24.777 777 777 777 777 757 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 757 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
24.777 777 777 777 777 757 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: 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: 0.777 777 777 777 777 757 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) 0.777 777 777 777 777 757 2 × 2 = 1 + 0.555 555 555 555 555 514 4;
  • 2) 0.555 555 555 555 555 514 4 × 2 = 1 + 0.111 111 111 111 111 028 8;
  • 3) 0.111 111 111 111 111 028 8 × 2 = 0 + 0.222 222 222 222 222 057 6;
  • 4) 0.222 222 222 222 222 057 6 × 2 = 0 + 0.444 444 444 444 444 115 2;
  • 5) 0.444 444 444 444 444 115 2 × 2 = 0 + 0.888 888 888 888 888 230 4;
  • 6) 0.888 888 888 888 888 230 4 × 2 = 1 + 0.777 777 777 777 776 460 8;
  • 7) 0.777 777 777 777 776 460 8 × 2 = 1 + 0.555 555 555 555 552 921 6;
  • 8) 0.555 555 555 555 552 921 6 × 2 = 1 + 0.111 111 111 111 105 843 2;
  • 9) 0.111 111 111 111 105 843 2 × 2 = 0 + 0.222 222 222 222 211 686 4;
  • 10) 0.222 222 222 222 211 686 4 × 2 = 0 + 0.444 444 444 444 423 372 8;
  • 11) 0.444 444 444 444 423 372 8 × 2 = 0 + 0.888 888 888 888 846 745 6;
  • 12) 0.888 888 888 888 846 745 6 × 2 = 1 + 0.777 777 777 777 693 491 2;
  • 13) 0.777 777 777 777 693 491 2 × 2 = 1 + 0.555 555 555 555 386 982 4;
  • 14) 0.555 555 555 555 386 982 4 × 2 = 1 + 0.111 111 111 110 773 964 8;
  • 15) 0.111 111 111 110 773 964 8 × 2 = 0 + 0.222 222 222 221 547 929 6;
  • 16) 0.222 222 222 221 547 929 6 × 2 = 0 + 0.444 444 444 443 095 859 2;
  • 17) 0.444 444 444 443 095 859 2 × 2 = 0 + 0.888 888 888 886 191 718 4;
  • 18) 0.888 888 888 886 191 718 4 × 2 = 1 + 0.777 777 777 772 383 436 8;
  • 19) 0.777 777 777 772 383 436 8 × 2 = 1 + 0.555 555 555 544 766 873 6;
  • 20) 0.555 555 555 544 766 873 6 × 2 = 1 + 0.111 111 111 089 533 747 2;
  • 21) 0.111 111 111 089 533 747 2 × 2 = 0 + 0.222 222 222 179 067 494 4;
  • 22) 0.222 222 222 179 067 494 4 × 2 = 0 + 0.444 444 444 358 134 988 8;
  • 23) 0.444 444 444 358 134 988 8 × 2 = 0 + 0.888 888 888 716 269 977 6;
  • 24) 0.888 888 888 716 269 977 6 × 2 = 1 + 0.777 777 777 432 539 955 2;
  • 25) 0.777 777 777 432 539 955 2 × 2 = 1 + 0.555 555 554 865 079 910 4;
  • 26) 0.555 555 554 865 079 910 4 × 2 = 1 + 0.111 111 109 730 159 820 8;
  • 27) 0.111 111 109 730 159 820 8 × 2 = 0 + 0.222 222 219 460 319 641 6;
  • 28) 0.222 222 219 460 319 641 6 × 2 = 0 + 0.444 444 438 920 639 283 2;
  • 29) 0.444 444 438 920 639 283 2 × 2 = 0 + 0.888 888 877 841 278 566 4;
  • 30) 0.888 888 877 841 278 566 4 × 2 = 1 + 0.777 777 755 682 557 132 8;
  • 31) 0.777 777 755 682 557 132 8 × 2 = 1 + 0.555 555 511 365 114 265 6;
  • 32) 0.555 555 511 365 114 265 6 × 2 = 1 + 0.111 111 022 730 228 531 2;
  • 33) 0.111 111 022 730 228 531 2 × 2 = 0 + 0.222 222 045 460 457 062 4;
  • 34) 0.222 222 045 460 457 062 4 × 2 = 0 + 0.444 444 090 920 914 124 8;
  • 35) 0.444 444 090 920 914 124 8 × 2 = 0 + 0.888 888 181 841 828 249 6;
  • 36) 0.888 888 181 841 828 249 6 × 2 = 1 + 0.777 776 363 683 656 499 2;
  • 37) 0.777 776 363 683 656 499 2 × 2 = 1 + 0.555 552 727 367 312 998 4;
  • 38) 0.555 552 727 367 312 998 4 × 2 = 1 + 0.111 105 454 734 625 996 8;
  • 39) 0.111 105 454 734 625 996 8 × 2 = 0 + 0.222 210 909 469 251 993 6;
  • 40) 0.222 210 909 469 251 993 6 × 2 = 0 + 0.444 421 818 938 503 987 2;
  • 41) 0.444 421 818 938 503 987 2 × 2 = 0 + 0.888 843 637 877 007 974 4;
  • 42) 0.888 843 637 877 007 974 4 × 2 = 1 + 0.777 687 275 754 015 948 8;
  • 43) 0.777 687 275 754 015 948 8 × 2 = 1 + 0.555 374 551 508 031 897 6;
  • 44) 0.555 374 551 508 031 897 6 × 2 = 1 + 0.110 749 103 016 063 795 2;
  • 45) 0.110 749 103 016 063 795 2 × 2 = 0 + 0.221 498 206 032 127 590 4;
  • 46) 0.221 498 206 032 127 590 4 × 2 = 0 + 0.442 996 412 064 255 180 8;
  • 47) 0.442 996 412 064 255 180 8 × 2 = 0 + 0.885 992 824 128 510 361 6;
  • 48) 0.885 992 824 128 510 361 6 × 2 = 1 + 0.771 985 648 257 020 723 2;
  • 49) 0.771 985 648 257 020 723 2 × 2 = 1 + 0.543 971 296 514 041 446 4;
  • 50) 0.543 971 296 514 041 446 4 × 2 = 1 + 0.087 942 593 028 082 892 8;
  • 51) 0.087 942 593 028 082 892 8 × 2 = 0 + 0.175 885 186 056 165 785 6;
  • 52) 0.175 885 186 056 165 785 6 × 2 = 0 + 0.351 770 372 112 331 571 2;
  • 53) 0.351 770 372 112 331 571 2 × 2 = 0 + 0.703 540 744 224 663 142 4;

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:


0.777 777 777 777 777 757 2(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 757 2(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:


24.777 777 777 777 777 757 2(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 24.777 777 777 777 777 757 2 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