-0.000 000 000 000 293 915 427 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 293 915 427 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
-0.000 000 000 000 293 915 427 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:

|-0.000 000 000 000 293 915 427 8| = 0.000 000 000 000 293 915 427 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: 0.000 000 000 000 293 915 427 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) 0.000 000 000 000 293 915 427 8 × 2 = 0 + 0.000 000 000 000 587 830 855 6;
  • 2) 0.000 000 000 000 587 830 855 6 × 2 = 0 + 0.000 000 000 001 175 661 711 2;
  • 3) 0.000 000 000 001 175 661 711 2 × 2 = 0 + 0.000 000 000 002 351 323 422 4;
  • 4) 0.000 000 000 002 351 323 422 4 × 2 = 0 + 0.000 000 000 004 702 646 844 8;
  • 5) 0.000 000 000 004 702 646 844 8 × 2 = 0 + 0.000 000 000 009 405 293 689 6;
  • 6) 0.000 000 000 009 405 293 689 6 × 2 = 0 + 0.000 000 000 018 810 587 379 2;
  • 7) 0.000 000 000 018 810 587 379 2 × 2 = 0 + 0.000 000 000 037 621 174 758 4;
  • 8) 0.000 000 000 037 621 174 758 4 × 2 = 0 + 0.000 000 000 075 242 349 516 8;
  • 9) 0.000 000 000 075 242 349 516 8 × 2 = 0 + 0.000 000 000 150 484 699 033 6;
  • 10) 0.000 000 000 150 484 699 033 6 × 2 = 0 + 0.000 000 000 300 969 398 067 2;
  • 11) 0.000 000 000 300 969 398 067 2 × 2 = 0 + 0.000 000 000 601 938 796 134 4;
  • 12) 0.000 000 000 601 938 796 134 4 × 2 = 0 + 0.000 000 001 203 877 592 268 8;
  • 13) 0.000 000 001 203 877 592 268 8 × 2 = 0 + 0.000 000 002 407 755 184 537 6;
  • 14) 0.000 000 002 407 755 184 537 6 × 2 = 0 + 0.000 000 004 815 510 369 075 2;
  • 15) 0.000 000 004 815 510 369 075 2 × 2 = 0 + 0.000 000 009 631 020 738 150 4;
  • 16) 0.000 000 009 631 020 738 150 4 × 2 = 0 + 0.000 000 019 262 041 476 300 8;
  • 17) 0.000 000 019 262 041 476 300 8 × 2 = 0 + 0.000 000 038 524 082 952 601 6;
  • 18) 0.000 000 038 524 082 952 601 6 × 2 = 0 + 0.000 000 077 048 165 905 203 2;
  • 19) 0.000 000 077 048 165 905 203 2 × 2 = 0 + 0.000 000 154 096 331 810 406 4;
  • 20) 0.000 000 154 096 331 810 406 4 × 2 = 0 + 0.000 000 308 192 663 620 812 8;
  • 21) 0.000 000 308 192 663 620 812 8 × 2 = 0 + 0.000 000 616 385 327 241 625 6;
  • 22) 0.000 000 616 385 327 241 625 6 × 2 = 0 + 0.000 001 232 770 654 483 251 2;
  • 23) 0.000 001 232 770 654 483 251 2 × 2 = 0 + 0.000 002 465 541 308 966 502 4;
  • 24) 0.000 002 465 541 308 966 502 4 × 2 = 0 + 0.000 004 931 082 617 933 004 8;
  • 25) 0.000 004 931 082 617 933 004 8 × 2 = 0 + 0.000 009 862 165 235 866 009 6;
  • 26) 0.000 009 862 165 235 866 009 6 × 2 = 0 + 0.000 019 724 330 471 732 019 2;
  • 27) 0.000 019 724 330 471 732 019 2 × 2 = 0 + 0.000 039 448 660 943 464 038 4;
  • 28) 0.000 039 448 660 943 464 038 4 × 2 = 0 + 0.000 078 897 321 886 928 076 8;
  • 29) 0.000 078 897 321 886 928 076 8 × 2 = 0 + 0.000 157 794 643 773 856 153 6;
  • 30) 0.000 157 794 643 773 856 153 6 × 2 = 0 + 0.000 315 589 287 547 712 307 2;
  • 31) 0.000 315 589 287 547 712 307 2 × 2 = 0 + 0.000 631 178 575 095 424 614 4;
  • 32) 0.000 631 178 575 095 424 614 4 × 2 = 0 + 0.001 262 357 150 190 849 228 8;
  • 33) 0.001 262 357 150 190 849 228 8 × 2 = 0 + 0.002 524 714 300 381 698 457 6;
  • 34) 0.002 524 714 300 381 698 457 6 × 2 = 0 + 0.005 049 428 600 763 396 915 2;
  • 35) 0.005 049 428 600 763 396 915 2 × 2 = 0 + 0.010 098 857 201 526 793 830 4;
  • 36) 0.010 098 857 201 526 793 830 4 × 2 = 0 + 0.020 197 714 403 053 587 660 8;
  • 37) 0.020 197 714 403 053 587 660 8 × 2 = 0 + 0.040 395 428 806 107 175 321 6;
  • 38) 0.040 395 428 806 107 175 321 6 × 2 = 0 + 0.080 790 857 612 214 350 643 2;
  • 39) 0.080 790 857 612 214 350 643 2 × 2 = 0 + 0.161 581 715 224 428 701 286 4;
  • 40) 0.161 581 715 224 428 701 286 4 × 2 = 0 + 0.323 163 430 448 857 402 572 8;
  • 41) 0.323 163 430 448 857 402 572 8 × 2 = 0 + 0.646 326 860 897 714 805 145 6;
  • 42) 0.646 326 860 897 714 805 145 6 × 2 = 1 + 0.292 653 721 795 429 610 291 2;
  • 43) 0.292 653 721 795 429 610 291 2 × 2 = 0 + 0.585 307 443 590 859 220 582 4;
  • 44) 0.585 307 443 590 859 220 582 4 × 2 = 1 + 0.170 614 887 181 718 441 164 8;
  • 45) 0.170 614 887 181 718 441 164 8 × 2 = 0 + 0.341 229 774 363 436 882 329 6;
  • 46) 0.341 229 774 363 436 882 329 6 × 2 = 0 + 0.682 459 548 726 873 764 659 2;
  • 47) 0.682 459 548 726 873 764 659 2 × 2 = 1 + 0.364 919 097 453 747 529 318 4;
  • 48) 0.364 919 097 453 747 529 318 4 × 2 = 0 + 0.729 838 194 907 495 058 636 8;
  • 49) 0.729 838 194 907 495 058 636 8 × 2 = 1 + 0.459 676 389 814 990 117 273 6;
  • 50) 0.459 676 389 814 990 117 273 6 × 2 = 0 + 0.919 352 779 629 980 234 547 2;
  • 51) 0.919 352 779 629 980 234 547 2 × 2 = 1 + 0.838 705 559 259 960 469 094 4;
  • 52) 0.838 705 559 259 960 469 094 4 × 2 = 1 + 0.677 411 118 519 920 938 188 8;
  • 53) 0.677 411 118 519 920 938 188 8 × 2 = 1 + 0.354 822 237 039 841 876 377 6;
  • 54) 0.354 822 237 039 841 876 377 6 × 2 = 0 + 0.709 644 474 079 683 752 755 2;
  • 55) 0.709 644 474 079 683 752 755 2 × 2 = 1 + 0.419 288 948 159 367 505 510 4;
  • 56) 0.419 288 948 159 367 505 510 4 × 2 = 0 + 0.838 577 896 318 735 011 020 8;
  • 57) 0.838 577 896 318 735 011 020 8 × 2 = 1 + 0.677 155 792 637 470 022 041 6;
  • 58) 0.677 155 792 637 470 022 041 6 × 2 = 1 + 0.354 311 585 274 940 044 083 2;
  • 59) 0.354 311 585 274 940 044 083 2 × 2 = 0 + 0.708 623 170 549 880 088 166 4;
  • 60) 0.708 623 170 549 880 088 166 4 × 2 = 1 + 0.417 246 341 099 760 176 332 8;
  • 61) 0.417 246 341 099 760 176 332 8 × 2 = 0 + 0.834 492 682 199 520 352 665 6;
  • 62) 0.834 492 682 199 520 352 665 6 × 2 = 1 + 0.668 985 364 399 040 705 331 2;
  • 63) 0.668 985 364 399 040 705 331 2 × 2 = 1 + 0.337 970 728 798 081 410 662 4;
  • 64) 0.337 970 728 798 081 410 662 4 × 2 = 0 + 0.675 941 457 596 162 821 324 8;
  • 65) 0.675 941 457 596 162 821 324 8 × 2 = 1 + 0.351 882 915 192 325 642 649 6;
  • 66) 0.351 882 915 192 325 642 649 6 × 2 = 0 + 0.703 765 830 384 651 285 299 2;
  • 67) 0.703 765 830 384 651 285 299 2 × 2 = 1 + 0.407 531 660 769 302 570 598 4;
  • 68) 0.407 531 660 769 302 570 598 4 × 2 = 0 + 0.815 063 321 538 605 141 196 8;
  • 69) 0.815 063 321 538 605 141 196 8 × 2 = 1 + 0.630 126 643 077 210 282 393 6;
  • 70) 0.630 126 643 077 210 282 393 6 × 2 = 1 + 0.260 253 286 154 420 564 787 2;
  • 71) 0.260 253 286 154 420 564 787 2 × 2 = 0 + 0.520 506 572 308 841 129 574 4;
  • 72) 0.520 506 572 308 841 129 574 4 × 2 = 1 + 0.041 013 144 617 682 259 148 8;
  • 73) 0.041 013 144 617 682 259 148 8 × 2 = 0 + 0.082 026 289 235 364 518 297 6;
  • 74) 0.082 026 289 235 364 518 297 6 × 2 = 0 + 0.164 052 578 470 729 036 595 2;
  • 75) 0.164 052 578 470 729 036 595 2 × 2 = 0 + 0.328 105 156 941 458 073 190 4;
  • 76) 0.328 105 156 941 458 073 190 4 × 2 = 0 + 0.656 210 313 882 916 146 380 8;
  • 77) 0.656 210 313 882 916 146 380 8 × 2 = 1 + 0.312 420 627 765 832 292 761 6;
  • 78) 0.312 420 627 765 832 292 761 6 × 2 = 0 + 0.624 841 255 531 664 585 523 2;
  • 79) 0.624 841 255 531 664 585 523 2 × 2 = 1 + 0.249 682 511 063 329 171 046 4;
  • 80) 0.249 682 511 063 329 171 046 4 × 2 = 0 + 0.499 365 022 126 658 342 092 8;
  • 81) 0.499 365 022 126 658 342 092 8 × 2 = 0 + 0.998 730 044 253 316 684 185 6;
  • 82) 0.998 730 044 253 316 684 185 6 × 2 = 1 + 0.997 460 088 506 633 368 371 2;
  • 83) 0.997 460 088 506 633 368 371 2 × 2 = 1 + 0.994 920 177 013 266 736 742 4;
  • 84) 0.994 920 177 013 266 736 742 4 × 2 = 1 + 0.989 840 354 026 533 473 484 8;
  • 85) 0.989 840 354 026 533 473 484 8 × 2 = 1 + 0.979 680 708 053 066 946 969 6;
  • 86) 0.979 680 708 053 066 946 969 6 × 2 = 1 + 0.959 361 416 106 133 893 939 2;
  • 87) 0.959 361 416 106 133 893 939 2 × 2 = 1 + 0.918 722 832 212 267 787 878 4;
  • 88) 0.918 722 832 212 267 787 878 4 × 2 = 1 + 0.837 445 664 424 535 575 756 8;
  • 89) 0.837 445 664 424 535 575 756 8 × 2 = 1 + 0.674 891 328 849 071 151 513 6;
  • 90) 0.674 891 328 849 071 151 513 6 × 2 = 1 + 0.349 782 657 698 142 303 027 2;
  • 91) 0.349 782 657 698 142 303 027 2 × 2 = 0 + 0.699 565 315 396 284 606 054 4;
  • 92) 0.699 565 315 396 284 606 054 4 × 2 = 1 + 0.399 130 630 792 569 212 108 8;
  • 93) 0.399 130 630 792 569 212 108 8 × 2 = 0 + 0.798 261 261 585 138 424 217 6;
  • 94) 0.798 261 261 585 138 424 217 6 × 2 = 1 + 0.596 522 523 170 276 848 435 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:


0.000 000 000 000 293 915 427 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1010 1101 0110 1010 1101 0000 1010 0111 1111 1101 01(2)

6. Positive number before normalization:

0.000 000 000 000 293 915 427 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1010 1101 0110 1010 1101 0000 1010 0111 1111 1101 01(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:


0.000 000 000 000 293 915 427 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1010 1101 0110 1010 1101 0000 1010 0111 1111 1101 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0010 1011 1010 1101 0110 1010 1101 0000 1010 0111 1111 1101 01(2) × 20 =

1.0100 1010 1110 1011 0101 1010 1011 0100 0010 1001 1111 1111 0101(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 1011 0100 0010 1001 1111 1111 0101


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 1011 0100 0010 1001 1111 1111 0101 =

0100 1010 1110 1011 0101 1010 1011 0100 0010 1001 1111 1111 0101


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 1011 0100 0010 1001 1111 1111 0101


Decimal number -0.000 000 000 000 293 915 427 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0101 - 0100 1010 1110 1011 0101 1010 1011 0100 0010 1001 1111 1111 0101

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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