1.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 051 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 051 15(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
1.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 051 15(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: 1.
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;
  • 1 ÷ 2 = 0 + 1;

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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 051 15.

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) 0.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 051 15 × 2 = 0 + 0.323 668 065 571 546 190 334 601 063 681 219 292 072 823 032 950 102 3;
  • 2) 0.323 668 065 571 546 190 334 601 063 681 219 292 072 823 032 950 102 3 × 2 = 0 + 0.647 336 131 143 092 380 669 202 127 362 438 584 145 646 065 900 204 6;
  • 3) 0.647 336 131 143 092 380 669 202 127 362 438 584 145 646 065 900 204 6 × 2 = 1 + 0.294 672 262 286 184 761 338 404 254 724 877 168 291 292 131 800 409 2;
  • 4) 0.294 672 262 286 184 761 338 404 254 724 877 168 291 292 131 800 409 2 × 2 = 0 + 0.589 344 524 572 369 522 676 808 509 449 754 336 582 584 263 600 818 4;
  • 5) 0.589 344 524 572 369 522 676 808 509 449 754 336 582 584 263 600 818 4 × 2 = 1 + 0.178 689 049 144 739 045 353 617 018 899 508 673 165 168 527 201 636 8;
  • 6) 0.178 689 049 144 739 045 353 617 018 899 508 673 165 168 527 201 636 8 × 2 = 0 + 0.357 378 098 289 478 090 707 234 037 799 017 346 330 337 054 403 273 6;
  • 7) 0.357 378 098 289 478 090 707 234 037 799 017 346 330 337 054 403 273 6 × 2 = 0 + 0.714 756 196 578 956 181 414 468 075 598 034 692 660 674 108 806 547 2;
  • 8) 0.714 756 196 578 956 181 414 468 075 598 034 692 660 674 108 806 547 2 × 2 = 1 + 0.429 512 393 157 912 362 828 936 151 196 069 385 321 348 217 613 094 4;
  • 9) 0.429 512 393 157 912 362 828 936 151 196 069 385 321 348 217 613 094 4 × 2 = 0 + 0.859 024 786 315 824 725 657 872 302 392 138 770 642 696 435 226 188 8;
  • 10) 0.859 024 786 315 824 725 657 872 302 392 138 770 642 696 435 226 188 8 × 2 = 1 + 0.718 049 572 631 649 451 315 744 604 784 277 541 285 392 870 452 377 6;
  • 11) 0.718 049 572 631 649 451 315 744 604 784 277 541 285 392 870 452 377 6 × 2 = 1 + 0.436 099 145 263 298 902 631 489 209 568 555 082 570 785 740 904 755 2;
  • 12) 0.436 099 145 263 298 902 631 489 209 568 555 082 570 785 740 904 755 2 × 2 = 0 + 0.872 198 290 526 597 805 262 978 419 137 110 165 141 571 481 809 510 4;
  • 13) 0.872 198 290 526 597 805 262 978 419 137 110 165 141 571 481 809 510 4 × 2 = 1 + 0.744 396 581 053 195 610 525 956 838 274 220 330 283 142 963 619 020 8;
  • 14) 0.744 396 581 053 195 610 525 956 838 274 220 330 283 142 963 619 020 8 × 2 = 1 + 0.488 793 162 106 391 221 051 913 676 548 440 660 566 285 927 238 041 6;
  • 15) 0.488 793 162 106 391 221 051 913 676 548 440 660 566 285 927 238 041 6 × 2 = 0 + 0.977 586 324 212 782 442 103 827 353 096 881 321 132 571 854 476 083 2;
  • 16) 0.977 586 324 212 782 442 103 827 353 096 881 321 132 571 854 476 083 2 × 2 = 1 + 0.955 172 648 425 564 884 207 654 706 193 762 642 265 143 708 952 166 4;
  • 17) 0.955 172 648 425 564 884 207 654 706 193 762 642 265 143 708 952 166 4 × 2 = 1 + 0.910 345 296 851 129 768 415 309 412 387 525 284 530 287 417 904 332 8;
  • 18) 0.910 345 296 851 129 768 415 309 412 387 525 284 530 287 417 904 332 8 × 2 = 1 + 0.820 690 593 702 259 536 830 618 824 775 050 569 060 574 835 808 665 6;
  • 19) 0.820 690 593 702 259 536 830 618 824 775 050 569 060 574 835 808 665 6 × 2 = 1 + 0.641 381 187 404 519 073 661 237 649 550 101 138 121 149 671 617 331 2;
  • 20) 0.641 381 187 404 519 073 661 237 649 550 101 138 121 149 671 617 331 2 × 2 = 1 + 0.282 762 374 809 038 147 322 475 299 100 202 276 242 299 343 234 662 4;
  • 21) 0.282 762 374 809 038 147 322 475 299 100 202 276 242 299 343 234 662 4 × 2 = 0 + 0.565 524 749 618 076 294 644 950 598 200 404 552 484 598 686 469 324 8;
  • 22) 0.565 524 749 618 076 294 644 950 598 200 404 552 484 598 686 469 324 8 × 2 = 1 + 0.131 049 499 236 152 589 289 901 196 400 809 104 969 197 372 938 649 6;
  • 23) 0.131 049 499 236 152 589 289 901 196 400 809 104 969 197 372 938 649 6 × 2 = 0 + 0.262 098 998 472 305 178 579 802 392 801 618 209 938 394 745 877 299 2;
  • 24) 0.262 098 998 472 305 178 579 802 392 801 618 209 938 394 745 877 299 2 × 2 = 0 + 0.524 197 996 944 610 357 159 604 785 603 236 419 876 789 491 754 598 4;
  • 25) 0.524 197 996 944 610 357 159 604 785 603 236 419 876 789 491 754 598 4 × 2 = 1 + 0.048 395 993 889 220 714 319 209 571 206 472 839 753 578 983 509 196 8;
  • 26) 0.048 395 993 889 220 714 319 209 571 206 472 839 753 578 983 509 196 8 × 2 = 0 + 0.096 791 987 778 441 428 638 419 142 412 945 679 507 157 967 018 393 6;
  • 27) 0.096 791 987 778 441 428 638 419 142 412 945 679 507 157 967 018 393 6 × 2 = 0 + 0.193 583 975 556 882 857 276 838 284 825 891 359 014 315 934 036 787 2;
  • 28) 0.193 583 975 556 882 857 276 838 284 825 891 359 014 315 934 036 787 2 × 2 = 0 + 0.387 167 951 113 765 714 553 676 569 651 782 718 028 631 868 073 574 4;
  • 29) 0.387 167 951 113 765 714 553 676 569 651 782 718 028 631 868 073 574 4 × 2 = 0 + 0.774 335 902 227 531 429 107 353 139 303 565 436 057 263 736 147 148 8;
  • 30) 0.774 335 902 227 531 429 107 353 139 303 565 436 057 263 736 147 148 8 × 2 = 1 + 0.548 671 804 455 062 858 214 706 278 607 130 872 114 527 472 294 297 6;
  • 31) 0.548 671 804 455 062 858 214 706 278 607 130 872 114 527 472 294 297 6 × 2 = 1 + 0.097 343 608 910 125 716 429 412 557 214 261 744 229 054 944 588 595 2;
  • 32) 0.097 343 608 910 125 716 429 412 557 214 261 744 229 054 944 588 595 2 × 2 = 0 + 0.194 687 217 820 251 432 858 825 114 428 523 488 458 109 889 177 190 4;
  • 33) 0.194 687 217 820 251 432 858 825 114 428 523 488 458 109 889 177 190 4 × 2 = 0 + 0.389 374 435 640 502 865 717 650 228 857 046 976 916 219 778 354 380 8;
  • 34) 0.389 374 435 640 502 865 717 650 228 857 046 976 916 219 778 354 380 8 × 2 = 0 + 0.778 748 871 281 005 731 435 300 457 714 093 953 832 439 556 708 761 6;
  • 35) 0.778 748 871 281 005 731 435 300 457 714 093 953 832 439 556 708 761 6 × 2 = 1 + 0.557 497 742 562 011 462 870 600 915 428 187 907 664 879 113 417 523 2;
  • 36) 0.557 497 742 562 011 462 870 600 915 428 187 907 664 879 113 417 523 2 × 2 = 1 + 0.114 995 485 124 022 925 741 201 830 856 375 815 329 758 226 835 046 4;
  • 37) 0.114 995 485 124 022 925 741 201 830 856 375 815 329 758 226 835 046 4 × 2 = 0 + 0.229 990 970 248 045 851 482 403 661 712 751 630 659 516 453 670 092 8;
  • 38) 0.229 990 970 248 045 851 482 403 661 712 751 630 659 516 453 670 092 8 × 2 = 0 + 0.459 981 940 496 091 702 964 807 323 425 503 261 319 032 907 340 185 6;
  • 39) 0.459 981 940 496 091 702 964 807 323 425 503 261 319 032 907 340 185 6 × 2 = 0 + 0.919 963 880 992 183 405 929 614 646 851 006 522 638 065 814 680 371 2;
  • 40) 0.919 963 880 992 183 405 929 614 646 851 006 522 638 065 814 680 371 2 × 2 = 1 + 0.839 927 761 984 366 811 859 229 293 702 013 045 276 131 629 360 742 4;
  • 41) 0.839 927 761 984 366 811 859 229 293 702 013 045 276 131 629 360 742 4 × 2 = 1 + 0.679 855 523 968 733 623 718 458 587 404 026 090 552 263 258 721 484 8;
  • 42) 0.679 855 523 968 733 623 718 458 587 404 026 090 552 263 258 721 484 8 × 2 = 1 + 0.359 711 047 937 467 247 436 917 174 808 052 181 104 526 517 442 969 6;
  • 43) 0.359 711 047 937 467 247 436 917 174 808 052 181 104 526 517 442 969 6 × 2 = 0 + 0.719 422 095 874 934 494 873 834 349 616 104 362 209 053 034 885 939 2;
  • 44) 0.719 422 095 874 934 494 873 834 349 616 104 362 209 053 034 885 939 2 × 2 = 1 + 0.438 844 191 749 868 989 747 668 699 232 208 724 418 106 069 771 878 4;
  • 45) 0.438 844 191 749 868 989 747 668 699 232 208 724 418 106 069 771 878 4 × 2 = 0 + 0.877 688 383 499 737 979 495 337 398 464 417 448 836 212 139 543 756 8;
  • 46) 0.877 688 383 499 737 979 495 337 398 464 417 448 836 212 139 543 756 8 × 2 = 1 + 0.755 376 766 999 475 958 990 674 796 928 834 897 672 424 279 087 513 6;
  • 47) 0.755 376 766 999 475 958 990 674 796 928 834 897 672 424 279 087 513 6 × 2 = 1 + 0.510 753 533 998 951 917 981 349 593 857 669 795 344 848 558 175 027 2;
  • 48) 0.510 753 533 998 951 917 981 349 593 857 669 795 344 848 558 175 027 2 × 2 = 1 + 0.021 507 067 997 903 835 962 699 187 715 339 590 689 697 116 350 054 4;
  • 49) 0.021 507 067 997 903 835 962 699 187 715 339 590 689 697 116 350 054 4 × 2 = 0 + 0.043 014 135 995 807 671 925 398 375 430 679 181 379 394 232 700 108 8;
  • 50) 0.043 014 135 995 807 671 925 398 375 430 679 181 379 394 232 700 108 8 × 2 = 0 + 0.086 028 271 991 615 343 850 796 750 861 358 362 758 788 465 400 217 6;
  • 51) 0.086 028 271 991 615 343 850 796 750 861 358 362 758 788 465 400 217 6 × 2 = 0 + 0.172 056 543 983 230 687 701 593 501 722 716 725 517 576 930 800 435 2;
  • 52) 0.172 056 543 983 230 687 701 593 501 722 716 725 517 576 930 800 435 2 × 2 = 0 + 0.344 113 087 966 461 375 403 187 003 445 433 451 035 153 861 600 870 4;
  • 53) 0.344 113 087 966 461 375 403 187 003 445 433 451 035 153 861 600 870 4 × 2 = 0 + 0.688 226 175 932 922 750 806 374 006 890 866 902 070 307 723 201 740 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:


0.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 051 15(10) =

0.0010 1001 0110 1101 1111 0100 1000 0110 0011 0001 1101 0111 0000 0(2)

5. Positive number before normalization:

1.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 051 15(10) =

1.0010 1001 0110 1101 1111 0100 1000 0110 0011 0001 1101 0111 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 051 15(10) =

1.0010 1001 0110 1101 1111 0100 1000 0110 0011 0001 1101 0111 0000 0(2) =

1.0010 1001 0110 1101 1111 0100 1000 0110 0011 0001 1101 0111 0000 0(2) × 20


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): 0

Mantissa (not normalized):
1.0010 1001 0110 1101 1111 0100 1000 0110 0011 0001 1101 0111 0000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1023(10) =

011 1111 1111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0010 1001 0110 1101 1111 0100 1000 0110 0011 0001 1101 0111 0000 0 =

0010 1001 0110 1101 1111 0100 1000 0110 0011 0001 1101 0111 0000


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 1111

Mantissa (52 bits) =
0010 1001 0110 1101 1111 0100 1000 0110 0011 0001 1101 0111 0000


Decimal number 1.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 051 15 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0010 1001 0110 1101 1111 0100 1000 0110 0011 0001 1101 0111 0000

Share

» Convert decimal 1.161 834 032 785 773 095 167 300 531 840 609 646 036 411 516 475 052 07 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: My Manager Only Says “We” When Referring to “My Work”. NotebookLM YouTube Video of 11.13 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