1.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 094 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 094 1(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
1.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 094 1(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 094 1.

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) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 094 1 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 188 2;
  • 2) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 188 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 376 4;
  • 3) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 376 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 752 8;
  • 4) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 752 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 505 6;
  • 5) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 505 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 011 2;
  • 6) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 011 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 022 4;
  • 7) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 110 022 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 220 044 8;
  • 8) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 220 044 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 440 089 6;
  • 9) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 440 089 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 880 179 2;
  • 10) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 880 179 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 760 358 4;
  • 11) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 760 358 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 520 716 8;
  • 12) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 520 716 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 041 433 6;
  • 13) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 041 433 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 082 867 2;
  • 14) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 082 867 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 165 734 4;
  • 15) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 165 734 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 331 468 8;
  • 16) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 331 468 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 776 662 937 6;
  • 17) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 776 662 937 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 553 325 875 2;
  • 18) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 553 325 875 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 106 651 750 4;
  • 19) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 106 651 750 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 213 303 500 8;
  • 20) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 213 303 500 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 426 607 001 6;
  • 21) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 426 607 001 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 853 214 003 2;
  • 22) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 853 214 003 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 706 428 006 4;
  • 23) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 706 428 006 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 412 856 012 8;
  • 24) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 412 856 012 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 110 825 712 025 6;
  • 25) 0.111 111 111 111 111 111 111 111 111 111 111 111 110 825 712 025 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 221 651 424 051 2;
  • 26) 0.222 222 222 222 222 222 222 222 222 222 222 222 221 651 424 051 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 443 302 848 102 4;
  • 27) 0.444 444 444 444 444 444 444 444 444 444 444 444 443 302 848 102 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 886 605 696 204 8;
  • 28) 0.888 888 888 888 888 888 888 888 888 888 888 888 886 605 696 204 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 773 211 392 409 6;
  • 29) 0.777 777 777 777 777 777 777 777 777 777 777 777 773 211 392 409 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 546 422 784 819 2;
  • 30) 0.555 555 555 555 555 555 555 555 555 555 555 555 546 422 784 819 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 092 845 569 638 4;
  • 31) 0.111 111 111 111 111 111 111 111 111 111 111 111 092 845 569 638 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 185 691 139 276 8;
  • 32) 0.222 222 222 222 222 222 222 222 222 222 222 222 185 691 139 276 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 371 382 278 553 6;
  • 33) 0.444 444 444 444 444 444 444 444 444 444 444 444 371 382 278 553 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 742 764 557 107 2;
  • 34) 0.888 888 888 888 888 888 888 888 888 888 888 888 742 764 557 107 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 485 529 114 214 4;
  • 35) 0.777 777 777 777 777 777 777 777 777 777 777 777 485 529 114 214 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 554 971 058 228 428 8;
  • 36) 0.555 555 555 555 555 555 555 555 555 555 555 554 971 058 228 428 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 109 942 116 456 857 6;
  • 37) 0.111 111 111 111 111 111 111 111 111 111 111 109 942 116 456 857 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 219 884 232 913 715 2;
  • 38) 0.222 222 222 222 222 222 222 222 222 222 222 219 884 232 913 715 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 439 768 465 827 430 4;
  • 39) 0.444 444 444 444 444 444 444 444 444 444 444 439 768 465 827 430 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 879 536 931 654 860 8;
  • 40) 0.888 888 888 888 888 888 888 888 888 888 888 879 536 931 654 860 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 759 073 863 309 721 6;
  • 41) 0.777 777 777 777 777 777 777 777 777 777 777 759 073 863 309 721 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 518 147 726 619 443 2;
  • 42) 0.555 555 555 555 555 555 555 555 555 555 555 518 147 726 619 443 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 036 295 453 238 886 4;
  • 43) 0.111 111 111 111 111 111 111 111 111 111 111 036 295 453 238 886 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 072 590 906 477 772 8;
  • 44) 0.222 222 222 222 222 222 222 222 222 222 222 072 590 906 477 772 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 145 181 812 955 545 6;
  • 45) 0.444 444 444 444 444 444 444 444 444 444 444 145 181 812 955 545 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 290 363 625 911 091 2;
  • 46) 0.888 888 888 888 888 888 888 888 888 888 888 290 363 625 911 091 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 776 580 727 251 822 182 4;
  • 47) 0.777 777 777 777 777 777 777 777 777 777 776 580 727 251 822 182 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 553 161 454 503 644 364 8;
  • 48) 0.555 555 555 555 555 555 555 555 555 555 553 161 454 503 644 364 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 106 322 909 007 288 729 6;
  • 49) 0.111 111 111 111 111 111 111 111 111 111 106 322 909 007 288 729 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 212 645 818 014 577 459 2;
  • 50) 0.222 222 222 222 222 222 222 222 222 222 212 645 818 014 577 459 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 425 291 636 029 154 918 4;
  • 51) 0.444 444 444 444 444 444 444 444 444 444 425 291 636 029 154 918 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 850 583 272 058 309 836 8;
  • 52) 0.888 888 888 888 888 888 888 888 888 888 850 583 272 058 309 836 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 701 166 544 116 619 673 6;
  • 53) 0.777 777 777 777 777 777 777 777 777 777 701 166 544 116 619 673 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 402 333 088 233 239 347 2;

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:


0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 094 1(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2)

5. Positive number before normalization:

1.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 094 1(10) =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 094 1(10) =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2) =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2) × 20


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): 0

Mantissa (not normalized):
1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1 =

0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


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 1111

Mantissa (52 bits) =
0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


Decimal number 1.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 094 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001

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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