-0.000 000 000 000 293 915 430 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 293 915 430 8(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
-0.000 000 000 000 293 915 430 8(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:

|-0.000 000 000 000 293 915 430 8| = 0.000 000 000 000 293 915 430 8


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: 0.000 000 000 000 293 915 430 8.

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) 0.000 000 000 000 293 915 430 8 × 2 = 0 + 0.000 000 000 000 587 830 861 6;
  • 2) 0.000 000 000 000 587 830 861 6 × 2 = 0 + 0.000 000 000 001 175 661 723 2;
  • 3) 0.000 000 000 001 175 661 723 2 × 2 = 0 + 0.000 000 000 002 351 323 446 4;
  • 4) 0.000 000 000 002 351 323 446 4 × 2 = 0 + 0.000 000 000 004 702 646 892 8;
  • 5) 0.000 000 000 004 702 646 892 8 × 2 = 0 + 0.000 000 000 009 405 293 785 6;
  • 6) 0.000 000 000 009 405 293 785 6 × 2 = 0 + 0.000 000 000 018 810 587 571 2;
  • 7) 0.000 000 000 018 810 587 571 2 × 2 = 0 + 0.000 000 000 037 621 175 142 4;
  • 8) 0.000 000 000 037 621 175 142 4 × 2 = 0 + 0.000 000 000 075 242 350 284 8;
  • 9) 0.000 000 000 075 242 350 284 8 × 2 = 0 + 0.000 000 000 150 484 700 569 6;
  • 10) 0.000 000 000 150 484 700 569 6 × 2 = 0 + 0.000 000 000 300 969 401 139 2;
  • 11) 0.000 000 000 300 969 401 139 2 × 2 = 0 + 0.000 000 000 601 938 802 278 4;
  • 12) 0.000 000 000 601 938 802 278 4 × 2 = 0 + 0.000 000 001 203 877 604 556 8;
  • 13) 0.000 000 001 203 877 604 556 8 × 2 = 0 + 0.000 000 002 407 755 209 113 6;
  • 14) 0.000 000 002 407 755 209 113 6 × 2 = 0 + 0.000 000 004 815 510 418 227 2;
  • 15) 0.000 000 004 815 510 418 227 2 × 2 = 0 + 0.000 000 009 631 020 836 454 4;
  • 16) 0.000 000 009 631 020 836 454 4 × 2 = 0 + 0.000 000 019 262 041 672 908 8;
  • 17) 0.000 000 019 262 041 672 908 8 × 2 = 0 + 0.000 000 038 524 083 345 817 6;
  • 18) 0.000 000 038 524 083 345 817 6 × 2 = 0 + 0.000 000 077 048 166 691 635 2;
  • 19) 0.000 000 077 048 166 691 635 2 × 2 = 0 + 0.000 000 154 096 333 383 270 4;
  • 20) 0.000 000 154 096 333 383 270 4 × 2 = 0 + 0.000 000 308 192 666 766 540 8;
  • 21) 0.000 000 308 192 666 766 540 8 × 2 = 0 + 0.000 000 616 385 333 533 081 6;
  • 22) 0.000 000 616 385 333 533 081 6 × 2 = 0 + 0.000 001 232 770 667 066 163 2;
  • 23) 0.000 001 232 770 667 066 163 2 × 2 = 0 + 0.000 002 465 541 334 132 326 4;
  • 24) 0.000 002 465 541 334 132 326 4 × 2 = 0 + 0.000 004 931 082 668 264 652 8;
  • 25) 0.000 004 931 082 668 264 652 8 × 2 = 0 + 0.000 009 862 165 336 529 305 6;
  • 26) 0.000 009 862 165 336 529 305 6 × 2 = 0 + 0.000 019 724 330 673 058 611 2;
  • 27) 0.000 019 724 330 673 058 611 2 × 2 = 0 + 0.000 039 448 661 346 117 222 4;
  • 28) 0.000 039 448 661 346 117 222 4 × 2 = 0 + 0.000 078 897 322 692 234 444 8;
  • 29) 0.000 078 897 322 692 234 444 8 × 2 = 0 + 0.000 157 794 645 384 468 889 6;
  • 30) 0.000 157 794 645 384 468 889 6 × 2 = 0 + 0.000 315 589 290 768 937 779 2;
  • 31) 0.000 315 589 290 768 937 779 2 × 2 = 0 + 0.000 631 178 581 537 875 558 4;
  • 32) 0.000 631 178 581 537 875 558 4 × 2 = 0 + 0.001 262 357 163 075 751 116 8;
  • 33) 0.001 262 357 163 075 751 116 8 × 2 = 0 + 0.002 524 714 326 151 502 233 6;
  • 34) 0.002 524 714 326 151 502 233 6 × 2 = 0 + 0.005 049 428 652 303 004 467 2;
  • 35) 0.005 049 428 652 303 004 467 2 × 2 = 0 + 0.010 098 857 304 606 008 934 4;
  • 36) 0.010 098 857 304 606 008 934 4 × 2 = 0 + 0.020 197 714 609 212 017 868 8;
  • 37) 0.020 197 714 609 212 017 868 8 × 2 = 0 + 0.040 395 429 218 424 035 737 6;
  • 38) 0.040 395 429 218 424 035 737 6 × 2 = 0 + 0.080 790 858 436 848 071 475 2;
  • 39) 0.080 790 858 436 848 071 475 2 × 2 = 0 + 0.161 581 716 873 696 142 950 4;
  • 40) 0.161 581 716 873 696 142 950 4 × 2 = 0 + 0.323 163 433 747 392 285 900 8;
  • 41) 0.323 163 433 747 392 285 900 8 × 2 = 0 + 0.646 326 867 494 784 571 801 6;
  • 42) 0.646 326 867 494 784 571 801 6 × 2 = 1 + 0.292 653 734 989 569 143 603 2;
  • 43) 0.292 653 734 989 569 143 603 2 × 2 = 0 + 0.585 307 469 979 138 287 206 4;
  • 44) 0.585 307 469 979 138 287 206 4 × 2 = 1 + 0.170 614 939 958 276 574 412 8;
  • 45) 0.170 614 939 958 276 574 412 8 × 2 = 0 + 0.341 229 879 916 553 148 825 6;
  • 46) 0.341 229 879 916 553 148 825 6 × 2 = 0 + 0.682 459 759 833 106 297 651 2;
  • 47) 0.682 459 759 833 106 297 651 2 × 2 = 1 + 0.364 919 519 666 212 595 302 4;
  • 48) 0.364 919 519 666 212 595 302 4 × 2 = 0 + 0.729 839 039 332 425 190 604 8;
  • 49) 0.729 839 039 332 425 190 604 8 × 2 = 1 + 0.459 678 078 664 850 381 209 6;
  • 50) 0.459 678 078 664 850 381 209 6 × 2 = 0 + 0.919 356 157 329 700 762 419 2;
  • 51) 0.919 356 157 329 700 762 419 2 × 2 = 1 + 0.838 712 314 659 401 524 838 4;
  • 52) 0.838 712 314 659 401 524 838 4 × 2 = 1 + 0.677 424 629 318 803 049 676 8;
  • 53) 0.677 424 629 318 803 049 676 8 × 2 = 1 + 0.354 849 258 637 606 099 353 6;
  • 54) 0.354 849 258 637 606 099 353 6 × 2 = 0 + 0.709 698 517 275 212 198 707 2;
  • 55) 0.709 698 517 275 212 198 707 2 × 2 = 1 + 0.419 397 034 550 424 397 414 4;
  • 56) 0.419 397 034 550 424 397 414 4 × 2 = 0 + 0.838 794 069 100 848 794 828 8;
  • 57) 0.838 794 069 100 848 794 828 8 × 2 = 1 + 0.677 588 138 201 697 589 657 6;
  • 58) 0.677 588 138 201 697 589 657 6 × 2 = 1 + 0.355 176 276 403 395 179 315 2;
  • 59) 0.355 176 276 403 395 179 315 2 × 2 = 0 + 0.710 352 552 806 790 358 630 4;
  • 60) 0.710 352 552 806 790 358 630 4 × 2 = 1 + 0.420 705 105 613 580 717 260 8;
  • 61) 0.420 705 105 613 580 717 260 8 × 2 = 0 + 0.841 410 211 227 161 434 521 6;
  • 62) 0.841 410 211 227 161 434 521 6 × 2 = 1 + 0.682 820 422 454 322 869 043 2;
  • 63) 0.682 820 422 454 322 869 043 2 × 2 = 1 + 0.365 640 844 908 645 738 086 4;
  • 64) 0.365 640 844 908 645 738 086 4 × 2 = 0 + 0.731 281 689 817 291 476 172 8;
  • 65) 0.731 281 689 817 291 476 172 8 × 2 = 1 + 0.462 563 379 634 582 952 345 6;
  • 66) 0.462 563 379 634 582 952 345 6 × 2 = 0 + 0.925 126 759 269 165 904 691 2;
  • 67) 0.925 126 759 269 165 904 691 2 × 2 = 1 + 0.850 253 518 538 331 809 382 4;
  • 68) 0.850 253 518 538 331 809 382 4 × 2 = 1 + 0.700 507 037 076 663 618 764 8;
  • 69) 0.700 507 037 076 663 618 764 8 × 2 = 1 + 0.401 014 074 153 327 237 529 6;
  • 70) 0.401 014 074 153 327 237 529 6 × 2 = 0 + 0.802 028 148 306 654 475 059 2;
  • 71) 0.802 028 148 306 654 475 059 2 × 2 = 1 + 0.604 056 296 613 308 950 118 4;
  • 72) 0.604 056 296 613 308 950 118 4 × 2 = 1 + 0.208 112 593 226 617 900 236 8;
  • 73) 0.208 112 593 226 617 900 236 8 × 2 = 0 + 0.416 225 186 453 235 800 473 6;
  • 74) 0.416 225 186 453 235 800 473 6 × 2 = 0 + 0.832 450 372 906 471 600 947 2;
  • 75) 0.832 450 372 906 471 600 947 2 × 2 = 1 + 0.664 900 745 812 943 201 894 4;
  • 76) 0.664 900 745 812 943 201 894 4 × 2 = 1 + 0.329 801 491 625 886 403 788 8;
  • 77) 0.329 801 491 625 886 403 788 8 × 2 = 0 + 0.659 602 983 251 772 807 577 6;
  • 78) 0.659 602 983 251 772 807 577 6 × 2 = 1 + 0.319 205 966 503 545 615 155 2;
  • 79) 0.319 205 966 503 545 615 155 2 × 2 = 0 + 0.638 411 933 007 091 230 310 4;
  • 80) 0.638 411 933 007 091 230 310 4 × 2 = 1 + 0.276 823 866 014 182 460 620 8;
  • 81) 0.276 823 866 014 182 460 620 8 × 2 = 0 + 0.553 647 732 028 364 921 241 6;
  • 82) 0.553 647 732 028 364 921 241 6 × 2 = 1 + 0.107 295 464 056 729 842 483 2;
  • 83) 0.107 295 464 056 729 842 483 2 × 2 = 0 + 0.214 590 928 113 459 684 966 4;
  • 84) 0.214 590 928 113 459 684 966 4 × 2 = 0 + 0.429 181 856 226 919 369 932 8;
  • 85) 0.429 181 856 226 919 369 932 8 × 2 = 0 + 0.858 363 712 453 838 739 865 6;
  • 86) 0.858 363 712 453 838 739 865 6 × 2 = 1 + 0.716 727 424 907 677 479 731 2;
  • 87) 0.716 727 424 907 677 479 731 2 × 2 = 1 + 0.433 454 849 815 354 959 462 4;
  • 88) 0.433 454 849 815 354 959 462 4 × 2 = 0 + 0.866 909 699 630 709 918 924 8;
  • 89) 0.866 909 699 630 709 918 924 8 × 2 = 1 + 0.733 819 399 261 419 837 849 6;
  • 90) 0.733 819 399 261 419 837 849 6 × 2 = 1 + 0.467 638 798 522 839 675 699 2;
  • 91) 0.467 638 798 522 839 675 699 2 × 2 = 0 + 0.935 277 597 045 679 351 398 4;
  • 92) 0.935 277 597 045 679 351 398 4 × 2 = 1 + 0.870 555 194 091 358 702 796 8;
  • 93) 0.870 555 194 091 358 702 796 8 × 2 = 1 + 0.741 110 388 182 717 405 593 6;
  • 94) 0.741 110 388 182 717 405 593 6 × 2 = 1 + 0.482 220 776 365 434 811 187 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:


0.000 000 000 000 293 915 430 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1010 1101 0110 1011 1011 0011 0101 0100 0110 1101 11(2)

6. Positive number before normalization:

0.000 000 000 000 293 915 430 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1010 1101 0110 1011 1011 0011 0101 0100 0110 1101 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 42 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 293 915 430 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1010 1101 0110 1011 1011 0011 0101 0100 0110 1101 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1010 1101 0110 1011 1011 0011 0101 0100 0110 1101 11(2) × 20 =

1.0100 1010 1110 1011 0101 1010 1110 1100 1101 0101 0001 1011 0111(2) × 2-42


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): -42

Mantissa (not normalized):
1.0100 1010 1110 1011 0101 1010 1110 1100 1101 0101 0001 1011 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-42 + 2(11-1) - 1 =

(-42 + 1 023)(10) =

981(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;
  • 981 ÷ 2 = 490 + 1;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

981(10) =

011 1101 0101(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. 0100 1010 1110 1011 0101 1010 1110 1100 1101 0101 0001 1011 0111 =

0100 1010 1110 1011 0101 1010 1110 1100 1101 0101 0001 1011 0111


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 1101 0101

Mantissa (52 bits) =
0100 1010 1110 1011 0101 1010 1110 1100 1101 0101 0001 1011 0111


Decimal number -0.000 000 000 000 293 915 430 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0101 - 0100 1010 1110 1011 0101 1010 1110 1100 1101 0101 0001 1011 0111

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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