24.777 777 777 777 777 779 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 779 37(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
24.777 777 777 777 777 779 37(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: 0.777 777 777 777 777 779 37.

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) 0.777 777 777 777 777 779 37 × 2 = 1 + 0.555 555 555 555 555 558 74;
  • 2) 0.555 555 555 555 555 558 74 × 2 = 1 + 0.111 111 111 111 111 117 48;
  • 3) 0.111 111 111 111 111 117 48 × 2 = 0 + 0.222 222 222 222 222 234 96;
  • 4) 0.222 222 222 222 222 234 96 × 2 = 0 + 0.444 444 444 444 444 469 92;
  • 5) 0.444 444 444 444 444 469 92 × 2 = 0 + 0.888 888 888 888 888 939 84;
  • 6) 0.888 888 888 888 888 939 84 × 2 = 1 + 0.777 777 777 777 777 879 68;
  • 7) 0.777 777 777 777 777 879 68 × 2 = 1 + 0.555 555 555 555 555 759 36;
  • 8) 0.555 555 555 555 555 759 36 × 2 = 1 + 0.111 111 111 111 111 518 72;
  • 9) 0.111 111 111 111 111 518 72 × 2 = 0 + 0.222 222 222 222 223 037 44;
  • 10) 0.222 222 222 222 223 037 44 × 2 = 0 + 0.444 444 444 444 446 074 88;
  • 11) 0.444 444 444 444 446 074 88 × 2 = 0 + 0.888 888 888 888 892 149 76;
  • 12) 0.888 888 888 888 892 149 76 × 2 = 1 + 0.777 777 777 777 784 299 52;
  • 13) 0.777 777 777 777 784 299 52 × 2 = 1 + 0.555 555 555 555 568 599 04;
  • 14) 0.555 555 555 555 568 599 04 × 2 = 1 + 0.111 111 111 111 137 198 08;
  • 15) 0.111 111 111 111 137 198 08 × 2 = 0 + 0.222 222 222 222 274 396 16;
  • 16) 0.222 222 222 222 274 396 16 × 2 = 0 + 0.444 444 444 444 548 792 32;
  • 17) 0.444 444 444 444 548 792 32 × 2 = 0 + 0.888 888 888 889 097 584 64;
  • 18) 0.888 888 888 889 097 584 64 × 2 = 1 + 0.777 777 777 778 195 169 28;
  • 19) 0.777 777 777 778 195 169 28 × 2 = 1 + 0.555 555 555 556 390 338 56;
  • 20) 0.555 555 555 556 390 338 56 × 2 = 1 + 0.111 111 111 112 780 677 12;
  • 21) 0.111 111 111 112 780 677 12 × 2 = 0 + 0.222 222 222 225 561 354 24;
  • 22) 0.222 222 222 225 561 354 24 × 2 = 0 + 0.444 444 444 451 122 708 48;
  • 23) 0.444 444 444 451 122 708 48 × 2 = 0 + 0.888 888 888 902 245 416 96;
  • 24) 0.888 888 888 902 245 416 96 × 2 = 1 + 0.777 777 777 804 490 833 92;
  • 25) 0.777 777 777 804 490 833 92 × 2 = 1 + 0.555 555 555 608 981 667 84;
  • 26) 0.555 555 555 608 981 667 84 × 2 = 1 + 0.111 111 111 217 963 335 68;
  • 27) 0.111 111 111 217 963 335 68 × 2 = 0 + 0.222 222 222 435 926 671 36;
  • 28) 0.222 222 222 435 926 671 36 × 2 = 0 + 0.444 444 444 871 853 342 72;
  • 29) 0.444 444 444 871 853 342 72 × 2 = 0 + 0.888 888 889 743 706 685 44;
  • 30) 0.888 888 889 743 706 685 44 × 2 = 1 + 0.777 777 779 487 413 370 88;
  • 31) 0.777 777 779 487 413 370 88 × 2 = 1 + 0.555 555 558 974 826 741 76;
  • 32) 0.555 555 558 974 826 741 76 × 2 = 1 + 0.111 111 117 949 653 483 52;
  • 33) 0.111 111 117 949 653 483 52 × 2 = 0 + 0.222 222 235 899 306 967 04;
  • 34) 0.222 222 235 899 306 967 04 × 2 = 0 + 0.444 444 471 798 613 934 08;
  • 35) 0.444 444 471 798 613 934 08 × 2 = 0 + 0.888 888 943 597 227 868 16;
  • 36) 0.888 888 943 597 227 868 16 × 2 = 1 + 0.777 777 887 194 455 736 32;
  • 37) 0.777 777 887 194 455 736 32 × 2 = 1 + 0.555 555 774 388 911 472 64;
  • 38) 0.555 555 774 388 911 472 64 × 2 = 1 + 0.111 111 548 777 822 945 28;
  • 39) 0.111 111 548 777 822 945 28 × 2 = 0 + 0.222 223 097 555 645 890 56;
  • 40) 0.222 223 097 555 645 890 56 × 2 = 0 + 0.444 446 195 111 291 781 12;
  • 41) 0.444 446 195 111 291 781 12 × 2 = 0 + 0.888 892 390 222 583 562 24;
  • 42) 0.888 892 390 222 583 562 24 × 2 = 1 + 0.777 784 780 445 167 124 48;
  • 43) 0.777 784 780 445 167 124 48 × 2 = 1 + 0.555 569 560 890 334 248 96;
  • 44) 0.555 569 560 890 334 248 96 × 2 = 1 + 0.111 139 121 780 668 497 92;
  • 45) 0.111 139 121 780 668 497 92 × 2 = 0 + 0.222 278 243 561 336 995 84;
  • 46) 0.222 278 243 561 336 995 84 × 2 = 0 + 0.444 556 487 122 673 991 68;
  • 47) 0.444 556 487 122 673 991 68 × 2 = 0 + 0.889 112 974 245 347 983 36;
  • 48) 0.889 112 974 245 347 983 36 × 2 = 1 + 0.778 225 948 490 695 966 72;
  • 49) 0.778 225 948 490 695 966 72 × 2 = 1 + 0.556 451 896 981 391 933 44;
  • 50) 0.556 451 896 981 391 933 44 × 2 = 1 + 0.112 903 793 962 783 866 88;
  • 51) 0.112 903 793 962 783 866 88 × 2 = 0 + 0.225 807 587 925 567 733 76;
  • 52) 0.225 807 587 925 567 733 76 × 2 = 0 + 0.451 615 175 851 135 467 52;
  • 53) 0.451 615 175 851 135 467 52 × 2 = 0 + 0.903 230 351 702 270 935 04;

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:


0.777 777 777 777 777 779 37(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 779 37(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:


24.777 777 777 777 777 779 37(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 24.777 777 777 777 777 779 37 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