24.777 777 777 777 777 777 442 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 442(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
24.777 777 777 777 777 777 442(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: 0.777 777 777 777 777 777 442.

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) 0.777 777 777 777 777 777 442 × 2 = 1 + 0.555 555 555 555 555 554 884;
  • 2) 0.555 555 555 555 555 554 884 × 2 = 1 + 0.111 111 111 111 111 109 768;
  • 3) 0.111 111 111 111 111 109 768 × 2 = 0 + 0.222 222 222 222 222 219 536;
  • 4) 0.222 222 222 222 222 219 536 × 2 = 0 + 0.444 444 444 444 444 439 072;
  • 5) 0.444 444 444 444 444 439 072 × 2 = 0 + 0.888 888 888 888 888 878 144;
  • 6) 0.888 888 888 888 888 878 144 × 2 = 1 + 0.777 777 777 777 777 756 288;
  • 7) 0.777 777 777 777 777 756 288 × 2 = 1 + 0.555 555 555 555 555 512 576;
  • 8) 0.555 555 555 555 555 512 576 × 2 = 1 + 0.111 111 111 111 111 025 152;
  • 9) 0.111 111 111 111 111 025 152 × 2 = 0 + 0.222 222 222 222 222 050 304;
  • 10) 0.222 222 222 222 222 050 304 × 2 = 0 + 0.444 444 444 444 444 100 608;
  • 11) 0.444 444 444 444 444 100 608 × 2 = 0 + 0.888 888 888 888 888 201 216;
  • 12) 0.888 888 888 888 888 201 216 × 2 = 1 + 0.777 777 777 777 776 402 432;
  • 13) 0.777 777 777 777 776 402 432 × 2 = 1 + 0.555 555 555 555 552 804 864;
  • 14) 0.555 555 555 555 552 804 864 × 2 = 1 + 0.111 111 111 111 105 609 728;
  • 15) 0.111 111 111 111 105 609 728 × 2 = 0 + 0.222 222 222 222 211 219 456;
  • 16) 0.222 222 222 222 211 219 456 × 2 = 0 + 0.444 444 444 444 422 438 912;
  • 17) 0.444 444 444 444 422 438 912 × 2 = 0 + 0.888 888 888 888 844 877 824;
  • 18) 0.888 888 888 888 844 877 824 × 2 = 1 + 0.777 777 777 777 689 755 648;
  • 19) 0.777 777 777 777 689 755 648 × 2 = 1 + 0.555 555 555 555 379 511 296;
  • 20) 0.555 555 555 555 379 511 296 × 2 = 1 + 0.111 111 111 110 759 022 592;
  • 21) 0.111 111 111 110 759 022 592 × 2 = 0 + 0.222 222 222 221 518 045 184;
  • 22) 0.222 222 222 221 518 045 184 × 2 = 0 + 0.444 444 444 443 036 090 368;
  • 23) 0.444 444 444 443 036 090 368 × 2 = 0 + 0.888 888 888 886 072 180 736;
  • 24) 0.888 888 888 886 072 180 736 × 2 = 1 + 0.777 777 777 772 144 361 472;
  • 25) 0.777 777 777 772 144 361 472 × 2 = 1 + 0.555 555 555 544 288 722 944;
  • 26) 0.555 555 555 544 288 722 944 × 2 = 1 + 0.111 111 111 088 577 445 888;
  • 27) 0.111 111 111 088 577 445 888 × 2 = 0 + 0.222 222 222 177 154 891 776;
  • 28) 0.222 222 222 177 154 891 776 × 2 = 0 + 0.444 444 444 354 309 783 552;
  • 29) 0.444 444 444 354 309 783 552 × 2 = 0 + 0.888 888 888 708 619 567 104;
  • 30) 0.888 888 888 708 619 567 104 × 2 = 1 + 0.777 777 777 417 239 134 208;
  • 31) 0.777 777 777 417 239 134 208 × 2 = 1 + 0.555 555 554 834 478 268 416;
  • 32) 0.555 555 554 834 478 268 416 × 2 = 1 + 0.111 111 109 668 956 536 832;
  • 33) 0.111 111 109 668 956 536 832 × 2 = 0 + 0.222 222 219 337 913 073 664;
  • 34) 0.222 222 219 337 913 073 664 × 2 = 0 + 0.444 444 438 675 826 147 328;
  • 35) 0.444 444 438 675 826 147 328 × 2 = 0 + 0.888 888 877 351 652 294 656;
  • 36) 0.888 888 877 351 652 294 656 × 2 = 1 + 0.777 777 754 703 304 589 312;
  • 37) 0.777 777 754 703 304 589 312 × 2 = 1 + 0.555 555 509 406 609 178 624;
  • 38) 0.555 555 509 406 609 178 624 × 2 = 1 + 0.111 111 018 813 218 357 248;
  • 39) 0.111 111 018 813 218 357 248 × 2 = 0 + 0.222 222 037 626 436 714 496;
  • 40) 0.222 222 037 626 436 714 496 × 2 = 0 + 0.444 444 075 252 873 428 992;
  • 41) 0.444 444 075 252 873 428 992 × 2 = 0 + 0.888 888 150 505 746 857 984;
  • 42) 0.888 888 150 505 746 857 984 × 2 = 1 + 0.777 776 301 011 493 715 968;
  • 43) 0.777 776 301 011 493 715 968 × 2 = 1 + 0.555 552 602 022 987 431 936;
  • 44) 0.555 552 602 022 987 431 936 × 2 = 1 + 0.111 105 204 045 974 863 872;
  • 45) 0.111 105 204 045 974 863 872 × 2 = 0 + 0.222 210 408 091 949 727 744;
  • 46) 0.222 210 408 091 949 727 744 × 2 = 0 + 0.444 420 816 183 899 455 488;
  • 47) 0.444 420 816 183 899 455 488 × 2 = 0 + 0.888 841 632 367 798 910 976;
  • 48) 0.888 841 632 367 798 910 976 × 2 = 1 + 0.777 683 264 735 597 821 952;
  • 49) 0.777 683 264 735 597 821 952 × 2 = 1 + 0.555 366 529 471 195 643 904;
  • 50) 0.555 366 529 471 195 643 904 × 2 = 1 + 0.110 733 058 942 391 287 808;
  • 51) 0.110 733 058 942 391 287 808 × 2 = 0 + 0.221 466 117 884 782 575 616;
  • 52) 0.221 466 117 884 782 575 616 × 2 = 0 + 0.442 932 235 769 565 151 232;
  • 53) 0.442 932 235 769 565 151 232 × 2 = 0 + 0.885 864 471 539 130 302 464;

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:


0.777 777 777 777 777 777 442(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 442(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:


24.777 777 777 777 777 777 442(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 24.777 777 777 777 777 777 442 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