0.000 000 000 000 000 471 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 471 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
0.000 000 000 000 000 471 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: 0.000 000 000 000 000 471 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) 0.000 000 000 000 000 471 2 × 2 = 0 + 0.000 000 000 000 000 942 4;
  • 2) 0.000 000 000 000 000 942 4 × 2 = 0 + 0.000 000 000 000 001 884 8;
  • 3) 0.000 000 000 000 001 884 8 × 2 = 0 + 0.000 000 000 000 003 769 6;
  • 4) 0.000 000 000 000 003 769 6 × 2 = 0 + 0.000 000 000 000 007 539 2;
  • 5) 0.000 000 000 000 007 539 2 × 2 = 0 + 0.000 000 000 000 015 078 4;
  • 6) 0.000 000 000 000 015 078 4 × 2 = 0 + 0.000 000 000 000 030 156 8;
  • 7) 0.000 000 000 000 030 156 8 × 2 = 0 + 0.000 000 000 000 060 313 6;
  • 8) 0.000 000 000 000 060 313 6 × 2 = 0 + 0.000 000 000 000 120 627 2;
  • 9) 0.000 000 000 000 120 627 2 × 2 = 0 + 0.000 000 000 000 241 254 4;
  • 10) 0.000 000 000 000 241 254 4 × 2 = 0 + 0.000 000 000 000 482 508 8;
  • 11) 0.000 000 000 000 482 508 8 × 2 = 0 + 0.000 000 000 000 965 017 6;
  • 12) 0.000 000 000 000 965 017 6 × 2 = 0 + 0.000 000 000 001 930 035 2;
  • 13) 0.000 000 000 001 930 035 2 × 2 = 0 + 0.000 000 000 003 860 070 4;
  • 14) 0.000 000 000 003 860 070 4 × 2 = 0 + 0.000 000 000 007 720 140 8;
  • 15) 0.000 000 000 007 720 140 8 × 2 = 0 + 0.000 000 000 015 440 281 6;
  • 16) 0.000 000 000 015 440 281 6 × 2 = 0 + 0.000 000 000 030 880 563 2;
  • 17) 0.000 000 000 030 880 563 2 × 2 = 0 + 0.000 000 000 061 761 126 4;
  • 18) 0.000 000 000 061 761 126 4 × 2 = 0 + 0.000 000 000 123 522 252 8;
  • 19) 0.000 000 000 123 522 252 8 × 2 = 0 + 0.000 000 000 247 044 505 6;
  • 20) 0.000 000 000 247 044 505 6 × 2 = 0 + 0.000 000 000 494 089 011 2;
  • 21) 0.000 000 000 494 089 011 2 × 2 = 0 + 0.000 000 000 988 178 022 4;
  • 22) 0.000 000 000 988 178 022 4 × 2 = 0 + 0.000 000 001 976 356 044 8;
  • 23) 0.000 000 001 976 356 044 8 × 2 = 0 + 0.000 000 003 952 712 089 6;
  • 24) 0.000 000 003 952 712 089 6 × 2 = 0 + 0.000 000 007 905 424 179 2;
  • 25) 0.000 000 007 905 424 179 2 × 2 = 0 + 0.000 000 015 810 848 358 4;
  • 26) 0.000 000 015 810 848 358 4 × 2 = 0 + 0.000 000 031 621 696 716 8;
  • 27) 0.000 000 031 621 696 716 8 × 2 = 0 + 0.000 000 063 243 393 433 6;
  • 28) 0.000 000 063 243 393 433 6 × 2 = 0 + 0.000 000 126 486 786 867 2;
  • 29) 0.000 000 126 486 786 867 2 × 2 = 0 + 0.000 000 252 973 573 734 4;
  • 30) 0.000 000 252 973 573 734 4 × 2 = 0 + 0.000 000 505 947 147 468 8;
  • 31) 0.000 000 505 947 147 468 8 × 2 = 0 + 0.000 001 011 894 294 937 6;
  • 32) 0.000 001 011 894 294 937 6 × 2 = 0 + 0.000 002 023 788 589 875 2;
  • 33) 0.000 002 023 788 589 875 2 × 2 = 0 + 0.000 004 047 577 179 750 4;
  • 34) 0.000 004 047 577 179 750 4 × 2 = 0 + 0.000 008 095 154 359 500 8;
  • 35) 0.000 008 095 154 359 500 8 × 2 = 0 + 0.000 016 190 308 719 001 6;
  • 36) 0.000 016 190 308 719 001 6 × 2 = 0 + 0.000 032 380 617 438 003 2;
  • 37) 0.000 032 380 617 438 003 2 × 2 = 0 + 0.000 064 761 234 876 006 4;
  • 38) 0.000 064 761 234 876 006 4 × 2 = 0 + 0.000 129 522 469 752 012 8;
  • 39) 0.000 129 522 469 752 012 8 × 2 = 0 + 0.000 259 044 939 504 025 6;
  • 40) 0.000 259 044 939 504 025 6 × 2 = 0 + 0.000 518 089 879 008 051 2;
  • 41) 0.000 518 089 879 008 051 2 × 2 = 0 + 0.001 036 179 758 016 102 4;
  • 42) 0.001 036 179 758 016 102 4 × 2 = 0 + 0.002 072 359 516 032 204 8;
  • 43) 0.002 072 359 516 032 204 8 × 2 = 0 + 0.004 144 719 032 064 409 6;
  • 44) 0.004 144 719 032 064 409 6 × 2 = 0 + 0.008 289 438 064 128 819 2;
  • 45) 0.008 289 438 064 128 819 2 × 2 = 0 + 0.016 578 876 128 257 638 4;
  • 46) 0.016 578 876 128 257 638 4 × 2 = 0 + 0.033 157 752 256 515 276 8;
  • 47) 0.033 157 752 256 515 276 8 × 2 = 0 + 0.066 315 504 513 030 553 6;
  • 48) 0.066 315 504 513 030 553 6 × 2 = 0 + 0.132 631 009 026 061 107 2;
  • 49) 0.132 631 009 026 061 107 2 × 2 = 0 + 0.265 262 018 052 122 214 4;
  • 50) 0.265 262 018 052 122 214 4 × 2 = 0 + 0.530 524 036 104 244 428 8;
  • 51) 0.530 524 036 104 244 428 8 × 2 = 1 + 0.061 048 072 208 488 857 6;
  • 52) 0.061 048 072 208 488 857 6 × 2 = 0 + 0.122 096 144 416 977 715 2;
  • 53) 0.122 096 144 416 977 715 2 × 2 = 0 + 0.244 192 288 833 955 430 4;
  • 54) 0.244 192 288 833 955 430 4 × 2 = 0 + 0.488 384 577 667 910 860 8;
  • 55) 0.488 384 577 667 910 860 8 × 2 = 0 + 0.976 769 155 335 821 721 6;
  • 56) 0.976 769 155 335 821 721 6 × 2 = 1 + 0.953 538 310 671 643 443 2;
  • 57) 0.953 538 310 671 643 443 2 × 2 = 1 + 0.907 076 621 343 286 886 4;
  • 58) 0.907 076 621 343 286 886 4 × 2 = 1 + 0.814 153 242 686 573 772 8;
  • 59) 0.814 153 242 686 573 772 8 × 2 = 1 + 0.628 306 485 373 147 545 6;
  • 60) 0.628 306 485 373 147 545 6 × 2 = 1 + 0.256 612 970 746 295 091 2;
  • 61) 0.256 612 970 746 295 091 2 × 2 = 0 + 0.513 225 941 492 590 182 4;
  • 62) 0.513 225 941 492 590 182 4 × 2 = 1 + 0.026 451 882 985 180 364 8;
  • 63) 0.026 451 882 985 180 364 8 × 2 = 0 + 0.052 903 765 970 360 729 6;
  • 64) 0.052 903 765 970 360 729 6 × 2 = 0 + 0.105 807 531 940 721 459 2;
  • 65) 0.105 807 531 940 721 459 2 × 2 = 0 + 0.211 615 063 881 442 918 4;
  • 66) 0.211 615 063 881 442 918 4 × 2 = 0 + 0.423 230 127 762 885 836 8;
  • 67) 0.423 230 127 762 885 836 8 × 2 = 0 + 0.846 460 255 525 771 673 6;
  • 68) 0.846 460 255 525 771 673 6 × 2 = 1 + 0.692 920 511 051 543 347 2;
  • 69) 0.692 920 511 051 543 347 2 × 2 = 1 + 0.385 841 022 103 086 694 4;
  • 70) 0.385 841 022 103 086 694 4 × 2 = 0 + 0.771 682 044 206 173 388 8;
  • 71) 0.771 682 044 206 173 388 8 × 2 = 1 + 0.543 364 088 412 346 777 6;
  • 72) 0.543 364 088 412 346 777 6 × 2 = 1 + 0.086 728 176 824 693 555 2;
  • 73) 0.086 728 176 824 693 555 2 × 2 = 0 + 0.173 456 353 649 387 110 4;
  • 74) 0.173 456 353 649 387 110 4 × 2 = 0 + 0.346 912 707 298 774 220 8;
  • 75) 0.346 912 707 298 774 220 8 × 2 = 0 + 0.693 825 414 597 548 441 6;
  • 76) 0.693 825 414 597 548 441 6 × 2 = 1 + 0.387 650 829 195 096 883 2;
  • 77) 0.387 650 829 195 096 883 2 × 2 = 0 + 0.775 301 658 390 193 766 4;
  • 78) 0.775 301 658 390 193 766 4 × 2 = 1 + 0.550 603 316 780 387 532 8;
  • 79) 0.550 603 316 780 387 532 8 × 2 = 1 + 0.101 206 633 560 775 065 6;
  • 80) 0.101 206 633 560 775 065 6 × 2 = 0 + 0.202 413 267 121 550 131 2;
  • 81) 0.202 413 267 121 550 131 2 × 2 = 0 + 0.404 826 534 243 100 262 4;
  • 82) 0.404 826 534 243 100 262 4 × 2 = 0 + 0.809 653 068 486 200 524 8;
  • 83) 0.809 653 068 486 200 524 8 × 2 = 1 + 0.619 306 136 972 401 049 6;
  • 84) 0.619 306 136 972 401 049 6 × 2 = 1 + 0.238 612 273 944 802 099 2;
  • 85) 0.238 612 273 944 802 099 2 × 2 = 0 + 0.477 224 547 889 604 198 4;
  • 86) 0.477 224 547 889 604 198 4 × 2 = 0 + 0.954 449 095 779 208 396 8;
  • 87) 0.954 449 095 779 208 396 8 × 2 = 1 + 0.908 898 191 558 416 793 6;
  • 88) 0.908 898 191 558 416 793 6 × 2 = 1 + 0.817 796 383 116 833 587 2;
  • 89) 0.817 796 383 116 833 587 2 × 2 = 1 + 0.635 592 766 233 667 174 4;
  • 90) 0.635 592 766 233 667 174 4 × 2 = 1 + 0.271 185 532 467 334 348 8;
  • 91) 0.271 185 532 467 334 348 8 × 2 = 0 + 0.542 371 064 934 668 697 6;
  • 92) 0.542 371 064 934 668 697 6 × 2 = 1 + 0.084 742 129 869 337 395 2;
  • 93) 0.084 742 129 869 337 395 2 × 2 = 0 + 0.169 484 259 738 674 790 4;
  • 94) 0.169 484 259 738 674 790 4 × 2 = 0 + 0.338 968 519 477 349 580 8;
  • 95) 0.338 968 519 477 349 580 8 × 2 = 0 + 0.677 937 038 954 699 161 6;
  • 96) 0.677 937 038 954 699 161 6 × 2 = 1 + 0.355 874 077 909 398 323 2;
  • 97) 0.355 874 077 909 398 323 2 × 2 = 0 + 0.711 748 155 818 796 646 4;
  • 98) 0.711 748 155 818 796 646 4 × 2 = 1 + 0.423 496 311 637 593 292 8;
  • 99) 0.423 496 311 637 593 292 8 × 2 = 0 + 0.846 992 623 275 186 585 6;
  • 100) 0.846 992 623 275 186 585 6 × 2 = 1 + 0.693 985 246 550 373 171 2;
  • 101) 0.693 985 246 550 373 171 2 × 2 = 1 + 0.387 970 493 100 746 342 4;
  • 102) 0.387 970 493 100 746 342 4 × 2 = 0 + 0.775 940 986 201 492 684 8;
  • 103) 0.775 940 986 201 492 684 8 × 2 = 1 + 0.551 881 972 402 985 369 6;

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:


0.000 000 000 000 000 471 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 1111 0100 0001 1011 0001 0110 0011 0011 1101 0001 0101 101(2)

5. Positive number before normalization:

0.000 000 000 000 000 471 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 1111 0100 0001 1011 0001 0110 0011 0011 1101 0001 0101 101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 51 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 471 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 1111 0100 0001 1011 0001 0110 0011 0011 1101 0001 0101 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 1111 0100 0001 1011 0001 0110 0011 0011 1101 0001 0101 101(2) × 20 =

1.0000 1111 1010 0000 1101 1000 1011 0001 1001 1110 1000 1010 1101(2) × 2-51


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): -51

Mantissa (not normalized):
1.0000 1111 1010 0000 1101 1000 1011 0001 1001 1110 1000 1010 1101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-51 + 2(11-1) - 1 =

(-51 + 1 023)(10) =

972(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;
  • 972 ÷ 2 = 486 + 0;
  • 486 ÷ 2 = 243 + 0;
  • 243 ÷ 2 = 121 + 1;
  • 121 ÷ 2 = 60 + 1;
  • 60 ÷ 2 = 30 + 0;
  • 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) =

972(10) =

011 1100 1100(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 1111 1010 0000 1101 1000 1011 0001 1001 1110 1000 1010 1101 =

0000 1111 1010 0000 1101 1000 1011 0001 1001 1110 1000 1010 1101


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 1100 1100

Mantissa (52 bits) =
0000 1111 1010 0000 1101 1000 1011 0001 1001 1110 1000 1010 1101


Decimal number 0.000 000 000 000 000 471 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1100 1100 - 0000 1111 1010 0000 1101 1000 1011 0001 1001 1110 1000 1010 1101

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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