24.777 777 777 777 777 773 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 773 9(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
24.777 777 777 777 777 773 9(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: 0.777 777 777 777 777 773 9.

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) 0.777 777 777 777 777 773 9 × 2 = 1 + 0.555 555 555 555 555 547 8;
  • 2) 0.555 555 555 555 555 547 8 × 2 = 1 + 0.111 111 111 111 111 095 6;
  • 3) 0.111 111 111 111 111 095 6 × 2 = 0 + 0.222 222 222 222 222 191 2;
  • 4) 0.222 222 222 222 222 191 2 × 2 = 0 + 0.444 444 444 444 444 382 4;
  • 5) 0.444 444 444 444 444 382 4 × 2 = 0 + 0.888 888 888 888 888 764 8;
  • 6) 0.888 888 888 888 888 764 8 × 2 = 1 + 0.777 777 777 777 777 529 6;
  • 7) 0.777 777 777 777 777 529 6 × 2 = 1 + 0.555 555 555 555 555 059 2;
  • 8) 0.555 555 555 555 555 059 2 × 2 = 1 + 0.111 111 111 111 110 118 4;
  • 9) 0.111 111 111 111 110 118 4 × 2 = 0 + 0.222 222 222 222 220 236 8;
  • 10) 0.222 222 222 222 220 236 8 × 2 = 0 + 0.444 444 444 444 440 473 6;
  • 11) 0.444 444 444 444 440 473 6 × 2 = 0 + 0.888 888 888 888 880 947 2;
  • 12) 0.888 888 888 888 880 947 2 × 2 = 1 + 0.777 777 777 777 761 894 4;
  • 13) 0.777 777 777 777 761 894 4 × 2 = 1 + 0.555 555 555 555 523 788 8;
  • 14) 0.555 555 555 555 523 788 8 × 2 = 1 + 0.111 111 111 111 047 577 6;
  • 15) 0.111 111 111 111 047 577 6 × 2 = 0 + 0.222 222 222 222 095 155 2;
  • 16) 0.222 222 222 222 095 155 2 × 2 = 0 + 0.444 444 444 444 190 310 4;
  • 17) 0.444 444 444 444 190 310 4 × 2 = 0 + 0.888 888 888 888 380 620 8;
  • 18) 0.888 888 888 888 380 620 8 × 2 = 1 + 0.777 777 777 776 761 241 6;
  • 19) 0.777 777 777 776 761 241 6 × 2 = 1 + 0.555 555 555 553 522 483 2;
  • 20) 0.555 555 555 553 522 483 2 × 2 = 1 + 0.111 111 111 107 044 966 4;
  • 21) 0.111 111 111 107 044 966 4 × 2 = 0 + 0.222 222 222 214 089 932 8;
  • 22) 0.222 222 222 214 089 932 8 × 2 = 0 + 0.444 444 444 428 179 865 6;
  • 23) 0.444 444 444 428 179 865 6 × 2 = 0 + 0.888 888 888 856 359 731 2;
  • 24) 0.888 888 888 856 359 731 2 × 2 = 1 + 0.777 777 777 712 719 462 4;
  • 25) 0.777 777 777 712 719 462 4 × 2 = 1 + 0.555 555 555 425 438 924 8;
  • 26) 0.555 555 555 425 438 924 8 × 2 = 1 + 0.111 111 110 850 877 849 6;
  • 27) 0.111 111 110 850 877 849 6 × 2 = 0 + 0.222 222 221 701 755 699 2;
  • 28) 0.222 222 221 701 755 699 2 × 2 = 0 + 0.444 444 443 403 511 398 4;
  • 29) 0.444 444 443 403 511 398 4 × 2 = 0 + 0.888 888 886 807 022 796 8;
  • 30) 0.888 888 886 807 022 796 8 × 2 = 1 + 0.777 777 773 614 045 593 6;
  • 31) 0.777 777 773 614 045 593 6 × 2 = 1 + 0.555 555 547 228 091 187 2;
  • 32) 0.555 555 547 228 091 187 2 × 2 = 1 + 0.111 111 094 456 182 374 4;
  • 33) 0.111 111 094 456 182 374 4 × 2 = 0 + 0.222 222 188 912 364 748 8;
  • 34) 0.222 222 188 912 364 748 8 × 2 = 0 + 0.444 444 377 824 729 497 6;
  • 35) 0.444 444 377 824 729 497 6 × 2 = 0 + 0.888 888 755 649 458 995 2;
  • 36) 0.888 888 755 649 458 995 2 × 2 = 1 + 0.777 777 511 298 917 990 4;
  • 37) 0.777 777 511 298 917 990 4 × 2 = 1 + 0.555 555 022 597 835 980 8;
  • 38) 0.555 555 022 597 835 980 8 × 2 = 1 + 0.111 110 045 195 671 961 6;
  • 39) 0.111 110 045 195 671 961 6 × 2 = 0 + 0.222 220 090 391 343 923 2;
  • 40) 0.222 220 090 391 343 923 2 × 2 = 0 + 0.444 440 180 782 687 846 4;
  • 41) 0.444 440 180 782 687 846 4 × 2 = 0 + 0.888 880 361 565 375 692 8;
  • 42) 0.888 880 361 565 375 692 8 × 2 = 1 + 0.777 760 723 130 751 385 6;
  • 43) 0.777 760 723 130 751 385 6 × 2 = 1 + 0.555 521 446 261 502 771 2;
  • 44) 0.555 521 446 261 502 771 2 × 2 = 1 + 0.111 042 892 523 005 542 4;
  • 45) 0.111 042 892 523 005 542 4 × 2 = 0 + 0.222 085 785 046 011 084 8;
  • 46) 0.222 085 785 046 011 084 8 × 2 = 0 + 0.444 171 570 092 022 169 6;
  • 47) 0.444 171 570 092 022 169 6 × 2 = 0 + 0.888 343 140 184 044 339 2;
  • 48) 0.888 343 140 184 044 339 2 × 2 = 1 + 0.776 686 280 368 088 678 4;
  • 49) 0.776 686 280 368 088 678 4 × 2 = 1 + 0.553 372 560 736 177 356 8;
  • 50) 0.553 372 560 736 177 356 8 × 2 = 1 + 0.106 745 121 472 354 713 6;
  • 51) 0.106 745 121 472 354 713 6 × 2 = 0 + 0.213 490 242 944 709 427 2;
  • 52) 0.213 490 242 944 709 427 2 × 2 = 0 + 0.426 980 485 889 418 854 4;
  • 53) 0.426 980 485 889 418 854 4 × 2 = 0 + 0.853 960 971 778 837 708 8;

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:


0.777 777 777 777 777 773 9(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 773 9(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:


24.777 777 777 777 777 773 9(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 24.777 777 777 777 777 773 9 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