-0.000 000 000 000 176 557 65 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 65(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 65(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 65| = 0.000 000 000 000 176 557 65


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 65.

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 65 × 2 = 0 + 0.000 000 000 000 353 115 3;
  • 2) 0.000 000 000 000 353 115 3 × 2 = 0 + 0.000 000 000 000 706 230 6;
  • 3) 0.000 000 000 000 706 230 6 × 2 = 0 + 0.000 000 000 001 412 461 2;
  • 4) 0.000 000 000 001 412 461 2 × 2 = 0 + 0.000 000 000 002 824 922 4;
  • 5) 0.000 000 000 002 824 922 4 × 2 = 0 + 0.000 000 000 005 649 844 8;
  • 6) 0.000 000 000 005 649 844 8 × 2 = 0 + 0.000 000 000 011 299 689 6;
  • 7) 0.000 000 000 011 299 689 6 × 2 = 0 + 0.000 000 000 022 599 379 2;
  • 8) 0.000 000 000 022 599 379 2 × 2 = 0 + 0.000 000 000 045 198 758 4;
  • 9) 0.000 000 000 045 198 758 4 × 2 = 0 + 0.000 000 000 090 397 516 8;
  • 10) 0.000 000 000 090 397 516 8 × 2 = 0 + 0.000 000 000 180 795 033 6;
  • 11) 0.000 000 000 180 795 033 6 × 2 = 0 + 0.000 000 000 361 590 067 2;
  • 12) 0.000 000 000 361 590 067 2 × 2 = 0 + 0.000 000 000 723 180 134 4;
  • 13) 0.000 000 000 723 180 134 4 × 2 = 0 + 0.000 000 001 446 360 268 8;
  • 14) 0.000 000 001 446 360 268 8 × 2 = 0 + 0.000 000 002 892 720 537 6;
  • 15) 0.000 000 002 892 720 537 6 × 2 = 0 + 0.000 000 005 785 441 075 2;
  • 16) 0.000 000 005 785 441 075 2 × 2 = 0 + 0.000 000 011 570 882 150 4;
  • 17) 0.000 000 011 570 882 150 4 × 2 = 0 + 0.000 000 023 141 764 300 8;
  • 18) 0.000 000 023 141 764 300 8 × 2 = 0 + 0.000 000 046 283 528 601 6;
  • 19) 0.000 000 046 283 528 601 6 × 2 = 0 + 0.000 000 092 567 057 203 2;
  • 20) 0.000 000 092 567 057 203 2 × 2 = 0 + 0.000 000 185 134 114 406 4;
  • 21) 0.000 000 185 134 114 406 4 × 2 = 0 + 0.000 000 370 268 228 812 8;
  • 22) 0.000 000 370 268 228 812 8 × 2 = 0 + 0.000 000 740 536 457 625 6;
  • 23) 0.000 000 740 536 457 625 6 × 2 = 0 + 0.000 001 481 072 915 251 2;
  • 24) 0.000 001 481 072 915 251 2 × 2 = 0 + 0.000 002 962 145 830 502 4;
  • 25) 0.000 002 962 145 830 502 4 × 2 = 0 + 0.000 005 924 291 661 004 8;
  • 26) 0.000 005 924 291 661 004 8 × 2 = 0 + 0.000 011 848 583 322 009 6;
  • 27) 0.000 011 848 583 322 009 6 × 2 = 0 + 0.000 023 697 166 644 019 2;
  • 28) 0.000 023 697 166 644 019 2 × 2 = 0 + 0.000 047 394 333 288 038 4;
  • 29) 0.000 047 394 333 288 038 4 × 2 = 0 + 0.000 094 788 666 576 076 8;
  • 30) 0.000 094 788 666 576 076 8 × 2 = 0 + 0.000 189 577 333 152 153 6;
  • 31) 0.000 189 577 333 152 153 6 × 2 = 0 + 0.000 379 154 666 304 307 2;
  • 32) 0.000 379 154 666 304 307 2 × 2 = 0 + 0.000 758 309 332 608 614 4;
  • 33) 0.000 758 309 332 608 614 4 × 2 = 0 + 0.001 516 618 665 217 228 8;
  • 34) 0.001 516 618 665 217 228 8 × 2 = 0 + 0.003 033 237 330 434 457 6;
  • 35) 0.003 033 237 330 434 457 6 × 2 = 0 + 0.006 066 474 660 868 915 2;
  • 36) 0.006 066 474 660 868 915 2 × 2 = 0 + 0.012 132 949 321 737 830 4;
  • 37) 0.012 132 949 321 737 830 4 × 2 = 0 + 0.024 265 898 643 475 660 8;
  • 38) 0.024 265 898 643 475 660 8 × 2 = 0 + 0.048 531 797 286 951 321 6;
  • 39) 0.048 531 797 286 951 321 6 × 2 = 0 + 0.097 063 594 573 902 643 2;
  • 40) 0.097 063 594 573 902 643 2 × 2 = 0 + 0.194 127 189 147 805 286 4;
  • 41) 0.194 127 189 147 805 286 4 × 2 = 0 + 0.388 254 378 295 610 572 8;
  • 42) 0.388 254 378 295 610 572 8 × 2 = 0 + 0.776 508 756 591 221 145 6;
  • 43) 0.776 508 756 591 221 145 6 × 2 = 1 + 0.553 017 513 182 442 291 2;
  • 44) 0.553 017 513 182 442 291 2 × 2 = 1 + 0.106 035 026 364 884 582 4;
  • 45) 0.106 035 026 364 884 582 4 × 2 = 0 + 0.212 070 052 729 769 164 8;
  • 46) 0.212 070 052 729 769 164 8 × 2 = 0 + 0.424 140 105 459 538 329 6;
  • 47) 0.424 140 105 459 538 329 6 × 2 = 0 + 0.848 280 210 919 076 659 2;
  • 48) 0.848 280 210 919 076 659 2 × 2 = 1 + 0.696 560 421 838 153 318 4;
  • 49) 0.696 560 421 838 153 318 4 × 2 = 1 + 0.393 120 843 676 306 636 8;
  • 50) 0.393 120 843 676 306 636 8 × 2 = 0 + 0.786 241 687 352 613 273 6;
  • 51) 0.786 241 687 352 613 273 6 × 2 = 1 + 0.572 483 374 705 226 547 2;
  • 52) 0.572 483 374 705 226 547 2 × 2 = 1 + 0.144 966 749 410 453 094 4;
  • 53) 0.144 966 749 410 453 094 4 × 2 = 0 + 0.289 933 498 820 906 188 8;
  • 54) 0.289 933 498 820 906 188 8 × 2 = 0 + 0.579 866 997 641 812 377 6;
  • 55) 0.579 866 997 641 812 377 6 × 2 = 1 + 0.159 733 995 283 624 755 2;
  • 56) 0.159 733 995 283 624 755 2 × 2 = 0 + 0.319 467 990 567 249 510 4;
  • 57) 0.319 467 990 567 249 510 4 × 2 = 0 + 0.638 935 981 134 499 020 8;
  • 58) 0.638 935 981 134 499 020 8 × 2 = 1 + 0.277 871 962 268 998 041 6;
  • 59) 0.277 871 962 268 998 041 6 × 2 = 0 + 0.555 743 924 537 996 083 2;
  • 60) 0.555 743 924 537 996 083 2 × 2 = 1 + 0.111 487 849 075 992 166 4;
  • 61) 0.111 487 849 075 992 166 4 × 2 = 0 + 0.222 975 698 151 984 332 8;
  • 62) 0.222 975 698 151 984 332 8 × 2 = 0 + 0.445 951 396 303 968 665 6;
  • 63) 0.445 951 396 303 968 665 6 × 2 = 0 + 0.891 902 792 607 937 331 2;
  • 64) 0.891 902 792 607 937 331 2 × 2 = 1 + 0.783 805 585 215 874 662 4;
  • 65) 0.783 805 585 215 874 662 4 × 2 = 1 + 0.567 611 170 431 749 324 8;
  • 66) 0.567 611 170 431 749 324 8 × 2 = 1 + 0.135 222 340 863 498 649 6;
  • 67) 0.135 222 340 863 498 649 6 × 2 = 0 + 0.270 444 681 726 997 299 2;
  • 68) 0.270 444 681 726 997 299 2 × 2 = 0 + 0.540 889 363 453 994 598 4;
  • 69) 0.540 889 363 453 994 598 4 × 2 = 1 + 0.081 778 726 907 989 196 8;
  • 70) 0.081 778 726 907 989 196 8 × 2 = 0 + 0.163 557 453 815 978 393 6;
  • 71) 0.163 557 453 815 978 393 6 × 2 = 0 + 0.327 114 907 631 956 787 2;
  • 72) 0.327 114 907 631 956 787 2 × 2 = 0 + 0.654 229 815 263 913 574 4;
  • 73) 0.654 229 815 263 913 574 4 × 2 = 1 + 0.308 459 630 527 827 148 8;
  • 74) 0.308 459 630 527 827 148 8 × 2 = 0 + 0.616 919 261 055 654 297 6;
  • 75) 0.616 919 261 055 654 297 6 × 2 = 1 + 0.233 838 522 111 308 595 2;
  • 76) 0.233 838 522 111 308 595 2 × 2 = 0 + 0.467 677 044 222 617 190 4;
  • 77) 0.467 677 044 222 617 190 4 × 2 = 0 + 0.935 354 088 445 234 380 8;
  • 78) 0.935 354 088 445 234 380 8 × 2 = 1 + 0.870 708 176 890 468 761 6;
  • 79) 0.870 708 176 890 468 761 6 × 2 = 1 + 0.741 416 353 780 937 523 2;
  • 80) 0.741 416 353 780 937 523 2 × 2 = 1 + 0.482 832 707 561 875 046 4;
  • 81) 0.482 832 707 561 875 046 4 × 2 = 0 + 0.965 665 415 123 750 092 8;
  • 82) 0.965 665 415 123 750 092 8 × 2 = 1 + 0.931 330 830 247 500 185 6;
  • 83) 0.931 330 830 247 500 185 6 × 2 = 1 + 0.862 661 660 495 000 371 2;
  • 84) 0.862 661 660 495 000 371 2 × 2 = 1 + 0.725 323 320 990 000 742 4;
  • 85) 0.725 323 320 990 000 742 4 × 2 = 1 + 0.450 646 641 980 001 484 8;
  • 86) 0.450 646 641 980 001 484 8 × 2 = 0 + 0.901 293 283 960 002 969 6;
  • 87) 0.901 293 283 960 002 969 6 × 2 = 1 + 0.802 586 567 920 005 939 2;
  • 88) 0.802 586 567 920 005 939 2 × 2 = 1 + 0.605 173 135 840 011 878 4;
  • 89) 0.605 173 135 840 011 878 4 × 2 = 1 + 0.210 346 271 680 023 756 8;
  • 90) 0.210 346 271 680 023 756 8 × 2 = 0 + 0.420 692 543 360 047 513 6;
  • 91) 0.420 692 543 360 047 513 6 × 2 = 0 + 0.841 385 086 720 095 027 2;
  • 92) 0.841 385 086 720 095 027 2 × 2 = 1 + 0.682 770 173 440 190 054 4;
  • 93) 0.682 770 173 440 190 054 4 × 2 = 1 + 0.365 540 346 880 380 108 8;
  • 94) 0.365 540 346 880 380 108 8 × 2 = 0 + 0.731 080 693 760 760 217 6;
  • 95) 0.731 080 693 760 760 217 6 × 2 = 1 + 0.462 161 387 521 520 435 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:


0.000 000 000 000 176 557 65(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 1100 1000 1010 0111 0111 1011 1001 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 65(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 1100 1000 1010 0111 0111 1011 1001 101(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 65(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 1100 1000 1010 0111 0111 1011 1001 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 1100 1000 1010 0111 0111 1011 1001 101(2) × 20 =

1.1000 1101 1001 0010 1000 1110 0100 0101 0011 1011 1101 1100 1101(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 1110 0100 0101 0011 1011 1101 1100 1101


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 1110 0100 0101 0011 1011 1101 1100 1101 =

1000 1101 1001 0010 1000 1110 0100 0101 0011 1011 1101 1100 1101


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 1110 0100 0101 0011 1011 1101 1100 1101


Decimal number -0.000 000 000 000 176 557 65 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1000 1110 0100 0101 0011 1011 1101 1100 1101

Share

» Convert decimal -0.000 000 000 000 176 557 58 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: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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