0.000 999 999 999 999 999 803 976 247 214 620 809 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 999 999 999 999 999 803 976 247 214 620 809 2(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
0.000 999 999 999 999 999 803 976 247 214 620 809 2(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 999 999 999 999 999 803 976 247 214 620 809 2.

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) 0.000 999 999 999 999 999 803 976 247 214 620 809 2 × 2 = 0 + 0.001 999 999 999 999 999 607 952 494 429 241 618 4;
  • 2) 0.001 999 999 999 999 999 607 952 494 429 241 618 4 × 2 = 0 + 0.003 999 999 999 999 999 215 904 988 858 483 236 8;
  • 3) 0.003 999 999 999 999 999 215 904 988 858 483 236 8 × 2 = 0 + 0.007 999 999 999 999 998 431 809 977 716 966 473 6;
  • 4) 0.007 999 999 999 999 998 431 809 977 716 966 473 6 × 2 = 0 + 0.015 999 999 999 999 996 863 619 955 433 932 947 2;
  • 5) 0.015 999 999 999 999 996 863 619 955 433 932 947 2 × 2 = 0 + 0.031 999 999 999 999 993 727 239 910 867 865 894 4;
  • 6) 0.031 999 999 999 999 993 727 239 910 867 865 894 4 × 2 = 0 + 0.063 999 999 999 999 987 454 479 821 735 731 788 8;
  • 7) 0.063 999 999 999 999 987 454 479 821 735 731 788 8 × 2 = 0 + 0.127 999 999 999 999 974 908 959 643 471 463 577 6;
  • 8) 0.127 999 999 999 999 974 908 959 643 471 463 577 6 × 2 = 0 + 0.255 999 999 999 999 949 817 919 286 942 927 155 2;
  • 9) 0.255 999 999 999 999 949 817 919 286 942 927 155 2 × 2 = 0 + 0.511 999 999 999 999 899 635 838 573 885 854 310 4;
  • 10) 0.511 999 999 999 999 899 635 838 573 885 854 310 4 × 2 = 1 + 0.023 999 999 999 999 799 271 677 147 771 708 620 8;
  • 11) 0.023 999 999 999 999 799 271 677 147 771 708 620 8 × 2 = 0 + 0.047 999 999 999 999 598 543 354 295 543 417 241 6;
  • 12) 0.047 999 999 999 999 598 543 354 295 543 417 241 6 × 2 = 0 + 0.095 999 999 999 999 197 086 708 591 086 834 483 2;
  • 13) 0.095 999 999 999 999 197 086 708 591 086 834 483 2 × 2 = 0 + 0.191 999 999 999 998 394 173 417 182 173 668 966 4;
  • 14) 0.191 999 999 999 998 394 173 417 182 173 668 966 4 × 2 = 0 + 0.383 999 999 999 996 788 346 834 364 347 337 932 8;
  • 15) 0.383 999 999 999 996 788 346 834 364 347 337 932 8 × 2 = 0 + 0.767 999 999 999 993 576 693 668 728 694 675 865 6;
  • 16) 0.767 999 999 999 993 576 693 668 728 694 675 865 6 × 2 = 1 + 0.535 999 999 999 987 153 387 337 457 389 351 731 2;
  • 17) 0.535 999 999 999 987 153 387 337 457 389 351 731 2 × 2 = 1 + 0.071 999 999 999 974 306 774 674 914 778 703 462 4;
  • 18) 0.071 999 999 999 974 306 774 674 914 778 703 462 4 × 2 = 0 + 0.143 999 999 999 948 613 549 349 829 557 406 924 8;
  • 19) 0.143 999 999 999 948 613 549 349 829 557 406 924 8 × 2 = 0 + 0.287 999 999 999 897 227 098 699 659 114 813 849 6;
  • 20) 0.287 999 999 999 897 227 098 699 659 114 813 849 6 × 2 = 0 + 0.575 999 999 999 794 454 197 399 318 229 627 699 2;
  • 21) 0.575 999 999 999 794 454 197 399 318 229 627 699 2 × 2 = 1 + 0.151 999 999 999 588 908 394 798 636 459 255 398 4;
  • 22) 0.151 999 999 999 588 908 394 798 636 459 255 398 4 × 2 = 0 + 0.303 999 999 999 177 816 789 597 272 918 510 796 8;
  • 23) 0.303 999 999 999 177 816 789 597 272 918 510 796 8 × 2 = 0 + 0.607 999 999 998 355 633 579 194 545 837 021 593 6;
  • 24) 0.607 999 999 998 355 633 579 194 545 837 021 593 6 × 2 = 1 + 0.215 999 999 996 711 267 158 389 091 674 043 187 2;
  • 25) 0.215 999 999 996 711 267 158 389 091 674 043 187 2 × 2 = 0 + 0.431 999 999 993 422 534 316 778 183 348 086 374 4;
  • 26) 0.431 999 999 993 422 534 316 778 183 348 086 374 4 × 2 = 0 + 0.863 999 999 986 845 068 633 556 366 696 172 748 8;
  • 27) 0.863 999 999 986 845 068 633 556 366 696 172 748 8 × 2 = 1 + 0.727 999 999 973 690 137 267 112 733 392 345 497 6;
  • 28) 0.727 999 999 973 690 137 267 112 733 392 345 497 6 × 2 = 1 + 0.455 999 999 947 380 274 534 225 466 784 690 995 2;
  • 29) 0.455 999 999 947 380 274 534 225 466 784 690 995 2 × 2 = 0 + 0.911 999 999 894 760 549 068 450 933 569 381 990 4;
  • 30) 0.911 999 999 894 760 549 068 450 933 569 381 990 4 × 2 = 1 + 0.823 999 999 789 521 098 136 901 867 138 763 980 8;
  • 31) 0.823 999 999 789 521 098 136 901 867 138 763 980 8 × 2 = 1 + 0.647 999 999 579 042 196 273 803 734 277 527 961 6;
  • 32) 0.647 999 999 579 042 196 273 803 734 277 527 961 6 × 2 = 1 + 0.295 999 999 158 084 392 547 607 468 555 055 923 2;
  • 33) 0.295 999 999 158 084 392 547 607 468 555 055 923 2 × 2 = 0 + 0.591 999 998 316 168 785 095 214 937 110 111 846 4;
  • 34) 0.591 999 998 316 168 785 095 214 937 110 111 846 4 × 2 = 1 + 0.183 999 996 632 337 570 190 429 874 220 223 692 8;
  • 35) 0.183 999 996 632 337 570 190 429 874 220 223 692 8 × 2 = 0 + 0.367 999 993 264 675 140 380 859 748 440 447 385 6;
  • 36) 0.367 999 993 264 675 140 380 859 748 440 447 385 6 × 2 = 0 + 0.735 999 986 529 350 280 761 719 496 880 894 771 2;
  • 37) 0.735 999 986 529 350 280 761 719 496 880 894 771 2 × 2 = 1 + 0.471 999 973 058 700 561 523 438 993 761 789 542 4;
  • 38) 0.471 999 973 058 700 561 523 438 993 761 789 542 4 × 2 = 0 + 0.943 999 946 117 401 123 046 877 987 523 579 084 8;
  • 39) 0.943 999 946 117 401 123 046 877 987 523 579 084 8 × 2 = 1 + 0.887 999 892 234 802 246 093 755 975 047 158 169 6;
  • 40) 0.887 999 892 234 802 246 093 755 975 047 158 169 6 × 2 = 1 + 0.775 999 784 469 604 492 187 511 950 094 316 339 2;
  • 41) 0.775 999 784 469 604 492 187 511 950 094 316 339 2 × 2 = 1 + 0.551 999 568 939 208 984 375 023 900 188 632 678 4;
  • 42) 0.551 999 568 939 208 984 375 023 900 188 632 678 4 × 2 = 1 + 0.103 999 137 878 417 968 750 047 800 377 265 356 8;
  • 43) 0.103 999 137 878 417 968 750 047 800 377 265 356 8 × 2 = 0 + 0.207 998 275 756 835 937 500 095 600 754 530 713 6;
  • 44) 0.207 998 275 756 835 937 500 095 600 754 530 713 6 × 2 = 0 + 0.415 996 551 513 671 875 000 191 201 509 061 427 2;
  • 45) 0.415 996 551 513 671 875 000 191 201 509 061 427 2 × 2 = 0 + 0.831 993 103 027 343 750 000 382 403 018 122 854 4;
  • 46) 0.831 993 103 027 343 750 000 382 403 018 122 854 4 × 2 = 1 + 0.663 986 206 054 687 500 000 764 806 036 245 708 8;
  • 47) 0.663 986 206 054 687 500 000 764 806 036 245 708 8 × 2 = 1 + 0.327 972 412 109 375 000 001 529 612 072 491 417 6;
  • 48) 0.327 972 412 109 375 000 001 529 612 072 491 417 6 × 2 = 0 + 0.655 944 824 218 750 000 003 059 224 144 982 835 2;
  • 49) 0.655 944 824 218 750 000 003 059 224 144 982 835 2 × 2 = 1 + 0.311 889 648 437 500 000 006 118 448 289 965 670 4;
  • 50) 0.311 889 648 437 500 000 006 118 448 289 965 670 4 × 2 = 0 + 0.623 779 296 875 000 000 012 236 896 579 931 340 8;
  • 51) 0.623 779 296 875 000 000 012 236 896 579 931 340 8 × 2 = 1 + 0.247 558 593 750 000 000 024 473 793 159 862 681 6;
  • 52) 0.247 558 593 750 000 000 024 473 793 159 862 681 6 × 2 = 0 + 0.495 117 187 500 000 000 048 947 586 319 725 363 2;
  • 53) 0.495 117 187 500 000 000 048 947 586 319 725 363 2 × 2 = 0 + 0.990 234 375 000 000 000 097 895 172 639 450 726 4;
  • 54) 0.990 234 375 000 000 000 097 895 172 639 450 726 4 × 2 = 1 + 0.980 468 750 000 000 000 195 790 345 278 901 452 8;
  • 55) 0.980 468 750 000 000 000 195 790 345 278 901 452 8 × 2 = 1 + 0.960 937 500 000 000 000 391 580 690 557 802 905 6;
  • 56) 0.960 937 500 000 000 000 391 580 690 557 802 905 6 × 2 = 1 + 0.921 875 000 000 000 000 783 161 381 115 605 811 2;
  • 57) 0.921 875 000 000 000 000 783 161 381 115 605 811 2 × 2 = 1 + 0.843 750 000 000 000 001 566 322 762 231 211 622 4;
  • 58) 0.843 750 000 000 000 001 566 322 762 231 211 622 4 × 2 = 1 + 0.687 500 000 000 000 003 132 645 524 462 423 244 8;
  • 59) 0.687 500 000 000 000 003 132 645 524 462 423 244 8 × 2 = 1 + 0.375 000 000 000 000 006 265 291 048 924 846 489 6;
  • 60) 0.375 000 000 000 000 006 265 291 048 924 846 489 6 × 2 = 0 + 0.750 000 000 000 000 012 530 582 097 849 692 979 2;
  • 61) 0.750 000 000 000 000 012 530 582 097 849 692 979 2 × 2 = 1 + 0.500 000 000 000 000 025 061 164 195 699 385 958 4;
  • 62) 0.500 000 000 000 000 025 061 164 195 699 385 958 4 × 2 = 1 + 0.000 000 000 000 000 050 122 328 391 398 771 916 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).


4. 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:


0.000 999 999 999 999 999 803 976 247 214 620 809 2(10) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 11(2)

5. Positive number before normalization:

0.000 999 999 999 999 999 803 976 247 214 620 809 2(10) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.000 999 999 999 999 999 803 976 247 214 620 809 2(10) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 11(2) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 11(2) × 20 =

1.0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011(2) × 2-10


7. 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 0 (a positive number)

Exponent (unadjusted): -10

Mantissa (not normalized):
1.0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-10 + 2(11-1) - 1 =

(-10 + 1 023)(10) =

1 013(10)


9. 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;
  • 1 013 ÷ 2 = 506 + 1;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1013(10) =

011 1111 0101(2)


11. 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. 0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011 =

0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 0101

Mantissa (52 bits) =
0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011


Decimal number 0.000 999 999 999 999 999 803 976 247 214 620 809 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0101 - 0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1011

Share

» Convert decimal 0.000 999 999 999 999 999 803 976 247 214 620 802 6 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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