-0.000 806 264 623 585 362 514 063 654 156 856 105 041 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 806 264 623 585 362 514 063 654 156 856 105 041 9(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
-0.000 806 264 623 585 362 514 063 654 156 856 105 041 9(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:

|-0.000 806 264 623 585 362 514 063 654 156 856 105 041 9| = 0.000 806 264 623 585 362 514 063 654 156 856 105 041 9


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: 0.000 806 264 623 585 362 514 063 654 156 856 105 041 9.

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) 0.000 806 264 623 585 362 514 063 654 156 856 105 041 9 × 2 = 0 + 0.001 612 529 247 170 725 028 127 308 313 712 210 083 8;
  • 2) 0.001 612 529 247 170 725 028 127 308 313 712 210 083 8 × 2 = 0 + 0.003 225 058 494 341 450 056 254 616 627 424 420 167 6;
  • 3) 0.003 225 058 494 341 450 056 254 616 627 424 420 167 6 × 2 = 0 + 0.006 450 116 988 682 900 112 509 233 254 848 840 335 2;
  • 4) 0.006 450 116 988 682 900 112 509 233 254 848 840 335 2 × 2 = 0 + 0.012 900 233 977 365 800 225 018 466 509 697 680 670 4;
  • 5) 0.012 900 233 977 365 800 225 018 466 509 697 680 670 4 × 2 = 0 + 0.025 800 467 954 731 600 450 036 933 019 395 361 340 8;
  • 6) 0.025 800 467 954 731 600 450 036 933 019 395 361 340 8 × 2 = 0 + 0.051 600 935 909 463 200 900 073 866 038 790 722 681 6;
  • 7) 0.051 600 935 909 463 200 900 073 866 038 790 722 681 6 × 2 = 0 + 0.103 201 871 818 926 401 800 147 732 077 581 445 363 2;
  • 8) 0.103 201 871 818 926 401 800 147 732 077 581 445 363 2 × 2 = 0 + 0.206 403 743 637 852 803 600 295 464 155 162 890 726 4;
  • 9) 0.206 403 743 637 852 803 600 295 464 155 162 890 726 4 × 2 = 0 + 0.412 807 487 275 705 607 200 590 928 310 325 781 452 8;
  • 10) 0.412 807 487 275 705 607 200 590 928 310 325 781 452 8 × 2 = 0 + 0.825 614 974 551 411 214 401 181 856 620 651 562 905 6;
  • 11) 0.825 614 974 551 411 214 401 181 856 620 651 562 905 6 × 2 = 1 + 0.651 229 949 102 822 428 802 363 713 241 303 125 811 2;
  • 12) 0.651 229 949 102 822 428 802 363 713 241 303 125 811 2 × 2 = 1 + 0.302 459 898 205 644 857 604 727 426 482 606 251 622 4;
  • 13) 0.302 459 898 205 644 857 604 727 426 482 606 251 622 4 × 2 = 0 + 0.604 919 796 411 289 715 209 454 852 965 212 503 244 8;
  • 14) 0.604 919 796 411 289 715 209 454 852 965 212 503 244 8 × 2 = 1 + 0.209 839 592 822 579 430 418 909 705 930 425 006 489 6;
  • 15) 0.209 839 592 822 579 430 418 909 705 930 425 006 489 6 × 2 = 0 + 0.419 679 185 645 158 860 837 819 411 860 850 012 979 2;
  • 16) 0.419 679 185 645 158 860 837 819 411 860 850 012 979 2 × 2 = 0 + 0.839 358 371 290 317 721 675 638 823 721 700 025 958 4;
  • 17) 0.839 358 371 290 317 721 675 638 823 721 700 025 958 4 × 2 = 1 + 0.678 716 742 580 635 443 351 277 647 443 400 051 916 8;
  • 18) 0.678 716 742 580 635 443 351 277 647 443 400 051 916 8 × 2 = 1 + 0.357 433 485 161 270 886 702 555 294 886 800 103 833 6;
  • 19) 0.357 433 485 161 270 886 702 555 294 886 800 103 833 6 × 2 = 0 + 0.714 866 970 322 541 773 405 110 589 773 600 207 667 2;
  • 20) 0.714 866 970 322 541 773 405 110 589 773 600 207 667 2 × 2 = 1 + 0.429 733 940 645 083 546 810 221 179 547 200 415 334 4;
  • 21) 0.429 733 940 645 083 546 810 221 179 547 200 415 334 4 × 2 = 0 + 0.859 467 881 290 167 093 620 442 359 094 400 830 668 8;
  • 22) 0.859 467 881 290 167 093 620 442 359 094 400 830 668 8 × 2 = 1 + 0.718 935 762 580 334 187 240 884 718 188 801 661 337 6;
  • 23) 0.718 935 762 580 334 187 240 884 718 188 801 661 337 6 × 2 = 1 + 0.437 871 525 160 668 374 481 769 436 377 603 322 675 2;
  • 24) 0.437 871 525 160 668 374 481 769 436 377 603 322 675 2 × 2 = 0 + 0.875 743 050 321 336 748 963 538 872 755 206 645 350 4;
  • 25) 0.875 743 050 321 336 748 963 538 872 755 206 645 350 4 × 2 = 1 + 0.751 486 100 642 673 497 927 077 745 510 413 290 700 8;
  • 26) 0.751 486 100 642 673 497 927 077 745 510 413 290 700 8 × 2 = 1 + 0.502 972 201 285 346 995 854 155 491 020 826 581 401 6;
  • 27) 0.502 972 201 285 346 995 854 155 491 020 826 581 401 6 × 2 = 1 + 0.005 944 402 570 693 991 708 310 982 041 653 162 803 2;
  • 28) 0.005 944 402 570 693 991 708 310 982 041 653 162 803 2 × 2 = 0 + 0.011 888 805 141 387 983 416 621 964 083 306 325 606 4;
  • 29) 0.011 888 805 141 387 983 416 621 964 083 306 325 606 4 × 2 = 0 + 0.023 777 610 282 775 966 833 243 928 166 612 651 212 8;
  • 30) 0.023 777 610 282 775 966 833 243 928 166 612 651 212 8 × 2 = 0 + 0.047 555 220 565 551 933 666 487 856 333 225 302 425 6;
  • 31) 0.047 555 220 565 551 933 666 487 856 333 225 302 425 6 × 2 = 0 + 0.095 110 441 131 103 867 332 975 712 666 450 604 851 2;
  • 32) 0.095 110 441 131 103 867 332 975 712 666 450 604 851 2 × 2 = 0 + 0.190 220 882 262 207 734 665 951 425 332 901 209 702 4;
  • 33) 0.190 220 882 262 207 734 665 951 425 332 901 209 702 4 × 2 = 0 + 0.380 441 764 524 415 469 331 902 850 665 802 419 404 8;
  • 34) 0.380 441 764 524 415 469 331 902 850 665 802 419 404 8 × 2 = 0 + 0.760 883 529 048 830 938 663 805 701 331 604 838 809 6;
  • 35) 0.760 883 529 048 830 938 663 805 701 331 604 838 809 6 × 2 = 1 + 0.521 767 058 097 661 877 327 611 402 663 209 677 619 2;
  • 36) 0.521 767 058 097 661 877 327 611 402 663 209 677 619 2 × 2 = 1 + 0.043 534 116 195 323 754 655 222 805 326 419 355 238 4;
  • 37) 0.043 534 116 195 323 754 655 222 805 326 419 355 238 4 × 2 = 0 + 0.087 068 232 390 647 509 310 445 610 652 838 710 476 8;
  • 38) 0.087 068 232 390 647 509 310 445 610 652 838 710 476 8 × 2 = 0 + 0.174 136 464 781 295 018 620 891 221 305 677 420 953 6;
  • 39) 0.174 136 464 781 295 018 620 891 221 305 677 420 953 6 × 2 = 0 + 0.348 272 929 562 590 037 241 782 442 611 354 841 907 2;
  • 40) 0.348 272 929 562 590 037 241 782 442 611 354 841 907 2 × 2 = 0 + 0.696 545 859 125 180 074 483 564 885 222 709 683 814 4;
  • 41) 0.696 545 859 125 180 074 483 564 885 222 709 683 814 4 × 2 = 1 + 0.393 091 718 250 360 148 967 129 770 445 419 367 628 8;
  • 42) 0.393 091 718 250 360 148 967 129 770 445 419 367 628 8 × 2 = 0 + 0.786 183 436 500 720 297 934 259 540 890 838 735 257 6;
  • 43) 0.786 183 436 500 720 297 934 259 540 890 838 735 257 6 × 2 = 1 + 0.572 366 873 001 440 595 868 519 081 781 677 470 515 2;
  • 44) 0.572 366 873 001 440 595 868 519 081 781 677 470 515 2 × 2 = 1 + 0.144 733 746 002 881 191 737 038 163 563 354 941 030 4;
  • 45) 0.144 733 746 002 881 191 737 038 163 563 354 941 030 4 × 2 = 0 + 0.289 467 492 005 762 383 474 076 327 126 709 882 060 8;
  • 46) 0.289 467 492 005 762 383 474 076 327 126 709 882 060 8 × 2 = 0 + 0.578 934 984 011 524 766 948 152 654 253 419 764 121 6;
  • 47) 0.578 934 984 011 524 766 948 152 654 253 419 764 121 6 × 2 = 1 + 0.157 869 968 023 049 533 896 305 308 506 839 528 243 2;
  • 48) 0.157 869 968 023 049 533 896 305 308 506 839 528 243 2 × 2 = 0 + 0.315 739 936 046 099 067 792 610 617 013 679 056 486 4;
  • 49) 0.315 739 936 046 099 067 792 610 617 013 679 056 486 4 × 2 = 0 + 0.631 479 872 092 198 135 585 221 234 027 358 112 972 8;
  • 50) 0.631 479 872 092 198 135 585 221 234 027 358 112 972 8 × 2 = 1 + 0.262 959 744 184 396 271 170 442 468 054 716 225 945 6;
  • 51) 0.262 959 744 184 396 271 170 442 468 054 716 225 945 6 × 2 = 0 + 0.525 919 488 368 792 542 340 884 936 109 432 451 891 2;
  • 52) 0.525 919 488 368 792 542 340 884 936 109 432 451 891 2 × 2 = 1 + 0.051 838 976 737 585 084 681 769 872 218 864 903 782 4;
  • 53) 0.051 838 976 737 585 084 681 769 872 218 864 903 782 4 × 2 = 0 + 0.103 677 953 475 170 169 363 539 744 437 729 807 564 8;
  • 54) 0.103 677 953 475 170 169 363 539 744 437 729 807 564 8 × 2 = 0 + 0.207 355 906 950 340 338 727 079 488 875 459 615 129 6;
  • 55) 0.207 355 906 950 340 338 727 079 488 875 459 615 129 6 × 2 = 0 + 0.414 711 813 900 680 677 454 158 977 750 919 230 259 2;
  • 56) 0.414 711 813 900 680 677 454 158 977 750 919 230 259 2 × 2 = 0 + 0.829 423 627 801 361 354 908 317 955 501 838 460 518 4;
  • 57) 0.829 423 627 801 361 354 908 317 955 501 838 460 518 4 × 2 = 1 + 0.658 847 255 602 722 709 816 635 911 003 676 921 036 8;
  • 58) 0.658 847 255 602 722 709 816 635 911 003 676 921 036 8 × 2 = 1 + 0.317 694 511 205 445 419 633 271 822 007 353 842 073 6;
  • 59) 0.317 694 511 205 445 419 633 271 822 007 353 842 073 6 × 2 = 0 + 0.635 389 022 410 890 839 266 543 644 014 707 684 147 2;
  • 60) 0.635 389 022 410 890 839 266 543 644 014 707 684 147 2 × 2 = 1 + 0.270 778 044 821 781 678 533 087 288 029 415 368 294 4;
  • 61) 0.270 778 044 821 781 678 533 087 288 029 415 368 294 4 × 2 = 0 + 0.541 556 089 643 563 357 066 174 576 058 830 736 588 8;
  • 62) 0.541 556 089 643 563 357 066 174 576 058 830 736 588 8 × 2 = 1 + 0.083 112 179 287 126 714 132 349 152 117 661 473 177 6;
  • 63) 0.083 112 179 287 126 714 132 349 152 117 661 473 177 6 × 2 = 0 + 0.166 224 358 574 253 428 264 698 304 235 322 946 355 2;

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:


0.000 806 264 623 585 362 514 063 654 156 856 105 041 9(10) =

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

6. Positive number before normalization:

0.000 806 264 623 585 362 514 063 654 156 856 105 041 9(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:


0.000 806 264 623 585 362 514 063 654 156 856 105 041 9(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 -0.000 806 264 623 585 362 514 063 654 156 856 105 041 9 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

Share

» Convert decimal -0.000 806 264 623 585 362 514 063 654 156 856 105 051 6 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 07, 2026 [July]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: Why Do Colleagues From Other Teams Have Better Results?. NotebookLM YouTube Video of 11.31 minutes

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