623 585 362 514 063 654 156 856 085 3(10) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2) × 20 =

1.1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010(2) × 2-11


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): -11

Mantissa (not normalized):
1.1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-11 + 1 023)(10) =

1 012(10)


10. 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;
  • 1 012 ÷ 2 = 506 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1012(10) =

011 1111 0100(2)


12. 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. 1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010 =

1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0100

Mantissa (52 bits) =
1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010


Decimal number -0.000 806 264 623 585 362 514 063 654 156 856 085 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0100 - 1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010

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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