24.777 777 777 777 778 044 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 778 044(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
24.777 777 777 777 778 044(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: 0.777 777 777 777 778 044.

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) 0.777 777 777 777 778 044 × 2 = 1 + 0.555 555 555 555 556 088;
  • 2) 0.555 555 555 555 556 088 × 2 = 1 + 0.111 111 111 111 112 176;
  • 3) 0.111 111 111 111 112 176 × 2 = 0 + 0.222 222 222 222 224 352;
  • 4) 0.222 222 222 222 224 352 × 2 = 0 + 0.444 444 444 444 448 704;
  • 5) 0.444 444 444 444 448 704 × 2 = 0 + 0.888 888 888 888 897 408;
  • 6) 0.888 888 888 888 897 408 × 2 = 1 + 0.777 777 777 777 794 816;
  • 7) 0.777 777 777 777 794 816 × 2 = 1 + 0.555 555 555 555 589 632;
  • 8) 0.555 555 555 555 589 632 × 2 = 1 + 0.111 111 111 111 179 264;
  • 9) 0.111 111 111 111 179 264 × 2 = 0 + 0.222 222 222 222 358 528;
  • 10) 0.222 222 222 222 358 528 × 2 = 0 + 0.444 444 444 444 717 056;
  • 11) 0.444 444 444 444 717 056 × 2 = 0 + 0.888 888 888 889 434 112;
  • 12) 0.888 888 888 889 434 112 × 2 = 1 + 0.777 777 777 778 868 224;
  • 13) 0.777 777 777 778 868 224 × 2 = 1 + 0.555 555 555 557 736 448;
  • 14) 0.555 555 555 557 736 448 × 2 = 1 + 0.111 111 111 115 472 896;
  • 15) 0.111 111 111 115 472 896 × 2 = 0 + 0.222 222 222 230 945 792;
  • 16) 0.222 222 222 230 945 792 × 2 = 0 + 0.444 444 444 461 891 584;
  • 17) 0.444 444 444 461 891 584 × 2 = 0 + 0.888 888 888 923 783 168;
  • 18) 0.888 888 888 923 783 168 × 2 = 1 + 0.777 777 777 847 566 336;
  • 19) 0.777 777 777 847 566 336 × 2 = 1 + 0.555 555 555 695 132 672;
  • 20) 0.555 555 555 695 132 672 × 2 = 1 + 0.111 111 111 390 265 344;
  • 21) 0.111 111 111 390 265 344 × 2 = 0 + 0.222 222 222 780 530 688;
  • 22) 0.222 222 222 780 530 688 × 2 = 0 + 0.444 444 445 561 061 376;
  • 23) 0.444 444 445 561 061 376 × 2 = 0 + 0.888 888 891 122 122 752;
  • 24) 0.888 888 891 122 122 752 × 2 = 1 + 0.777 777 782 244 245 504;
  • 25) 0.777 777 782 244 245 504 × 2 = 1 + 0.555 555 564 488 491 008;
  • 26) 0.555 555 564 488 491 008 × 2 = 1 + 0.111 111 128 976 982 016;
  • 27) 0.111 111 128 976 982 016 × 2 = 0 + 0.222 222 257 953 964 032;
  • 28) 0.222 222 257 953 964 032 × 2 = 0 + 0.444 444 515 907 928 064;
  • 29) 0.444 444 515 907 928 064 × 2 = 0 + 0.888 889 031 815 856 128;
  • 30) 0.888 889 031 815 856 128 × 2 = 1 + 0.777 778 063 631 712 256;
  • 31) 0.777 778 063 631 712 256 × 2 = 1 + 0.555 556 127 263 424 512;
  • 32) 0.555 556 127 263 424 512 × 2 = 1 + 0.111 112 254 526 849 024;
  • 33) 0.111 112 254 526 849 024 × 2 = 0 + 0.222 224 509 053 698 048;
  • 34) 0.222 224 509 053 698 048 × 2 = 0 + 0.444 449 018 107 396 096;
  • 35) 0.444 449 018 107 396 096 × 2 = 0 + 0.888 898 036 214 792 192;
  • 36) 0.888 898 036 214 792 192 × 2 = 1 + 0.777 796 072 429 584 384;
  • 37) 0.777 796 072 429 584 384 × 2 = 1 + 0.555 592 144 859 168 768;
  • 38) 0.555 592 144 859 168 768 × 2 = 1 + 0.111 184 289 718 337 536;
  • 39) 0.111 184 289 718 337 536 × 2 = 0 + 0.222 368 579 436 675 072;
  • 40) 0.222 368 579 436 675 072 × 2 = 0 + 0.444 737 158 873 350 144;
  • 41) 0.444 737 158 873 350 144 × 2 = 0 + 0.889 474 317 746 700 288;
  • 42) 0.889 474 317 746 700 288 × 2 = 1 + 0.778 948 635 493 400 576;
  • 43) 0.778 948 635 493 400 576 × 2 = 1 + 0.557 897 270 986 801 152;
  • 44) 0.557 897 270 986 801 152 × 2 = 1 + 0.115 794 541 973 602 304;
  • 45) 0.115 794 541 973 602 304 × 2 = 0 + 0.231 589 083 947 204 608;
  • 46) 0.231 589 083 947 204 608 × 2 = 0 + 0.463 178 167 894 409 216;
  • 47) 0.463 178 167 894 409 216 × 2 = 0 + 0.926 356 335 788 818 432;
  • 48) 0.926 356 335 788 818 432 × 2 = 1 + 0.852 712 671 577 636 864;
  • 49) 0.852 712 671 577 636 864 × 2 = 1 + 0.705 425 343 155 273 728;
  • 50) 0.705 425 343 155 273 728 × 2 = 1 + 0.410 850 686 310 547 456;
  • 51) 0.410 850 686 310 547 456 × 2 = 0 + 0.821 701 372 621 094 912;
  • 52) 0.821 701 372 621 094 912 × 2 = 1 + 0.643 402 745 242 189 824;
  • 53) 0.643 402 745 242 189 824 × 2 = 1 + 0.286 805 490 484 379 648;

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:


0.777 777 777 777 778 044(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1(2)

5. Positive number before normalization:

24.777 777 777 777 778 044(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1(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:


24.777 777 777 777 778 044(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1101 1(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 1101 1


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 1011 =

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 24.777 777 777 777 778 044 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