24.777 777 777 777 601 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 601(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
24.777 777 777 777 601(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: 0.777 777 777 777 601.

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) 0.777 777 777 777 601 × 2 = 1 + 0.555 555 555 555 202;
  • 2) 0.555 555 555 555 202 × 2 = 1 + 0.111 111 111 110 404;
  • 3) 0.111 111 111 110 404 × 2 = 0 + 0.222 222 222 220 808;
  • 4) 0.222 222 222 220 808 × 2 = 0 + 0.444 444 444 441 616;
  • 5) 0.444 444 444 441 616 × 2 = 0 + 0.888 888 888 883 232;
  • 6) 0.888 888 888 883 232 × 2 = 1 + 0.777 777 777 766 464;
  • 7) 0.777 777 777 766 464 × 2 = 1 + 0.555 555 555 532 928;
  • 8) 0.555 555 555 532 928 × 2 = 1 + 0.111 111 111 065 856;
  • 9) 0.111 111 111 065 856 × 2 = 0 + 0.222 222 222 131 712;
  • 10) 0.222 222 222 131 712 × 2 = 0 + 0.444 444 444 263 424;
  • 11) 0.444 444 444 263 424 × 2 = 0 + 0.888 888 888 526 848;
  • 12) 0.888 888 888 526 848 × 2 = 1 + 0.777 777 777 053 696;
  • 13) 0.777 777 777 053 696 × 2 = 1 + 0.555 555 554 107 392;
  • 14) 0.555 555 554 107 392 × 2 = 1 + 0.111 111 108 214 784;
  • 15) 0.111 111 108 214 784 × 2 = 0 + 0.222 222 216 429 568;
  • 16) 0.222 222 216 429 568 × 2 = 0 + 0.444 444 432 859 136;
  • 17) 0.444 444 432 859 136 × 2 = 0 + 0.888 888 865 718 272;
  • 18) 0.888 888 865 718 272 × 2 = 1 + 0.777 777 731 436 544;
  • 19) 0.777 777 731 436 544 × 2 = 1 + 0.555 555 462 873 088;
  • 20) 0.555 555 462 873 088 × 2 = 1 + 0.111 110 925 746 176;
  • 21) 0.111 110 925 746 176 × 2 = 0 + 0.222 221 851 492 352;
  • 22) 0.222 221 851 492 352 × 2 = 0 + 0.444 443 702 984 704;
  • 23) 0.444 443 702 984 704 × 2 = 0 + 0.888 887 405 969 408;
  • 24) 0.888 887 405 969 408 × 2 = 1 + 0.777 774 811 938 816;
  • 25) 0.777 774 811 938 816 × 2 = 1 + 0.555 549 623 877 632;
  • 26) 0.555 549 623 877 632 × 2 = 1 + 0.111 099 247 755 264;
  • 27) 0.111 099 247 755 264 × 2 = 0 + 0.222 198 495 510 528;
  • 28) 0.222 198 495 510 528 × 2 = 0 + 0.444 396 991 021 056;
  • 29) 0.444 396 991 021 056 × 2 = 0 + 0.888 793 982 042 112;
  • 30) 0.888 793 982 042 112 × 2 = 1 + 0.777 587 964 084 224;
  • 31) 0.777 587 964 084 224 × 2 = 1 + 0.555 175 928 168 448;
  • 32) 0.555 175 928 168 448 × 2 = 1 + 0.110 351 856 336 896;
  • 33) 0.110 351 856 336 896 × 2 = 0 + 0.220 703 712 673 792;
  • 34) 0.220 703 712 673 792 × 2 = 0 + 0.441 407 425 347 584;
  • 35) 0.441 407 425 347 584 × 2 = 0 + 0.882 814 850 695 168;
  • 36) 0.882 814 850 695 168 × 2 = 1 + 0.765 629 701 390 336;
  • 37) 0.765 629 701 390 336 × 2 = 1 + 0.531 259 402 780 672;
  • 38) 0.531 259 402 780 672 × 2 = 1 + 0.062 518 805 561 344;
  • 39) 0.062 518 805 561 344 × 2 = 0 + 0.125 037 611 122 688;
  • 40) 0.125 037 611 122 688 × 2 = 0 + 0.250 075 222 245 376;
  • 41) 0.250 075 222 245 376 × 2 = 0 + 0.500 150 444 490 752;
  • 42) 0.500 150 444 490 752 × 2 = 1 + 0.000 300 888 981 504;
  • 43) 0.000 300 888 981 504 × 2 = 0 + 0.000 601 777 963 008;
  • 44) 0.000 601 777 963 008 × 2 = 0 + 0.001 203 555 926 016;
  • 45) 0.001 203 555 926 016 × 2 = 0 + 0.002 407 111 852 032;
  • 46) 0.002 407 111 852 032 × 2 = 0 + 0.004 814 223 704 064;
  • 47) 0.004 814 223 704 064 × 2 = 0 + 0.009 628 447 408 128;
  • 48) 0.009 628 447 408 128 × 2 = 0 + 0.019 256 894 816 256;
  • 49) 0.019 256 894 816 256 × 2 = 0 + 0.038 513 789 632 512;
  • 50) 0.038 513 789 632 512 × 2 = 0 + 0.077 027 579 265 024;
  • 51) 0.077 027 579 265 024 × 2 = 0 + 0.154 055 158 530 048;
  • 52) 0.154 055 158 530 048 × 2 = 0 + 0.308 110 317 060 096;
  • 53) 0.308 110 317 060 096 × 2 = 0 + 0.616 220 634 120 192;

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:


0.777 777 777 777 601(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0100 0000 0000 0(2)

5. Positive number before normalization:

24.777 777 777 777 601(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0100 0000 0000 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:


24.777 777 777 777 601(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0100 0000 0000 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0100 0000 0000 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0100 0000 0000 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 0100 0000 0000 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 0100 0000 0 0000 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0100 0000


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 0100 0000


Decimal number 24.777 777 777 777 601 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 0100 0000

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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