-0.000 000 000 000 176 562 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 562 4(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
-0.000 000 000 000 176 562 4(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:

|-0.000 000 000 000 176 562 4| = 0.000 000 000 000 176 562 4


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: 0.000 000 000 000 176 562 4.

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) 0.000 000 000 000 176 562 4 × 2 = 0 + 0.000 000 000 000 353 124 8;
  • 2) 0.000 000 000 000 353 124 8 × 2 = 0 + 0.000 000 000 000 706 249 6;
  • 3) 0.000 000 000 000 706 249 6 × 2 = 0 + 0.000 000 000 001 412 499 2;
  • 4) 0.000 000 000 001 412 499 2 × 2 = 0 + 0.000 000 000 002 824 998 4;
  • 5) 0.000 000 000 002 824 998 4 × 2 = 0 + 0.000 000 000 005 649 996 8;
  • 6) 0.000 000 000 005 649 996 8 × 2 = 0 + 0.000 000 000 011 299 993 6;
  • 7) 0.000 000 000 011 299 993 6 × 2 = 0 + 0.000 000 000 022 599 987 2;
  • 8) 0.000 000 000 022 599 987 2 × 2 = 0 + 0.000 000 000 045 199 974 4;
  • 9) 0.000 000 000 045 199 974 4 × 2 = 0 + 0.000 000 000 090 399 948 8;
  • 10) 0.000 000 000 090 399 948 8 × 2 = 0 + 0.000 000 000 180 799 897 6;
  • 11) 0.000 000 000 180 799 897 6 × 2 = 0 + 0.000 000 000 361 599 795 2;
  • 12) 0.000 000 000 361 599 795 2 × 2 = 0 + 0.000 000 000 723 199 590 4;
  • 13) 0.000 000 000 723 199 590 4 × 2 = 0 + 0.000 000 001 446 399 180 8;
  • 14) 0.000 000 001 446 399 180 8 × 2 = 0 + 0.000 000 002 892 798 361 6;
  • 15) 0.000 000 002 892 798 361 6 × 2 = 0 + 0.000 000 005 785 596 723 2;
  • 16) 0.000 000 005 785 596 723 2 × 2 = 0 + 0.000 000 011 571 193 446 4;
  • 17) 0.000 000 011 571 193 446 4 × 2 = 0 + 0.000 000 023 142 386 892 8;
  • 18) 0.000 000 023 142 386 892 8 × 2 = 0 + 0.000 000 046 284 773 785 6;
  • 19) 0.000 000 046 284 773 785 6 × 2 = 0 + 0.000 000 092 569 547 571 2;
  • 20) 0.000 000 092 569 547 571 2 × 2 = 0 + 0.000 000 185 139 095 142 4;
  • 21) 0.000 000 185 139 095 142 4 × 2 = 0 + 0.000 000 370 278 190 284 8;
  • 22) 0.000 000 370 278 190 284 8 × 2 = 0 + 0.000 000 740 556 380 569 6;
  • 23) 0.000 000 740 556 380 569 6 × 2 = 0 + 0.000 001 481 112 761 139 2;
  • 24) 0.000 001 481 112 761 139 2 × 2 = 0 + 0.000 002 962 225 522 278 4;
  • 25) 0.000 002 962 225 522 278 4 × 2 = 0 + 0.000 005 924 451 044 556 8;
  • 26) 0.000 005 924 451 044 556 8 × 2 = 0 + 0.000 011 848 902 089 113 6;
  • 27) 0.000 011 848 902 089 113 6 × 2 = 0 + 0.000 023 697 804 178 227 2;
  • 28) 0.000 023 697 804 178 227 2 × 2 = 0 + 0.000 047 395 608 356 454 4;
  • 29) 0.000 047 395 608 356 454 4 × 2 = 0 + 0.000 094 791 216 712 908 8;
  • 30) 0.000 094 791 216 712 908 8 × 2 = 0 + 0.000 189 582 433 425 817 6;
  • 31) 0.000 189 582 433 425 817 6 × 2 = 0 + 0.000 379 164 866 851 635 2;
  • 32) 0.000 379 164 866 851 635 2 × 2 = 0 + 0.000 758 329 733 703 270 4;
  • 33) 0.000 758 329 733 703 270 4 × 2 = 0 + 0.001 516 659 467 406 540 8;
  • 34) 0.001 516 659 467 406 540 8 × 2 = 0 + 0.003 033 318 934 813 081 6;
  • 35) 0.003 033 318 934 813 081 6 × 2 = 0 + 0.006 066 637 869 626 163 2;
  • 36) 0.006 066 637 869 626 163 2 × 2 = 0 + 0.012 133 275 739 252 326 4;
  • 37) 0.012 133 275 739 252 326 4 × 2 = 0 + 0.024 266 551 478 504 652 8;
  • 38) 0.024 266 551 478 504 652 8 × 2 = 0 + 0.048 533 102 957 009 305 6;
  • 39) 0.048 533 102 957 009 305 6 × 2 = 0 + 0.097 066 205 914 018 611 2;
  • 40) 0.097 066 205 914 018 611 2 × 2 = 0 + 0.194 132 411 828 037 222 4;
  • 41) 0.194 132 411 828 037 222 4 × 2 = 0 + 0.388 264 823 656 074 444 8;
  • 42) 0.388 264 823 656 074 444 8 × 2 = 0 + 0.776 529 647 312 148 889 6;
  • 43) 0.776 529 647 312 148 889 6 × 2 = 1 + 0.553 059 294 624 297 779 2;
  • 44) 0.553 059 294 624 297 779 2 × 2 = 1 + 0.106 118 589 248 595 558 4;
  • 45) 0.106 118 589 248 595 558 4 × 2 = 0 + 0.212 237 178 497 191 116 8;
  • 46) 0.212 237 178 497 191 116 8 × 2 = 0 + 0.424 474 356 994 382 233 6;
  • 47) 0.424 474 356 994 382 233 6 × 2 = 0 + 0.848 948 713 988 764 467 2;
  • 48) 0.848 948 713 988 764 467 2 × 2 = 1 + 0.697 897 427 977 528 934 4;
  • 49) 0.697 897 427 977 528 934 4 × 2 = 1 + 0.395 794 855 955 057 868 8;
  • 50) 0.395 794 855 955 057 868 8 × 2 = 0 + 0.791 589 711 910 115 737 6;
  • 51) 0.791 589 711 910 115 737 6 × 2 = 1 + 0.583 179 423 820 231 475 2;
  • 52) 0.583 179 423 820 231 475 2 × 2 = 1 + 0.166 358 847 640 462 950 4;
  • 53) 0.166 358 847 640 462 950 4 × 2 = 0 + 0.332 717 695 280 925 900 8;
  • 54) 0.332 717 695 280 925 900 8 × 2 = 0 + 0.665 435 390 561 851 801 6;
  • 55) 0.665 435 390 561 851 801 6 × 2 = 1 + 0.330 870 781 123 703 603 2;
  • 56) 0.330 870 781 123 703 603 2 × 2 = 0 + 0.661 741 562 247 407 206 4;
  • 57) 0.661 741 562 247 407 206 4 × 2 = 1 + 0.323 483 124 494 814 412 8;
  • 58) 0.323 483 124 494 814 412 8 × 2 = 0 + 0.646 966 248 989 628 825 6;
  • 59) 0.646 966 248 989 628 825 6 × 2 = 1 + 0.293 932 497 979 257 651 2;
  • 60) 0.293 932 497 979 257 651 2 × 2 = 0 + 0.587 864 995 958 515 302 4;
  • 61) 0.587 864 995 958 515 302 4 × 2 = 1 + 0.175 729 991 917 030 604 8;
  • 62) 0.175 729 991 917 030 604 8 × 2 = 0 + 0.351 459 983 834 061 209 6;
  • 63) 0.351 459 983 834 061 209 6 × 2 = 0 + 0.702 919 967 668 122 419 2;
  • 64) 0.702 919 967 668 122 419 2 × 2 = 1 + 0.405 839 935 336 244 838 4;
  • 65) 0.405 839 935 336 244 838 4 × 2 = 0 + 0.811 679 870 672 489 676 8;
  • 66) 0.811 679 870 672 489 676 8 × 2 = 1 + 0.623 359 741 344 979 353 6;
  • 67) 0.623 359 741 344 979 353 6 × 2 = 1 + 0.246 719 482 689 958 707 2;
  • 68) 0.246 719 482 689 958 707 2 × 2 = 0 + 0.493 438 965 379 917 414 4;
  • 69) 0.493 438 965 379 917 414 4 × 2 = 0 + 0.986 877 930 759 834 828 8;
  • 70) 0.986 877 930 759 834 828 8 × 2 = 1 + 0.973 755 861 519 669 657 6;
  • 71) 0.973 755 861 519 669 657 6 × 2 = 1 + 0.947 511 723 039 339 315 2;
  • 72) 0.947 511 723 039 339 315 2 × 2 = 1 + 0.895 023 446 078 678 630 4;
  • 73) 0.895 023 446 078 678 630 4 × 2 = 1 + 0.790 046 892 157 357 260 8;
  • 74) 0.790 046 892 157 357 260 8 × 2 = 1 + 0.580 093 784 314 714 521 6;
  • 75) 0.580 093 784 314 714 521 6 × 2 = 1 + 0.160 187 568 629 429 043 2;
  • 76) 0.160 187 568 629 429 043 2 × 2 = 0 + 0.320 375 137 258 858 086 4;
  • 77) 0.320 375 137 258 858 086 4 × 2 = 0 + 0.640 750 274 517 716 172 8;
  • 78) 0.640 750 274 517 716 172 8 × 2 = 1 + 0.281 500 549 035 432 345 6;
  • 79) 0.281 500 549 035 432 345 6 × 2 = 0 + 0.563 001 098 070 864 691 2;
  • 80) 0.563 001 098 070 864 691 2 × 2 = 1 + 0.126 002 196 141 729 382 4;
  • 81) 0.126 002 196 141 729 382 4 × 2 = 0 + 0.252 004 392 283 458 764 8;
  • 82) 0.252 004 392 283 458 764 8 × 2 = 0 + 0.504 008 784 566 917 529 6;
  • 83) 0.504 008 784 566 917 529 6 × 2 = 1 + 0.008 017 569 133 835 059 2;
  • 84) 0.008 017 569 133 835 059 2 × 2 = 0 + 0.016 035 138 267 670 118 4;
  • 85) 0.016 035 138 267 670 118 4 × 2 = 0 + 0.032 070 276 535 340 236 8;
  • 86) 0.032 070 276 535 340 236 8 × 2 = 0 + 0.064 140 553 070 680 473 6;
  • 87) 0.064 140 553 070 680 473 6 × 2 = 0 + 0.128 281 106 141 360 947 2;
  • 88) 0.128 281 106 141 360 947 2 × 2 = 0 + 0.256 562 212 282 721 894 4;
  • 89) 0.256 562 212 282 721 894 4 × 2 = 0 + 0.513 124 424 565 443 788 8;
  • 90) 0.513 124 424 565 443 788 8 × 2 = 1 + 0.026 248 849 130 887 577 6;
  • 91) 0.026 248 849 130 887 577 6 × 2 = 0 + 0.052 497 698 261 775 155 2;
  • 92) 0.052 497 698 261 775 155 2 × 2 = 0 + 0.104 995 396 523 550 310 4;
  • 93) 0.104 995 396 523 550 310 4 × 2 = 0 + 0.209 990 793 047 100 620 8;
  • 94) 0.209 990 793 047 100 620 8 × 2 = 0 + 0.419 981 586 094 201 241 6;
  • 95) 0.419 981 586 094 201 241 6 × 2 = 0 + 0.839 963 172 188 402 483 2;

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:


0.000 000 000 000 176 562 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1010 1001 0110 0111 1110 0101 0010 0000 0100 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 562 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1010 1001 0110 0111 1110 0101 0010 0000 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:


0.000 000 000 000 176 562 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1010 1001 0110 0111 1110 0101 0010 0000 0100 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1010 1001 0110 0111 1110 0101 0010 0000 0100 000(2) × 20 =

1.1000 1101 1001 0101 0100 1011 0011 1111 0010 1001 0000 0010 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 0101 0100 1011 0011 1111 0010 1001 0000 0010 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 0101 0100 1011 0011 1111 0010 1001 0000 0010 0000 =

1000 1101 1001 0101 0100 1011 0011 1111 0010 1001 0000 0010 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 0101 0100 1011 0011 1111 0010 1001 0000 0010 0000


Decimal number -0.000 000 000 000 176 562 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0101 0100 1011 0011 1111 0010 1001 0000 0010 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