Convert Decimal 24.777 777 777 777 777 777 777 777 777 713 1 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 777 777 777 713 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
24.777 777 777 777 777 777 777 777 777 713 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: 24.
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;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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.

24(10) =

1 1000(2)


3. Convert to binary (base 2) the fractional part: 0.777 777 777 777 777 777 777 777 777 713 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) 0.777 777 777 777 777 777 777 777 777 713 1 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 426 2;
  • 2) 0.555 555 555 555 555 555 555 555 555 426 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 110 852 4;
  • 3) 0.111 111 111 111 111 111 111 111 110 852 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 221 704 8;
  • 4) 0.222 222 222 222 222 222 222 222 221 704 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 443 409 6;
  • 5) 0.444 444 444 444 444 444 444 444 443 409 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 886 819 2;
  • 6) 0.888 888 888 888 888 888 888 888 886 819 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 773 638 4;
  • 7) 0.777 777 777 777 777 777 777 777 773 638 4 × 2 = 1 + 0.555 555 555 555 555 555 555 555 547 276 8;
  • 8) 0.555 555 555 555 555 555 555 555 547 276 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 094 553 6;
  • 9) 0.111 111 111 111 111 111 111 111 094 553 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 189 107 2;
  • 10) 0.222 222 222 222 222 222 222 222 189 107 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 378 214 4;
  • 11) 0.444 444 444 444 444 444 444 444 378 214 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 756 428 8;
  • 12) 0.888 888 888 888 888 888 888 888 756 428 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 512 857 6;
  • 13) 0.777 777 777 777 777 777 777 777 512 857 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 025 715 2;
  • 14) 0.555 555 555 555 555 555 555 555 025 715 2 × 2 = 1 + 0.111 111 111 111 111 111 111 110 051 430 4;
  • 15) 0.111 111 111 111 111 111 111 110 051 430 4 × 2 = 0 + 0.222 222 222 222 222 222 222 220 102 860 8;
  • 16) 0.222 222 222 222 222 222 222 220 102 860 8 × 2 = 0 + 0.444 444 444 444 444 444 444 440 205 721 6;
  • 17) 0.444 444 444 444 444 444 444 440 205 721 6 × 2 = 0 + 0.888 888 888 888 888 888 888 880 411 443 2;
  • 18) 0.888 888 888 888 888 888 888 880 411 443 2 × 2 = 1 + 0.777 777 777 777 777 777 777 760 822 886 4;
  • 19) 0.777 777 777 777 777 777 777 760 822 886 4 × 2 = 1 + 0.555 555 555 555 555 555 555 521 645 772 8;
  • 20) 0.555 555 555 555 555 555 555 521 645 772 8 × 2 = 1 + 0.111 111 111 111 111 111 111 043 291 545 6;
  • 21) 0.111 111 111 111 111 111 111 043 291 545 6 × 2 = 0 + 0.222 222 222 222 222 222 222 086 583 091 2;
  • 22) 0.222 222 222 222 222 222 222 086 583 091 2 × 2 = 0 + 0.444 444 444 444 444 444 444 173 166 182 4;
  • 23) 0.444 444 444 444 444 444 444 173 166 182 4 × 2 = 0 + 0.888 888 888 888 888 888 888 346 332 364 8;
  • 24) 0.888 888 888 888 888 888 888 346 332 364 8 × 2 = 1 + 0.777 777 777 777 777 777 776 692 664 729 6;
  • 25) 0.777 777 777 777 777 777 776 692 664 729 6 × 2 = 1 + 0.555 555 555 555 555 555 553 385 329 459 2;
  • 26) 0.555 555 555 555 555 555 553 385 329 459 2 × 2 = 1 + 0.111 111 111 111 111 111 106 770 658 918 4;
  • 27) 0.111 111 111 111 111 111 106 770 658 918 4 × 2 = 0 + 0.222 222 222 222 222 222 213 541 317 836 8;
  • 28) 0.222 222 222 222 222 222 213 541 317 836 8 × 2 = 0 + 0.444 444 444 444 444 444 427 082 635 673 6;
  • 29) 0.444 444 444 444 444 444 427 082 635 673 6 × 2 = 0 + 0.888 888 888 888 888 888 854 165 271 347 2;
  • 30) 0.888 888 888 888 888 888 854 165 271 347 2 × 2 = 1 + 0.777 777 777 777 777 777 708 330 542 694 4;
  • 31) 0.777 777 777 777 777 777 708 330 542 694 4 × 2 = 1 + 0.555 555 555 555 555 555 416 661 085 388 8;
  • 32) 0.555 555 555 555 555 555 416 661 085 388 8 × 2 = 1 + 0.111 111 111 111 111 110 833 322 170 777 6;
  • 33) 0.111 111 111 111 111 110 833 322 170 777 6 × 2 = 0 + 0.222 222 222 222 222 221 666 644 341 555 2;
  • 34) 0.222 222 222 222 222 221 666 644 341 555 2 × 2 = 0 + 0.444 444 444 444 444 443 333 288 683 110 4;
  • 35) 0.444 444 444 444 444 443 333 288 683 110 4 × 2 = 0 + 0.888 888 888 888 888 886 666 577 366 220 8;
  • 36) 0.888 888 888 888 888 886 666 577 366 220 8 × 2 = 1 + 0.777 777 777 777 777 773 333 154 732 441 6;
  • 37) 0.777 777 777 777 777 773 333 154 732 441 6 × 2 = 1 + 0.555 555 555 555 555 546 666 309 464 883 2;
  • 38) 0.555 555 555 555 555 546 666 309 464 883 2 × 2 = 1 + 0.111 111 111 111 111 093 332 618 929 766 4;
  • 39) 0.111 111 111 111 111 093 332 618 929 766 4 × 2 = 0 + 0.222 222 222 222 222 186 665 237 859 532 8;
  • 40) 0.222 222 222 222 222 186 665 237 859 532 8 × 2 = 0 + 0.444 444 444 444 444 373 330 475 719 065 6;
  • 41) 0.444 444 444 444 444 373 330 475 719 065 6 × 2 = 0 + 0.888 888 888 888 888 746 660 951 438 131 2;
  • 42) 0.888 888 888 888 888 746 660 951 438 131 2 × 2 = 1 + 0.777 777 777 777 777 493 321 902 876 262 4;
  • 43) 0.777 777 777 777 777 493 321 902 876 262 4 × 2 = 1 + 0.555 555 555 555 554 986 643 805 752 524 8;
  • 44) 0.555 555 555 555 554 986 643 805 752 524 8 × 2 = 1 + 0.111 111 111 111 109 973 287 611 505 049 6;
  • 45) 0.111 111 111 111 109 973 287 611 505 049 6 × 2 = 0 + 0.222 222 222 222 219 946 575 223 010 099 2;
  • 46) 0.222 222 222 222 219 946 575 223 010 099 2 × 2 = 0 + 0.444 444 444 444 439 893 150 446 020 198 4;
  • 47) 0.444 444 444 444 439 893 150 446 020 198 4 × 2 = 0 + 0.888 888 888 888 879 786 300 892 040 396 8;
  • 48) 0.888 888 888 888 879 786 300 892 040 396 8 × 2 = 1 + 0.777 777 777 777 759 572 601 784 080 793 6;
  • 49) 0.777 777 777 777 759 572 601 784 080 793 6 × 2 = 1 + 0.555 555 555 555 519 145 203 568 161 587 2;
  • 50) 0.555 555 555 555 519 145 203 568 161 587 2 × 2 = 1 + 0.111 111 111 111 038 290 407 136 323 174 4;
  • 51) 0.111 111 111 111 038 290 407 136 323 174 4 × 2 = 0 + 0.222 222 222 222 076 580 814 272 646 348 8;
  • 52) 0.222 222 222 222 076 580 814 272 646 348 8 × 2 = 0 + 0.444 444 444 444 153 161 628 545 292 697 6;
  • 53) 0.444 444 444 444 153 161 628 545 292 697 6 × 2 = 0 + 0.888 888 888 888 306 323 257 090 585 395 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:


0.777 777 777 777 777 777 777 777 777 713 1(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2)

5. Positive number before normalization:

24.777 777 777 777 777 777 777 777 777 713 1(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


24.777 777 777 777 777 777 777 777 777 713 1(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2) × 24


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): 4

Mantissa (not normalized):
1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

4 + 2(11-1) - 1 =

(4 + 1 023)(10) =

1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 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) =

1027(10) =

100 0000 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, 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. 1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1 1000 =

1000 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) =
100 0000 0011

Mantissa (52 bits) =
1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


Decimal number 24.777 777 777 777 777 777 777 777 777 713 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 1000 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