24.777 777 777 778 011 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 778 011(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
24.777 777 777 778 011(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: 0.777 777 777 778 011.

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) 0.777 777 777 778 011 × 2 = 1 + 0.555 555 555 556 022;
  • 2) 0.555 555 555 556 022 × 2 = 1 + 0.111 111 111 112 044;
  • 3) 0.111 111 111 112 044 × 2 = 0 + 0.222 222 222 224 088;
  • 4) 0.222 222 222 224 088 × 2 = 0 + 0.444 444 444 448 176;
  • 5) 0.444 444 444 448 176 × 2 = 0 + 0.888 888 888 896 352;
  • 6) 0.888 888 888 896 352 × 2 = 1 + 0.777 777 777 792 704;
  • 7) 0.777 777 777 792 704 × 2 = 1 + 0.555 555 555 585 408;
  • 8) 0.555 555 555 585 408 × 2 = 1 + 0.111 111 111 170 816;
  • 9) 0.111 111 111 170 816 × 2 = 0 + 0.222 222 222 341 632;
  • 10) 0.222 222 222 341 632 × 2 = 0 + 0.444 444 444 683 264;
  • 11) 0.444 444 444 683 264 × 2 = 0 + 0.888 888 889 366 528;
  • 12) 0.888 888 889 366 528 × 2 = 1 + 0.777 777 778 733 056;
  • 13) 0.777 777 778 733 056 × 2 = 1 + 0.555 555 557 466 112;
  • 14) 0.555 555 557 466 112 × 2 = 1 + 0.111 111 114 932 224;
  • 15) 0.111 111 114 932 224 × 2 = 0 + 0.222 222 229 864 448;
  • 16) 0.222 222 229 864 448 × 2 = 0 + 0.444 444 459 728 896;
  • 17) 0.444 444 459 728 896 × 2 = 0 + 0.888 888 919 457 792;
  • 18) 0.888 888 919 457 792 × 2 = 1 + 0.777 777 838 915 584;
  • 19) 0.777 777 838 915 584 × 2 = 1 + 0.555 555 677 831 168;
  • 20) 0.555 555 677 831 168 × 2 = 1 + 0.111 111 355 662 336;
  • 21) 0.111 111 355 662 336 × 2 = 0 + 0.222 222 711 324 672;
  • 22) 0.222 222 711 324 672 × 2 = 0 + 0.444 445 422 649 344;
  • 23) 0.444 445 422 649 344 × 2 = 0 + 0.888 890 845 298 688;
  • 24) 0.888 890 845 298 688 × 2 = 1 + 0.777 781 690 597 376;
  • 25) 0.777 781 690 597 376 × 2 = 1 + 0.555 563 381 194 752;
  • 26) 0.555 563 381 194 752 × 2 = 1 + 0.111 126 762 389 504;
  • 27) 0.111 126 762 389 504 × 2 = 0 + 0.222 253 524 779 008;
  • 28) 0.222 253 524 779 008 × 2 = 0 + 0.444 507 049 558 016;
  • 29) 0.444 507 049 558 016 × 2 = 0 + 0.889 014 099 116 032;
  • 30) 0.889 014 099 116 032 × 2 = 1 + 0.778 028 198 232 064;
  • 31) 0.778 028 198 232 064 × 2 = 1 + 0.556 056 396 464 128;
  • 32) 0.556 056 396 464 128 × 2 = 1 + 0.112 112 792 928 256;
  • 33) 0.112 112 792 928 256 × 2 = 0 + 0.224 225 585 856 512;
  • 34) 0.224 225 585 856 512 × 2 = 0 + 0.448 451 171 713 024;
  • 35) 0.448 451 171 713 024 × 2 = 0 + 0.896 902 343 426 048;
  • 36) 0.896 902 343 426 048 × 2 = 1 + 0.793 804 686 852 096;
  • 37) 0.793 804 686 852 096 × 2 = 1 + 0.587 609 373 704 192;
  • 38) 0.587 609 373 704 192 × 2 = 1 + 0.175 218 747 408 384;
  • 39) 0.175 218 747 408 384 × 2 = 0 + 0.350 437 494 816 768;
  • 40) 0.350 437 494 816 768 × 2 = 0 + 0.700 874 989 633 536;
  • 41) 0.700 874 989 633 536 × 2 = 1 + 0.401 749 979 267 072;
  • 42) 0.401 749 979 267 072 × 2 = 0 + 0.803 499 958 534 144;
  • 43) 0.803 499 958 534 144 × 2 = 1 + 0.606 999 917 068 288;
  • 44) 0.606 999 917 068 288 × 2 = 1 + 0.213 999 834 136 576;
  • 45) 0.213 999 834 136 576 × 2 = 0 + 0.427 999 668 273 152;
  • 46) 0.427 999 668 273 152 × 2 = 0 + 0.855 999 336 546 304;
  • 47) 0.855 999 336 546 304 × 2 = 1 + 0.711 998 673 092 608;
  • 48) 0.711 998 673 092 608 × 2 = 1 + 0.423 997 346 185 216;
  • 49) 0.423 997 346 185 216 × 2 = 0 + 0.847 994 692 370 432;
  • 50) 0.847 994 692 370 432 × 2 = 1 + 0.695 989 384 740 864;
  • 51) 0.695 989 384 740 864 × 2 = 1 + 0.391 978 769 481 728;
  • 52) 0.391 978 769 481 728 × 2 = 0 + 0.783 957 538 963 456;
  • 53) 0.783 957 538 963 456 × 2 = 1 + 0.567 915 077 926 912;

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:


0.777 777 777 778 011(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1011 0011 0110 1(2)

5. Positive number before normalization:

24.777 777 777 778 011(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1011 0011 0110 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:


24.777 777 777 778 011(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1011 0011 0110 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1011 0011 0110 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1011 0011 0110 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 1011 0011 0110 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 1011 0011 0 1101 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1011 0011


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 1011 0011


Decimal number 24.777 777 777 778 011 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 1011 0011

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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