-0.000 000 000 000 176 557 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 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
-0.000 000 000 000 176 557 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:

|-0.000 000 000 000 176 557 12| = 0.000 000 000 000 176 557 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: 0.000 000 000 000 176 557 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) 0.000 000 000 000 176 557 12 × 2 = 0 + 0.000 000 000 000 353 114 24;
  • 2) 0.000 000 000 000 353 114 24 × 2 = 0 + 0.000 000 000 000 706 228 48;
  • 3) 0.000 000 000 000 706 228 48 × 2 = 0 + 0.000 000 000 001 412 456 96;
  • 4) 0.000 000 000 001 412 456 96 × 2 = 0 + 0.000 000 000 002 824 913 92;
  • 5) 0.000 000 000 002 824 913 92 × 2 = 0 + 0.000 000 000 005 649 827 84;
  • 6) 0.000 000 000 005 649 827 84 × 2 = 0 + 0.000 000 000 011 299 655 68;
  • 7) 0.000 000 000 011 299 655 68 × 2 = 0 + 0.000 000 000 022 599 311 36;
  • 8) 0.000 000 000 022 599 311 36 × 2 = 0 + 0.000 000 000 045 198 622 72;
  • 9) 0.000 000 000 045 198 622 72 × 2 = 0 + 0.000 000 000 090 397 245 44;
  • 10) 0.000 000 000 090 397 245 44 × 2 = 0 + 0.000 000 000 180 794 490 88;
  • 11) 0.000 000 000 180 794 490 88 × 2 = 0 + 0.000 000 000 361 588 981 76;
  • 12) 0.000 000 000 361 588 981 76 × 2 = 0 + 0.000 000 000 723 177 963 52;
  • 13) 0.000 000 000 723 177 963 52 × 2 = 0 + 0.000 000 001 446 355 927 04;
  • 14) 0.000 000 001 446 355 927 04 × 2 = 0 + 0.000 000 002 892 711 854 08;
  • 15) 0.000 000 002 892 711 854 08 × 2 = 0 + 0.000 000 005 785 423 708 16;
  • 16) 0.000 000 005 785 423 708 16 × 2 = 0 + 0.000 000 011 570 847 416 32;
  • 17) 0.000 000 011 570 847 416 32 × 2 = 0 + 0.000 000 023 141 694 832 64;
  • 18) 0.000 000 023 141 694 832 64 × 2 = 0 + 0.000 000 046 283 389 665 28;
  • 19) 0.000 000 046 283 389 665 28 × 2 = 0 + 0.000 000 092 566 779 330 56;
  • 20) 0.000 000 092 566 779 330 56 × 2 = 0 + 0.000 000 185 133 558 661 12;
  • 21) 0.000 000 185 133 558 661 12 × 2 = 0 + 0.000 000 370 267 117 322 24;
  • 22) 0.000 000 370 267 117 322 24 × 2 = 0 + 0.000 000 740 534 234 644 48;
  • 23) 0.000 000 740 534 234 644 48 × 2 = 0 + 0.000 001 481 068 469 288 96;
  • 24) 0.000 001 481 068 469 288 96 × 2 = 0 + 0.000 002 962 136 938 577 92;
  • 25) 0.000 002 962 136 938 577 92 × 2 = 0 + 0.000 005 924 273 877 155 84;
  • 26) 0.000 005 924 273 877 155 84 × 2 = 0 + 0.000 011 848 547 754 311 68;
  • 27) 0.000 011 848 547 754 311 68 × 2 = 0 + 0.000 023 697 095 508 623 36;
  • 28) 0.000 023 697 095 508 623 36 × 2 = 0 + 0.000 047 394 191 017 246 72;
  • 29) 0.000 047 394 191 017 246 72 × 2 = 0 + 0.000 094 788 382 034 493 44;
  • 30) 0.000 094 788 382 034 493 44 × 2 = 0 + 0.000 189 576 764 068 986 88;
  • 31) 0.000 189 576 764 068 986 88 × 2 = 0 + 0.000 379 153 528 137 973 76;
  • 32) 0.000 379 153 528 137 973 76 × 2 = 0 + 0.000 758 307 056 275 947 52;
  • 33) 0.000 758 307 056 275 947 52 × 2 = 0 + 0.001 516 614 112 551 895 04;
  • 34) 0.001 516 614 112 551 895 04 × 2 = 0 + 0.003 033 228 225 103 790 08;
  • 35) 0.003 033 228 225 103 790 08 × 2 = 0 + 0.006 066 456 450 207 580 16;
  • 36) 0.006 066 456 450 207 580 16 × 2 = 0 + 0.012 132 912 900 415 160 32;
  • 37) 0.012 132 912 900 415 160 32 × 2 = 0 + 0.024 265 825 800 830 320 64;
  • 38) 0.024 265 825 800 830 320 64 × 2 = 0 + 0.048 531 651 601 660 641 28;
  • 39) 0.048 531 651 601 660 641 28 × 2 = 0 + 0.097 063 303 203 321 282 56;
  • 40) 0.097 063 303 203 321 282 56 × 2 = 0 + 0.194 126 606 406 642 565 12;
  • 41) 0.194 126 606 406 642 565 12 × 2 = 0 + 0.388 253 212 813 285 130 24;
  • 42) 0.388 253 212 813 285 130 24 × 2 = 0 + 0.776 506 425 626 570 260 48;
  • 43) 0.776 506 425 626 570 260 48 × 2 = 1 + 0.553 012 851 253 140 520 96;
  • 44) 0.553 012 851 253 140 520 96 × 2 = 1 + 0.106 025 702 506 281 041 92;
  • 45) 0.106 025 702 506 281 041 92 × 2 = 0 + 0.212 051 405 012 562 083 84;
  • 46) 0.212 051 405 012 562 083 84 × 2 = 0 + 0.424 102 810 025 124 167 68;
  • 47) 0.424 102 810 025 124 167 68 × 2 = 0 + 0.848 205 620 050 248 335 36;
  • 48) 0.848 205 620 050 248 335 36 × 2 = 1 + 0.696 411 240 100 496 670 72;
  • 49) 0.696 411 240 100 496 670 72 × 2 = 1 + 0.392 822 480 200 993 341 44;
  • 50) 0.392 822 480 200 993 341 44 × 2 = 0 + 0.785 644 960 401 986 682 88;
  • 51) 0.785 644 960 401 986 682 88 × 2 = 1 + 0.571 289 920 803 973 365 76;
  • 52) 0.571 289 920 803 973 365 76 × 2 = 1 + 0.142 579 841 607 946 731 52;
  • 53) 0.142 579 841 607 946 731 52 × 2 = 0 + 0.285 159 683 215 893 463 04;
  • 54) 0.285 159 683 215 893 463 04 × 2 = 0 + 0.570 319 366 431 786 926 08;
  • 55) 0.570 319 366 431 786 926 08 × 2 = 1 + 0.140 638 732 863 573 852 16;
  • 56) 0.140 638 732 863 573 852 16 × 2 = 0 + 0.281 277 465 727 147 704 32;
  • 57) 0.281 277 465 727 147 704 32 × 2 = 0 + 0.562 554 931 454 295 408 64;
  • 58) 0.562 554 931 454 295 408 64 × 2 = 1 + 0.125 109 862 908 590 817 28;
  • 59) 0.125 109 862 908 590 817 28 × 2 = 0 + 0.250 219 725 817 181 634 56;
  • 60) 0.250 219 725 817 181 634 56 × 2 = 0 + 0.500 439 451 634 363 269 12;
  • 61) 0.500 439 451 634 363 269 12 × 2 = 1 + 0.000 878 903 268 726 538 24;
  • 62) 0.000 878 903 268 726 538 24 × 2 = 0 + 0.001 757 806 537 453 076 48;
  • 63) 0.001 757 806 537 453 076 48 × 2 = 0 + 0.003 515 613 074 906 152 96;
  • 64) 0.003 515 613 074 906 152 96 × 2 = 0 + 0.007 031 226 149 812 305 92;
  • 65) 0.007 031 226 149 812 305 92 × 2 = 0 + 0.014 062 452 299 624 611 84;
  • 66) 0.014 062 452 299 624 611 84 × 2 = 0 + 0.028 124 904 599 249 223 68;
  • 67) 0.028 124 904 599 249 223 68 × 2 = 0 + 0.056 249 809 198 498 447 36;
  • 68) 0.056 249 809 198 498 447 36 × 2 = 0 + 0.112 499 618 396 996 894 72;
  • 69) 0.112 499 618 396 996 894 72 × 2 = 0 + 0.224 999 236 793 993 789 44;
  • 70) 0.224 999 236 793 993 789 44 × 2 = 0 + 0.449 998 473 587 987 578 88;
  • 71) 0.449 998 473 587 987 578 88 × 2 = 0 + 0.899 996 947 175 975 157 76;
  • 72) 0.899 996 947 175 975 157 76 × 2 = 1 + 0.799 993 894 351 950 315 52;
  • 73) 0.799 993 894 351 950 315 52 × 2 = 1 + 0.599 987 788 703 900 631 04;
  • 74) 0.599 987 788 703 900 631 04 × 2 = 1 + 0.199 975 577 407 801 262 08;
  • 75) 0.199 975 577 407 801 262 08 × 2 = 0 + 0.399 951 154 815 602 524 16;
  • 76) 0.399 951 154 815 602 524 16 × 2 = 0 + 0.799 902 309 631 205 048 32;
  • 77) 0.799 902 309 631 205 048 32 × 2 = 1 + 0.599 804 619 262 410 096 64;
  • 78) 0.599 804 619 262 410 096 64 × 2 = 1 + 0.199 609 238 524 820 193 28;
  • 79) 0.199 609 238 524 820 193 28 × 2 = 0 + 0.399 218 477 049 640 386 56;
  • 80) 0.399 218 477 049 640 386 56 × 2 = 0 + 0.798 436 954 099 280 773 12;
  • 81) 0.798 436 954 099 280 773 12 × 2 = 1 + 0.596 873 908 198 561 546 24;
  • 82) 0.596 873 908 198 561 546 24 × 2 = 1 + 0.193 747 816 397 123 092 48;
  • 83) 0.193 747 816 397 123 092 48 × 2 = 0 + 0.387 495 632 794 246 184 96;
  • 84) 0.387 495 632 794 246 184 96 × 2 = 0 + 0.774 991 265 588 492 369 92;
  • 85) 0.774 991 265 588 492 369 92 × 2 = 1 + 0.549 982 531 176 984 739 84;
  • 86) 0.549 982 531 176 984 739 84 × 2 = 1 + 0.099 965 062 353 969 479 68;
  • 87) 0.099 965 062 353 969 479 68 × 2 = 0 + 0.199 930 124 707 938 959 36;
  • 88) 0.199 930 124 707 938 959 36 × 2 = 0 + 0.399 860 249 415 877 918 72;
  • 89) 0.399 860 249 415 877 918 72 × 2 = 0 + 0.799 720 498 831 755 837 44;
  • 90) 0.799 720 498 831 755 837 44 × 2 = 1 + 0.599 440 997 663 511 674 88;
  • 91) 0.599 440 997 663 511 674 88 × 2 = 1 + 0.198 881 995 327 023 349 76;
  • 92) 0.198 881 995 327 023 349 76 × 2 = 0 + 0.397 763 990 654 046 699 52;
  • 93) 0.397 763 990 654 046 699 52 × 2 = 0 + 0.795 527 981 308 093 399 04;
  • 94) 0.795 527 981 308 093 399 04 × 2 = 1 + 0.591 055 962 616 186 798 08;
  • 95) 0.591 055 962 616 186 798 08 × 2 = 1 + 0.182 111 925 232 373 596 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:


0.000 000 000 000 176 557 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1000 0000 0001 1100 1100 1100 1100 0110 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1000 0000 0001 1100 1100 1100 1100 0110 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 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1000 0000 0001 1100 1100 1100 1100 0110 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1000 0000 0001 1100 1100 1100 1100 0110 011(2) × 20 =

1.1000 1101 1001 0010 0100 0000 0000 1110 0110 0110 0110 0011 0011(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 0000 0000 1110 0110 0110 0110 0011 0011


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 0000 0000 1110 0110 0110 0110 0011 0011 =

1000 1101 1001 0010 0100 0000 0000 1110 0110 0110 0110 0011 0011


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 0000 0000 1110 0110 0110 0110 0011 0011


Decimal number -0.000 000 000 000 176 557 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0100 0000 0000 1110 0110 0110 0110 0011 0011

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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