-0.000 000 000 000 176 556 81 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 81(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
-0.000 000 000 000 176 556 81(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:

|-0.000 000 000 000 176 556 81| = 0.000 000 000 000 176 556 81


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: 0.000 000 000 000 176 556 81.

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) 0.000 000 000 000 176 556 81 × 2 = 0 + 0.000 000 000 000 353 113 62;
  • 2) 0.000 000 000 000 353 113 62 × 2 = 0 + 0.000 000 000 000 706 227 24;
  • 3) 0.000 000 000 000 706 227 24 × 2 = 0 + 0.000 000 000 001 412 454 48;
  • 4) 0.000 000 000 001 412 454 48 × 2 = 0 + 0.000 000 000 002 824 908 96;
  • 5) 0.000 000 000 002 824 908 96 × 2 = 0 + 0.000 000 000 005 649 817 92;
  • 6) 0.000 000 000 005 649 817 92 × 2 = 0 + 0.000 000 000 011 299 635 84;
  • 7) 0.000 000 000 011 299 635 84 × 2 = 0 + 0.000 000 000 022 599 271 68;
  • 8) 0.000 000 000 022 599 271 68 × 2 = 0 + 0.000 000 000 045 198 543 36;
  • 9) 0.000 000 000 045 198 543 36 × 2 = 0 + 0.000 000 000 090 397 086 72;
  • 10) 0.000 000 000 090 397 086 72 × 2 = 0 + 0.000 000 000 180 794 173 44;
  • 11) 0.000 000 000 180 794 173 44 × 2 = 0 + 0.000 000 000 361 588 346 88;
  • 12) 0.000 000 000 361 588 346 88 × 2 = 0 + 0.000 000 000 723 176 693 76;
  • 13) 0.000 000 000 723 176 693 76 × 2 = 0 + 0.000 000 001 446 353 387 52;
  • 14) 0.000 000 001 446 353 387 52 × 2 = 0 + 0.000 000 002 892 706 775 04;
  • 15) 0.000 000 002 892 706 775 04 × 2 = 0 + 0.000 000 005 785 413 550 08;
  • 16) 0.000 000 005 785 413 550 08 × 2 = 0 + 0.000 000 011 570 827 100 16;
  • 17) 0.000 000 011 570 827 100 16 × 2 = 0 + 0.000 000 023 141 654 200 32;
  • 18) 0.000 000 023 141 654 200 32 × 2 = 0 + 0.000 000 046 283 308 400 64;
  • 19) 0.000 000 046 283 308 400 64 × 2 = 0 + 0.000 000 092 566 616 801 28;
  • 20) 0.000 000 092 566 616 801 28 × 2 = 0 + 0.000 000 185 133 233 602 56;
  • 21) 0.000 000 185 133 233 602 56 × 2 = 0 + 0.000 000 370 266 467 205 12;
  • 22) 0.000 000 370 266 467 205 12 × 2 = 0 + 0.000 000 740 532 934 410 24;
  • 23) 0.000 000 740 532 934 410 24 × 2 = 0 + 0.000 001 481 065 868 820 48;
  • 24) 0.000 001 481 065 868 820 48 × 2 = 0 + 0.000 002 962 131 737 640 96;
  • 25) 0.000 002 962 131 737 640 96 × 2 = 0 + 0.000 005 924 263 475 281 92;
  • 26) 0.000 005 924 263 475 281 92 × 2 = 0 + 0.000 011 848 526 950 563 84;
  • 27) 0.000 011 848 526 950 563 84 × 2 = 0 + 0.000 023 697 053 901 127 68;
  • 28) 0.000 023 697 053 901 127 68 × 2 = 0 + 0.000 047 394 107 802 255 36;
  • 29) 0.000 047 394 107 802 255 36 × 2 = 0 + 0.000 094 788 215 604 510 72;
  • 30) 0.000 094 788 215 604 510 72 × 2 = 0 + 0.000 189 576 431 209 021 44;
  • 31) 0.000 189 576 431 209 021 44 × 2 = 0 + 0.000 379 152 862 418 042 88;
  • 32) 0.000 379 152 862 418 042 88 × 2 = 0 + 0.000 758 305 724 836 085 76;
  • 33) 0.000 758 305 724 836 085 76 × 2 = 0 + 0.001 516 611 449 672 171 52;
  • 34) 0.001 516 611 449 672 171 52 × 2 = 0 + 0.003 033 222 899 344 343 04;
  • 35) 0.003 033 222 899 344 343 04 × 2 = 0 + 0.006 066 445 798 688 686 08;
  • 36) 0.006 066 445 798 688 686 08 × 2 = 0 + 0.012 132 891 597 377 372 16;
  • 37) 0.012 132 891 597 377 372 16 × 2 = 0 + 0.024 265 783 194 754 744 32;
  • 38) 0.024 265 783 194 754 744 32 × 2 = 0 + 0.048 531 566 389 509 488 64;
  • 39) 0.048 531 566 389 509 488 64 × 2 = 0 + 0.097 063 132 779 018 977 28;
  • 40) 0.097 063 132 779 018 977 28 × 2 = 0 + 0.194 126 265 558 037 954 56;
  • 41) 0.194 126 265 558 037 954 56 × 2 = 0 + 0.388 252 531 116 075 909 12;
  • 42) 0.388 252 531 116 075 909 12 × 2 = 0 + 0.776 505 062 232 151 818 24;
  • 43) 0.776 505 062 232 151 818 24 × 2 = 1 + 0.553 010 124 464 303 636 48;
  • 44) 0.553 010 124 464 303 636 48 × 2 = 1 + 0.106 020 248 928 607 272 96;
  • 45) 0.106 020 248 928 607 272 96 × 2 = 0 + 0.212 040 497 857 214 545 92;
  • 46) 0.212 040 497 857 214 545 92 × 2 = 0 + 0.424 080 995 714 429 091 84;
  • 47) 0.424 080 995 714 429 091 84 × 2 = 0 + 0.848 161 991 428 858 183 68;
  • 48) 0.848 161 991 428 858 183 68 × 2 = 1 + 0.696 323 982 857 716 367 36;
  • 49) 0.696 323 982 857 716 367 36 × 2 = 1 + 0.392 647 965 715 432 734 72;
  • 50) 0.392 647 965 715 432 734 72 × 2 = 0 + 0.785 295 931 430 865 469 44;
  • 51) 0.785 295 931 430 865 469 44 × 2 = 1 + 0.570 591 862 861 730 938 88;
  • 52) 0.570 591 862 861 730 938 88 × 2 = 1 + 0.141 183 725 723 461 877 76;
  • 53) 0.141 183 725 723 461 877 76 × 2 = 0 + 0.282 367 451 446 923 755 52;
  • 54) 0.282 367 451 446 923 755 52 × 2 = 0 + 0.564 734 902 893 847 511 04;
  • 55) 0.564 734 902 893 847 511 04 × 2 = 1 + 0.129 469 805 787 695 022 08;
  • 56) 0.129 469 805 787 695 022 08 × 2 = 0 + 0.258 939 611 575 390 044 16;
  • 57) 0.258 939 611 575 390 044 16 × 2 = 0 + 0.517 879 223 150 780 088 32;
  • 58) 0.517 879 223 150 780 088 32 × 2 = 1 + 0.035 758 446 301 560 176 64;
  • 59) 0.035 758 446 301 560 176 64 × 2 = 0 + 0.071 516 892 603 120 353 28;
  • 60) 0.071 516 892 603 120 353 28 × 2 = 0 + 0.143 033 785 206 240 706 56;
  • 61) 0.143 033 785 206 240 706 56 × 2 = 0 + 0.286 067 570 412 481 413 12;
  • 62) 0.286 067 570 412 481 413 12 × 2 = 0 + 0.572 135 140 824 962 826 24;
  • 63) 0.572 135 140 824 962 826 24 × 2 = 1 + 0.144 270 281 649 925 652 48;
  • 64) 0.144 270 281 649 925 652 48 × 2 = 0 + 0.288 540 563 299 851 304 96;
  • 65) 0.288 540 563 299 851 304 96 × 2 = 0 + 0.577 081 126 599 702 609 92;
  • 66) 0.577 081 126 599 702 609 92 × 2 = 1 + 0.154 162 253 199 405 219 84;
  • 67) 0.154 162 253 199 405 219 84 × 2 = 0 + 0.308 324 506 398 810 439 68;
  • 68) 0.308 324 506 398 810 439 68 × 2 = 0 + 0.616 649 012 797 620 879 36;
  • 69) 0.616 649 012 797 620 879 36 × 2 = 1 + 0.233 298 025 595 241 758 72;
  • 70) 0.233 298 025 595 241 758 72 × 2 = 0 + 0.466 596 051 190 483 517 44;
  • 71) 0.466 596 051 190 483 517 44 × 2 = 0 + 0.933 192 102 380 967 034 88;
  • 72) 0.933 192 102 380 967 034 88 × 2 = 1 + 0.866 384 204 761 934 069 76;
  • 73) 0.866 384 204 761 934 069 76 × 2 = 1 + 0.732 768 409 523 868 139 52;
  • 74) 0.732 768 409 523 868 139 52 × 2 = 1 + 0.465 536 819 047 736 279 04;
  • 75) 0.465 536 819 047 736 279 04 × 2 = 0 + 0.931 073 638 095 472 558 08;
  • 76) 0.931 073 638 095 472 558 08 × 2 = 1 + 0.862 147 276 190 945 116 16;
  • 77) 0.862 147 276 190 945 116 16 × 2 = 1 + 0.724 294 552 381 890 232 32;
  • 78) 0.724 294 552 381 890 232 32 × 2 = 1 + 0.448 589 104 763 780 464 64;
  • 79) 0.448 589 104 763 780 464 64 × 2 = 0 + 0.897 178 209 527 560 929 28;
  • 80) 0.897 178 209 527 560 929 28 × 2 = 1 + 0.794 356 419 055 121 858 56;
  • 81) 0.794 356 419 055 121 858 56 × 2 = 1 + 0.588 712 838 110 243 717 12;
  • 82) 0.588 712 838 110 243 717 12 × 2 = 1 + 0.177 425 676 220 487 434 24;
  • 83) 0.177 425 676 220 487 434 24 × 2 = 0 + 0.354 851 352 440 974 868 48;
  • 84) 0.354 851 352 440 974 868 48 × 2 = 0 + 0.709 702 704 881 949 736 96;
  • 85) 0.709 702 704 881 949 736 96 × 2 = 1 + 0.419 405 409 763 899 473 92;
  • 86) 0.419 405 409 763 899 473 92 × 2 = 0 + 0.838 810 819 527 798 947 84;
  • 87) 0.838 810 819 527 798 947 84 × 2 = 1 + 0.677 621 639 055 597 895 68;
  • 88) 0.677 621 639 055 597 895 68 × 2 = 1 + 0.355 243 278 111 195 791 36;
  • 89) 0.355 243 278 111 195 791 36 × 2 = 0 + 0.710 486 556 222 391 582 72;
  • 90) 0.710 486 556 222 391 582 72 × 2 = 1 + 0.420 973 112 444 783 165 44;
  • 91) 0.420 973 112 444 783 165 44 × 2 = 0 + 0.841 946 224 889 566 330 88;
  • 92) 0.841 946 224 889 566 330 88 × 2 = 1 + 0.683 892 449 779 132 661 76;
  • 93) 0.683 892 449 779 132 661 76 × 2 = 1 + 0.367 784 899 558 265 323 52;
  • 94) 0.367 784 899 558 265 323 52 × 2 = 0 + 0.735 569 799 116 530 647 04;
  • 95) 0.735 569 799 116 530 647 04 × 2 = 1 + 0.471 139 598 233 061 294 08;

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:


0.000 000 000 000 176 556 81(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0010 0100 1001 1101 1101 1100 1011 0101 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 81(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0010 0100 1001 1101 1101 1100 1011 0101 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:


0.000 000 000 000 176 556 81(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0010 0100 1001 1101 1101 1100 1011 0101 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0010 0100 1001 1101 1101 1100 1011 0101 101(2) × 20 =

1.1000 1101 1001 0010 0001 0010 0100 1110 1110 1110 0101 1010 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 0001 0010 0100 1110 1110 1110 0101 1010 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 0001 0010 0100 1110 1110 1110 0101 1010 1101 =

1000 1101 1001 0010 0001 0010 0100 1110 1110 1110 0101 1010 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 0001 0010 0100 1110 1110 1110 0101 1010 1101


Decimal number -0.000 000 000 000 176 556 81 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0001 0010 0100 1110 1110 1110 0101 1010 1101

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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