24.777 777 777 777 594 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 594(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
24.777 777 777 777 594(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: 0.777 777 777 777 594.

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) 0.777 777 777 777 594 × 2 = 1 + 0.555 555 555 555 188;
  • 2) 0.555 555 555 555 188 × 2 = 1 + 0.111 111 111 110 376;
  • 3) 0.111 111 111 110 376 × 2 = 0 + 0.222 222 222 220 752;
  • 4) 0.222 222 222 220 752 × 2 = 0 + 0.444 444 444 441 504;
  • 5) 0.444 444 444 441 504 × 2 = 0 + 0.888 888 888 883 008;
  • 6) 0.888 888 888 883 008 × 2 = 1 + 0.777 777 777 766 016;
  • 7) 0.777 777 777 766 016 × 2 = 1 + 0.555 555 555 532 032;
  • 8) 0.555 555 555 532 032 × 2 = 1 + 0.111 111 111 064 064;
  • 9) 0.111 111 111 064 064 × 2 = 0 + 0.222 222 222 128 128;
  • 10) 0.222 222 222 128 128 × 2 = 0 + 0.444 444 444 256 256;
  • 11) 0.444 444 444 256 256 × 2 = 0 + 0.888 888 888 512 512;
  • 12) 0.888 888 888 512 512 × 2 = 1 + 0.777 777 777 025 024;
  • 13) 0.777 777 777 025 024 × 2 = 1 + 0.555 555 554 050 048;
  • 14) 0.555 555 554 050 048 × 2 = 1 + 0.111 111 108 100 096;
  • 15) 0.111 111 108 100 096 × 2 = 0 + 0.222 222 216 200 192;
  • 16) 0.222 222 216 200 192 × 2 = 0 + 0.444 444 432 400 384;
  • 17) 0.444 444 432 400 384 × 2 = 0 + 0.888 888 864 800 768;
  • 18) 0.888 888 864 800 768 × 2 = 1 + 0.777 777 729 601 536;
  • 19) 0.777 777 729 601 536 × 2 = 1 + 0.555 555 459 203 072;
  • 20) 0.555 555 459 203 072 × 2 = 1 + 0.111 110 918 406 144;
  • 21) 0.111 110 918 406 144 × 2 = 0 + 0.222 221 836 812 288;
  • 22) 0.222 221 836 812 288 × 2 = 0 + 0.444 443 673 624 576;
  • 23) 0.444 443 673 624 576 × 2 = 0 + 0.888 887 347 249 152;
  • 24) 0.888 887 347 249 152 × 2 = 1 + 0.777 774 694 498 304;
  • 25) 0.777 774 694 498 304 × 2 = 1 + 0.555 549 388 996 608;
  • 26) 0.555 549 388 996 608 × 2 = 1 + 0.111 098 777 993 216;
  • 27) 0.111 098 777 993 216 × 2 = 0 + 0.222 197 555 986 432;
  • 28) 0.222 197 555 986 432 × 2 = 0 + 0.444 395 111 972 864;
  • 29) 0.444 395 111 972 864 × 2 = 0 + 0.888 790 223 945 728;
  • 30) 0.888 790 223 945 728 × 2 = 1 + 0.777 580 447 891 456;
  • 31) 0.777 580 447 891 456 × 2 = 1 + 0.555 160 895 782 912;
  • 32) 0.555 160 895 782 912 × 2 = 1 + 0.110 321 791 565 824;
  • 33) 0.110 321 791 565 824 × 2 = 0 + 0.220 643 583 131 648;
  • 34) 0.220 643 583 131 648 × 2 = 0 + 0.441 287 166 263 296;
  • 35) 0.441 287 166 263 296 × 2 = 0 + 0.882 574 332 526 592;
  • 36) 0.882 574 332 526 592 × 2 = 1 + 0.765 148 665 053 184;
  • 37) 0.765 148 665 053 184 × 2 = 1 + 0.530 297 330 106 368;
  • 38) 0.530 297 330 106 368 × 2 = 1 + 0.060 594 660 212 736;
  • 39) 0.060 594 660 212 736 × 2 = 0 + 0.121 189 320 425 472;
  • 40) 0.121 189 320 425 472 × 2 = 0 + 0.242 378 640 850 944;
  • 41) 0.242 378 640 850 944 × 2 = 0 + 0.484 757 281 701 888;
  • 42) 0.484 757 281 701 888 × 2 = 0 + 0.969 514 563 403 776;
  • 43) 0.969 514 563 403 776 × 2 = 1 + 0.939 029 126 807 552;
  • 44) 0.939 029 126 807 552 × 2 = 1 + 0.878 058 253 615 104;
  • 45) 0.878 058 253 615 104 × 2 = 1 + 0.756 116 507 230 208;
  • 46) 0.756 116 507 230 208 × 2 = 1 + 0.512 233 014 460 416;
  • 47) 0.512 233 014 460 416 × 2 = 1 + 0.024 466 028 920 832;
  • 48) 0.024 466 028 920 832 × 2 = 0 + 0.048 932 057 841 664;
  • 49) 0.048 932 057 841 664 × 2 = 0 + 0.097 864 115 683 328;
  • 50) 0.097 864 115 683 328 × 2 = 0 + 0.195 728 231 366 656;
  • 51) 0.195 728 231 366 656 × 2 = 0 + 0.391 456 462 733 312;
  • 52) 0.391 456 462 733 312 × 2 = 0 + 0.782 912 925 466 624;
  • 53) 0.782 912 925 466 624 × 2 = 1 + 0.565 825 850 933 248;

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:


0.777 777 777 777 594(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0011 1110 0000 1(2)

5. Positive number before normalization:

24.777 777 777 777 594(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0011 1110 0000 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:


24.777 777 777 777 594(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0011 1110 0000 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0011 1110 0000 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0011 1110 0000 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 0011 1110 0000 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 0011 1110 0 0001 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0011 1110


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 0011 1110


Decimal number 24.777 777 777 777 594 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 0011 1110

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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