-0.000 000 000 000 176 547 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 547 6(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
-0.000 000 000 000 176 547 6(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:

|-0.000 000 000 000 176 547 6| = 0.000 000 000 000 176 547 6


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: 0.000 000 000 000 176 547 6.

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) 0.000 000 000 000 176 547 6 × 2 = 0 + 0.000 000 000 000 353 095 2;
  • 2) 0.000 000 000 000 353 095 2 × 2 = 0 + 0.000 000 000 000 706 190 4;
  • 3) 0.000 000 000 000 706 190 4 × 2 = 0 + 0.000 000 000 001 412 380 8;
  • 4) 0.000 000 000 001 412 380 8 × 2 = 0 + 0.000 000 000 002 824 761 6;
  • 5) 0.000 000 000 002 824 761 6 × 2 = 0 + 0.000 000 000 005 649 523 2;
  • 6) 0.000 000 000 005 649 523 2 × 2 = 0 + 0.000 000 000 011 299 046 4;
  • 7) 0.000 000 000 011 299 046 4 × 2 = 0 + 0.000 000 000 022 598 092 8;
  • 8) 0.000 000 000 022 598 092 8 × 2 = 0 + 0.000 000 000 045 196 185 6;
  • 9) 0.000 000 000 045 196 185 6 × 2 = 0 + 0.000 000 000 090 392 371 2;
  • 10) 0.000 000 000 090 392 371 2 × 2 = 0 + 0.000 000 000 180 784 742 4;
  • 11) 0.000 000 000 180 784 742 4 × 2 = 0 + 0.000 000 000 361 569 484 8;
  • 12) 0.000 000 000 361 569 484 8 × 2 = 0 + 0.000 000 000 723 138 969 6;
  • 13) 0.000 000 000 723 138 969 6 × 2 = 0 + 0.000 000 001 446 277 939 2;
  • 14) 0.000 000 001 446 277 939 2 × 2 = 0 + 0.000 000 002 892 555 878 4;
  • 15) 0.000 000 002 892 555 878 4 × 2 = 0 + 0.000 000 005 785 111 756 8;
  • 16) 0.000 000 005 785 111 756 8 × 2 = 0 + 0.000 000 011 570 223 513 6;
  • 17) 0.000 000 011 570 223 513 6 × 2 = 0 + 0.000 000 023 140 447 027 2;
  • 18) 0.000 000 023 140 447 027 2 × 2 = 0 + 0.000 000 046 280 894 054 4;
  • 19) 0.000 000 046 280 894 054 4 × 2 = 0 + 0.000 000 092 561 788 108 8;
  • 20) 0.000 000 092 561 788 108 8 × 2 = 0 + 0.000 000 185 123 576 217 6;
  • 21) 0.000 000 185 123 576 217 6 × 2 = 0 + 0.000 000 370 247 152 435 2;
  • 22) 0.000 000 370 247 152 435 2 × 2 = 0 + 0.000 000 740 494 304 870 4;
  • 23) 0.000 000 740 494 304 870 4 × 2 = 0 + 0.000 001 480 988 609 740 8;
  • 24) 0.000 001 480 988 609 740 8 × 2 = 0 + 0.000 002 961 977 219 481 6;
  • 25) 0.000 002 961 977 219 481 6 × 2 = 0 + 0.000 005 923 954 438 963 2;
  • 26) 0.000 005 923 954 438 963 2 × 2 = 0 + 0.000 011 847 908 877 926 4;
  • 27) 0.000 011 847 908 877 926 4 × 2 = 0 + 0.000 023 695 817 755 852 8;
  • 28) 0.000 023 695 817 755 852 8 × 2 = 0 + 0.000 047 391 635 511 705 6;
  • 29) 0.000 047 391 635 511 705 6 × 2 = 0 + 0.000 094 783 271 023 411 2;
  • 30) 0.000 094 783 271 023 411 2 × 2 = 0 + 0.000 189 566 542 046 822 4;
  • 31) 0.000 189 566 542 046 822 4 × 2 = 0 + 0.000 379 133 084 093 644 8;
  • 32) 0.000 379 133 084 093 644 8 × 2 = 0 + 0.000 758 266 168 187 289 6;
  • 33) 0.000 758 266 168 187 289 6 × 2 = 0 + 0.001 516 532 336 374 579 2;
  • 34) 0.001 516 532 336 374 579 2 × 2 = 0 + 0.003 033 064 672 749 158 4;
  • 35) 0.003 033 064 672 749 158 4 × 2 = 0 + 0.006 066 129 345 498 316 8;
  • 36) 0.006 066 129 345 498 316 8 × 2 = 0 + 0.012 132 258 690 996 633 6;
  • 37) 0.012 132 258 690 996 633 6 × 2 = 0 + 0.024 264 517 381 993 267 2;
  • 38) 0.024 264 517 381 993 267 2 × 2 = 0 + 0.048 529 034 763 986 534 4;
  • 39) 0.048 529 034 763 986 534 4 × 2 = 0 + 0.097 058 069 527 973 068 8;
  • 40) 0.097 058 069 527 973 068 8 × 2 = 0 + 0.194 116 139 055 946 137 6;
  • 41) 0.194 116 139 055 946 137 6 × 2 = 0 + 0.388 232 278 111 892 275 2;
  • 42) 0.388 232 278 111 892 275 2 × 2 = 0 + 0.776 464 556 223 784 550 4;
  • 43) 0.776 464 556 223 784 550 4 × 2 = 1 + 0.552 929 112 447 569 100 8;
  • 44) 0.552 929 112 447 569 100 8 × 2 = 1 + 0.105 858 224 895 138 201 6;
  • 45) 0.105 858 224 895 138 201 6 × 2 = 0 + 0.211 716 449 790 276 403 2;
  • 46) 0.211 716 449 790 276 403 2 × 2 = 0 + 0.423 432 899 580 552 806 4;
  • 47) 0.423 432 899 580 552 806 4 × 2 = 0 + 0.846 865 799 161 105 612 8;
  • 48) 0.846 865 799 161 105 612 8 × 2 = 1 + 0.693 731 598 322 211 225 6;
  • 49) 0.693 731 598 322 211 225 6 × 2 = 1 + 0.387 463 196 644 422 451 2;
  • 50) 0.387 463 196 644 422 451 2 × 2 = 0 + 0.774 926 393 288 844 902 4;
  • 51) 0.774 926 393 288 844 902 4 × 2 = 1 + 0.549 852 786 577 689 804 8;
  • 52) 0.549 852 786 577 689 804 8 × 2 = 1 + 0.099 705 573 155 379 609 6;
  • 53) 0.099 705 573 155 379 609 6 × 2 = 0 + 0.199 411 146 310 759 219 2;
  • 54) 0.199 411 146 310 759 219 2 × 2 = 0 + 0.398 822 292 621 518 438 4;
  • 55) 0.398 822 292 621 518 438 4 × 2 = 0 + 0.797 644 585 243 036 876 8;
  • 56) 0.797 644 585 243 036 876 8 × 2 = 1 + 0.595 289 170 486 073 753 6;
  • 57) 0.595 289 170 486 073 753 6 × 2 = 1 + 0.190 578 340 972 147 507 2;
  • 58) 0.190 578 340 972 147 507 2 × 2 = 0 + 0.381 156 681 944 295 014 4;
  • 59) 0.381 156 681 944 295 014 4 × 2 = 0 + 0.762 313 363 888 590 028 8;
  • 60) 0.762 313 363 888 590 028 8 × 2 = 1 + 0.524 626 727 777 180 057 6;
  • 61) 0.524 626 727 777 180 057 6 × 2 = 1 + 0.049 253 455 554 360 115 2;
  • 62) 0.049 253 455 554 360 115 2 × 2 = 0 + 0.098 506 911 108 720 230 4;
  • 63) 0.098 506 911 108 720 230 4 × 2 = 0 + 0.197 013 822 217 440 460 8;
  • 64) 0.197 013 822 217 440 460 8 × 2 = 0 + 0.394 027 644 434 880 921 6;
  • 65) 0.394 027 644 434 880 921 6 × 2 = 0 + 0.788 055 288 869 761 843 2;
  • 66) 0.788 055 288 869 761 843 2 × 2 = 1 + 0.576 110 577 739 523 686 4;
  • 67) 0.576 110 577 739 523 686 4 × 2 = 1 + 0.152 221 155 479 047 372 8;
  • 68) 0.152 221 155 479 047 372 8 × 2 = 0 + 0.304 442 310 958 094 745 6;
  • 69) 0.304 442 310 958 094 745 6 × 2 = 0 + 0.608 884 621 916 189 491 2;
  • 70) 0.608 884 621 916 189 491 2 × 2 = 1 + 0.217 769 243 832 378 982 4;
  • 71) 0.217 769 243 832 378 982 4 × 2 = 0 + 0.435 538 487 664 757 964 8;
  • 72) 0.435 538 487 664 757 964 8 × 2 = 0 + 0.871 076 975 329 515 929 6;
  • 73) 0.871 076 975 329 515 929 6 × 2 = 1 + 0.742 153 950 659 031 859 2;
  • 74) 0.742 153 950 659 031 859 2 × 2 = 1 + 0.484 307 901 318 063 718 4;
  • 75) 0.484 307 901 318 063 718 4 × 2 = 0 + 0.968 615 802 636 127 436 8;
  • 76) 0.968 615 802 636 127 436 8 × 2 = 1 + 0.937 231 605 272 254 873 6;
  • 77) 0.937 231 605 272 254 873 6 × 2 = 1 + 0.874 463 210 544 509 747 2;
  • 78) 0.874 463 210 544 509 747 2 × 2 = 1 + 0.748 926 421 089 019 494 4;
  • 79) 0.748 926 421 089 019 494 4 × 2 = 1 + 0.497 852 842 178 038 988 8;
  • 80) 0.497 852 842 178 038 988 8 × 2 = 0 + 0.995 705 684 356 077 977 6;
  • 81) 0.995 705 684 356 077 977 6 × 2 = 1 + 0.991 411 368 712 155 955 2;
  • 82) 0.991 411 368 712 155 955 2 × 2 = 1 + 0.982 822 737 424 311 910 4;
  • 83) 0.982 822 737 424 311 910 4 × 2 = 1 + 0.965 645 474 848 623 820 8;
  • 84) 0.965 645 474 848 623 820 8 × 2 = 1 + 0.931 290 949 697 247 641 6;
  • 85) 0.931 290 949 697 247 641 6 × 2 = 1 + 0.862 581 899 394 495 283 2;
  • 86) 0.862 581 899 394 495 283 2 × 2 = 1 + 0.725 163 798 788 990 566 4;
  • 87) 0.725 163 798 788 990 566 4 × 2 = 1 + 0.450 327 597 577 981 132 8;
  • 88) 0.450 327 597 577 981 132 8 × 2 = 0 + 0.900 655 195 155 962 265 6;
  • 89) 0.900 655 195 155 962 265 6 × 2 = 1 + 0.801 310 390 311 924 531 2;
  • 90) 0.801 310 390 311 924 531 2 × 2 = 1 + 0.602 620 780 623 849 062 4;
  • 91) 0.602 620 780 623 849 062 4 × 2 = 1 + 0.205 241 561 247 698 124 8;
  • 92) 0.205 241 561 247 698 124 8 × 2 = 0 + 0.410 483 122 495 396 249 6;
  • 93) 0.410 483 122 495 396 249 6 × 2 = 0 + 0.820 966 244 990 792 499 2;
  • 94) 0.820 966 244 990 792 499 2 × 2 = 1 + 0.641 932 489 981 584 998 4;
  • 95) 0.641 932 489 981 584 998 4 × 2 = 1 + 0.283 864 979 963 169 996 8;

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:


0.000 000 000 000 176 547 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1001 1000 0110 0100 1101 1110 1111 1110 1110 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 547 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1001 1000 0110 0100 1101 1110 1111 1110 1110 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:


0.000 000 000 000 176 547 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1001 1000 0110 0100 1101 1110 1111 1110 1110 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1001 1000 0110 0100 1101 1110 1111 1110 1110 011(2) × 20 =

1.1000 1101 1000 1100 1100 0011 0010 0110 1111 0111 1111 0111 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 1000 1100 1100 0011 0010 0110 1111 0111 1111 0111 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 1000 1100 1100 0011 0010 0110 1111 0111 1111 0111 0011 =

1000 1101 1000 1100 1100 0011 0010 0110 1111 0111 1111 0111 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 1000 1100 1100 0011 0010 0110 1111 0111 1111 0111 0011


Decimal number -0.000 000 000 000 176 547 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1100 1100 0011 0010 0110 1111 0111 1111 0111 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