-0.000 000 000 000 176 556 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 8(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 8(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 8| = 0.000 000 000 000 176 556 8


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 8.

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 8 × 2 = 0 + 0.000 000 000 000 353 113 6;
  • 2) 0.000 000 000 000 353 113 6 × 2 = 0 + 0.000 000 000 000 706 227 2;
  • 3) 0.000 000 000 000 706 227 2 × 2 = 0 + 0.000 000 000 001 412 454 4;
  • 4) 0.000 000 000 001 412 454 4 × 2 = 0 + 0.000 000 000 002 824 908 8;
  • 5) 0.000 000 000 002 824 908 8 × 2 = 0 + 0.000 000 000 005 649 817 6;
  • 6) 0.000 000 000 005 649 817 6 × 2 = 0 + 0.000 000 000 011 299 635 2;
  • 7) 0.000 000 000 011 299 635 2 × 2 = 0 + 0.000 000 000 022 599 270 4;
  • 8) 0.000 000 000 022 599 270 4 × 2 = 0 + 0.000 000 000 045 198 540 8;
  • 9) 0.000 000 000 045 198 540 8 × 2 = 0 + 0.000 000 000 090 397 081 6;
  • 10) 0.000 000 000 090 397 081 6 × 2 = 0 + 0.000 000 000 180 794 163 2;
  • 11) 0.000 000 000 180 794 163 2 × 2 = 0 + 0.000 000 000 361 588 326 4;
  • 12) 0.000 000 000 361 588 326 4 × 2 = 0 + 0.000 000 000 723 176 652 8;
  • 13) 0.000 000 000 723 176 652 8 × 2 = 0 + 0.000 000 001 446 353 305 6;
  • 14) 0.000 000 001 446 353 305 6 × 2 = 0 + 0.000 000 002 892 706 611 2;
  • 15) 0.000 000 002 892 706 611 2 × 2 = 0 + 0.000 000 005 785 413 222 4;
  • 16) 0.000 000 005 785 413 222 4 × 2 = 0 + 0.000 000 011 570 826 444 8;
  • 17) 0.000 000 011 570 826 444 8 × 2 = 0 + 0.000 000 023 141 652 889 6;
  • 18) 0.000 000 023 141 652 889 6 × 2 = 0 + 0.000 000 046 283 305 779 2;
  • 19) 0.000 000 046 283 305 779 2 × 2 = 0 + 0.000 000 092 566 611 558 4;
  • 20) 0.000 000 092 566 611 558 4 × 2 = 0 + 0.000 000 185 133 223 116 8;
  • 21) 0.000 000 185 133 223 116 8 × 2 = 0 + 0.000 000 370 266 446 233 6;
  • 22) 0.000 000 370 266 446 233 6 × 2 = 0 + 0.000 000 740 532 892 467 2;
  • 23) 0.000 000 740 532 892 467 2 × 2 = 0 + 0.000 001 481 065 784 934 4;
  • 24) 0.000 001 481 065 784 934 4 × 2 = 0 + 0.000 002 962 131 569 868 8;
  • 25) 0.000 002 962 131 569 868 8 × 2 = 0 + 0.000 005 924 263 139 737 6;
  • 26) 0.000 005 924 263 139 737 6 × 2 = 0 + 0.000 011 848 526 279 475 2;
  • 27) 0.000 011 848 526 279 475 2 × 2 = 0 + 0.000 023 697 052 558 950 4;
  • 28) 0.000 023 697 052 558 950 4 × 2 = 0 + 0.000 047 394 105 117 900 8;
  • 29) 0.000 047 394 105 117 900 8 × 2 = 0 + 0.000 094 788 210 235 801 6;
  • 30) 0.000 094 788 210 235 801 6 × 2 = 0 + 0.000 189 576 420 471 603 2;
  • 31) 0.000 189 576 420 471 603 2 × 2 = 0 + 0.000 379 152 840 943 206 4;
  • 32) 0.000 379 152 840 943 206 4 × 2 = 0 + 0.000 758 305 681 886 412 8;
  • 33) 0.000 758 305 681 886 412 8 × 2 = 0 + 0.001 516 611 363 772 825 6;
  • 34) 0.001 516 611 363 772 825 6 × 2 = 0 + 0.003 033 222 727 545 651 2;
  • 35) 0.003 033 222 727 545 651 2 × 2 = 0 + 0.006 066 445 455 091 302 4;
  • 36) 0.006 066 445 455 091 302 4 × 2 = 0 + 0.012 132 890 910 182 604 8;
  • 37) 0.012 132 890 910 182 604 8 × 2 = 0 + 0.024 265 781 820 365 209 6;
  • 38) 0.024 265 781 820 365 209 6 × 2 = 0 + 0.048 531 563 640 730 419 2;
  • 39) 0.048 531 563 640 730 419 2 × 2 = 0 + 0.097 063 127 281 460 838 4;
  • 40) 0.097 063 127 281 460 838 4 × 2 = 0 + 0.194 126 254 562 921 676 8;
  • 41) 0.194 126 254 562 921 676 8 × 2 = 0 + 0.388 252 509 125 843 353 6;
  • 42) 0.388 252 509 125 843 353 6 × 2 = 0 + 0.776 505 018 251 686 707 2;
  • 43) 0.776 505 018 251 686 707 2 × 2 = 1 + 0.553 010 036 503 373 414 4;
  • 44) 0.553 010 036 503 373 414 4 × 2 = 1 + 0.106 020 073 006 746 828 8;
  • 45) 0.106 020 073 006 746 828 8 × 2 = 0 + 0.212 040 146 013 493 657 6;
  • 46) 0.212 040 146 013 493 657 6 × 2 = 0 + 0.424 080 292 026 987 315 2;
  • 47) 0.424 080 292 026 987 315 2 × 2 = 0 + 0.848 160 584 053 974 630 4;
  • 48) 0.848 160 584 053 974 630 4 × 2 = 1 + 0.696 321 168 107 949 260 8;
  • 49) 0.696 321 168 107 949 260 8 × 2 = 1 + 0.392 642 336 215 898 521 6;
  • 50) 0.392 642 336 215 898 521 6 × 2 = 0 + 0.785 284 672 431 797 043 2;
  • 51) 0.785 284 672 431 797 043 2 × 2 = 1 + 0.570 569 344 863 594 086 4;
  • 52) 0.570 569 344 863 594 086 4 × 2 = 1 + 0.141 138 689 727 188 172 8;
  • 53) 0.141 138 689 727 188 172 8 × 2 = 0 + 0.282 277 379 454 376 345 6;
  • 54) 0.282 277 379 454 376 345 6 × 2 = 0 + 0.564 554 758 908 752 691 2;
  • 55) 0.564 554 758 908 752 691 2 × 2 = 1 + 0.129 109 517 817 505 382 4;
  • 56) 0.129 109 517 817 505 382 4 × 2 = 0 + 0.258 219 035 635 010 764 8;
  • 57) 0.258 219 035 635 010 764 8 × 2 = 0 + 0.516 438 071 270 021 529 6;
  • 58) 0.516 438 071 270 021 529 6 × 2 = 1 + 0.032 876 142 540 043 059 2;
  • 59) 0.032 876 142 540 043 059 2 × 2 = 0 + 0.065 752 285 080 086 118 4;
  • 60) 0.065 752 285 080 086 118 4 × 2 = 0 + 0.131 504 570 160 172 236 8;
  • 61) 0.131 504 570 160 172 236 8 × 2 = 0 + 0.263 009 140 320 344 473 6;
  • 62) 0.263 009 140 320 344 473 6 × 2 = 0 + 0.526 018 280 640 688 947 2;
  • 63) 0.526 018 280 640 688 947 2 × 2 = 1 + 0.052 036 561 281 377 894 4;
  • 64) 0.052 036 561 281 377 894 4 × 2 = 0 + 0.104 073 122 562 755 788 8;
  • 65) 0.104 073 122 562 755 788 8 × 2 = 0 + 0.208 146 245 125 511 577 6;
  • 66) 0.208 146 245 125 511 577 6 × 2 = 0 + 0.416 292 490 251 023 155 2;
  • 67) 0.416 292 490 251 023 155 2 × 2 = 0 + 0.832 584 980 502 046 310 4;
  • 68) 0.832 584 980 502 046 310 4 × 2 = 1 + 0.665 169 961 004 092 620 8;
  • 69) 0.665 169 961 004 092 620 8 × 2 = 1 + 0.330 339 922 008 185 241 6;
  • 70) 0.330 339 922 008 185 241 6 × 2 = 0 + 0.660 679 844 016 370 483 2;
  • 71) 0.660 679 844 016 370 483 2 × 2 = 1 + 0.321 359 688 032 740 966 4;
  • 72) 0.321 359 688 032 740 966 4 × 2 = 0 + 0.642 719 376 065 481 932 8;
  • 73) 0.642 719 376 065 481 932 8 × 2 = 1 + 0.285 438 752 130 963 865 6;
  • 74) 0.285 438 752 130 963 865 6 × 2 = 0 + 0.570 877 504 261 927 731 2;
  • 75) 0.570 877 504 261 927 731 2 × 2 = 1 + 0.141 755 008 523 855 462 4;
  • 76) 0.141 755 008 523 855 462 4 × 2 = 0 + 0.283 510 017 047 710 924 8;
  • 77) 0.283 510 017 047 710 924 8 × 2 = 0 + 0.567 020 034 095 421 849 6;
  • 78) 0.567 020 034 095 421 849 6 × 2 = 1 + 0.134 040 068 190 843 699 2;
  • 79) 0.134 040 068 190 843 699 2 × 2 = 0 + 0.268 080 136 381 687 398 4;
  • 80) 0.268 080 136 381 687 398 4 × 2 = 0 + 0.536 160 272 763 374 796 8;
  • 81) 0.536 160 272 763 374 796 8 × 2 = 1 + 0.072 320 545 526 749 593 6;
  • 82) 0.072 320 545 526 749 593 6 × 2 = 0 + 0.144 641 091 053 499 187 2;
  • 83) 0.144 641 091 053 499 187 2 × 2 = 0 + 0.289 282 182 106 998 374 4;
  • 84) 0.289 282 182 106 998 374 4 × 2 = 0 + 0.578 564 364 213 996 748 8;
  • 85) 0.578 564 364 213 996 748 8 × 2 = 1 + 0.157 128 728 427 993 497 6;
  • 86) 0.157 128 728 427 993 497 6 × 2 = 0 + 0.314 257 456 855 986 995 2;
  • 87) 0.314 257 456 855 986 995 2 × 2 = 0 + 0.628 514 913 711 973 990 4;
  • 88) 0.628 514 913 711 973 990 4 × 2 = 1 + 0.257 029 827 423 947 980 8;
  • 89) 0.257 029 827 423 947 980 8 × 2 = 0 + 0.514 059 654 847 895 961 6;
  • 90) 0.514 059 654 847 895 961 6 × 2 = 1 + 0.028 119 309 695 791 923 2;
  • 91) 0.028 119 309 695 791 923 2 × 2 = 0 + 0.056 238 619 391 583 846 4;
  • 92) 0.056 238 619 391 583 846 4 × 2 = 0 + 0.112 477 238 783 167 692 8;
  • 93) 0.112 477 238 783 167 692 8 × 2 = 0 + 0.224 954 477 566 335 385 6;
  • 94) 0.224 954 477 566 335 385 6 × 2 = 0 + 0.449 908 955 132 670 771 2;
  • 95) 0.449 908 955 132 670 771 2 × 2 = 0 + 0.899 817 910 265 341 542 4;

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 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0010 0001 1010 1010 0100 1000 1001 0100 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0010 0001 1010 1010 0100 1000 1001 0100 000(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 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0010 0001 1010 1010 0100 1000 1001 0100 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0010 0001 1010 1010 0100 1000 1001 0100 000(2) × 20 =

1.1000 1101 1001 0010 0001 0000 1101 0101 0010 0100 0100 1010 0000(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 0000 1101 0101 0010 0100 0100 1010 0000


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 0000 1101 0101 0010 0100 0100 1010 0000 =

1000 1101 1001 0010 0001 0000 1101 0101 0010 0100 0100 1010 0000


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 0000 1101 0101 0010 0100 0100 1010 0000


Decimal number -0.000 000 000 000 176 556 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0001 0000 1101 0101 0010 0100 0100 1010 0000

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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