-0.000 000 000 000 176 557 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 19(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 19(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 19| = 0.000 000 000 000 176 557 19


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 19.

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 19 × 2 = 0 + 0.000 000 000 000 353 114 38;
  • 2) 0.000 000 000 000 353 114 38 × 2 = 0 + 0.000 000 000 000 706 228 76;
  • 3) 0.000 000 000 000 706 228 76 × 2 = 0 + 0.000 000 000 001 412 457 52;
  • 4) 0.000 000 000 001 412 457 52 × 2 = 0 + 0.000 000 000 002 824 915 04;
  • 5) 0.000 000 000 002 824 915 04 × 2 = 0 + 0.000 000 000 005 649 830 08;
  • 6) 0.000 000 000 005 649 830 08 × 2 = 0 + 0.000 000 000 011 299 660 16;
  • 7) 0.000 000 000 011 299 660 16 × 2 = 0 + 0.000 000 000 022 599 320 32;
  • 8) 0.000 000 000 022 599 320 32 × 2 = 0 + 0.000 000 000 045 198 640 64;
  • 9) 0.000 000 000 045 198 640 64 × 2 = 0 + 0.000 000 000 090 397 281 28;
  • 10) 0.000 000 000 090 397 281 28 × 2 = 0 + 0.000 000 000 180 794 562 56;
  • 11) 0.000 000 000 180 794 562 56 × 2 = 0 + 0.000 000 000 361 589 125 12;
  • 12) 0.000 000 000 361 589 125 12 × 2 = 0 + 0.000 000 000 723 178 250 24;
  • 13) 0.000 000 000 723 178 250 24 × 2 = 0 + 0.000 000 001 446 356 500 48;
  • 14) 0.000 000 001 446 356 500 48 × 2 = 0 + 0.000 000 002 892 713 000 96;
  • 15) 0.000 000 002 892 713 000 96 × 2 = 0 + 0.000 000 005 785 426 001 92;
  • 16) 0.000 000 005 785 426 001 92 × 2 = 0 + 0.000 000 011 570 852 003 84;
  • 17) 0.000 000 011 570 852 003 84 × 2 = 0 + 0.000 000 023 141 704 007 68;
  • 18) 0.000 000 023 141 704 007 68 × 2 = 0 + 0.000 000 046 283 408 015 36;
  • 19) 0.000 000 046 283 408 015 36 × 2 = 0 + 0.000 000 092 566 816 030 72;
  • 20) 0.000 000 092 566 816 030 72 × 2 = 0 + 0.000 000 185 133 632 061 44;
  • 21) 0.000 000 185 133 632 061 44 × 2 = 0 + 0.000 000 370 267 264 122 88;
  • 22) 0.000 000 370 267 264 122 88 × 2 = 0 + 0.000 000 740 534 528 245 76;
  • 23) 0.000 000 740 534 528 245 76 × 2 = 0 + 0.000 001 481 069 056 491 52;
  • 24) 0.000 001 481 069 056 491 52 × 2 = 0 + 0.000 002 962 138 112 983 04;
  • 25) 0.000 002 962 138 112 983 04 × 2 = 0 + 0.000 005 924 276 225 966 08;
  • 26) 0.000 005 924 276 225 966 08 × 2 = 0 + 0.000 011 848 552 451 932 16;
  • 27) 0.000 011 848 552 451 932 16 × 2 = 0 + 0.000 023 697 104 903 864 32;
  • 28) 0.000 023 697 104 903 864 32 × 2 = 0 + 0.000 047 394 209 807 728 64;
  • 29) 0.000 047 394 209 807 728 64 × 2 = 0 + 0.000 094 788 419 615 457 28;
  • 30) 0.000 094 788 419 615 457 28 × 2 = 0 + 0.000 189 576 839 230 914 56;
  • 31) 0.000 189 576 839 230 914 56 × 2 = 0 + 0.000 379 153 678 461 829 12;
  • 32) 0.000 379 153 678 461 829 12 × 2 = 0 + 0.000 758 307 356 923 658 24;
  • 33) 0.000 758 307 356 923 658 24 × 2 = 0 + 0.001 516 614 713 847 316 48;
  • 34) 0.001 516 614 713 847 316 48 × 2 = 0 + 0.003 033 229 427 694 632 96;
  • 35) 0.003 033 229 427 694 632 96 × 2 = 0 + 0.006 066 458 855 389 265 92;
  • 36) 0.006 066 458 855 389 265 92 × 2 = 0 + 0.012 132 917 710 778 531 84;
  • 37) 0.012 132 917 710 778 531 84 × 2 = 0 + 0.024 265 835 421 557 063 68;
  • 38) 0.024 265 835 421 557 063 68 × 2 = 0 + 0.048 531 670 843 114 127 36;
  • 39) 0.048 531 670 843 114 127 36 × 2 = 0 + 0.097 063 341 686 228 254 72;
  • 40) 0.097 063 341 686 228 254 72 × 2 = 0 + 0.194 126 683 372 456 509 44;
  • 41) 0.194 126 683 372 456 509 44 × 2 = 0 + 0.388 253 366 744 913 018 88;
  • 42) 0.388 253 366 744 913 018 88 × 2 = 0 + 0.776 506 733 489 826 037 76;
  • 43) 0.776 506 733 489 826 037 76 × 2 = 1 + 0.553 013 466 979 652 075 52;
  • 44) 0.553 013 466 979 652 075 52 × 2 = 1 + 0.106 026 933 959 304 151 04;
  • 45) 0.106 026 933 959 304 151 04 × 2 = 0 + 0.212 053 867 918 608 302 08;
  • 46) 0.212 053 867 918 608 302 08 × 2 = 0 + 0.424 107 735 837 216 604 16;
  • 47) 0.424 107 735 837 216 604 16 × 2 = 0 + 0.848 215 471 674 433 208 32;
  • 48) 0.848 215 471 674 433 208 32 × 2 = 1 + 0.696 430 943 348 866 416 64;
  • 49) 0.696 430 943 348 866 416 64 × 2 = 1 + 0.392 861 886 697 732 833 28;
  • 50) 0.392 861 886 697 732 833 28 × 2 = 0 + 0.785 723 773 395 465 666 56;
  • 51) 0.785 723 773 395 465 666 56 × 2 = 1 + 0.571 447 546 790 931 333 12;
  • 52) 0.571 447 546 790 931 333 12 × 2 = 1 + 0.142 895 093 581 862 666 24;
  • 53) 0.142 895 093 581 862 666 24 × 2 = 0 + 0.285 790 187 163 725 332 48;
  • 54) 0.285 790 187 163 725 332 48 × 2 = 0 + 0.571 580 374 327 450 664 96;
  • 55) 0.571 580 374 327 450 664 96 × 2 = 1 + 0.143 160 748 654 901 329 92;
  • 56) 0.143 160 748 654 901 329 92 × 2 = 0 + 0.286 321 497 309 802 659 84;
  • 57) 0.286 321 497 309 802 659 84 × 2 = 0 + 0.572 642 994 619 605 319 68;
  • 58) 0.572 642 994 619 605 319 68 × 2 = 1 + 0.145 285 989 239 210 639 36;
  • 59) 0.145 285 989 239 210 639 36 × 2 = 0 + 0.290 571 978 478 421 278 72;
  • 60) 0.290 571 978 478 421 278 72 × 2 = 0 + 0.581 143 956 956 842 557 44;
  • 61) 0.581 143 956 956 842 557 44 × 2 = 1 + 0.162 287 913 913 685 114 88;
  • 62) 0.162 287 913 913 685 114 88 × 2 = 0 + 0.324 575 827 827 370 229 76;
  • 63) 0.324 575 827 827 370 229 76 × 2 = 0 + 0.649 151 655 654 740 459 52;
  • 64) 0.649 151 655 654 740 459 52 × 2 = 1 + 0.298 303 311 309 480 919 04;
  • 65) 0.298 303 311 309 480 919 04 × 2 = 0 + 0.596 606 622 618 961 838 08;
  • 66) 0.596 606 622 618 961 838 08 × 2 = 1 + 0.193 213 245 237 923 676 16;
  • 67) 0.193 213 245 237 923 676 16 × 2 = 0 + 0.386 426 490 475 847 352 32;
  • 68) 0.386 426 490 475 847 352 32 × 2 = 0 + 0.772 852 980 951 694 704 64;
  • 69) 0.772 852 980 951 694 704 64 × 2 = 1 + 0.545 705 961 903 389 409 28;
  • 70) 0.545 705 961 903 389 409 28 × 2 = 1 + 0.091 411 923 806 778 818 56;
  • 71) 0.091 411 923 806 778 818 56 × 2 = 0 + 0.182 823 847 613 557 637 12;
  • 72) 0.182 823 847 613 557 637 12 × 2 = 0 + 0.365 647 695 227 115 274 24;
  • 73) 0.365 647 695 227 115 274 24 × 2 = 0 + 0.731 295 390 454 230 548 48;
  • 74) 0.731 295 390 454 230 548 48 × 2 = 1 + 0.462 590 780 908 461 096 96;
  • 75) 0.462 590 780 908 461 096 96 × 2 = 0 + 0.925 181 561 816 922 193 92;
  • 76) 0.925 181 561 816 922 193 92 × 2 = 1 + 0.850 363 123 633 844 387 84;
  • 77) 0.850 363 123 633 844 387 84 × 2 = 1 + 0.700 726 247 267 688 775 68;
  • 78) 0.700 726 247 267 688 775 68 × 2 = 1 + 0.401 452 494 535 377 551 36;
  • 79) 0.401 452 494 535 377 551 36 × 2 = 0 + 0.802 904 989 070 755 102 72;
  • 80) 0.802 904 989 070 755 102 72 × 2 = 1 + 0.605 809 978 141 510 205 44;
  • 81) 0.605 809 978 141 510 205 44 × 2 = 1 + 0.211 619 956 283 020 410 88;
  • 82) 0.211 619 956 283 020 410 88 × 2 = 0 + 0.423 239 912 566 040 821 76;
  • 83) 0.423 239 912 566 040 821 76 × 2 = 0 + 0.846 479 825 132 081 643 52;
  • 84) 0.846 479 825 132 081 643 52 × 2 = 1 + 0.692 959 650 264 163 287 04;
  • 85) 0.692 959 650 264 163 287 04 × 2 = 1 + 0.385 919 300 528 326 574 08;
  • 86) 0.385 919 300 528 326 574 08 × 2 = 0 + 0.771 838 601 056 653 148 16;
  • 87) 0.771 838 601 056 653 148 16 × 2 = 1 + 0.543 677 202 113 306 296 32;
  • 88) 0.543 677 202 113 306 296 32 × 2 = 1 + 0.087 354 404 226 612 592 64;
  • 89) 0.087 354 404 226 612 592 64 × 2 = 0 + 0.174 708 808 453 225 185 28;
  • 90) 0.174 708 808 453 225 185 28 × 2 = 0 + 0.349 417 616 906 450 370 56;
  • 91) 0.349 417 616 906 450 370 56 × 2 = 0 + 0.698 835 233 812 900 741 12;
  • 92) 0.698 835 233 812 900 741 12 × 2 = 1 + 0.397 670 467 625 801 482 24;
  • 93) 0.397 670 467 625 801 482 24 × 2 = 0 + 0.795 340 935 251 602 964 48;
  • 94) 0.795 340 935 251 602 964 48 × 2 = 1 + 0.590 681 870 503 205 928 96;
  • 95) 0.590 681 870 503 205 928 96 × 2 = 1 + 0.181 363 741 006 411 857 92;

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 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0100 1100 0101 1101 1001 1011 0001 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0100 1100 0101 1101 1001 1011 0001 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 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0100 1100 0101 1101 1001 1011 0001 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0100 1100 0101 1101 1001 1011 0001 011(2) × 20 =

1.1000 1101 1001 0010 0100 1010 0110 0010 1110 1100 1101 1000 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 0100 1010 0110 0010 1110 1100 1101 1000 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 0100 1010 0110 0010 1110 1100 1101 1000 1011 =

1000 1101 1001 0010 0100 1010 0110 0010 1110 1100 1101 1000 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 0100 1010 0110 0010 1110 1100 1101 1000 1011


Decimal number -0.000 000 000 000 176 557 19 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0100 1010 0110 0010 1110 1100 1101 1000 1011

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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