-0.000 000 000 000 176 557 623 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 623(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
-0.000 000 000 000 176 557 623(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:

|-0.000 000 000 000 176 557 623| = 0.000 000 000 000 176 557 623


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: 0.000 000 000 000 176 557 623.

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) 0.000 000 000 000 176 557 623 × 2 = 0 + 0.000 000 000 000 353 115 246;
  • 2) 0.000 000 000 000 353 115 246 × 2 = 0 + 0.000 000 000 000 706 230 492;
  • 3) 0.000 000 000 000 706 230 492 × 2 = 0 + 0.000 000 000 001 412 460 984;
  • 4) 0.000 000 000 001 412 460 984 × 2 = 0 + 0.000 000 000 002 824 921 968;
  • 5) 0.000 000 000 002 824 921 968 × 2 = 0 + 0.000 000 000 005 649 843 936;
  • 6) 0.000 000 000 005 649 843 936 × 2 = 0 + 0.000 000 000 011 299 687 872;
  • 7) 0.000 000 000 011 299 687 872 × 2 = 0 + 0.000 000 000 022 599 375 744;
  • 8) 0.000 000 000 022 599 375 744 × 2 = 0 + 0.000 000 000 045 198 751 488;
  • 9) 0.000 000 000 045 198 751 488 × 2 = 0 + 0.000 000 000 090 397 502 976;
  • 10) 0.000 000 000 090 397 502 976 × 2 = 0 + 0.000 000 000 180 795 005 952;
  • 11) 0.000 000 000 180 795 005 952 × 2 = 0 + 0.000 000 000 361 590 011 904;
  • 12) 0.000 000 000 361 590 011 904 × 2 = 0 + 0.000 000 000 723 180 023 808;
  • 13) 0.000 000 000 723 180 023 808 × 2 = 0 + 0.000 000 001 446 360 047 616;
  • 14) 0.000 000 001 446 360 047 616 × 2 = 0 + 0.000 000 002 892 720 095 232;
  • 15) 0.000 000 002 892 720 095 232 × 2 = 0 + 0.000 000 005 785 440 190 464;
  • 16) 0.000 000 005 785 440 190 464 × 2 = 0 + 0.000 000 011 570 880 380 928;
  • 17) 0.000 000 011 570 880 380 928 × 2 = 0 + 0.000 000 023 141 760 761 856;
  • 18) 0.000 000 023 141 760 761 856 × 2 = 0 + 0.000 000 046 283 521 523 712;
  • 19) 0.000 000 046 283 521 523 712 × 2 = 0 + 0.000 000 092 567 043 047 424;
  • 20) 0.000 000 092 567 043 047 424 × 2 = 0 + 0.000 000 185 134 086 094 848;
  • 21) 0.000 000 185 134 086 094 848 × 2 = 0 + 0.000 000 370 268 172 189 696;
  • 22) 0.000 000 370 268 172 189 696 × 2 = 0 + 0.000 000 740 536 344 379 392;
  • 23) 0.000 000 740 536 344 379 392 × 2 = 0 + 0.000 001 481 072 688 758 784;
  • 24) 0.000 001 481 072 688 758 784 × 2 = 0 + 0.000 002 962 145 377 517 568;
  • 25) 0.000 002 962 145 377 517 568 × 2 = 0 + 0.000 005 924 290 755 035 136;
  • 26) 0.000 005 924 290 755 035 136 × 2 = 0 + 0.000 011 848 581 510 070 272;
  • 27) 0.000 011 848 581 510 070 272 × 2 = 0 + 0.000 023 697 163 020 140 544;
  • 28) 0.000 023 697 163 020 140 544 × 2 = 0 + 0.000 047 394 326 040 281 088;
  • 29) 0.000 047 394 326 040 281 088 × 2 = 0 + 0.000 094 788 652 080 562 176;
  • 30) 0.000 094 788 652 080 562 176 × 2 = 0 + 0.000 189 577 304 161 124 352;
  • 31) 0.000 189 577 304 161 124 352 × 2 = 0 + 0.000 379 154 608 322 248 704;
  • 32) 0.000 379 154 608 322 248 704 × 2 = 0 + 0.000 758 309 216 644 497 408;
  • 33) 0.000 758 309 216 644 497 408 × 2 = 0 + 0.001 516 618 433 288 994 816;
  • 34) 0.001 516 618 433 288 994 816 × 2 = 0 + 0.003 033 236 866 577 989 632;
  • 35) 0.003 033 236 866 577 989 632 × 2 = 0 + 0.006 066 473 733 155 979 264;
  • 36) 0.006 066 473 733 155 979 264 × 2 = 0 + 0.012 132 947 466 311 958 528;
  • 37) 0.012 132 947 466 311 958 528 × 2 = 0 + 0.024 265 894 932 623 917 056;
  • 38) 0.024 265 894 932 623 917 056 × 2 = 0 + 0.048 531 789 865 247 834 112;
  • 39) 0.048 531 789 865 247 834 112 × 2 = 0 + 0.097 063 579 730 495 668 224;
  • 40) 0.097 063 579 730 495 668 224 × 2 = 0 + 0.194 127 159 460 991 336 448;
  • 41) 0.194 127 159 460 991 336 448 × 2 = 0 + 0.388 254 318 921 982 672 896;
  • 42) 0.388 254 318 921 982 672 896 × 2 = 0 + 0.776 508 637 843 965 345 792;
  • 43) 0.776 508 637 843 965 345 792 × 2 = 1 + 0.553 017 275 687 930 691 584;
  • 44) 0.553 017 275 687 930 691 584 × 2 = 1 + 0.106 034 551 375 861 383 168;
  • 45) 0.106 034 551 375 861 383 168 × 2 = 0 + 0.212 069 102 751 722 766 336;
  • 46) 0.212 069 102 751 722 766 336 × 2 = 0 + 0.424 138 205 503 445 532 672;
  • 47) 0.424 138 205 503 445 532 672 × 2 = 0 + 0.848 276 411 006 891 065 344;
  • 48) 0.848 276 411 006 891 065 344 × 2 = 1 + 0.696 552 822 013 782 130 688;
  • 49) 0.696 552 822 013 782 130 688 × 2 = 1 + 0.393 105 644 027 564 261 376;
  • 50) 0.393 105 644 027 564 261 376 × 2 = 0 + 0.786 211 288 055 128 522 752;
  • 51) 0.786 211 288 055 128 522 752 × 2 = 1 + 0.572 422 576 110 257 045 504;
  • 52) 0.572 422 576 110 257 045 504 × 2 = 1 + 0.144 845 152 220 514 091 008;
  • 53) 0.144 845 152 220 514 091 008 × 2 = 0 + 0.289 690 304 441 028 182 016;
  • 54) 0.289 690 304 441 028 182 016 × 2 = 0 + 0.579 380 608 882 056 364 032;
  • 55) 0.579 380 608 882 056 364 032 × 2 = 1 + 0.158 761 217 764 112 728 064;
  • 56) 0.158 761 217 764 112 728 064 × 2 = 0 + 0.317 522 435 528 225 456 128;
  • 57) 0.317 522 435 528 225 456 128 × 2 = 0 + 0.635 044 871 056 450 912 256;
  • 58) 0.635 044 871 056 450 912 256 × 2 = 1 + 0.270 089 742 112 901 824 512;
  • 59) 0.270 089 742 112 901 824 512 × 2 = 0 + 0.540 179 484 225 803 649 024;
  • 60) 0.540 179 484 225 803 649 024 × 2 = 1 + 0.080 358 968 451 607 298 048;
  • 61) 0.080 358 968 451 607 298 048 × 2 = 0 + 0.160 717 936 903 214 596 096;
  • 62) 0.160 717 936 903 214 596 096 × 2 = 0 + 0.321 435 873 806 429 192 192;
  • 63) 0.321 435 873 806 429 192 192 × 2 = 0 + 0.642 871 747 612 858 384 384;
  • 64) 0.642 871 747 612 858 384 384 × 2 = 1 + 0.285 743 495 225 716 768 768;
  • 65) 0.285 743 495 225 716 768 768 × 2 = 0 + 0.571 486 990 451 433 537 536;
  • 66) 0.571 486 990 451 433 537 536 × 2 = 1 + 0.142 973 980 902 867 075 072;
  • 67) 0.142 973 980 902 867 075 072 × 2 = 0 + 0.285 947 961 805 734 150 144;
  • 68) 0.285 947 961 805 734 150 144 × 2 = 0 + 0.571 895 923 611 468 300 288;
  • 69) 0.571 895 923 611 468 300 288 × 2 = 1 + 0.143 791 847 222 936 600 576;
  • 70) 0.143 791 847 222 936 600 576 × 2 = 0 + 0.287 583 694 445 873 201 152;
  • 71) 0.287 583 694 445 873 201 152 × 2 = 0 + 0.575 167 388 891 746 402 304;
  • 72) 0.575 167 388 891 746 402 304 × 2 = 1 + 0.150 334 777 783 492 804 608;
  • 73) 0.150 334 777 783 492 804 608 × 2 = 0 + 0.300 669 555 566 985 609 216;
  • 74) 0.300 669 555 566 985 609 216 × 2 = 0 + 0.601 339 111 133 971 218 432;
  • 75) 0.601 339 111 133 971 218 432 × 2 = 1 + 0.202 678 222 267 942 436 864;
  • 76) 0.202 678 222 267 942 436 864 × 2 = 0 + 0.405 356 444 535 884 873 728;
  • 77) 0.405 356 444 535 884 873 728 × 2 = 0 + 0.810 712 889 071 769 747 456;
  • 78) 0.810 712 889 071 769 747 456 × 2 = 1 + 0.621 425 778 143 539 494 912;
  • 79) 0.621 425 778 143 539 494 912 × 2 = 1 + 0.242 851 556 287 078 989 824;
  • 80) 0.242 851 556 287 078 989 824 × 2 = 0 + 0.485 703 112 574 157 979 648;
  • 81) 0.485 703 112 574 157 979 648 × 2 = 0 + 0.971 406 225 148 315 959 296;
  • 82) 0.971 406 225 148 315 959 296 × 2 = 1 + 0.942 812 450 296 631 918 592;
  • 83) 0.942 812 450 296 631 918 592 × 2 = 1 + 0.885 624 900 593 263 837 184;
  • 84) 0.885 624 900 593 263 837 184 × 2 = 1 + 0.771 249 801 186 527 674 368;
  • 85) 0.771 249 801 186 527 674 368 × 2 = 1 + 0.542 499 602 373 055 348 736;
  • 86) 0.542 499 602 373 055 348 736 × 2 = 1 + 0.084 999 204 746 110 697 472;
  • 87) 0.084 999 204 746 110 697 472 × 2 = 0 + 0.169 998 409 492 221 394 944;
  • 88) 0.169 998 409 492 221 394 944 × 2 = 0 + 0.339 996 818 984 442 789 888;
  • 89) 0.339 996 818 984 442 789 888 × 2 = 0 + 0.679 993 637 968 885 579 776;
  • 90) 0.679 993 637 968 885 579 776 × 2 = 1 + 0.359 987 275 937 771 159 552;
  • 91) 0.359 987 275 937 771 159 552 × 2 = 0 + 0.719 974 551 875 542 319 104;
  • 92) 0.719 974 551 875 542 319 104 × 2 = 1 + 0.439 949 103 751 084 638 208;
  • 93) 0.439 949 103 751 084 638 208 × 2 = 0 + 0.879 898 207 502 169 276 416;
  • 94) 0.879 898 207 502 169 276 416 × 2 = 1 + 0.759 796 415 004 338 552 832;
  • 95) 0.759 796 415 004 338 552 832 × 2 = 1 + 0.519 592 830 008 677 105 664;

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:


0.000 000 000 000 176 557 623(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 0100 1001 0010 0110 0111 1100 0101 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 623(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 0100 1001 0010 0110 0111 1100 0101 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 557 623(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 0100 1001 0010 0110 0111 1100 0101 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 0100 1001 0010 0110 0111 1100 0101 011(2) × 20 =

1.1000 1101 1001 0010 1000 1010 0100 1001 0011 0011 1110 0010 1011(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0010 1000 1010 0100 1001 0011 0011 1110 0010 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 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) =

980(10) =

011 1101 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. 1000 1101 1001 0010 1000 1010 0100 1001 0011 0011 1110 0010 1011 =

1000 1101 1001 0010 1000 1010 0100 1001 0011 0011 1110 0010 1011


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 0100

Mantissa (52 bits) =
1000 1101 1001 0010 1000 1010 0100 1001 0011 0011 1110 0010 1011


Decimal number -0.000 000 000 000 176 557 623 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1000 1010 0100 1001 0011 0011 1110 0010 1011

Share

» Convert decimal -0.000 000 000 000 176 557 607 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 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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