-0.000 000 000 000 176 557 24 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 24(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 24(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 24| = 0.000 000 000 000 176 557 24


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 24.

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 24 × 2 = 0 + 0.000 000 000 000 353 114 48;
  • 2) 0.000 000 000 000 353 114 48 × 2 = 0 + 0.000 000 000 000 706 228 96;
  • 3) 0.000 000 000 000 706 228 96 × 2 = 0 + 0.000 000 000 001 412 457 92;
  • 4) 0.000 000 000 001 412 457 92 × 2 = 0 + 0.000 000 000 002 824 915 84;
  • 5) 0.000 000 000 002 824 915 84 × 2 = 0 + 0.000 000 000 005 649 831 68;
  • 6) 0.000 000 000 005 649 831 68 × 2 = 0 + 0.000 000 000 011 299 663 36;
  • 7) 0.000 000 000 011 299 663 36 × 2 = 0 + 0.000 000 000 022 599 326 72;
  • 8) 0.000 000 000 022 599 326 72 × 2 = 0 + 0.000 000 000 045 198 653 44;
  • 9) 0.000 000 000 045 198 653 44 × 2 = 0 + 0.000 000 000 090 397 306 88;
  • 10) 0.000 000 000 090 397 306 88 × 2 = 0 + 0.000 000 000 180 794 613 76;
  • 11) 0.000 000 000 180 794 613 76 × 2 = 0 + 0.000 000 000 361 589 227 52;
  • 12) 0.000 000 000 361 589 227 52 × 2 = 0 + 0.000 000 000 723 178 455 04;
  • 13) 0.000 000 000 723 178 455 04 × 2 = 0 + 0.000 000 001 446 356 910 08;
  • 14) 0.000 000 001 446 356 910 08 × 2 = 0 + 0.000 000 002 892 713 820 16;
  • 15) 0.000 000 002 892 713 820 16 × 2 = 0 + 0.000 000 005 785 427 640 32;
  • 16) 0.000 000 005 785 427 640 32 × 2 = 0 + 0.000 000 011 570 855 280 64;
  • 17) 0.000 000 011 570 855 280 64 × 2 = 0 + 0.000 000 023 141 710 561 28;
  • 18) 0.000 000 023 141 710 561 28 × 2 = 0 + 0.000 000 046 283 421 122 56;
  • 19) 0.000 000 046 283 421 122 56 × 2 = 0 + 0.000 000 092 566 842 245 12;
  • 20) 0.000 000 092 566 842 245 12 × 2 = 0 + 0.000 000 185 133 684 490 24;
  • 21) 0.000 000 185 133 684 490 24 × 2 = 0 + 0.000 000 370 267 368 980 48;
  • 22) 0.000 000 370 267 368 980 48 × 2 = 0 + 0.000 000 740 534 737 960 96;
  • 23) 0.000 000 740 534 737 960 96 × 2 = 0 + 0.000 001 481 069 475 921 92;
  • 24) 0.000 001 481 069 475 921 92 × 2 = 0 + 0.000 002 962 138 951 843 84;
  • 25) 0.000 002 962 138 951 843 84 × 2 = 0 + 0.000 005 924 277 903 687 68;
  • 26) 0.000 005 924 277 903 687 68 × 2 = 0 + 0.000 011 848 555 807 375 36;
  • 27) 0.000 011 848 555 807 375 36 × 2 = 0 + 0.000 023 697 111 614 750 72;
  • 28) 0.000 023 697 111 614 750 72 × 2 = 0 + 0.000 047 394 223 229 501 44;
  • 29) 0.000 047 394 223 229 501 44 × 2 = 0 + 0.000 094 788 446 459 002 88;
  • 30) 0.000 094 788 446 459 002 88 × 2 = 0 + 0.000 189 576 892 918 005 76;
  • 31) 0.000 189 576 892 918 005 76 × 2 = 0 + 0.000 379 153 785 836 011 52;
  • 32) 0.000 379 153 785 836 011 52 × 2 = 0 + 0.000 758 307 571 672 023 04;
  • 33) 0.000 758 307 571 672 023 04 × 2 = 0 + 0.001 516 615 143 344 046 08;
  • 34) 0.001 516 615 143 344 046 08 × 2 = 0 + 0.003 033 230 286 688 092 16;
  • 35) 0.003 033 230 286 688 092 16 × 2 = 0 + 0.006 066 460 573 376 184 32;
  • 36) 0.006 066 460 573 376 184 32 × 2 = 0 + 0.012 132 921 146 752 368 64;
  • 37) 0.012 132 921 146 752 368 64 × 2 = 0 + 0.024 265 842 293 504 737 28;
  • 38) 0.024 265 842 293 504 737 28 × 2 = 0 + 0.048 531 684 587 009 474 56;
  • 39) 0.048 531 684 587 009 474 56 × 2 = 0 + 0.097 063 369 174 018 949 12;
  • 40) 0.097 063 369 174 018 949 12 × 2 = 0 + 0.194 126 738 348 037 898 24;
  • 41) 0.194 126 738 348 037 898 24 × 2 = 0 + 0.388 253 476 696 075 796 48;
  • 42) 0.388 253 476 696 075 796 48 × 2 = 0 + 0.776 506 953 392 151 592 96;
  • 43) 0.776 506 953 392 151 592 96 × 2 = 1 + 0.553 013 906 784 303 185 92;
  • 44) 0.553 013 906 784 303 185 92 × 2 = 1 + 0.106 027 813 568 606 371 84;
  • 45) 0.106 027 813 568 606 371 84 × 2 = 0 + 0.212 055 627 137 212 743 68;
  • 46) 0.212 055 627 137 212 743 68 × 2 = 0 + 0.424 111 254 274 425 487 36;
  • 47) 0.424 111 254 274 425 487 36 × 2 = 0 + 0.848 222 508 548 850 974 72;
  • 48) 0.848 222 508 548 850 974 72 × 2 = 1 + 0.696 445 017 097 701 949 44;
  • 49) 0.696 445 017 097 701 949 44 × 2 = 1 + 0.392 890 034 195 403 898 88;
  • 50) 0.392 890 034 195 403 898 88 × 2 = 0 + 0.785 780 068 390 807 797 76;
  • 51) 0.785 780 068 390 807 797 76 × 2 = 1 + 0.571 560 136 781 615 595 52;
  • 52) 0.571 560 136 781 615 595 52 × 2 = 1 + 0.143 120 273 563 231 191 04;
  • 53) 0.143 120 273 563 231 191 04 × 2 = 0 + 0.286 240 547 126 462 382 08;
  • 54) 0.286 240 547 126 462 382 08 × 2 = 0 + 0.572 481 094 252 924 764 16;
  • 55) 0.572 481 094 252 924 764 16 × 2 = 1 + 0.144 962 188 505 849 528 32;
  • 56) 0.144 962 188 505 849 528 32 × 2 = 0 + 0.289 924 377 011 699 056 64;
  • 57) 0.289 924 377 011 699 056 64 × 2 = 0 + 0.579 848 754 023 398 113 28;
  • 58) 0.579 848 754 023 398 113 28 × 2 = 1 + 0.159 697 508 046 796 226 56;
  • 59) 0.159 697 508 046 796 226 56 × 2 = 0 + 0.319 395 016 093 592 453 12;
  • 60) 0.319 395 016 093 592 453 12 × 2 = 0 + 0.638 790 032 187 184 906 24;
  • 61) 0.638 790 032 187 184 906 24 × 2 = 1 + 0.277 580 064 374 369 812 48;
  • 62) 0.277 580 064 374 369 812 48 × 2 = 0 + 0.555 160 128 748 739 624 96;
  • 63) 0.555 160 128 748 739 624 96 × 2 = 1 + 0.110 320 257 497 479 249 92;
  • 64) 0.110 320 257 497 479 249 92 × 2 = 0 + 0.220 640 514 994 958 499 84;
  • 65) 0.220 640 514 994 958 499 84 × 2 = 0 + 0.441 281 029 989 916 999 68;
  • 66) 0.441 281 029 989 916 999 68 × 2 = 0 + 0.882 562 059 979 833 999 36;
  • 67) 0.882 562 059 979 833 999 36 × 2 = 1 + 0.765 124 119 959 667 998 72;
  • 68) 0.765 124 119 959 667 998 72 × 2 = 1 + 0.530 248 239 919 335 997 44;
  • 69) 0.530 248 239 919 335 997 44 × 2 = 1 + 0.060 496 479 838 671 994 88;
  • 70) 0.060 496 479 838 671 994 88 × 2 = 0 + 0.120 992 959 677 343 989 76;
  • 71) 0.120 992 959 677 343 989 76 × 2 = 0 + 0.241 985 919 354 687 979 52;
  • 72) 0.241 985 919 354 687 979 52 × 2 = 0 + 0.483 971 838 709 375 959 04;
  • 73) 0.483 971 838 709 375 959 04 × 2 = 0 + 0.967 943 677 418 751 918 08;
  • 74) 0.967 943 677 418 751 918 08 × 2 = 1 + 0.935 887 354 837 503 836 16;
  • 75) 0.935 887 354 837 503 836 16 × 2 = 1 + 0.871 774 709 675 007 672 32;
  • 76) 0.871 774 709 675 007 672 32 × 2 = 1 + 0.743 549 419 350 015 344 64;
  • 77) 0.743 549 419 350 015 344 64 × 2 = 1 + 0.487 098 838 700 030 689 28;
  • 78) 0.487 098 838 700 030 689 28 × 2 = 0 + 0.974 197 677 400 061 378 56;
  • 79) 0.974 197 677 400 061 378 56 × 2 = 1 + 0.948 395 354 800 122 757 12;
  • 80) 0.948 395 354 800 122 757 12 × 2 = 1 + 0.896 790 709 600 245 514 24;
  • 81) 0.896 790 709 600 245 514 24 × 2 = 1 + 0.793 581 419 200 491 028 48;
  • 82) 0.793 581 419 200 491 028 48 × 2 = 1 + 0.587 162 838 400 982 056 96;
  • 83) 0.587 162 838 400 982 056 96 × 2 = 1 + 0.174 325 676 801 964 113 92;
  • 84) 0.174 325 676 801 964 113 92 × 2 = 0 + 0.348 651 353 603 928 227 84;
  • 85) 0.348 651 353 603 928 227 84 × 2 = 0 + 0.697 302 707 207 856 455 68;
  • 86) 0.697 302 707 207 856 455 68 × 2 = 1 + 0.394 605 414 415 712 911 36;
  • 87) 0.394 605 414 415 712 911 36 × 2 = 0 + 0.789 210 828 831 425 822 72;
  • 88) 0.789 210 828 831 425 822 72 × 2 = 1 + 0.578 421 657 662 851 645 44;
  • 89) 0.578 421 657 662 851 645 44 × 2 = 1 + 0.156 843 315 325 703 290 88;
  • 90) 0.156 843 315 325 703 290 88 × 2 = 0 + 0.313 686 630 651 406 581 76;
  • 91) 0.313 686 630 651 406 581 76 × 2 = 0 + 0.627 373 261 302 813 163 52;
  • 92) 0.627 373 261 302 813 163 52 × 2 = 1 + 0.254 746 522 605 626 327 04;
  • 93) 0.254 746 522 605 626 327 04 × 2 = 0 + 0.509 493 045 211 252 654 08;
  • 94) 0.509 493 045 211 252 654 08 × 2 = 1 + 0.018 986 090 422 505 308 16;
  • 95) 0.018 986 090 422 505 308 16 × 2 = 0 + 0.037 972 180 845 010 616 32;

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 24(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0011 1000 0111 1011 1110 0101 1001 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 24(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0011 1000 0111 1011 1110 0101 1001 010(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 24(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0011 1000 0111 1011 1110 0101 1001 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0011 1000 0111 1011 1110 0101 1001 010(2) × 20 =

1.1000 1101 1001 0010 0101 0001 1100 0011 1101 1111 0010 1100 1010(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 0101 0001 1100 0011 1101 1111 0010 1100 1010


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 0101 0001 1100 0011 1101 1111 0010 1100 1010 =

1000 1101 1001 0010 0101 0001 1100 0011 1101 1111 0010 1100 1010


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 0101 0001 1100 0011 1101 1111 0010 1100 1010


Decimal number -0.000 000 000 000 176 557 24 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0101 0001 1100 0011 1101 1111 0010 1100 1010

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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