24.777 777 777 777 733 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 733(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
24.777 777 777 777 733(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: 0.777 777 777 777 733.

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) 0.777 777 777 777 733 × 2 = 1 + 0.555 555 555 555 466;
  • 2) 0.555 555 555 555 466 × 2 = 1 + 0.111 111 111 110 932;
  • 3) 0.111 111 111 110 932 × 2 = 0 + 0.222 222 222 221 864;
  • 4) 0.222 222 222 221 864 × 2 = 0 + 0.444 444 444 443 728;
  • 5) 0.444 444 444 443 728 × 2 = 0 + 0.888 888 888 887 456;
  • 6) 0.888 888 888 887 456 × 2 = 1 + 0.777 777 777 774 912;
  • 7) 0.777 777 777 774 912 × 2 = 1 + 0.555 555 555 549 824;
  • 8) 0.555 555 555 549 824 × 2 = 1 + 0.111 111 111 099 648;
  • 9) 0.111 111 111 099 648 × 2 = 0 + 0.222 222 222 199 296;
  • 10) 0.222 222 222 199 296 × 2 = 0 + 0.444 444 444 398 592;
  • 11) 0.444 444 444 398 592 × 2 = 0 + 0.888 888 888 797 184;
  • 12) 0.888 888 888 797 184 × 2 = 1 + 0.777 777 777 594 368;
  • 13) 0.777 777 777 594 368 × 2 = 1 + 0.555 555 555 188 736;
  • 14) 0.555 555 555 188 736 × 2 = 1 + 0.111 111 110 377 472;
  • 15) 0.111 111 110 377 472 × 2 = 0 + 0.222 222 220 754 944;
  • 16) 0.222 222 220 754 944 × 2 = 0 + 0.444 444 441 509 888;
  • 17) 0.444 444 441 509 888 × 2 = 0 + 0.888 888 883 019 776;
  • 18) 0.888 888 883 019 776 × 2 = 1 + 0.777 777 766 039 552;
  • 19) 0.777 777 766 039 552 × 2 = 1 + 0.555 555 532 079 104;
  • 20) 0.555 555 532 079 104 × 2 = 1 + 0.111 111 064 158 208;
  • 21) 0.111 111 064 158 208 × 2 = 0 + 0.222 222 128 316 416;
  • 22) 0.222 222 128 316 416 × 2 = 0 + 0.444 444 256 632 832;
  • 23) 0.444 444 256 632 832 × 2 = 0 + 0.888 888 513 265 664;
  • 24) 0.888 888 513 265 664 × 2 = 1 + 0.777 777 026 531 328;
  • 25) 0.777 777 026 531 328 × 2 = 1 + 0.555 554 053 062 656;
  • 26) 0.555 554 053 062 656 × 2 = 1 + 0.111 108 106 125 312;
  • 27) 0.111 108 106 125 312 × 2 = 0 + 0.222 216 212 250 624;
  • 28) 0.222 216 212 250 624 × 2 = 0 + 0.444 432 424 501 248;
  • 29) 0.444 432 424 501 248 × 2 = 0 + 0.888 864 849 002 496;
  • 30) 0.888 864 849 002 496 × 2 = 1 + 0.777 729 698 004 992;
  • 31) 0.777 729 698 004 992 × 2 = 1 + 0.555 459 396 009 984;
  • 32) 0.555 459 396 009 984 × 2 = 1 + 0.110 918 792 019 968;
  • 33) 0.110 918 792 019 968 × 2 = 0 + 0.221 837 584 039 936;
  • 34) 0.221 837 584 039 936 × 2 = 0 + 0.443 675 168 079 872;
  • 35) 0.443 675 168 079 872 × 2 = 0 + 0.887 350 336 159 744;
  • 36) 0.887 350 336 159 744 × 2 = 1 + 0.774 700 672 319 488;
  • 37) 0.774 700 672 319 488 × 2 = 1 + 0.549 401 344 638 976;
  • 38) 0.549 401 344 638 976 × 2 = 1 + 0.098 802 689 277 952;
  • 39) 0.098 802 689 277 952 × 2 = 0 + 0.197 605 378 555 904;
  • 40) 0.197 605 378 555 904 × 2 = 0 + 0.395 210 757 111 808;
  • 41) 0.395 210 757 111 808 × 2 = 0 + 0.790 421 514 223 616;
  • 42) 0.790 421 514 223 616 × 2 = 1 + 0.580 843 028 447 232;
  • 43) 0.580 843 028 447 232 × 2 = 1 + 0.161 686 056 894 464;
  • 44) 0.161 686 056 894 464 × 2 = 0 + 0.323 372 113 788 928;
  • 45) 0.323 372 113 788 928 × 2 = 0 + 0.646 744 227 577 856;
  • 46) 0.646 744 227 577 856 × 2 = 1 + 0.293 488 455 155 712;
  • 47) 0.293 488 455 155 712 × 2 = 0 + 0.586 976 910 311 424;
  • 48) 0.586 976 910 311 424 × 2 = 1 + 0.173 953 820 622 848;
  • 49) 0.173 953 820 622 848 × 2 = 0 + 0.347 907 641 245 696;
  • 50) 0.347 907 641 245 696 × 2 = 0 + 0.695 815 282 491 392;
  • 51) 0.695 815 282 491 392 × 2 = 1 + 0.391 630 564 982 784;
  • 52) 0.391 630 564 982 784 × 2 = 0 + 0.783 261 129 965 568;
  • 53) 0.783 261 129 965 568 × 2 = 1 + 0.566 522 259 931 136;

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:


0.777 777 777 777 733(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0101 0010 1(2)

5. Positive number before normalization:

24.777 777 777 777 733(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0101 0010 1(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:


24.777 777 777 777 733(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0101 0010 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0101 0010 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0101 0010 1(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 0110 0101 0010 1


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 0110 0101 0 0101 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0101


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 0110 0101


Decimal number 24.777 777 777 777 733 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 0110 0101

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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