24.777 777 777 777 777 777 918 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 918(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
24.777 777 777 777 777 777 918(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: 0.777 777 777 777 777 777 918.

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) 0.777 777 777 777 777 777 918 × 2 = 1 + 0.555 555 555 555 555 555 836;
  • 2) 0.555 555 555 555 555 555 836 × 2 = 1 + 0.111 111 111 111 111 111 672;
  • 3) 0.111 111 111 111 111 111 672 × 2 = 0 + 0.222 222 222 222 222 223 344;
  • 4) 0.222 222 222 222 222 223 344 × 2 = 0 + 0.444 444 444 444 444 446 688;
  • 5) 0.444 444 444 444 444 446 688 × 2 = 0 + 0.888 888 888 888 888 893 376;
  • 6) 0.888 888 888 888 888 893 376 × 2 = 1 + 0.777 777 777 777 777 786 752;
  • 7) 0.777 777 777 777 777 786 752 × 2 = 1 + 0.555 555 555 555 555 573 504;
  • 8) 0.555 555 555 555 555 573 504 × 2 = 1 + 0.111 111 111 111 111 147 008;
  • 9) 0.111 111 111 111 111 147 008 × 2 = 0 + 0.222 222 222 222 222 294 016;
  • 10) 0.222 222 222 222 222 294 016 × 2 = 0 + 0.444 444 444 444 444 588 032;
  • 11) 0.444 444 444 444 444 588 032 × 2 = 0 + 0.888 888 888 888 889 176 064;
  • 12) 0.888 888 888 888 889 176 064 × 2 = 1 + 0.777 777 777 777 778 352 128;
  • 13) 0.777 777 777 777 778 352 128 × 2 = 1 + 0.555 555 555 555 556 704 256;
  • 14) 0.555 555 555 555 556 704 256 × 2 = 1 + 0.111 111 111 111 113 408 512;
  • 15) 0.111 111 111 111 113 408 512 × 2 = 0 + 0.222 222 222 222 226 817 024;
  • 16) 0.222 222 222 222 226 817 024 × 2 = 0 + 0.444 444 444 444 453 634 048;
  • 17) 0.444 444 444 444 453 634 048 × 2 = 0 + 0.888 888 888 888 907 268 096;
  • 18) 0.888 888 888 888 907 268 096 × 2 = 1 + 0.777 777 777 777 814 536 192;
  • 19) 0.777 777 777 777 814 536 192 × 2 = 1 + 0.555 555 555 555 629 072 384;
  • 20) 0.555 555 555 555 629 072 384 × 2 = 1 + 0.111 111 111 111 258 144 768;
  • 21) 0.111 111 111 111 258 144 768 × 2 = 0 + 0.222 222 222 222 516 289 536;
  • 22) 0.222 222 222 222 516 289 536 × 2 = 0 + 0.444 444 444 445 032 579 072;
  • 23) 0.444 444 444 445 032 579 072 × 2 = 0 + 0.888 888 888 890 065 158 144;
  • 24) 0.888 888 888 890 065 158 144 × 2 = 1 + 0.777 777 777 780 130 316 288;
  • 25) 0.777 777 777 780 130 316 288 × 2 = 1 + 0.555 555 555 560 260 632 576;
  • 26) 0.555 555 555 560 260 632 576 × 2 = 1 + 0.111 111 111 120 521 265 152;
  • 27) 0.111 111 111 120 521 265 152 × 2 = 0 + 0.222 222 222 241 042 530 304;
  • 28) 0.222 222 222 241 042 530 304 × 2 = 0 + 0.444 444 444 482 085 060 608;
  • 29) 0.444 444 444 482 085 060 608 × 2 = 0 + 0.888 888 888 964 170 121 216;
  • 30) 0.888 888 888 964 170 121 216 × 2 = 1 + 0.777 777 777 928 340 242 432;
  • 31) 0.777 777 777 928 340 242 432 × 2 = 1 + 0.555 555 555 856 680 484 864;
  • 32) 0.555 555 555 856 680 484 864 × 2 = 1 + 0.111 111 111 713 360 969 728;
  • 33) 0.111 111 111 713 360 969 728 × 2 = 0 + 0.222 222 223 426 721 939 456;
  • 34) 0.222 222 223 426 721 939 456 × 2 = 0 + 0.444 444 446 853 443 878 912;
  • 35) 0.444 444 446 853 443 878 912 × 2 = 0 + 0.888 888 893 706 887 757 824;
  • 36) 0.888 888 893 706 887 757 824 × 2 = 1 + 0.777 777 787 413 775 515 648;
  • 37) 0.777 777 787 413 775 515 648 × 2 = 1 + 0.555 555 574 827 551 031 296;
  • 38) 0.555 555 574 827 551 031 296 × 2 = 1 + 0.111 111 149 655 102 062 592;
  • 39) 0.111 111 149 655 102 062 592 × 2 = 0 + 0.222 222 299 310 204 125 184;
  • 40) 0.222 222 299 310 204 125 184 × 2 = 0 + 0.444 444 598 620 408 250 368;
  • 41) 0.444 444 598 620 408 250 368 × 2 = 0 + 0.888 889 197 240 816 500 736;
  • 42) 0.888 889 197 240 816 500 736 × 2 = 1 + 0.777 778 394 481 633 001 472;
  • 43) 0.777 778 394 481 633 001 472 × 2 = 1 + 0.555 556 788 963 266 002 944;
  • 44) 0.555 556 788 963 266 002 944 × 2 = 1 + 0.111 113 577 926 532 005 888;
  • 45) 0.111 113 577 926 532 005 888 × 2 = 0 + 0.222 227 155 853 064 011 776;
  • 46) 0.222 227 155 853 064 011 776 × 2 = 0 + 0.444 454 311 706 128 023 552;
  • 47) 0.444 454 311 706 128 023 552 × 2 = 0 + 0.888 908 623 412 256 047 104;
  • 48) 0.888 908 623 412 256 047 104 × 2 = 1 + 0.777 817 246 824 512 094 208;
  • 49) 0.777 817 246 824 512 094 208 × 2 = 1 + 0.555 634 493 649 024 188 416;
  • 50) 0.555 634 493 649 024 188 416 × 2 = 1 + 0.111 268 987 298 048 376 832;
  • 51) 0.111 268 987 298 048 376 832 × 2 = 0 + 0.222 537 974 596 096 753 664;
  • 52) 0.222 537 974 596 096 753 664 × 2 = 0 + 0.445 075 949 192 193 507 328;
  • 53) 0.445 075 949 192 193 507 328 × 2 = 0 + 0.890 151 898 384 387 014 656;

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:


0.777 777 777 777 777 777 918(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 918(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:


24.777 777 777 777 777 777 918(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 24.777 777 777 777 777 777 918 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