-0.000 000 000 000 176 556 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 12(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 12(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 12| = 0.000 000 000 000 176 556 12


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 12.

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 12 × 2 = 0 + 0.000 000 000 000 353 112 24;
  • 2) 0.000 000 000 000 353 112 24 × 2 = 0 + 0.000 000 000 000 706 224 48;
  • 3) 0.000 000 000 000 706 224 48 × 2 = 0 + 0.000 000 000 001 412 448 96;
  • 4) 0.000 000 000 001 412 448 96 × 2 = 0 + 0.000 000 000 002 824 897 92;
  • 5) 0.000 000 000 002 824 897 92 × 2 = 0 + 0.000 000 000 005 649 795 84;
  • 6) 0.000 000 000 005 649 795 84 × 2 = 0 + 0.000 000 000 011 299 591 68;
  • 7) 0.000 000 000 011 299 591 68 × 2 = 0 + 0.000 000 000 022 599 183 36;
  • 8) 0.000 000 000 022 599 183 36 × 2 = 0 + 0.000 000 000 045 198 366 72;
  • 9) 0.000 000 000 045 198 366 72 × 2 = 0 + 0.000 000 000 090 396 733 44;
  • 10) 0.000 000 000 090 396 733 44 × 2 = 0 + 0.000 000 000 180 793 466 88;
  • 11) 0.000 000 000 180 793 466 88 × 2 = 0 + 0.000 000 000 361 586 933 76;
  • 12) 0.000 000 000 361 586 933 76 × 2 = 0 + 0.000 000 000 723 173 867 52;
  • 13) 0.000 000 000 723 173 867 52 × 2 = 0 + 0.000 000 001 446 347 735 04;
  • 14) 0.000 000 001 446 347 735 04 × 2 = 0 + 0.000 000 002 892 695 470 08;
  • 15) 0.000 000 002 892 695 470 08 × 2 = 0 + 0.000 000 005 785 390 940 16;
  • 16) 0.000 000 005 785 390 940 16 × 2 = 0 + 0.000 000 011 570 781 880 32;
  • 17) 0.000 000 011 570 781 880 32 × 2 = 0 + 0.000 000 023 141 563 760 64;
  • 18) 0.000 000 023 141 563 760 64 × 2 = 0 + 0.000 000 046 283 127 521 28;
  • 19) 0.000 000 046 283 127 521 28 × 2 = 0 + 0.000 000 092 566 255 042 56;
  • 20) 0.000 000 092 566 255 042 56 × 2 = 0 + 0.000 000 185 132 510 085 12;
  • 21) 0.000 000 185 132 510 085 12 × 2 = 0 + 0.000 000 370 265 020 170 24;
  • 22) 0.000 000 370 265 020 170 24 × 2 = 0 + 0.000 000 740 530 040 340 48;
  • 23) 0.000 000 740 530 040 340 48 × 2 = 0 + 0.000 001 481 060 080 680 96;
  • 24) 0.000 001 481 060 080 680 96 × 2 = 0 + 0.000 002 962 120 161 361 92;
  • 25) 0.000 002 962 120 161 361 92 × 2 = 0 + 0.000 005 924 240 322 723 84;
  • 26) 0.000 005 924 240 322 723 84 × 2 = 0 + 0.000 011 848 480 645 447 68;
  • 27) 0.000 011 848 480 645 447 68 × 2 = 0 + 0.000 023 696 961 290 895 36;
  • 28) 0.000 023 696 961 290 895 36 × 2 = 0 + 0.000 047 393 922 581 790 72;
  • 29) 0.000 047 393 922 581 790 72 × 2 = 0 + 0.000 094 787 845 163 581 44;
  • 30) 0.000 094 787 845 163 581 44 × 2 = 0 + 0.000 189 575 690 327 162 88;
  • 31) 0.000 189 575 690 327 162 88 × 2 = 0 + 0.000 379 151 380 654 325 76;
  • 32) 0.000 379 151 380 654 325 76 × 2 = 0 + 0.000 758 302 761 308 651 52;
  • 33) 0.000 758 302 761 308 651 52 × 2 = 0 + 0.001 516 605 522 617 303 04;
  • 34) 0.001 516 605 522 617 303 04 × 2 = 0 + 0.003 033 211 045 234 606 08;
  • 35) 0.003 033 211 045 234 606 08 × 2 = 0 + 0.006 066 422 090 469 212 16;
  • 36) 0.006 066 422 090 469 212 16 × 2 = 0 + 0.012 132 844 180 938 424 32;
  • 37) 0.012 132 844 180 938 424 32 × 2 = 0 + 0.024 265 688 361 876 848 64;
  • 38) 0.024 265 688 361 876 848 64 × 2 = 0 + 0.048 531 376 723 753 697 28;
  • 39) 0.048 531 376 723 753 697 28 × 2 = 0 + 0.097 062 753 447 507 394 56;
  • 40) 0.097 062 753 447 507 394 56 × 2 = 0 + 0.194 125 506 895 014 789 12;
  • 41) 0.194 125 506 895 014 789 12 × 2 = 0 + 0.388 251 013 790 029 578 24;
  • 42) 0.388 251 013 790 029 578 24 × 2 = 0 + 0.776 502 027 580 059 156 48;
  • 43) 0.776 502 027 580 059 156 48 × 2 = 1 + 0.553 004 055 160 118 312 96;
  • 44) 0.553 004 055 160 118 312 96 × 2 = 1 + 0.106 008 110 320 236 625 92;
  • 45) 0.106 008 110 320 236 625 92 × 2 = 0 + 0.212 016 220 640 473 251 84;
  • 46) 0.212 016 220 640 473 251 84 × 2 = 0 + 0.424 032 441 280 946 503 68;
  • 47) 0.424 032 441 280 946 503 68 × 2 = 0 + 0.848 064 882 561 893 007 36;
  • 48) 0.848 064 882 561 893 007 36 × 2 = 1 + 0.696 129 765 123 786 014 72;
  • 49) 0.696 129 765 123 786 014 72 × 2 = 1 + 0.392 259 530 247 572 029 44;
  • 50) 0.392 259 530 247 572 029 44 × 2 = 0 + 0.784 519 060 495 144 058 88;
  • 51) 0.784 519 060 495 144 058 88 × 2 = 1 + 0.569 038 120 990 288 117 76;
  • 52) 0.569 038 120 990 288 117 76 × 2 = 1 + 0.138 076 241 980 576 235 52;
  • 53) 0.138 076 241 980 576 235 52 × 2 = 0 + 0.276 152 483 961 152 471 04;
  • 54) 0.276 152 483 961 152 471 04 × 2 = 0 + 0.552 304 967 922 304 942 08;
  • 55) 0.552 304 967 922 304 942 08 × 2 = 1 + 0.104 609 935 844 609 884 16;
  • 56) 0.104 609 935 844 609 884 16 × 2 = 0 + 0.209 219 871 689 219 768 32;
  • 57) 0.209 219 871 689 219 768 32 × 2 = 0 + 0.418 439 743 378 439 536 64;
  • 58) 0.418 439 743 378 439 536 64 × 2 = 0 + 0.836 879 486 756 879 073 28;
  • 59) 0.836 879 486 756 879 073 28 × 2 = 1 + 0.673 758 973 513 758 146 56;
  • 60) 0.673 758 973 513 758 146 56 × 2 = 1 + 0.347 517 947 027 516 293 12;
  • 61) 0.347 517 947 027 516 293 12 × 2 = 0 + 0.695 035 894 055 032 586 24;
  • 62) 0.695 035 894 055 032 586 24 × 2 = 1 + 0.390 071 788 110 065 172 48;
  • 63) 0.390 071 788 110 065 172 48 × 2 = 0 + 0.780 143 576 220 130 344 96;
  • 64) 0.780 143 576 220 130 344 96 × 2 = 1 + 0.560 287 152 440 260 689 92;
  • 65) 0.560 287 152 440 260 689 92 × 2 = 1 + 0.120 574 304 880 521 379 84;
  • 66) 0.120 574 304 880 521 379 84 × 2 = 0 + 0.241 148 609 761 042 759 68;
  • 67) 0.241 148 609 761 042 759 68 × 2 = 0 + 0.482 297 219 522 085 519 36;
  • 68) 0.482 297 219 522 085 519 36 × 2 = 0 + 0.964 594 439 044 171 038 72;
  • 69) 0.964 594 439 044 171 038 72 × 2 = 1 + 0.929 188 878 088 342 077 44;
  • 70) 0.929 188 878 088 342 077 44 × 2 = 1 + 0.858 377 756 176 684 154 88;
  • 71) 0.858 377 756 176 684 154 88 × 2 = 1 + 0.716 755 512 353 368 309 76;
  • 72) 0.716 755 512 353 368 309 76 × 2 = 1 + 0.433 511 024 706 736 619 52;
  • 73) 0.433 511 024 706 736 619 52 × 2 = 0 + 0.867 022 049 413 473 239 04;
  • 74) 0.867 022 049 413 473 239 04 × 2 = 1 + 0.734 044 098 826 946 478 08;
  • 75) 0.734 044 098 826 946 478 08 × 2 = 1 + 0.468 088 197 653 892 956 16;
  • 76) 0.468 088 197 653 892 956 16 × 2 = 0 + 0.936 176 395 307 785 912 32;
  • 77) 0.936 176 395 307 785 912 32 × 2 = 1 + 0.872 352 790 615 571 824 64;
  • 78) 0.872 352 790 615 571 824 64 × 2 = 1 + 0.744 705 581 231 143 649 28;
  • 79) 0.744 705 581 231 143 649 28 × 2 = 1 + 0.489 411 162 462 287 298 56;
  • 80) 0.489 411 162 462 287 298 56 × 2 = 0 + 0.978 822 324 924 574 597 12;
  • 81) 0.978 822 324 924 574 597 12 × 2 = 1 + 0.957 644 649 849 149 194 24;
  • 82) 0.957 644 649 849 149 194 24 × 2 = 1 + 0.915 289 299 698 298 388 48;
  • 83) 0.915 289 299 698 298 388 48 × 2 = 1 + 0.830 578 599 396 596 776 96;
  • 84) 0.830 578 599 396 596 776 96 × 2 = 1 + 0.661 157 198 793 193 553 92;
  • 85) 0.661 157 198 793 193 553 92 × 2 = 1 + 0.322 314 397 586 387 107 84;
  • 86) 0.322 314 397 586 387 107 84 × 2 = 0 + 0.644 628 795 172 774 215 68;
  • 87) 0.644 628 795 172 774 215 68 × 2 = 1 + 0.289 257 590 345 548 431 36;
  • 88) 0.289 257 590 345 548 431 36 × 2 = 0 + 0.578 515 180 691 096 862 72;
  • 89) 0.578 515 180 691 096 862 72 × 2 = 1 + 0.157 030 361 382 193 725 44;
  • 90) 0.157 030 361 382 193 725 44 × 2 = 0 + 0.314 060 722 764 387 450 88;
  • 91) 0.314 060 722 764 387 450 88 × 2 = 0 + 0.628 121 445 528 774 901 76;
  • 92) 0.628 121 445 528 774 901 76 × 2 = 1 + 0.256 242 891 057 549 803 52;
  • 93) 0.256 242 891 057 549 803 52 × 2 = 0 + 0.512 485 782 115 099 607 04;
  • 94) 0.512 485 782 115 099 607 04 × 2 = 1 + 0.024 971 564 230 199 214 08;
  • 95) 0.024 971 564 230 199 214 08 × 2 = 0 + 0.049 943 128 460 398 428 16;

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 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0101 1000 1111 0110 1110 1111 1010 1001 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0101 1000 1111 0110 1110 1111 1010 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:


0.000 000 000 000 176 556 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0101 1000 1111 0110 1110 1111 1010 1001 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0101 1000 1111 0110 1110 1111 1010 1001 010(2) × 20 =

1.1000 1101 1001 0001 1010 1100 0111 1011 0111 0111 1101 0100 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 0001 1010 1100 0111 1011 0111 0111 1101 0100 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 0001 1010 1100 0111 1011 0111 0111 1101 0100 1010 =

1000 1101 1001 0001 1010 1100 0111 1011 0111 0111 1101 0100 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 0001 1010 1100 0111 1011 0111 0111 1101 0100 1010


Decimal number -0.000 000 000 000 176 556 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1010 1100 0111 1011 0111 0111 1101 0100 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