24.777 777 777 777 777 777 663 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 663(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
24.777 777 777 777 777 777 663(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: 0.777 777 777 777 777 777 663.

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) 0.777 777 777 777 777 777 663 × 2 = 1 + 0.555 555 555 555 555 555 326;
  • 2) 0.555 555 555 555 555 555 326 × 2 = 1 + 0.111 111 111 111 111 110 652;
  • 3) 0.111 111 111 111 111 110 652 × 2 = 0 + 0.222 222 222 222 222 221 304;
  • 4) 0.222 222 222 222 222 221 304 × 2 = 0 + 0.444 444 444 444 444 442 608;
  • 5) 0.444 444 444 444 444 442 608 × 2 = 0 + 0.888 888 888 888 888 885 216;
  • 6) 0.888 888 888 888 888 885 216 × 2 = 1 + 0.777 777 777 777 777 770 432;
  • 7) 0.777 777 777 777 777 770 432 × 2 = 1 + 0.555 555 555 555 555 540 864;
  • 8) 0.555 555 555 555 555 540 864 × 2 = 1 + 0.111 111 111 111 111 081 728;
  • 9) 0.111 111 111 111 111 081 728 × 2 = 0 + 0.222 222 222 222 222 163 456;
  • 10) 0.222 222 222 222 222 163 456 × 2 = 0 + 0.444 444 444 444 444 326 912;
  • 11) 0.444 444 444 444 444 326 912 × 2 = 0 + 0.888 888 888 888 888 653 824;
  • 12) 0.888 888 888 888 888 653 824 × 2 = 1 + 0.777 777 777 777 777 307 648;
  • 13) 0.777 777 777 777 777 307 648 × 2 = 1 + 0.555 555 555 555 554 615 296;
  • 14) 0.555 555 555 555 554 615 296 × 2 = 1 + 0.111 111 111 111 109 230 592;
  • 15) 0.111 111 111 111 109 230 592 × 2 = 0 + 0.222 222 222 222 218 461 184;
  • 16) 0.222 222 222 222 218 461 184 × 2 = 0 + 0.444 444 444 444 436 922 368;
  • 17) 0.444 444 444 444 436 922 368 × 2 = 0 + 0.888 888 888 888 873 844 736;
  • 18) 0.888 888 888 888 873 844 736 × 2 = 1 + 0.777 777 777 777 747 689 472;
  • 19) 0.777 777 777 777 747 689 472 × 2 = 1 + 0.555 555 555 555 495 378 944;
  • 20) 0.555 555 555 555 495 378 944 × 2 = 1 + 0.111 111 111 110 990 757 888;
  • 21) 0.111 111 111 110 990 757 888 × 2 = 0 + 0.222 222 222 221 981 515 776;
  • 22) 0.222 222 222 221 981 515 776 × 2 = 0 + 0.444 444 444 443 963 031 552;
  • 23) 0.444 444 444 443 963 031 552 × 2 = 0 + 0.888 888 888 887 926 063 104;
  • 24) 0.888 888 888 887 926 063 104 × 2 = 1 + 0.777 777 777 775 852 126 208;
  • 25) 0.777 777 777 775 852 126 208 × 2 = 1 + 0.555 555 555 551 704 252 416;
  • 26) 0.555 555 555 551 704 252 416 × 2 = 1 + 0.111 111 111 103 408 504 832;
  • 27) 0.111 111 111 103 408 504 832 × 2 = 0 + 0.222 222 222 206 817 009 664;
  • 28) 0.222 222 222 206 817 009 664 × 2 = 0 + 0.444 444 444 413 634 019 328;
  • 29) 0.444 444 444 413 634 019 328 × 2 = 0 + 0.888 888 888 827 268 038 656;
  • 30) 0.888 888 888 827 268 038 656 × 2 = 1 + 0.777 777 777 654 536 077 312;
  • 31) 0.777 777 777 654 536 077 312 × 2 = 1 + 0.555 555 555 309 072 154 624;
  • 32) 0.555 555 555 309 072 154 624 × 2 = 1 + 0.111 111 110 618 144 309 248;
  • 33) 0.111 111 110 618 144 309 248 × 2 = 0 + 0.222 222 221 236 288 618 496;
  • 34) 0.222 222 221 236 288 618 496 × 2 = 0 + 0.444 444 442 472 577 236 992;
  • 35) 0.444 444 442 472 577 236 992 × 2 = 0 + 0.888 888 884 945 154 473 984;
  • 36) 0.888 888 884 945 154 473 984 × 2 = 1 + 0.777 777 769 890 308 947 968;
  • 37) 0.777 777 769 890 308 947 968 × 2 = 1 + 0.555 555 539 780 617 895 936;
  • 38) 0.555 555 539 780 617 895 936 × 2 = 1 + 0.111 111 079 561 235 791 872;
  • 39) 0.111 111 079 561 235 791 872 × 2 = 0 + 0.222 222 159 122 471 583 744;
  • 40) 0.222 222 159 122 471 583 744 × 2 = 0 + 0.444 444 318 244 943 167 488;
  • 41) 0.444 444 318 244 943 167 488 × 2 = 0 + 0.888 888 636 489 886 334 976;
  • 42) 0.888 888 636 489 886 334 976 × 2 = 1 + 0.777 777 272 979 772 669 952;
  • 43) 0.777 777 272 979 772 669 952 × 2 = 1 + 0.555 554 545 959 545 339 904;
  • 44) 0.555 554 545 959 545 339 904 × 2 = 1 + 0.111 109 091 919 090 679 808;
  • 45) 0.111 109 091 919 090 679 808 × 2 = 0 + 0.222 218 183 838 181 359 616;
  • 46) 0.222 218 183 838 181 359 616 × 2 = 0 + 0.444 436 367 676 362 719 232;
  • 47) 0.444 436 367 676 362 719 232 × 2 = 0 + 0.888 872 735 352 725 438 464;
  • 48) 0.888 872 735 352 725 438 464 × 2 = 1 + 0.777 745 470 705 450 876 928;
  • 49) 0.777 745 470 705 450 876 928 × 2 = 1 + 0.555 490 941 410 901 753 856;
  • 50) 0.555 490 941 410 901 753 856 × 2 = 1 + 0.110 981 882 821 803 507 712;
  • 51) 0.110 981 882 821 803 507 712 × 2 = 0 + 0.221 963 765 643 607 015 424;
  • 52) 0.221 963 765 643 607 015 424 × 2 = 0 + 0.443 927 531 287 214 030 848;
  • 53) 0.443 927 531 287 214 030 848 × 2 = 0 + 0.887 855 062 574 428 061 696;

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:


0.777 777 777 777 777 777 663(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 663(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:


24.777 777 777 777 777 777 663(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 24.777 777 777 777 777 777 663 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