24.777 777 777 777 777 777 802 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 802 6(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
24.777 777 777 777 777 777 802 6(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: 0.777 777 777 777 777 777 802 6.

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) 0.777 777 777 777 777 777 802 6 × 2 = 1 + 0.555 555 555 555 555 555 605 2;
  • 2) 0.555 555 555 555 555 555 605 2 × 2 = 1 + 0.111 111 111 111 111 111 210 4;
  • 3) 0.111 111 111 111 111 111 210 4 × 2 = 0 + 0.222 222 222 222 222 222 420 8;
  • 4) 0.222 222 222 222 222 222 420 8 × 2 = 0 + 0.444 444 444 444 444 444 841 6;
  • 5) 0.444 444 444 444 444 444 841 6 × 2 = 0 + 0.888 888 888 888 888 889 683 2;
  • 6) 0.888 888 888 888 888 889 683 2 × 2 = 1 + 0.777 777 777 777 777 779 366 4;
  • 7) 0.777 777 777 777 777 779 366 4 × 2 = 1 + 0.555 555 555 555 555 558 732 8;
  • 8) 0.555 555 555 555 555 558 732 8 × 2 = 1 + 0.111 111 111 111 111 117 465 6;
  • 9) 0.111 111 111 111 111 117 465 6 × 2 = 0 + 0.222 222 222 222 222 234 931 2;
  • 10) 0.222 222 222 222 222 234 931 2 × 2 = 0 + 0.444 444 444 444 444 469 862 4;
  • 11) 0.444 444 444 444 444 469 862 4 × 2 = 0 + 0.888 888 888 888 888 939 724 8;
  • 12) 0.888 888 888 888 888 939 724 8 × 2 = 1 + 0.777 777 777 777 777 879 449 6;
  • 13) 0.777 777 777 777 777 879 449 6 × 2 = 1 + 0.555 555 555 555 555 758 899 2;
  • 14) 0.555 555 555 555 555 758 899 2 × 2 = 1 + 0.111 111 111 111 111 517 798 4;
  • 15) 0.111 111 111 111 111 517 798 4 × 2 = 0 + 0.222 222 222 222 223 035 596 8;
  • 16) 0.222 222 222 222 223 035 596 8 × 2 = 0 + 0.444 444 444 444 446 071 193 6;
  • 17) 0.444 444 444 444 446 071 193 6 × 2 = 0 + 0.888 888 888 888 892 142 387 2;
  • 18) 0.888 888 888 888 892 142 387 2 × 2 = 1 + 0.777 777 777 777 784 284 774 4;
  • 19) 0.777 777 777 777 784 284 774 4 × 2 = 1 + 0.555 555 555 555 568 569 548 8;
  • 20) 0.555 555 555 555 568 569 548 8 × 2 = 1 + 0.111 111 111 111 137 139 097 6;
  • 21) 0.111 111 111 111 137 139 097 6 × 2 = 0 + 0.222 222 222 222 274 278 195 2;
  • 22) 0.222 222 222 222 274 278 195 2 × 2 = 0 + 0.444 444 444 444 548 556 390 4;
  • 23) 0.444 444 444 444 548 556 390 4 × 2 = 0 + 0.888 888 888 889 097 112 780 8;
  • 24) 0.888 888 888 889 097 112 780 8 × 2 = 1 + 0.777 777 777 778 194 225 561 6;
  • 25) 0.777 777 777 778 194 225 561 6 × 2 = 1 + 0.555 555 555 556 388 451 123 2;
  • 26) 0.555 555 555 556 388 451 123 2 × 2 = 1 + 0.111 111 111 112 776 902 246 4;
  • 27) 0.111 111 111 112 776 902 246 4 × 2 = 0 + 0.222 222 222 225 553 804 492 8;
  • 28) 0.222 222 222 225 553 804 492 8 × 2 = 0 + 0.444 444 444 451 107 608 985 6;
  • 29) 0.444 444 444 451 107 608 985 6 × 2 = 0 + 0.888 888 888 902 215 217 971 2;
  • 30) 0.888 888 888 902 215 217 971 2 × 2 = 1 + 0.777 777 777 804 430 435 942 4;
  • 31) 0.777 777 777 804 430 435 942 4 × 2 = 1 + 0.555 555 555 608 860 871 884 8;
  • 32) 0.555 555 555 608 860 871 884 8 × 2 = 1 + 0.111 111 111 217 721 743 769 6;
  • 33) 0.111 111 111 217 721 743 769 6 × 2 = 0 + 0.222 222 222 435 443 487 539 2;
  • 34) 0.222 222 222 435 443 487 539 2 × 2 = 0 + 0.444 444 444 870 886 975 078 4;
  • 35) 0.444 444 444 870 886 975 078 4 × 2 = 0 + 0.888 888 889 741 773 950 156 8;
  • 36) 0.888 888 889 741 773 950 156 8 × 2 = 1 + 0.777 777 779 483 547 900 313 6;
  • 37) 0.777 777 779 483 547 900 313 6 × 2 = 1 + 0.555 555 558 967 095 800 627 2;
  • 38) 0.555 555 558 967 095 800 627 2 × 2 = 1 + 0.111 111 117 934 191 601 254 4;
  • 39) 0.111 111 117 934 191 601 254 4 × 2 = 0 + 0.222 222 235 868 383 202 508 8;
  • 40) 0.222 222 235 868 383 202 508 8 × 2 = 0 + 0.444 444 471 736 766 405 017 6;
  • 41) 0.444 444 471 736 766 405 017 6 × 2 = 0 + 0.888 888 943 473 532 810 035 2;
  • 42) 0.888 888 943 473 532 810 035 2 × 2 = 1 + 0.777 777 886 947 065 620 070 4;
  • 43) 0.777 777 886 947 065 620 070 4 × 2 = 1 + 0.555 555 773 894 131 240 140 8;
  • 44) 0.555 555 773 894 131 240 140 8 × 2 = 1 + 0.111 111 547 788 262 480 281 6;
  • 45) 0.111 111 547 788 262 480 281 6 × 2 = 0 + 0.222 223 095 576 524 960 563 2;
  • 46) 0.222 223 095 576 524 960 563 2 × 2 = 0 + 0.444 446 191 153 049 921 126 4;
  • 47) 0.444 446 191 153 049 921 126 4 × 2 = 0 + 0.888 892 382 306 099 842 252 8;
  • 48) 0.888 892 382 306 099 842 252 8 × 2 = 1 + 0.777 784 764 612 199 684 505 6;
  • 49) 0.777 784 764 612 199 684 505 6 × 2 = 1 + 0.555 569 529 224 399 369 011 2;
  • 50) 0.555 569 529 224 399 369 011 2 × 2 = 1 + 0.111 139 058 448 798 738 022 4;
  • 51) 0.111 139 058 448 798 738 022 4 × 2 = 0 + 0.222 278 116 897 597 476 044 8;
  • 52) 0.222 278 116 897 597 476 044 8 × 2 = 0 + 0.444 556 233 795 194 952 089 6;
  • 53) 0.444 556 233 795 194 952 089 6 × 2 = 0 + 0.889 112 467 590 389 904 179 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:


0.777 777 777 777 777 777 802 6(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 802 6(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:


24.777 777 777 777 777 777 802 6(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 24.777 777 777 777 777 777 802 6 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