-0.000 000 000 000 176 556 72 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 72(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 72(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 72| = 0.000 000 000 000 176 556 72


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 72.

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 72 × 2 = 0 + 0.000 000 000 000 353 113 44;
  • 2) 0.000 000 000 000 353 113 44 × 2 = 0 + 0.000 000 000 000 706 226 88;
  • 3) 0.000 000 000 000 706 226 88 × 2 = 0 + 0.000 000 000 001 412 453 76;
  • 4) 0.000 000 000 001 412 453 76 × 2 = 0 + 0.000 000 000 002 824 907 52;
  • 5) 0.000 000 000 002 824 907 52 × 2 = 0 + 0.000 000 000 005 649 815 04;
  • 6) 0.000 000 000 005 649 815 04 × 2 = 0 + 0.000 000 000 011 299 630 08;
  • 7) 0.000 000 000 011 299 630 08 × 2 = 0 + 0.000 000 000 022 599 260 16;
  • 8) 0.000 000 000 022 599 260 16 × 2 = 0 + 0.000 000 000 045 198 520 32;
  • 9) 0.000 000 000 045 198 520 32 × 2 = 0 + 0.000 000 000 090 397 040 64;
  • 10) 0.000 000 000 090 397 040 64 × 2 = 0 + 0.000 000 000 180 794 081 28;
  • 11) 0.000 000 000 180 794 081 28 × 2 = 0 + 0.000 000 000 361 588 162 56;
  • 12) 0.000 000 000 361 588 162 56 × 2 = 0 + 0.000 000 000 723 176 325 12;
  • 13) 0.000 000 000 723 176 325 12 × 2 = 0 + 0.000 000 001 446 352 650 24;
  • 14) 0.000 000 001 446 352 650 24 × 2 = 0 + 0.000 000 002 892 705 300 48;
  • 15) 0.000 000 002 892 705 300 48 × 2 = 0 + 0.000 000 005 785 410 600 96;
  • 16) 0.000 000 005 785 410 600 96 × 2 = 0 + 0.000 000 011 570 821 201 92;
  • 17) 0.000 000 011 570 821 201 92 × 2 = 0 + 0.000 000 023 141 642 403 84;
  • 18) 0.000 000 023 141 642 403 84 × 2 = 0 + 0.000 000 046 283 284 807 68;
  • 19) 0.000 000 046 283 284 807 68 × 2 = 0 + 0.000 000 092 566 569 615 36;
  • 20) 0.000 000 092 566 569 615 36 × 2 = 0 + 0.000 000 185 133 139 230 72;
  • 21) 0.000 000 185 133 139 230 72 × 2 = 0 + 0.000 000 370 266 278 461 44;
  • 22) 0.000 000 370 266 278 461 44 × 2 = 0 + 0.000 000 740 532 556 922 88;
  • 23) 0.000 000 740 532 556 922 88 × 2 = 0 + 0.000 001 481 065 113 845 76;
  • 24) 0.000 001 481 065 113 845 76 × 2 = 0 + 0.000 002 962 130 227 691 52;
  • 25) 0.000 002 962 130 227 691 52 × 2 = 0 + 0.000 005 924 260 455 383 04;
  • 26) 0.000 005 924 260 455 383 04 × 2 = 0 + 0.000 011 848 520 910 766 08;
  • 27) 0.000 011 848 520 910 766 08 × 2 = 0 + 0.000 023 697 041 821 532 16;
  • 28) 0.000 023 697 041 821 532 16 × 2 = 0 + 0.000 047 394 083 643 064 32;
  • 29) 0.000 047 394 083 643 064 32 × 2 = 0 + 0.000 094 788 167 286 128 64;
  • 30) 0.000 094 788 167 286 128 64 × 2 = 0 + 0.000 189 576 334 572 257 28;
  • 31) 0.000 189 576 334 572 257 28 × 2 = 0 + 0.000 379 152 669 144 514 56;
  • 32) 0.000 379 152 669 144 514 56 × 2 = 0 + 0.000 758 305 338 289 029 12;
  • 33) 0.000 758 305 338 289 029 12 × 2 = 0 + 0.001 516 610 676 578 058 24;
  • 34) 0.001 516 610 676 578 058 24 × 2 = 0 + 0.003 033 221 353 156 116 48;
  • 35) 0.003 033 221 353 156 116 48 × 2 = 0 + 0.006 066 442 706 312 232 96;
  • 36) 0.006 066 442 706 312 232 96 × 2 = 0 + 0.012 132 885 412 624 465 92;
  • 37) 0.012 132 885 412 624 465 92 × 2 = 0 + 0.024 265 770 825 248 931 84;
  • 38) 0.024 265 770 825 248 931 84 × 2 = 0 + 0.048 531 541 650 497 863 68;
  • 39) 0.048 531 541 650 497 863 68 × 2 = 0 + 0.097 063 083 300 995 727 36;
  • 40) 0.097 063 083 300 995 727 36 × 2 = 0 + 0.194 126 166 601 991 454 72;
  • 41) 0.194 126 166 601 991 454 72 × 2 = 0 + 0.388 252 333 203 982 909 44;
  • 42) 0.388 252 333 203 982 909 44 × 2 = 0 + 0.776 504 666 407 965 818 88;
  • 43) 0.776 504 666 407 965 818 88 × 2 = 1 + 0.553 009 332 815 931 637 76;
  • 44) 0.553 009 332 815 931 637 76 × 2 = 1 + 0.106 018 665 631 863 275 52;
  • 45) 0.106 018 665 631 863 275 52 × 2 = 0 + 0.212 037 331 263 726 551 04;
  • 46) 0.212 037 331 263 726 551 04 × 2 = 0 + 0.424 074 662 527 453 102 08;
  • 47) 0.424 074 662 527 453 102 08 × 2 = 0 + 0.848 149 325 054 906 204 16;
  • 48) 0.848 149 325 054 906 204 16 × 2 = 1 + 0.696 298 650 109 812 408 32;
  • 49) 0.696 298 650 109 812 408 32 × 2 = 1 + 0.392 597 300 219 624 816 64;
  • 50) 0.392 597 300 219 624 816 64 × 2 = 0 + 0.785 194 600 439 249 633 28;
  • 51) 0.785 194 600 439 249 633 28 × 2 = 1 + 0.570 389 200 878 499 266 56;
  • 52) 0.570 389 200 878 499 266 56 × 2 = 1 + 0.140 778 401 756 998 533 12;
  • 53) 0.140 778 401 756 998 533 12 × 2 = 0 + 0.281 556 803 513 997 066 24;
  • 54) 0.281 556 803 513 997 066 24 × 2 = 0 + 0.563 113 607 027 994 132 48;
  • 55) 0.563 113 607 027 994 132 48 × 2 = 1 + 0.126 227 214 055 988 264 96;
  • 56) 0.126 227 214 055 988 264 96 × 2 = 0 + 0.252 454 428 111 976 529 92;
  • 57) 0.252 454 428 111 976 529 92 × 2 = 0 + 0.504 908 856 223 953 059 84;
  • 58) 0.504 908 856 223 953 059 84 × 2 = 1 + 0.009 817 712 447 906 119 68;
  • 59) 0.009 817 712 447 906 119 68 × 2 = 0 + 0.019 635 424 895 812 239 36;
  • 60) 0.019 635 424 895 812 239 36 × 2 = 0 + 0.039 270 849 791 624 478 72;
  • 61) 0.039 270 849 791 624 478 72 × 2 = 0 + 0.078 541 699 583 248 957 44;
  • 62) 0.078 541 699 583 248 957 44 × 2 = 0 + 0.157 083 399 166 497 914 88;
  • 63) 0.157 083 399 166 497 914 88 × 2 = 0 + 0.314 166 798 332 995 829 76;
  • 64) 0.314 166 798 332 995 829 76 × 2 = 0 + 0.628 333 596 665 991 659 52;
  • 65) 0.628 333 596 665 991 659 52 × 2 = 1 + 0.256 667 193 331 983 319 04;
  • 66) 0.256 667 193 331 983 319 04 × 2 = 0 + 0.513 334 386 663 966 638 08;
  • 67) 0.513 334 386 663 966 638 08 × 2 = 1 + 0.026 668 773 327 933 276 16;
  • 68) 0.026 668 773 327 933 276 16 × 2 = 0 + 0.053 337 546 655 866 552 32;
  • 69) 0.053 337 546 655 866 552 32 × 2 = 0 + 0.106 675 093 311 733 104 64;
  • 70) 0.106 675 093 311 733 104 64 × 2 = 0 + 0.213 350 186 623 466 209 28;
  • 71) 0.213 350 186 623 466 209 28 × 2 = 0 + 0.426 700 373 246 932 418 56;
  • 72) 0.426 700 373 246 932 418 56 × 2 = 0 + 0.853 400 746 493 864 837 12;
  • 73) 0.853 400 746 493 864 837 12 × 2 = 1 + 0.706 801 492 987 729 674 24;
  • 74) 0.706 801 492 987 729 674 24 × 2 = 1 + 0.413 602 985 975 459 348 48;
  • 75) 0.413 602 985 975 459 348 48 × 2 = 0 + 0.827 205 971 950 918 696 96;
  • 76) 0.827 205 971 950 918 696 96 × 2 = 1 + 0.654 411 943 901 837 393 92;
  • 77) 0.654 411 943 901 837 393 92 × 2 = 1 + 0.308 823 887 803 674 787 84;
  • 78) 0.308 823 887 803 674 787 84 × 2 = 0 + 0.617 647 775 607 349 575 68;
  • 79) 0.617 647 775 607 349 575 68 × 2 = 1 + 0.235 295 551 214 699 151 36;
  • 80) 0.235 295 551 214 699 151 36 × 2 = 0 + 0.470 591 102 429 398 302 72;
  • 81) 0.470 591 102 429 398 302 72 × 2 = 0 + 0.941 182 204 858 796 605 44;
  • 82) 0.941 182 204 858 796 605 44 × 2 = 1 + 0.882 364 409 717 593 210 88;
  • 83) 0.882 364 409 717 593 210 88 × 2 = 1 + 0.764 728 819 435 186 421 76;
  • 84) 0.764 728 819 435 186 421 76 × 2 = 1 + 0.529 457 638 870 372 843 52;
  • 85) 0.529 457 638 870 372 843 52 × 2 = 1 + 0.058 915 277 740 745 687 04;
  • 86) 0.058 915 277 740 745 687 04 × 2 = 0 + 0.117 830 555 481 491 374 08;
  • 87) 0.117 830 555 481 491 374 08 × 2 = 0 + 0.235 661 110 962 982 748 16;
  • 88) 0.235 661 110 962 982 748 16 × 2 = 0 + 0.471 322 221 925 965 496 32;
  • 89) 0.471 322 221 925 965 496 32 × 2 = 0 + 0.942 644 443 851 930 992 64;
  • 90) 0.942 644 443 851 930 992 64 × 2 = 1 + 0.885 288 887 703 861 985 28;
  • 91) 0.885 288 887 703 861 985 28 × 2 = 1 + 0.770 577 775 407 723 970 56;
  • 92) 0.770 577 775 407 723 970 56 × 2 = 1 + 0.541 155 550 815 447 941 12;
  • 93) 0.541 155 550 815 447 941 12 × 2 = 1 + 0.082 311 101 630 895 882 24;
  • 94) 0.082 311 101 630 895 882 24 × 2 = 0 + 0.164 622 203 261 791 764 48;
  • 95) 0.164 622 203 261 791 764 48 × 2 = 0 + 0.329 244 406 523 583 528 96;

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 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0000 1010 0000 1101 1010 0111 1000 0111 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0000 1010 0000 1101 1010 0111 1000 0111 100(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 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0000 1010 0000 1101 1010 0111 1000 0111 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0000 1010 0000 1101 1010 0111 1000 0111 100(2) × 20 =

1.1000 1101 1001 0010 0000 0101 0000 0110 1101 0011 1100 0011 1100(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 0000 0101 0000 0110 1101 0011 1100 0011 1100


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 0000 0101 0000 0110 1101 0011 1100 0011 1100 =

1000 1101 1001 0010 0000 0101 0000 0110 1101 0011 1100 0011 1100


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 0000 0101 0000 0110 1101 0011 1100 0011 1100


Decimal number -0.000 000 000 000 176 556 72 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0000 0101 0000 0110 1101 0011 1100 0011 1100

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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