24.777 777 777 777 777 777 467 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 467(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
24.777 777 777 777 777 777 467(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: 0.777 777 777 777 777 777 467.

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) 0.777 777 777 777 777 777 467 × 2 = 1 + 0.555 555 555 555 555 554 934;
  • 2) 0.555 555 555 555 555 554 934 × 2 = 1 + 0.111 111 111 111 111 109 868;
  • 3) 0.111 111 111 111 111 109 868 × 2 = 0 + 0.222 222 222 222 222 219 736;
  • 4) 0.222 222 222 222 222 219 736 × 2 = 0 + 0.444 444 444 444 444 439 472;
  • 5) 0.444 444 444 444 444 439 472 × 2 = 0 + 0.888 888 888 888 888 878 944;
  • 6) 0.888 888 888 888 888 878 944 × 2 = 1 + 0.777 777 777 777 777 757 888;
  • 7) 0.777 777 777 777 777 757 888 × 2 = 1 + 0.555 555 555 555 555 515 776;
  • 8) 0.555 555 555 555 555 515 776 × 2 = 1 + 0.111 111 111 111 111 031 552;
  • 9) 0.111 111 111 111 111 031 552 × 2 = 0 + 0.222 222 222 222 222 063 104;
  • 10) 0.222 222 222 222 222 063 104 × 2 = 0 + 0.444 444 444 444 444 126 208;
  • 11) 0.444 444 444 444 444 126 208 × 2 = 0 + 0.888 888 888 888 888 252 416;
  • 12) 0.888 888 888 888 888 252 416 × 2 = 1 + 0.777 777 777 777 776 504 832;
  • 13) 0.777 777 777 777 776 504 832 × 2 = 1 + 0.555 555 555 555 553 009 664;
  • 14) 0.555 555 555 555 553 009 664 × 2 = 1 + 0.111 111 111 111 106 019 328;
  • 15) 0.111 111 111 111 106 019 328 × 2 = 0 + 0.222 222 222 222 212 038 656;
  • 16) 0.222 222 222 222 212 038 656 × 2 = 0 + 0.444 444 444 444 424 077 312;
  • 17) 0.444 444 444 444 424 077 312 × 2 = 0 + 0.888 888 888 888 848 154 624;
  • 18) 0.888 888 888 888 848 154 624 × 2 = 1 + 0.777 777 777 777 696 309 248;
  • 19) 0.777 777 777 777 696 309 248 × 2 = 1 + 0.555 555 555 555 392 618 496;
  • 20) 0.555 555 555 555 392 618 496 × 2 = 1 + 0.111 111 111 110 785 236 992;
  • 21) 0.111 111 111 110 785 236 992 × 2 = 0 + 0.222 222 222 221 570 473 984;
  • 22) 0.222 222 222 221 570 473 984 × 2 = 0 + 0.444 444 444 443 140 947 968;
  • 23) 0.444 444 444 443 140 947 968 × 2 = 0 + 0.888 888 888 886 281 895 936;
  • 24) 0.888 888 888 886 281 895 936 × 2 = 1 + 0.777 777 777 772 563 791 872;
  • 25) 0.777 777 777 772 563 791 872 × 2 = 1 + 0.555 555 555 545 127 583 744;
  • 26) 0.555 555 555 545 127 583 744 × 2 = 1 + 0.111 111 111 090 255 167 488;
  • 27) 0.111 111 111 090 255 167 488 × 2 = 0 + 0.222 222 222 180 510 334 976;
  • 28) 0.222 222 222 180 510 334 976 × 2 = 0 + 0.444 444 444 361 020 669 952;
  • 29) 0.444 444 444 361 020 669 952 × 2 = 0 + 0.888 888 888 722 041 339 904;
  • 30) 0.888 888 888 722 041 339 904 × 2 = 1 + 0.777 777 777 444 082 679 808;
  • 31) 0.777 777 777 444 082 679 808 × 2 = 1 + 0.555 555 554 888 165 359 616;
  • 32) 0.555 555 554 888 165 359 616 × 2 = 1 + 0.111 111 109 776 330 719 232;
  • 33) 0.111 111 109 776 330 719 232 × 2 = 0 + 0.222 222 219 552 661 438 464;
  • 34) 0.222 222 219 552 661 438 464 × 2 = 0 + 0.444 444 439 105 322 876 928;
  • 35) 0.444 444 439 105 322 876 928 × 2 = 0 + 0.888 888 878 210 645 753 856;
  • 36) 0.888 888 878 210 645 753 856 × 2 = 1 + 0.777 777 756 421 291 507 712;
  • 37) 0.777 777 756 421 291 507 712 × 2 = 1 + 0.555 555 512 842 583 015 424;
  • 38) 0.555 555 512 842 583 015 424 × 2 = 1 + 0.111 111 025 685 166 030 848;
  • 39) 0.111 111 025 685 166 030 848 × 2 = 0 + 0.222 222 051 370 332 061 696;
  • 40) 0.222 222 051 370 332 061 696 × 2 = 0 + 0.444 444 102 740 664 123 392;
  • 41) 0.444 444 102 740 664 123 392 × 2 = 0 + 0.888 888 205 481 328 246 784;
  • 42) 0.888 888 205 481 328 246 784 × 2 = 1 + 0.777 776 410 962 656 493 568;
  • 43) 0.777 776 410 962 656 493 568 × 2 = 1 + 0.555 552 821 925 312 987 136;
  • 44) 0.555 552 821 925 312 987 136 × 2 = 1 + 0.111 105 643 850 625 974 272;
  • 45) 0.111 105 643 850 625 974 272 × 2 = 0 + 0.222 211 287 701 251 948 544;
  • 46) 0.222 211 287 701 251 948 544 × 2 = 0 + 0.444 422 575 402 503 897 088;
  • 47) 0.444 422 575 402 503 897 088 × 2 = 0 + 0.888 845 150 805 007 794 176;
  • 48) 0.888 845 150 805 007 794 176 × 2 = 1 + 0.777 690 301 610 015 588 352;
  • 49) 0.777 690 301 610 015 588 352 × 2 = 1 + 0.555 380 603 220 031 176 704;
  • 50) 0.555 380 603 220 031 176 704 × 2 = 1 + 0.110 761 206 440 062 353 408;
  • 51) 0.110 761 206 440 062 353 408 × 2 = 0 + 0.221 522 412 880 124 706 816;
  • 52) 0.221 522 412 880 124 706 816 × 2 = 0 + 0.443 044 825 760 249 413 632;
  • 53) 0.443 044 825 760 249 413 632 × 2 = 0 + 0.886 089 651 520 498 827 264;

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:


0.777 777 777 777 777 777 467(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 467(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:


24.777 777 777 777 777 777 467(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 24.777 777 777 777 777 777 467 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