-0.000 000 000 000 176 557 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 9(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 9(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 9| = 0.000 000 000 000 176 557 9


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 9.

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 9 × 2 = 0 + 0.000 000 000 000 353 115 8;
  • 2) 0.000 000 000 000 353 115 8 × 2 = 0 + 0.000 000 000 000 706 231 6;
  • 3) 0.000 000 000 000 706 231 6 × 2 = 0 + 0.000 000 000 001 412 463 2;
  • 4) 0.000 000 000 001 412 463 2 × 2 = 0 + 0.000 000 000 002 824 926 4;
  • 5) 0.000 000 000 002 824 926 4 × 2 = 0 + 0.000 000 000 005 649 852 8;
  • 6) 0.000 000 000 005 649 852 8 × 2 = 0 + 0.000 000 000 011 299 705 6;
  • 7) 0.000 000 000 011 299 705 6 × 2 = 0 + 0.000 000 000 022 599 411 2;
  • 8) 0.000 000 000 022 599 411 2 × 2 = 0 + 0.000 000 000 045 198 822 4;
  • 9) 0.000 000 000 045 198 822 4 × 2 = 0 + 0.000 000 000 090 397 644 8;
  • 10) 0.000 000 000 090 397 644 8 × 2 = 0 + 0.000 000 000 180 795 289 6;
  • 11) 0.000 000 000 180 795 289 6 × 2 = 0 + 0.000 000 000 361 590 579 2;
  • 12) 0.000 000 000 361 590 579 2 × 2 = 0 + 0.000 000 000 723 181 158 4;
  • 13) 0.000 000 000 723 181 158 4 × 2 = 0 + 0.000 000 001 446 362 316 8;
  • 14) 0.000 000 001 446 362 316 8 × 2 = 0 + 0.000 000 002 892 724 633 6;
  • 15) 0.000 000 002 892 724 633 6 × 2 = 0 + 0.000 000 005 785 449 267 2;
  • 16) 0.000 000 005 785 449 267 2 × 2 = 0 + 0.000 000 011 570 898 534 4;
  • 17) 0.000 000 011 570 898 534 4 × 2 = 0 + 0.000 000 023 141 797 068 8;
  • 18) 0.000 000 023 141 797 068 8 × 2 = 0 + 0.000 000 046 283 594 137 6;
  • 19) 0.000 000 046 283 594 137 6 × 2 = 0 + 0.000 000 092 567 188 275 2;
  • 20) 0.000 000 092 567 188 275 2 × 2 = 0 + 0.000 000 185 134 376 550 4;
  • 21) 0.000 000 185 134 376 550 4 × 2 = 0 + 0.000 000 370 268 753 100 8;
  • 22) 0.000 000 370 268 753 100 8 × 2 = 0 + 0.000 000 740 537 506 201 6;
  • 23) 0.000 000 740 537 506 201 6 × 2 = 0 + 0.000 001 481 075 012 403 2;
  • 24) 0.000 001 481 075 012 403 2 × 2 = 0 + 0.000 002 962 150 024 806 4;
  • 25) 0.000 002 962 150 024 806 4 × 2 = 0 + 0.000 005 924 300 049 612 8;
  • 26) 0.000 005 924 300 049 612 8 × 2 = 0 + 0.000 011 848 600 099 225 6;
  • 27) 0.000 011 848 600 099 225 6 × 2 = 0 + 0.000 023 697 200 198 451 2;
  • 28) 0.000 023 697 200 198 451 2 × 2 = 0 + 0.000 047 394 400 396 902 4;
  • 29) 0.000 047 394 400 396 902 4 × 2 = 0 + 0.000 094 788 800 793 804 8;
  • 30) 0.000 094 788 800 793 804 8 × 2 = 0 + 0.000 189 577 601 587 609 6;
  • 31) 0.000 189 577 601 587 609 6 × 2 = 0 + 0.000 379 155 203 175 219 2;
  • 32) 0.000 379 155 203 175 219 2 × 2 = 0 + 0.000 758 310 406 350 438 4;
  • 33) 0.000 758 310 406 350 438 4 × 2 = 0 + 0.001 516 620 812 700 876 8;
  • 34) 0.001 516 620 812 700 876 8 × 2 = 0 + 0.003 033 241 625 401 753 6;
  • 35) 0.003 033 241 625 401 753 6 × 2 = 0 + 0.006 066 483 250 803 507 2;
  • 36) 0.006 066 483 250 803 507 2 × 2 = 0 + 0.012 132 966 501 607 014 4;
  • 37) 0.012 132 966 501 607 014 4 × 2 = 0 + 0.024 265 933 003 214 028 8;
  • 38) 0.024 265 933 003 214 028 8 × 2 = 0 + 0.048 531 866 006 428 057 6;
  • 39) 0.048 531 866 006 428 057 6 × 2 = 0 + 0.097 063 732 012 856 115 2;
  • 40) 0.097 063 732 012 856 115 2 × 2 = 0 + 0.194 127 464 025 712 230 4;
  • 41) 0.194 127 464 025 712 230 4 × 2 = 0 + 0.388 254 928 051 424 460 8;
  • 42) 0.388 254 928 051 424 460 8 × 2 = 0 + 0.776 509 856 102 848 921 6;
  • 43) 0.776 509 856 102 848 921 6 × 2 = 1 + 0.553 019 712 205 697 843 2;
  • 44) 0.553 019 712 205 697 843 2 × 2 = 1 + 0.106 039 424 411 395 686 4;
  • 45) 0.106 039 424 411 395 686 4 × 2 = 0 + 0.212 078 848 822 791 372 8;
  • 46) 0.212 078 848 822 791 372 8 × 2 = 0 + 0.424 157 697 645 582 745 6;
  • 47) 0.424 157 697 645 582 745 6 × 2 = 0 + 0.848 315 395 291 165 491 2;
  • 48) 0.848 315 395 291 165 491 2 × 2 = 1 + 0.696 630 790 582 330 982 4;
  • 49) 0.696 630 790 582 330 982 4 × 2 = 1 + 0.393 261 581 164 661 964 8;
  • 50) 0.393 261 581 164 661 964 8 × 2 = 0 + 0.786 523 162 329 323 929 6;
  • 51) 0.786 523 162 329 323 929 6 × 2 = 1 + 0.573 046 324 658 647 859 2;
  • 52) 0.573 046 324 658 647 859 2 × 2 = 1 + 0.146 092 649 317 295 718 4;
  • 53) 0.146 092 649 317 295 718 4 × 2 = 0 + 0.292 185 298 634 591 436 8;
  • 54) 0.292 185 298 634 591 436 8 × 2 = 0 + 0.584 370 597 269 182 873 6;
  • 55) 0.584 370 597 269 182 873 6 × 2 = 1 + 0.168 741 194 538 365 747 2;
  • 56) 0.168 741 194 538 365 747 2 × 2 = 0 + 0.337 482 389 076 731 494 4;
  • 57) 0.337 482 389 076 731 494 4 × 2 = 0 + 0.674 964 778 153 462 988 8;
  • 58) 0.674 964 778 153 462 988 8 × 2 = 1 + 0.349 929 556 306 925 977 6;
  • 59) 0.349 929 556 306 925 977 6 × 2 = 0 + 0.699 859 112 613 851 955 2;
  • 60) 0.699 859 112 613 851 955 2 × 2 = 1 + 0.399 718 225 227 703 910 4;
  • 61) 0.399 718 225 227 703 910 4 × 2 = 0 + 0.799 436 450 455 407 820 8;
  • 62) 0.799 436 450 455 407 820 8 × 2 = 1 + 0.598 872 900 910 815 641 6;
  • 63) 0.598 872 900 910 815 641 6 × 2 = 1 + 0.197 745 801 821 631 283 2;
  • 64) 0.197 745 801 821 631 283 2 × 2 = 0 + 0.395 491 603 643 262 566 4;
  • 65) 0.395 491 603 643 262 566 4 × 2 = 0 + 0.790 983 207 286 525 132 8;
  • 66) 0.790 983 207 286 525 132 8 × 2 = 1 + 0.581 966 414 573 050 265 6;
  • 67) 0.581 966 414 573 050 265 6 × 2 = 1 + 0.163 932 829 146 100 531 2;
  • 68) 0.163 932 829 146 100 531 2 × 2 = 0 + 0.327 865 658 292 201 062 4;
  • 69) 0.327 865 658 292 201 062 4 × 2 = 0 + 0.655 731 316 584 402 124 8;
  • 70) 0.655 731 316 584 402 124 8 × 2 = 1 + 0.311 462 633 168 804 249 6;
  • 71) 0.311 462 633 168 804 249 6 × 2 = 0 + 0.622 925 266 337 608 499 2;
  • 72) 0.622 925 266 337 608 499 2 × 2 = 1 + 0.245 850 532 675 216 998 4;
  • 73) 0.245 850 532 675 216 998 4 × 2 = 0 + 0.491 701 065 350 433 996 8;
  • 74) 0.491 701 065 350 433 996 8 × 2 = 0 + 0.983 402 130 700 867 993 6;
  • 75) 0.983 402 130 700 867 993 6 × 2 = 1 + 0.966 804 261 401 735 987 2;
  • 76) 0.966 804 261 401 735 987 2 × 2 = 1 + 0.933 608 522 803 471 974 4;
  • 77) 0.933 608 522 803 471 974 4 × 2 = 1 + 0.867 217 045 606 943 948 8;
  • 78) 0.867 217 045 606 943 948 8 × 2 = 1 + 0.734 434 091 213 887 897 6;
  • 79) 0.734 434 091 213 887 897 6 × 2 = 1 + 0.468 868 182 427 775 795 2;
  • 80) 0.468 868 182 427 775 795 2 × 2 = 0 + 0.937 736 364 855 551 590 4;
  • 81) 0.937 736 364 855 551 590 4 × 2 = 1 + 0.875 472 729 711 103 180 8;
  • 82) 0.875 472 729 711 103 180 8 × 2 = 1 + 0.750 945 459 422 206 361 6;
  • 83) 0.750 945 459 422 206 361 6 × 2 = 1 + 0.501 890 918 844 412 723 2;
  • 84) 0.501 890 918 844 412 723 2 × 2 = 1 + 0.003 781 837 688 825 446 4;
  • 85) 0.003 781 837 688 825 446 4 × 2 = 0 + 0.007 563 675 377 650 892 8;
  • 86) 0.007 563 675 377 650 892 8 × 2 = 0 + 0.015 127 350 755 301 785 6;
  • 87) 0.015 127 350 755 301 785 6 × 2 = 0 + 0.030 254 701 510 603 571 2;
  • 88) 0.030 254 701 510 603 571 2 × 2 = 0 + 0.060 509 403 021 207 142 4;
  • 89) 0.060 509 403 021 207 142 4 × 2 = 0 + 0.121 018 806 042 414 284 8;
  • 90) 0.121 018 806 042 414 284 8 × 2 = 0 + 0.242 037 612 084 828 569 6;
  • 91) 0.242 037 612 084 828 569 6 × 2 = 0 + 0.484 075 224 169 657 139 2;
  • 92) 0.484 075 224 169 657 139 2 × 2 = 0 + 0.968 150 448 339 314 278 4;
  • 93) 0.968 150 448 339 314 278 4 × 2 = 1 + 0.936 300 896 678 628 556 8;
  • 94) 0.936 300 896 678 628 556 8 × 2 = 1 + 0.872 601 793 357 257 113 6;
  • 95) 0.872 601 793 357 257 113 6 × 2 = 1 + 0.745 203 586 714 514 227 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 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0110 0110 0101 0011 1110 1111 0000 0000 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0110 0110 0101 0011 1110 1111 0000 0000 111(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 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0110 0110 0101 0011 1110 1111 0000 0000 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0110 0110 0101 0011 1110 1111 0000 0000 111(2) × 20 =

1.1000 1101 1001 0010 1011 0011 0010 1001 1111 0111 1000 0000 0111(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 1011 0011 0010 1001 1111 0111 1000 0000 0111


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 1011 0011 0010 1001 1111 0111 1000 0000 0111 =

1000 1101 1001 0010 1011 0011 0010 1001 1111 0111 1000 0000 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 0100

Mantissa (52 bits) =
1000 1101 1001 0010 1011 0011 0010 1001 1111 0111 1000 0000 0111


Decimal number -0.000 000 000 000 176 557 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1011 0011 0010 1001 1111 0111 1000 0000 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