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

Convert decimal 24.777 777 777 777 754(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
24.777 777 777 777 754(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: 0.777 777 777 777 754.

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) 0.777 777 777 777 754 × 2 = 1 + 0.555 555 555 555 508;
  • 2) 0.555 555 555 555 508 × 2 = 1 + 0.111 111 111 111 016;
  • 3) 0.111 111 111 111 016 × 2 = 0 + 0.222 222 222 222 032;
  • 4) 0.222 222 222 222 032 × 2 = 0 + 0.444 444 444 444 064;
  • 5) 0.444 444 444 444 064 × 2 = 0 + 0.888 888 888 888 128;
  • 6) 0.888 888 888 888 128 × 2 = 1 + 0.777 777 777 776 256;
  • 7) 0.777 777 777 776 256 × 2 = 1 + 0.555 555 555 552 512;
  • 8) 0.555 555 555 552 512 × 2 = 1 + 0.111 111 111 105 024;
  • 9) 0.111 111 111 105 024 × 2 = 0 + 0.222 222 222 210 048;
  • 10) 0.222 222 222 210 048 × 2 = 0 + 0.444 444 444 420 096;
  • 11) 0.444 444 444 420 096 × 2 = 0 + 0.888 888 888 840 192;
  • 12) 0.888 888 888 840 192 × 2 = 1 + 0.777 777 777 680 384;
  • 13) 0.777 777 777 680 384 × 2 = 1 + 0.555 555 555 360 768;
  • 14) 0.555 555 555 360 768 × 2 = 1 + 0.111 111 110 721 536;
  • 15) 0.111 111 110 721 536 × 2 = 0 + 0.222 222 221 443 072;
  • 16) 0.222 222 221 443 072 × 2 = 0 + 0.444 444 442 886 144;
  • 17) 0.444 444 442 886 144 × 2 = 0 + 0.888 888 885 772 288;
  • 18) 0.888 888 885 772 288 × 2 = 1 + 0.777 777 771 544 576;
  • 19) 0.777 777 771 544 576 × 2 = 1 + 0.555 555 543 089 152;
  • 20) 0.555 555 543 089 152 × 2 = 1 + 0.111 111 086 178 304;
  • 21) 0.111 111 086 178 304 × 2 = 0 + 0.222 222 172 356 608;
  • 22) 0.222 222 172 356 608 × 2 = 0 + 0.444 444 344 713 216;
  • 23) 0.444 444 344 713 216 × 2 = 0 + 0.888 888 689 426 432;
  • 24) 0.888 888 689 426 432 × 2 = 1 + 0.777 777 378 852 864;
  • 25) 0.777 777 378 852 864 × 2 = 1 + 0.555 554 757 705 728;
  • 26) 0.555 554 757 705 728 × 2 = 1 + 0.111 109 515 411 456;
  • 27) 0.111 109 515 411 456 × 2 = 0 + 0.222 219 030 822 912;
  • 28) 0.222 219 030 822 912 × 2 = 0 + 0.444 438 061 645 824;
  • 29) 0.444 438 061 645 824 × 2 = 0 + 0.888 876 123 291 648;
  • 30) 0.888 876 123 291 648 × 2 = 1 + 0.777 752 246 583 296;
  • 31) 0.777 752 246 583 296 × 2 = 1 + 0.555 504 493 166 592;
  • 32) 0.555 504 493 166 592 × 2 = 1 + 0.111 008 986 333 184;
  • 33) 0.111 008 986 333 184 × 2 = 0 + 0.222 017 972 666 368;
  • 34) 0.222 017 972 666 368 × 2 = 0 + 0.444 035 945 332 736;
  • 35) 0.444 035 945 332 736 × 2 = 0 + 0.888 071 890 665 472;
  • 36) 0.888 071 890 665 472 × 2 = 1 + 0.776 143 781 330 944;
  • 37) 0.776 143 781 330 944 × 2 = 1 + 0.552 287 562 661 888;
  • 38) 0.552 287 562 661 888 × 2 = 1 + 0.104 575 125 323 776;
  • 39) 0.104 575 125 323 776 × 2 = 0 + 0.209 150 250 647 552;
  • 40) 0.209 150 250 647 552 × 2 = 0 + 0.418 300 501 295 104;
  • 41) 0.418 300 501 295 104 × 2 = 0 + 0.836 601 002 590 208;
  • 42) 0.836 601 002 590 208 × 2 = 1 + 0.673 202 005 180 416;
  • 43) 0.673 202 005 180 416 × 2 = 1 + 0.346 404 010 360 832;
  • 44) 0.346 404 010 360 832 × 2 = 0 + 0.692 808 020 721 664;
  • 45) 0.692 808 020 721 664 × 2 = 1 + 0.385 616 041 443 328;
  • 46) 0.385 616 041 443 328 × 2 = 0 + 0.771 232 082 886 656;
  • 47) 0.771 232 082 886 656 × 2 = 1 + 0.542 464 165 773 312;
  • 48) 0.542 464 165 773 312 × 2 = 1 + 0.084 928 331 546 624;
  • 49) 0.084 928 331 546 624 × 2 = 0 + 0.169 856 663 093 248;
  • 50) 0.169 856 663 093 248 × 2 = 0 + 0.339 713 326 186 496;
  • 51) 0.339 713 326 186 496 × 2 = 0 + 0.679 426 652 372 992;
  • 52) 0.679 426 652 372 992 × 2 = 1 + 0.358 853 304 745 984;
  • 53) 0.358 853 304 745 984 × 2 = 0 + 0.717 706 609 491 968;

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:


0.777 777 777 777 754(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1011 0001 0(2)

5. Positive number before normalization:

24.777 777 777 777 754(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1011 0001 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:


24.777 777 777 777 754(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1011 0001 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1011 0001 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1011 0001 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 0110 1011 0001 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 0110 1011 0 0010 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 1011


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 1011


Decimal number 24.777 777 777 777 754 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 1011

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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