0.000 000 000 000 099 999 997 23 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 099 999 997 23(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
0.000 000 000 000 099 999 997 23(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: 0.000 000 000 000 099 999 997 23.

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) 0.000 000 000 000 099 999 997 23 × 2 = 0 + 0.000 000 000 000 199 999 994 46;
  • 2) 0.000 000 000 000 199 999 994 46 × 2 = 0 + 0.000 000 000 000 399 999 988 92;
  • 3) 0.000 000 000 000 399 999 988 92 × 2 = 0 + 0.000 000 000 000 799 999 977 84;
  • 4) 0.000 000 000 000 799 999 977 84 × 2 = 0 + 0.000 000 000 001 599 999 955 68;
  • 5) 0.000 000 000 001 599 999 955 68 × 2 = 0 + 0.000 000 000 003 199 999 911 36;
  • 6) 0.000 000 000 003 199 999 911 36 × 2 = 0 + 0.000 000 000 006 399 999 822 72;
  • 7) 0.000 000 000 006 399 999 822 72 × 2 = 0 + 0.000 000 000 012 799 999 645 44;
  • 8) 0.000 000 000 012 799 999 645 44 × 2 = 0 + 0.000 000 000 025 599 999 290 88;
  • 9) 0.000 000 000 025 599 999 290 88 × 2 = 0 + 0.000 000 000 051 199 998 581 76;
  • 10) 0.000 000 000 051 199 998 581 76 × 2 = 0 + 0.000 000 000 102 399 997 163 52;
  • 11) 0.000 000 000 102 399 997 163 52 × 2 = 0 + 0.000 000 000 204 799 994 327 04;
  • 12) 0.000 000 000 204 799 994 327 04 × 2 = 0 + 0.000 000 000 409 599 988 654 08;
  • 13) 0.000 000 000 409 599 988 654 08 × 2 = 0 + 0.000 000 000 819 199 977 308 16;
  • 14) 0.000 000 000 819 199 977 308 16 × 2 = 0 + 0.000 000 001 638 399 954 616 32;
  • 15) 0.000 000 001 638 399 954 616 32 × 2 = 0 + 0.000 000 003 276 799 909 232 64;
  • 16) 0.000 000 003 276 799 909 232 64 × 2 = 0 + 0.000 000 006 553 599 818 465 28;
  • 17) 0.000 000 006 553 599 818 465 28 × 2 = 0 + 0.000 000 013 107 199 636 930 56;
  • 18) 0.000 000 013 107 199 636 930 56 × 2 = 0 + 0.000 000 026 214 399 273 861 12;
  • 19) 0.000 000 026 214 399 273 861 12 × 2 = 0 + 0.000 000 052 428 798 547 722 24;
  • 20) 0.000 000 052 428 798 547 722 24 × 2 = 0 + 0.000 000 104 857 597 095 444 48;
  • 21) 0.000 000 104 857 597 095 444 48 × 2 = 0 + 0.000 000 209 715 194 190 888 96;
  • 22) 0.000 000 209 715 194 190 888 96 × 2 = 0 + 0.000 000 419 430 388 381 777 92;
  • 23) 0.000 000 419 430 388 381 777 92 × 2 = 0 + 0.000 000 838 860 776 763 555 84;
  • 24) 0.000 000 838 860 776 763 555 84 × 2 = 0 + 0.000 001 677 721 553 527 111 68;
  • 25) 0.000 001 677 721 553 527 111 68 × 2 = 0 + 0.000 003 355 443 107 054 223 36;
  • 26) 0.000 003 355 443 107 054 223 36 × 2 = 0 + 0.000 006 710 886 214 108 446 72;
  • 27) 0.000 006 710 886 214 108 446 72 × 2 = 0 + 0.000 013 421 772 428 216 893 44;
  • 28) 0.000 013 421 772 428 216 893 44 × 2 = 0 + 0.000 026 843 544 856 433 786 88;
  • 29) 0.000 026 843 544 856 433 786 88 × 2 = 0 + 0.000 053 687 089 712 867 573 76;
  • 30) 0.000 053 687 089 712 867 573 76 × 2 = 0 + 0.000 107 374 179 425 735 147 52;
  • 31) 0.000 107 374 179 425 735 147 52 × 2 = 0 + 0.000 214 748 358 851 470 295 04;
  • 32) 0.000 214 748 358 851 470 295 04 × 2 = 0 + 0.000 429 496 717 702 940 590 08;
  • 33) 0.000 429 496 717 702 940 590 08 × 2 = 0 + 0.000 858 993 435 405 881 180 16;
  • 34) 0.000 858 993 435 405 881 180 16 × 2 = 0 + 0.001 717 986 870 811 762 360 32;
  • 35) 0.001 717 986 870 811 762 360 32 × 2 = 0 + 0.003 435 973 741 623 524 720 64;
  • 36) 0.003 435 973 741 623 524 720 64 × 2 = 0 + 0.006 871 947 483 247 049 441 28;
  • 37) 0.006 871 947 483 247 049 441 28 × 2 = 0 + 0.013 743 894 966 494 098 882 56;
  • 38) 0.013 743 894 966 494 098 882 56 × 2 = 0 + 0.027 487 789 932 988 197 765 12;
  • 39) 0.027 487 789 932 988 197 765 12 × 2 = 0 + 0.054 975 579 865 976 395 530 24;
  • 40) 0.054 975 579 865 976 395 530 24 × 2 = 0 + 0.109 951 159 731 952 791 060 48;
  • 41) 0.109 951 159 731 952 791 060 48 × 2 = 0 + 0.219 902 319 463 905 582 120 96;
  • 42) 0.219 902 319 463 905 582 120 96 × 2 = 0 + 0.439 804 638 927 811 164 241 92;
  • 43) 0.439 804 638 927 811 164 241 92 × 2 = 0 + 0.879 609 277 855 622 328 483 84;
  • 44) 0.879 609 277 855 622 328 483 84 × 2 = 1 + 0.759 218 555 711 244 656 967 68;
  • 45) 0.759 218 555 711 244 656 967 68 × 2 = 1 + 0.518 437 111 422 489 313 935 36;
  • 46) 0.518 437 111 422 489 313 935 36 × 2 = 1 + 0.036 874 222 844 978 627 870 72;
  • 47) 0.036 874 222 844 978 627 870 72 × 2 = 0 + 0.073 748 445 689 957 255 741 44;
  • 48) 0.073 748 445 689 957 255 741 44 × 2 = 0 + 0.147 496 891 379 914 511 482 88;
  • 49) 0.147 496 891 379 914 511 482 88 × 2 = 0 + 0.294 993 782 759 829 022 965 76;
  • 50) 0.294 993 782 759 829 022 965 76 × 2 = 0 + 0.589 987 565 519 658 045 931 52;
  • 51) 0.589 987 565 519 658 045 931 52 × 2 = 1 + 0.179 975 131 039 316 091 863 04;
  • 52) 0.179 975 131 039 316 091 863 04 × 2 = 0 + 0.359 950 262 078 632 183 726 08;
  • 53) 0.359 950 262 078 632 183 726 08 × 2 = 0 + 0.719 900 524 157 264 367 452 16;
  • 54) 0.719 900 524 157 264 367 452 16 × 2 = 1 + 0.439 801 048 314 528 734 904 32;
  • 55) 0.439 801 048 314 528 734 904 32 × 2 = 0 + 0.879 602 096 629 057 469 808 64;
  • 56) 0.879 602 096 629 057 469 808 64 × 2 = 1 + 0.759 204 193 258 114 939 617 28;
  • 57) 0.759 204 193 258 114 939 617 28 × 2 = 1 + 0.518 408 386 516 229 879 234 56;
  • 58) 0.518 408 386 516 229 879 234 56 × 2 = 1 + 0.036 816 773 032 459 758 469 12;
  • 59) 0.036 816 773 032 459 758 469 12 × 2 = 0 + 0.073 633 546 064 919 516 938 24;
  • 60) 0.073 633 546 064 919 516 938 24 × 2 = 0 + 0.147 267 092 129 839 033 876 48;
  • 61) 0.147 267 092 129 839 033 876 48 × 2 = 0 + 0.294 534 184 259 678 067 752 96;
  • 62) 0.294 534 184 259 678 067 752 96 × 2 = 0 + 0.589 068 368 519 356 135 505 92;
  • 63) 0.589 068 368 519 356 135 505 92 × 2 = 1 + 0.178 136 737 038 712 271 011 84;
  • 64) 0.178 136 737 038 712 271 011 84 × 2 = 0 + 0.356 273 474 077 424 542 023 68;
  • 65) 0.356 273 474 077 424 542 023 68 × 2 = 0 + 0.712 546 948 154 849 084 047 36;
  • 66) 0.712 546 948 154 849 084 047 36 × 2 = 1 + 0.425 093 896 309 698 168 094 72;
  • 67) 0.425 093 896 309 698 168 094 72 × 2 = 0 + 0.850 187 792 619 396 336 189 44;
  • 68) 0.850 187 792 619 396 336 189 44 × 2 = 1 + 0.700 375 585 238 792 672 378 88;
  • 69) 0.700 375 585 238 792 672 378 88 × 2 = 1 + 0.400 751 170 477 585 344 757 76;
  • 70) 0.400 751 170 477 585 344 757 76 × 2 = 0 + 0.801 502 340 955 170 689 515 52;
  • 71) 0.801 502 340 955 170 689 515 52 × 2 = 1 + 0.603 004 681 910 341 379 031 04;
  • 72) 0.603 004 681 910 341 379 031 04 × 2 = 1 + 0.206 009 363 820 682 758 062 08;
  • 73) 0.206 009 363 820 682 758 062 08 × 2 = 0 + 0.412 018 727 641 365 516 124 16;
  • 74) 0.412 018 727 641 365 516 124 16 × 2 = 0 + 0.824 037 455 282 731 032 248 32;
  • 75) 0.824 037 455 282 731 032 248 32 × 2 = 1 + 0.648 074 910 565 462 064 496 64;
  • 76) 0.648 074 910 565 462 064 496 64 × 2 = 1 + 0.296 149 821 130 924 128 993 28;
  • 77) 0.296 149 821 130 924 128 993 28 × 2 = 0 + 0.592 299 642 261 848 257 986 56;
  • 78) 0.592 299 642 261 848 257 986 56 × 2 = 1 + 0.184 599 284 523 696 515 973 12;
  • 79) 0.184 599 284 523 696 515 973 12 × 2 = 0 + 0.369 198 569 047 393 031 946 24;
  • 80) 0.369 198 569 047 393 031 946 24 × 2 = 0 + 0.738 397 138 094 786 063 892 48;
  • 81) 0.738 397 138 094 786 063 892 48 × 2 = 1 + 0.476 794 276 189 572 127 784 96;
  • 82) 0.476 794 276 189 572 127 784 96 × 2 = 0 + 0.953 588 552 379 144 255 569 92;
  • 83) 0.953 588 552 379 144 255 569 92 × 2 = 1 + 0.907 177 104 758 288 511 139 84;
  • 84) 0.907 177 104 758 288 511 139 84 × 2 = 1 + 0.814 354 209 516 577 022 279 68;
  • 85) 0.814 354 209 516 577 022 279 68 × 2 = 1 + 0.628 708 419 033 154 044 559 36;
  • 86) 0.628 708 419 033 154 044 559 36 × 2 = 1 + 0.257 416 838 066 308 089 118 72;
  • 87) 0.257 416 838 066 308 089 118 72 × 2 = 0 + 0.514 833 676 132 616 178 237 44;
  • 88) 0.514 833 676 132 616 178 237 44 × 2 = 1 + 0.029 667 352 265 232 356 474 88;
  • 89) 0.029 667 352 265 232 356 474 88 × 2 = 0 + 0.059 334 704 530 464 712 949 76;
  • 90) 0.059 334 704 530 464 712 949 76 × 2 = 0 + 0.118 669 409 060 929 425 899 52;
  • 91) 0.118 669 409 060 929 425 899 52 × 2 = 0 + 0.237 338 818 121 858 851 799 04;
  • 92) 0.237 338 818 121 858 851 799 04 × 2 = 0 + 0.474 677 636 243 717 703 598 08;
  • 93) 0.474 677 636 243 717 703 598 08 × 2 = 0 + 0.949 355 272 487 435 407 196 16;
  • 94) 0.949 355 272 487 435 407 196 16 × 2 = 1 + 0.898 710 544 974 870 814 392 32;
  • 95) 0.898 710 544 974 870 814 392 32 × 2 = 1 + 0.797 421 089 949 741 628 784 64;
  • 96) 0.797 421 089 949 741 628 784 64 × 2 = 1 + 0.594 842 179 899 483 257 569 28;

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:


0.000 000 000 000 099 999 997 23(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111(2)

5. Positive number before normalization:

0.000 000 000 000 099 999 997 23(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 44 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 099 999 997 23(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111(2) × 20 =

1.1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111(2) × 2-44


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): -44

Mantissa (not normalized):
1.1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-44 + 2(11-1) - 1 =

(-44 + 1 023)(10) =

979(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;
  • 979 ÷ 2 = 489 + 1;
  • 489 ÷ 2 = 244 + 1;
  • 244 ÷ 2 = 122 + 0;
  • 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;

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) =

979(10) =

011 1101 0011(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. 1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111 =

1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111


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 1101 0011

Mantissa (52 bits) =
1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111


Decimal number 0.000 000 000 000 099 999 997 23 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0011 - 1100 0010 0101 1100 0010 0101 1011 0011 0100 1011 1101 0000 0111

» Convert decimal 0.000 000 000 000 099 999 998 17 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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