64bit IEEE 754: Decimal -> Double Precision Floating Point Binary: 654.599 999 999 999 909 050 529 822 77 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 654.599 999 999 999 909 050 529 822 77(10) converted and written in 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: 654.
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;
  • 654 ÷ 2 = 327 + 0;
  • 327 ÷ 2 = 163 + 1;
  • 163 ÷ 2 = 81 + 1;
  • 81 ÷ 2 = 40 + 1;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.


654(10) =


10 1000 1110(2)


3. Convert to binary (base 2) the fractional part: 0.599 999 999 999 909 050 529 822 77.

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) 0.599 999 999 999 909 050 529 822 77 × 2 = 1 + 0.199 999 999 999 818 101 059 645 54;
  • 2) 0.199 999 999 999 818 101 059 645 54 × 2 = 0 + 0.399 999 999 999 636 202 119 291 08;
  • 3) 0.399 999 999 999 636 202 119 291 08 × 2 = 0 + 0.799 999 999 999 272 404 238 582 16;
  • 4) 0.799 999 999 999 272 404 238 582 16 × 2 = 1 + 0.599 999 999 998 544 808 477 164 32;
  • 5) 0.599 999 999 998 544 808 477 164 32 × 2 = 1 + 0.199 999 999 997 089 616 954 328 64;
  • 6) 0.199 999 999 997 089 616 954 328 64 × 2 = 0 + 0.399 999 999 994 179 233 908 657 28;
  • 7) 0.399 999 999 994 179 233 908 657 28 × 2 = 0 + 0.799 999 999 988 358 467 817 314 56;
  • 8) 0.799 999 999 988 358 467 817 314 56 × 2 = 1 + 0.599 999 999 976 716 935 634 629 12;
  • 9) 0.599 999 999 976 716 935 634 629 12 × 2 = 1 + 0.199 999 999 953 433 871 269 258 24;
  • 10) 0.199 999 999 953 433 871 269 258 24 × 2 = 0 + 0.399 999 999 906 867 742 538 516 48;
  • 11) 0.399 999 999 906 867 742 538 516 48 × 2 = 0 + 0.799 999 999 813 735 485 077 032 96;
  • 12) 0.799 999 999 813 735 485 077 032 96 × 2 = 1 + 0.599 999 999 627 470 970 154 065 92;
  • 13) 0.599 999 999 627 470 970 154 065 92 × 2 = 1 + 0.199 999 999 254 941 940 308 131 84;
  • 14) 0.199 999 999 254 941 940 308 131 84 × 2 = 0 + 0.399 999 998 509 883 880 616 263 68;
  • 15) 0.399 999 998 509 883 880 616 263 68 × 2 = 0 + 0.799 999 997 019 767 761 232 527 36;
  • 16) 0.799 999 997 019 767 761 232 527 36 × 2 = 1 + 0.599 999 994 039 535 522 465 054 72;
  • 17) 0.599 999 994 039 535 522 465 054 72 × 2 = 1 + 0.199 999 988 079 071 044 930 109 44;
  • 18) 0.199 999 988 079 071 044 930 109 44 × 2 = 0 + 0.399 999 976 158 142 089 860 218 88;
  • 19) 0.399 999 976 158 142 089 860 218 88 × 2 = 0 + 0.799 999 952 316 284 179 720 437 76;
  • 20) 0.799 999 952 316 284 179 720 437 76 × 2 = 1 + 0.599 999 904 632 568 359 440 875 52;
  • 21) 0.599 999 904 632 568 359 440 875 52 × 2 = 1 + 0.199 999 809 265 136 718 881 751 04;
  • 22) 0.199 999 809 265 136 718 881 751 04 × 2 = 0 + 0.399 999 618 530 273 437 763 502 08;
  • 23) 0.399 999 618 530 273 437 763 502 08 × 2 = 0 + 0.799 999 237 060 546 875 527 004 16;
  • 24) 0.799 999 237 060 546 875 527 004 16 × 2 = 1 + 0.599 998 474 121 093 751 054 008 32;
  • 25) 0.599 998 474 121 093 751 054 008 32 × 2 = 1 + 0.199 996 948 242 187 502 108 016 64;
  • 26) 0.199 996 948 242 187 502 108 016 64 × 2 = 0 + 0.399 993 896 484 375 004 216 033 28;
  • 27) 0.399 993 896 484 375 004 216 033 28 × 2 = 0 + 0.799 987 792 968 750 008 432 066 56;
  • 28) 0.799 987 792 968 750 008 432 066 56 × 2 = 1 + 0.599 975 585 937 500 016 864 133 12;
  • 29) 0.599 975 585 937 500 016 864 133 12 × 2 = 1 + 0.199 951 171 875 000 033 728 266 24;
  • 30) 0.199 951 171 875 000 033 728 266 24 × 2 = 0 + 0.399 902 343 750 000 067 456 532 48;
  • 31) 0.399 902 343 750 000 067 456 532 48 × 2 = 0 + 0.799 804 687 500 000 134 913 064 96;
  • 32) 0.799 804 687 500 000 134 913 064 96 × 2 = 1 + 0.599 609 375 000 000 269 826 129 92;
  • 33) 0.599 609 375 000 000 269 826 129 92 × 2 = 1 + 0.199 218 750 000 000 539 652 259 84;
  • 34) 0.199 218 750 000 000 539 652 259 84 × 2 = 0 + 0.398 437 500 000 001 079 304 519 68;
  • 35) 0.398 437 500 000 001 079 304 519 68 × 2 = 0 + 0.796 875 000 000 002 158 609 039 36;
  • 36) 0.796 875 000 000 002 158 609 039 36 × 2 = 1 + 0.593 750 000 000 004 317 218 078 72;
  • 37) 0.593 750 000 000 004 317 218 078 72 × 2 = 1 + 0.187 500 000 000 008 634 436 157 44;
  • 38) 0.187 500 000 000 008 634 436 157 44 × 2 = 0 + 0.375 000 000 000 017 268 872 314 88;
  • 39) 0.375 000 000 000 017 268 872 314 88 × 2 = 0 + 0.750 000 000 000 034 537 744 629 76;
  • 40) 0.750 000 000 000 034 537 744 629 76 × 2 = 1 + 0.500 000 000 000 069 075 489 259 52;
  • 41) 0.500 000 000 000 069 075 489 259 52 × 2 = 1 + 0.000 000 000 000 138 150 978 519 04;
  • 42) 0.000 000 000 000 138 150 978 519 04 × 2 = 0 + 0.000 000 000 000 276 301 957 038 08;
  • 43) 0.000 000 000 000 276 301 957 038 08 × 2 = 0 + 0.000 000 000 000 552 603 914 076 16;
  • 44) 0.000 000 000 000 552 603 914 076 16 × 2 = 0 + 0.000 000 000 001 105 207 828 152 32;
  • 45) 0.000 000 000 001 105 207 828 152 32 × 2 = 0 + 0.000 000 000 002 210 415 656 304 64;
  • 46) 0.000 000 000 002 210 415 656 304 64 × 2 = 0 + 0.000 000 000 004 420 831 312 609 28;
  • 47) 0.000 000 000 004 420 831 312 609 28 × 2 = 0 + 0.000 000 000 008 841 662 625 218 56;
  • 48) 0.000 000 000 008 841 662 625 218 56 × 2 = 0 + 0.000 000 000 017 683 325 250 437 12;
  • 49) 0.000 000 000 017 683 325 250 437 12 × 2 = 0 + 0.000 000 000 035 366 650 500 874 24;
  • 50) 0.000 000 000 035 366 650 500 874 24 × 2 = 0 + 0.000 000 000 070 733 301 001 748 48;
  • 51) 0.000 000 000 070 733 301 001 748 48 × 2 = 0 + 0.000 000 000 141 466 602 003 496 96;
  • 52) 0.000 000 000 141 466 602 003 496 96 × 2 = 0 + 0.000 000 000 282 933 204 006 993 92;
  • 53) 0.000 000 000 282 933 204 006 993 92 × 2 = 0 + 0.000 000 000 565 866 408 013 987 84;

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...)


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:


0.599 999 999 999 909 050 529 822 77(10) =


0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2)


5. Positive number before normalization:

654.599 999 999 999 909 050 529 822 77(10) =


10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the left, so that only one non zero digit remains to the left of it:


654.599 999 999 999 909 050 529 822 77(10) =


10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2) =


10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2) × 20 =


1.0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0000 00(2) × 29


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): 9


Mantissa (not normalized):
1.0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0000 00


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


9 + 2(11-1) - 1 =


(9 + 1 023)(10) =


1 032(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;
  • 1 032 ÷ 2 = 516 + 0;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1032(10) =


100 0000 1000(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 00 0000 0000 =


0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100


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) =
100 0000 1000


Mantissa (52 bits) =
0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100


The base ten decimal number 654.599 999 999 999 909 050 529 822 77 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 1000 - 0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 1

      55
    • 0

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 1

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 1

      46
    • 1

      45
    • 1

      44
    • 0

      43
    • 1

      42
    • 0

      41
    • 0

      40
    • 1

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 1

      35
    • 1

      34
    • 0

      33
    • 0

      32
    • 1

      31
    • 1

      30
    • 0

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 1

      23
    • 1

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 0

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 0

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 0

      8
    • 1

      7
    • 1

      6
    • 0

      5
    • 0

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 0

      0

Convert to 64 bit double precision IEEE 754 binary floating point representation standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

Number 654.599 999 999 999 909 050 529 822 77 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
Number 102 001 200 112 120 020 020 012 110 000 100 112 000 010 120 011 100 010 012 000 120 111 112 012 201 000 011 012 201 202 200 210 100 189 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
Number 4 603 978 641 751 548 072 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
Number 1 886 674 808 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
Number 13 835 058 055 282 163 683 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
Number -6 917 529 027 641 081 804 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
Number 1 665 814 675 311 891 105 022 807 165 128 931 001 170 047 465 289 672 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
Number 20 547 673 299 877 883 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
Number -473.66 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
Number -0.000 000 000 000 000 222 044 604 925 031 308 084 9 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 28 09:04 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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