24.777 777 777 777 777 780 85 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 780 85(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
24.777 777 777 777 777 780 85(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: 0.777 777 777 777 777 780 85.

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) 0.777 777 777 777 777 780 85 × 2 = 1 + 0.555 555 555 555 555 561 7;
  • 2) 0.555 555 555 555 555 561 7 × 2 = 1 + 0.111 111 111 111 111 123 4;
  • 3) 0.111 111 111 111 111 123 4 × 2 = 0 + 0.222 222 222 222 222 246 8;
  • 4) 0.222 222 222 222 222 246 8 × 2 = 0 + 0.444 444 444 444 444 493 6;
  • 5) 0.444 444 444 444 444 493 6 × 2 = 0 + 0.888 888 888 888 888 987 2;
  • 6) 0.888 888 888 888 888 987 2 × 2 = 1 + 0.777 777 777 777 777 974 4;
  • 7) 0.777 777 777 777 777 974 4 × 2 = 1 + 0.555 555 555 555 555 948 8;
  • 8) 0.555 555 555 555 555 948 8 × 2 = 1 + 0.111 111 111 111 111 897 6;
  • 9) 0.111 111 111 111 111 897 6 × 2 = 0 + 0.222 222 222 222 223 795 2;
  • 10) 0.222 222 222 222 223 795 2 × 2 = 0 + 0.444 444 444 444 447 590 4;
  • 11) 0.444 444 444 444 447 590 4 × 2 = 0 + 0.888 888 888 888 895 180 8;
  • 12) 0.888 888 888 888 895 180 8 × 2 = 1 + 0.777 777 777 777 790 361 6;
  • 13) 0.777 777 777 777 790 361 6 × 2 = 1 + 0.555 555 555 555 580 723 2;
  • 14) 0.555 555 555 555 580 723 2 × 2 = 1 + 0.111 111 111 111 161 446 4;
  • 15) 0.111 111 111 111 161 446 4 × 2 = 0 + 0.222 222 222 222 322 892 8;
  • 16) 0.222 222 222 222 322 892 8 × 2 = 0 + 0.444 444 444 444 645 785 6;
  • 17) 0.444 444 444 444 645 785 6 × 2 = 0 + 0.888 888 888 889 291 571 2;
  • 18) 0.888 888 888 889 291 571 2 × 2 = 1 + 0.777 777 777 778 583 142 4;
  • 19) 0.777 777 777 778 583 142 4 × 2 = 1 + 0.555 555 555 557 166 284 8;
  • 20) 0.555 555 555 557 166 284 8 × 2 = 1 + 0.111 111 111 114 332 569 6;
  • 21) 0.111 111 111 114 332 569 6 × 2 = 0 + 0.222 222 222 228 665 139 2;
  • 22) 0.222 222 222 228 665 139 2 × 2 = 0 + 0.444 444 444 457 330 278 4;
  • 23) 0.444 444 444 457 330 278 4 × 2 = 0 + 0.888 888 888 914 660 556 8;
  • 24) 0.888 888 888 914 660 556 8 × 2 = 1 + 0.777 777 777 829 321 113 6;
  • 25) 0.777 777 777 829 321 113 6 × 2 = 1 + 0.555 555 555 658 642 227 2;
  • 26) 0.555 555 555 658 642 227 2 × 2 = 1 + 0.111 111 111 317 284 454 4;
  • 27) 0.111 111 111 317 284 454 4 × 2 = 0 + 0.222 222 222 634 568 908 8;
  • 28) 0.222 222 222 634 568 908 8 × 2 = 0 + 0.444 444 445 269 137 817 6;
  • 29) 0.444 444 445 269 137 817 6 × 2 = 0 + 0.888 888 890 538 275 635 2;
  • 30) 0.888 888 890 538 275 635 2 × 2 = 1 + 0.777 777 781 076 551 270 4;
  • 31) 0.777 777 781 076 551 270 4 × 2 = 1 + 0.555 555 562 153 102 540 8;
  • 32) 0.555 555 562 153 102 540 8 × 2 = 1 + 0.111 111 124 306 205 081 6;
  • 33) 0.111 111 124 306 205 081 6 × 2 = 0 + 0.222 222 248 612 410 163 2;
  • 34) 0.222 222 248 612 410 163 2 × 2 = 0 + 0.444 444 497 224 820 326 4;
  • 35) 0.444 444 497 224 820 326 4 × 2 = 0 + 0.888 888 994 449 640 652 8;
  • 36) 0.888 888 994 449 640 652 8 × 2 = 1 + 0.777 777 988 899 281 305 6;
  • 37) 0.777 777 988 899 281 305 6 × 2 = 1 + 0.555 555 977 798 562 611 2;
  • 38) 0.555 555 977 798 562 611 2 × 2 = 1 + 0.111 111 955 597 125 222 4;
  • 39) 0.111 111 955 597 125 222 4 × 2 = 0 + 0.222 223 911 194 250 444 8;
  • 40) 0.222 223 911 194 250 444 8 × 2 = 0 + 0.444 447 822 388 500 889 6;
  • 41) 0.444 447 822 388 500 889 6 × 2 = 0 + 0.888 895 644 777 001 779 2;
  • 42) 0.888 895 644 777 001 779 2 × 2 = 1 + 0.777 791 289 554 003 558 4;
  • 43) 0.777 791 289 554 003 558 4 × 2 = 1 + 0.555 582 579 108 007 116 8;
  • 44) 0.555 582 579 108 007 116 8 × 2 = 1 + 0.111 165 158 216 014 233 6;
  • 45) 0.111 165 158 216 014 233 6 × 2 = 0 + 0.222 330 316 432 028 467 2;
  • 46) 0.222 330 316 432 028 467 2 × 2 = 0 + 0.444 660 632 864 056 934 4;
  • 47) 0.444 660 632 864 056 934 4 × 2 = 0 + 0.889 321 265 728 113 868 8;
  • 48) 0.889 321 265 728 113 868 8 × 2 = 1 + 0.778 642 531 456 227 737 6;
  • 49) 0.778 642 531 456 227 737 6 × 2 = 1 + 0.557 285 062 912 455 475 2;
  • 50) 0.557 285 062 912 455 475 2 × 2 = 1 + 0.114 570 125 824 910 950 4;
  • 51) 0.114 570 125 824 910 950 4 × 2 = 0 + 0.229 140 251 649 821 900 8;
  • 52) 0.229 140 251 649 821 900 8 × 2 = 0 + 0.458 280 503 299 643 801 6;
  • 53) 0.458 280 503 299 643 801 6 × 2 = 0 + 0.916 561 006 599 287 603 2;

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:


0.777 777 777 777 777 780 85(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 780 85(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:


24.777 777 777 777 777 780 85(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 24.777 777 777 777 777 780 85 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