0.000 000 000 000 000 461 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 461 57(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
0.000 000 000 000 000 461 57(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: 0.000 000 000 000 000 461 57.

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) 0.000 000 000 000 000 461 57 × 2 = 0 + 0.000 000 000 000 000 923 14;
  • 2) 0.000 000 000 000 000 923 14 × 2 = 0 + 0.000 000 000 000 001 846 28;
  • 3) 0.000 000 000 000 001 846 28 × 2 = 0 + 0.000 000 000 000 003 692 56;
  • 4) 0.000 000 000 000 003 692 56 × 2 = 0 + 0.000 000 000 000 007 385 12;
  • 5) 0.000 000 000 000 007 385 12 × 2 = 0 + 0.000 000 000 000 014 770 24;
  • 6) 0.000 000 000 000 014 770 24 × 2 = 0 + 0.000 000 000 000 029 540 48;
  • 7) 0.000 000 000 000 029 540 48 × 2 = 0 + 0.000 000 000 000 059 080 96;
  • 8) 0.000 000 000 000 059 080 96 × 2 = 0 + 0.000 000 000 000 118 161 92;
  • 9) 0.000 000 000 000 118 161 92 × 2 = 0 + 0.000 000 000 000 236 323 84;
  • 10) 0.000 000 000 000 236 323 84 × 2 = 0 + 0.000 000 000 000 472 647 68;
  • 11) 0.000 000 000 000 472 647 68 × 2 = 0 + 0.000 000 000 000 945 295 36;
  • 12) 0.000 000 000 000 945 295 36 × 2 = 0 + 0.000 000 000 001 890 590 72;
  • 13) 0.000 000 000 001 890 590 72 × 2 = 0 + 0.000 000 000 003 781 181 44;
  • 14) 0.000 000 000 003 781 181 44 × 2 = 0 + 0.000 000 000 007 562 362 88;
  • 15) 0.000 000 000 007 562 362 88 × 2 = 0 + 0.000 000 000 015 124 725 76;
  • 16) 0.000 000 000 015 124 725 76 × 2 = 0 + 0.000 000 000 030 249 451 52;
  • 17) 0.000 000 000 030 249 451 52 × 2 = 0 + 0.000 000 000 060 498 903 04;
  • 18) 0.000 000 000 060 498 903 04 × 2 = 0 + 0.000 000 000 120 997 806 08;
  • 19) 0.000 000 000 120 997 806 08 × 2 = 0 + 0.000 000 000 241 995 612 16;
  • 20) 0.000 000 000 241 995 612 16 × 2 = 0 + 0.000 000 000 483 991 224 32;
  • 21) 0.000 000 000 483 991 224 32 × 2 = 0 + 0.000 000 000 967 982 448 64;
  • 22) 0.000 000 000 967 982 448 64 × 2 = 0 + 0.000 000 001 935 964 897 28;
  • 23) 0.000 000 001 935 964 897 28 × 2 = 0 + 0.000 000 003 871 929 794 56;
  • 24) 0.000 000 003 871 929 794 56 × 2 = 0 + 0.000 000 007 743 859 589 12;
  • 25) 0.000 000 007 743 859 589 12 × 2 = 0 + 0.000 000 015 487 719 178 24;
  • 26) 0.000 000 015 487 719 178 24 × 2 = 0 + 0.000 000 030 975 438 356 48;
  • 27) 0.000 000 030 975 438 356 48 × 2 = 0 + 0.000 000 061 950 876 712 96;
  • 28) 0.000 000 061 950 876 712 96 × 2 = 0 + 0.000 000 123 901 753 425 92;
  • 29) 0.000 000 123 901 753 425 92 × 2 = 0 + 0.000 000 247 803 506 851 84;
  • 30) 0.000 000 247 803 506 851 84 × 2 = 0 + 0.000 000 495 607 013 703 68;
  • 31) 0.000 000 495 607 013 703 68 × 2 = 0 + 0.000 000 991 214 027 407 36;
  • 32) 0.000 000 991 214 027 407 36 × 2 = 0 + 0.000 001 982 428 054 814 72;
  • 33) 0.000 001 982 428 054 814 72 × 2 = 0 + 0.000 003 964 856 109 629 44;
  • 34) 0.000 003 964 856 109 629 44 × 2 = 0 + 0.000 007 929 712 219 258 88;
  • 35) 0.000 007 929 712 219 258 88 × 2 = 0 + 0.000 015 859 424 438 517 76;
  • 36) 0.000 015 859 424 438 517 76 × 2 = 0 + 0.000 031 718 848 877 035 52;
  • 37) 0.000 031 718 848 877 035 52 × 2 = 0 + 0.000 063 437 697 754 071 04;
  • 38) 0.000 063 437 697 754 071 04 × 2 = 0 + 0.000 126 875 395 508 142 08;
  • 39) 0.000 126 875 395 508 142 08 × 2 = 0 + 0.000 253 750 791 016 284 16;
  • 40) 0.000 253 750 791 016 284 16 × 2 = 0 + 0.000 507 501 582 032 568 32;
  • 41) 0.000 507 501 582 032 568 32 × 2 = 0 + 0.001 015 003 164 065 136 64;
  • 42) 0.001 015 003 164 065 136 64 × 2 = 0 + 0.002 030 006 328 130 273 28;
  • 43) 0.002 030 006 328 130 273 28 × 2 = 0 + 0.004 060 012 656 260 546 56;
  • 44) 0.004 060 012 656 260 546 56 × 2 = 0 + 0.008 120 025 312 521 093 12;
  • 45) 0.008 120 025 312 521 093 12 × 2 = 0 + 0.016 240 050 625 042 186 24;
  • 46) 0.016 240 050 625 042 186 24 × 2 = 0 + 0.032 480 101 250 084 372 48;
  • 47) 0.032 480 101 250 084 372 48 × 2 = 0 + 0.064 960 202 500 168 744 96;
  • 48) 0.064 960 202 500 168 744 96 × 2 = 0 + 0.129 920 405 000 337 489 92;
  • 49) 0.129 920 405 000 337 489 92 × 2 = 0 + 0.259 840 810 000 674 979 84;
  • 50) 0.259 840 810 000 674 979 84 × 2 = 0 + 0.519 681 620 001 349 959 68;
  • 51) 0.519 681 620 001 349 959 68 × 2 = 1 + 0.039 363 240 002 699 919 36;
  • 52) 0.039 363 240 002 699 919 36 × 2 = 0 + 0.078 726 480 005 399 838 72;
  • 53) 0.078 726 480 005 399 838 72 × 2 = 0 + 0.157 452 960 010 799 677 44;
  • 54) 0.157 452 960 010 799 677 44 × 2 = 0 + 0.314 905 920 021 599 354 88;
  • 55) 0.314 905 920 021 599 354 88 × 2 = 0 + 0.629 811 840 043 198 709 76;
  • 56) 0.629 811 840 043 198 709 76 × 2 = 1 + 0.259 623 680 086 397 419 52;
  • 57) 0.259 623 680 086 397 419 52 × 2 = 0 + 0.519 247 360 172 794 839 04;
  • 58) 0.519 247 360 172 794 839 04 × 2 = 1 + 0.038 494 720 345 589 678 08;
  • 59) 0.038 494 720 345 589 678 08 × 2 = 0 + 0.076 989 440 691 179 356 16;
  • 60) 0.076 989 440 691 179 356 16 × 2 = 0 + 0.153 978 881 382 358 712 32;
  • 61) 0.153 978 881 382 358 712 32 × 2 = 0 + 0.307 957 762 764 717 424 64;
  • 62) 0.307 957 762 764 717 424 64 × 2 = 0 + 0.615 915 525 529 434 849 28;
  • 63) 0.615 915 525 529 434 849 28 × 2 = 1 + 0.231 831 051 058 869 698 56;
  • 64) 0.231 831 051 058 869 698 56 × 2 = 0 + 0.463 662 102 117 739 397 12;
  • 65) 0.463 662 102 117 739 397 12 × 2 = 0 + 0.927 324 204 235 478 794 24;
  • 66) 0.927 324 204 235 478 794 24 × 2 = 1 + 0.854 648 408 470 957 588 48;
  • 67) 0.854 648 408 470 957 588 48 × 2 = 1 + 0.709 296 816 941 915 176 96;
  • 68) 0.709 296 816 941 915 176 96 × 2 = 1 + 0.418 593 633 883 830 353 92;
  • 69) 0.418 593 633 883 830 353 92 × 2 = 0 + 0.837 187 267 767 660 707 84;
  • 70) 0.837 187 267 767 660 707 84 × 2 = 1 + 0.674 374 535 535 321 415 68;
  • 71) 0.674 374 535 535 321 415 68 × 2 = 1 + 0.348 749 071 070 642 831 36;
  • 72) 0.348 749 071 070 642 831 36 × 2 = 0 + 0.697 498 142 141 285 662 72;
  • 73) 0.697 498 142 141 285 662 72 × 2 = 1 + 0.394 996 284 282 571 325 44;
  • 74) 0.394 996 284 282 571 325 44 × 2 = 0 + 0.789 992 568 565 142 650 88;
  • 75) 0.789 992 568 565 142 650 88 × 2 = 1 + 0.579 985 137 130 285 301 76;
  • 76) 0.579 985 137 130 285 301 76 × 2 = 1 + 0.159 970 274 260 570 603 52;
  • 77) 0.159 970 274 260 570 603 52 × 2 = 0 + 0.319 940 548 521 141 207 04;
  • 78) 0.319 940 548 521 141 207 04 × 2 = 0 + 0.639 881 097 042 282 414 08;
  • 79) 0.639 881 097 042 282 414 08 × 2 = 1 + 0.279 762 194 084 564 828 16;
  • 80) 0.279 762 194 084 564 828 16 × 2 = 0 + 0.559 524 388 169 129 656 32;
  • 81) 0.559 524 388 169 129 656 32 × 2 = 1 + 0.119 048 776 338 259 312 64;
  • 82) 0.119 048 776 338 259 312 64 × 2 = 0 + 0.238 097 552 676 518 625 28;
  • 83) 0.238 097 552 676 518 625 28 × 2 = 0 + 0.476 195 105 353 037 250 56;
  • 84) 0.476 195 105 353 037 250 56 × 2 = 0 + 0.952 390 210 706 074 501 12;
  • 85) 0.952 390 210 706 074 501 12 × 2 = 1 + 0.904 780 421 412 149 002 24;
  • 86) 0.904 780 421 412 149 002 24 × 2 = 1 + 0.809 560 842 824 298 004 48;
  • 87) 0.809 560 842 824 298 004 48 × 2 = 1 + 0.619 121 685 648 596 008 96;
  • 88) 0.619 121 685 648 596 008 96 × 2 = 1 + 0.238 243 371 297 192 017 92;
  • 89) 0.238 243 371 297 192 017 92 × 2 = 0 + 0.476 486 742 594 384 035 84;
  • 90) 0.476 486 742 594 384 035 84 × 2 = 0 + 0.952 973 485 188 768 071 68;
  • 91) 0.952 973 485 188 768 071 68 × 2 = 1 + 0.905 946 970 377 536 143 36;
  • 92) 0.905 946 970 377 536 143 36 × 2 = 1 + 0.811 893 940 755 072 286 72;
  • 93) 0.811 893 940 755 072 286 72 × 2 = 1 + 0.623 787 881 510 144 573 44;
  • 94) 0.623 787 881 510 144 573 44 × 2 = 1 + 0.247 575 763 020 289 146 88;
  • 95) 0.247 575 763 020 289 146 88 × 2 = 0 + 0.495 151 526 040 578 293 76;
  • 96) 0.495 151 526 040 578 293 76 × 2 = 0 + 0.990 303 052 081 156 587 52;
  • 97) 0.990 303 052 081 156 587 52 × 2 = 1 + 0.980 606 104 162 313 175 04;
  • 98) 0.980 606 104 162 313 175 04 × 2 = 1 + 0.961 212 208 324 626 350 08;
  • 99) 0.961 212 208 324 626 350 08 × 2 = 1 + 0.922 424 416 649 252 700 16;
  • 100) 0.922 424 416 649 252 700 16 × 2 = 1 + 0.844 848 833 298 505 400 32;
  • 101) 0.844 848 833 298 505 400 32 × 2 = 1 + 0.689 697 666 597 010 800 64;
  • 102) 0.689 697 666 597 010 800 64 × 2 = 1 + 0.379 395 333 194 021 601 28;
  • 103) 0.379 395 333 194 021 601 28 × 2 = 0 + 0.758 790 666 388 043 202 56;

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:


0.000 000 000 000 000 461 57(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 0100 0010 0111 0110 1011 0010 1000 1111 0011 1100 1111 110(2)

5. Positive number before normalization:

0.000 000 000 000 000 461 57(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 0100 0010 0111 0110 1011 0010 1000 1111 0011 1100 1111 110(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:


0.000 000 000 000 000 461 57(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 0100 0010 0111 0110 1011 0010 1000 1111 0011 1100 1111 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 0100 0010 0111 0110 1011 0010 1000 1111 0011 1100 1111 110(2) × 20 =

1.0000 1010 0001 0011 1011 0101 1001 0100 0111 1001 1110 0111 1110(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 1010 0001 0011 1011 0101 1001 0100 0111 1001 1110 0111 1110


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 1010 0001 0011 1011 0101 1001 0100 0111 1001 1110 0111 1110 =

0000 1010 0001 0011 1011 0101 1001 0100 0111 1001 1110 0111 1110


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 1010 0001 0011 1011 0101 1001 0100 0111 1001 1110 0111 1110


Decimal number 0.000 000 000 000 000 461 57 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1100 1100 - 0000 1010 0001 0011 1011 0101 1001 0100 0111 1001 1110 0111 1110

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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