24.777 777 777 777 777 763 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 763 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
24.777 777 777 777 777 763 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: 0.777 777 777 777 777 763 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) 0.777 777 777 777 777 763 9 × 2 = 1 + 0.555 555 555 555 555 527 8;
  • 2) 0.555 555 555 555 555 527 8 × 2 = 1 + 0.111 111 111 111 111 055 6;
  • 3) 0.111 111 111 111 111 055 6 × 2 = 0 + 0.222 222 222 222 222 111 2;
  • 4) 0.222 222 222 222 222 111 2 × 2 = 0 + 0.444 444 444 444 444 222 4;
  • 5) 0.444 444 444 444 444 222 4 × 2 = 0 + 0.888 888 888 888 888 444 8;
  • 6) 0.888 888 888 888 888 444 8 × 2 = 1 + 0.777 777 777 777 776 889 6;
  • 7) 0.777 777 777 777 776 889 6 × 2 = 1 + 0.555 555 555 555 553 779 2;
  • 8) 0.555 555 555 555 553 779 2 × 2 = 1 + 0.111 111 111 111 107 558 4;
  • 9) 0.111 111 111 111 107 558 4 × 2 = 0 + 0.222 222 222 222 215 116 8;
  • 10) 0.222 222 222 222 215 116 8 × 2 = 0 + 0.444 444 444 444 430 233 6;
  • 11) 0.444 444 444 444 430 233 6 × 2 = 0 + 0.888 888 888 888 860 467 2;
  • 12) 0.888 888 888 888 860 467 2 × 2 = 1 + 0.777 777 777 777 720 934 4;
  • 13) 0.777 777 777 777 720 934 4 × 2 = 1 + 0.555 555 555 555 441 868 8;
  • 14) 0.555 555 555 555 441 868 8 × 2 = 1 + 0.111 111 111 110 883 737 6;
  • 15) 0.111 111 111 110 883 737 6 × 2 = 0 + 0.222 222 222 221 767 475 2;
  • 16) 0.222 222 222 221 767 475 2 × 2 = 0 + 0.444 444 444 443 534 950 4;
  • 17) 0.444 444 444 443 534 950 4 × 2 = 0 + 0.888 888 888 887 069 900 8;
  • 18) 0.888 888 888 887 069 900 8 × 2 = 1 + 0.777 777 777 774 139 801 6;
  • 19) 0.777 777 777 774 139 801 6 × 2 = 1 + 0.555 555 555 548 279 603 2;
  • 20) 0.555 555 555 548 279 603 2 × 2 = 1 + 0.111 111 111 096 559 206 4;
  • 21) 0.111 111 111 096 559 206 4 × 2 = 0 + 0.222 222 222 193 118 412 8;
  • 22) 0.222 222 222 193 118 412 8 × 2 = 0 + 0.444 444 444 386 236 825 6;
  • 23) 0.444 444 444 386 236 825 6 × 2 = 0 + 0.888 888 888 772 473 651 2;
  • 24) 0.888 888 888 772 473 651 2 × 2 = 1 + 0.777 777 777 544 947 302 4;
  • 25) 0.777 777 777 544 947 302 4 × 2 = 1 + 0.555 555 555 089 894 604 8;
  • 26) 0.555 555 555 089 894 604 8 × 2 = 1 + 0.111 111 110 179 789 209 6;
  • 27) 0.111 111 110 179 789 209 6 × 2 = 0 + 0.222 222 220 359 578 419 2;
  • 28) 0.222 222 220 359 578 419 2 × 2 = 0 + 0.444 444 440 719 156 838 4;
  • 29) 0.444 444 440 719 156 838 4 × 2 = 0 + 0.888 888 881 438 313 676 8;
  • 30) 0.888 888 881 438 313 676 8 × 2 = 1 + 0.777 777 762 876 627 353 6;
  • 31) 0.777 777 762 876 627 353 6 × 2 = 1 + 0.555 555 525 753 254 707 2;
  • 32) 0.555 555 525 753 254 707 2 × 2 = 1 + 0.111 111 051 506 509 414 4;
  • 33) 0.111 111 051 506 509 414 4 × 2 = 0 + 0.222 222 103 013 018 828 8;
  • 34) 0.222 222 103 013 018 828 8 × 2 = 0 + 0.444 444 206 026 037 657 6;
  • 35) 0.444 444 206 026 037 657 6 × 2 = 0 + 0.888 888 412 052 075 315 2;
  • 36) 0.888 888 412 052 075 315 2 × 2 = 1 + 0.777 776 824 104 150 630 4;
  • 37) 0.777 776 824 104 150 630 4 × 2 = 1 + 0.555 553 648 208 301 260 8;
  • 38) 0.555 553 648 208 301 260 8 × 2 = 1 + 0.111 107 296 416 602 521 6;
  • 39) 0.111 107 296 416 602 521 6 × 2 = 0 + 0.222 214 592 833 205 043 2;
  • 40) 0.222 214 592 833 205 043 2 × 2 = 0 + 0.444 429 185 666 410 086 4;
  • 41) 0.444 429 185 666 410 086 4 × 2 = 0 + 0.888 858 371 332 820 172 8;
  • 42) 0.888 858 371 332 820 172 8 × 2 = 1 + 0.777 716 742 665 640 345 6;
  • 43) 0.777 716 742 665 640 345 6 × 2 = 1 + 0.555 433 485 331 280 691 2;
  • 44) 0.555 433 485 331 280 691 2 × 2 = 1 + 0.110 866 970 662 561 382 4;
  • 45) 0.110 866 970 662 561 382 4 × 2 = 0 + 0.221 733 941 325 122 764 8;
  • 46) 0.221 733 941 325 122 764 8 × 2 = 0 + 0.443 467 882 650 245 529 6;
  • 47) 0.443 467 882 650 245 529 6 × 2 = 0 + 0.886 935 765 300 491 059 2;
  • 48) 0.886 935 765 300 491 059 2 × 2 = 1 + 0.773 871 530 600 982 118 4;
  • 49) 0.773 871 530 600 982 118 4 × 2 = 1 + 0.547 743 061 201 964 236 8;
  • 50) 0.547 743 061 201 964 236 8 × 2 = 1 + 0.095 486 122 403 928 473 6;
  • 51) 0.095 486 122 403 928 473 6 × 2 = 0 + 0.190 972 244 807 856 947 2;
  • 52) 0.190 972 244 807 856 947 2 × 2 = 0 + 0.381 944 489 615 713 894 4;
  • 53) 0.381 944 489 615 713 894 4 × 2 = 0 + 0.763 888 979 231 427 788 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:


0.777 777 777 777 777 763 9(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 763 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:


24.777 777 777 777 777 763 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 24.777 777 777 777 777 763 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