24.777 777 777 777 777 791 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 791 3(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
24.777 777 777 777 777 791 3(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: 0.777 777 777 777 777 791 3.

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) 0.777 777 777 777 777 791 3 × 2 = 1 + 0.555 555 555 555 555 582 6;
  • 2) 0.555 555 555 555 555 582 6 × 2 = 1 + 0.111 111 111 111 111 165 2;
  • 3) 0.111 111 111 111 111 165 2 × 2 = 0 + 0.222 222 222 222 222 330 4;
  • 4) 0.222 222 222 222 222 330 4 × 2 = 0 + 0.444 444 444 444 444 660 8;
  • 5) 0.444 444 444 444 444 660 8 × 2 = 0 + 0.888 888 888 888 889 321 6;
  • 6) 0.888 888 888 888 889 321 6 × 2 = 1 + 0.777 777 777 777 778 643 2;
  • 7) 0.777 777 777 777 778 643 2 × 2 = 1 + 0.555 555 555 555 557 286 4;
  • 8) 0.555 555 555 555 557 286 4 × 2 = 1 + 0.111 111 111 111 114 572 8;
  • 9) 0.111 111 111 111 114 572 8 × 2 = 0 + 0.222 222 222 222 229 145 6;
  • 10) 0.222 222 222 222 229 145 6 × 2 = 0 + 0.444 444 444 444 458 291 2;
  • 11) 0.444 444 444 444 458 291 2 × 2 = 0 + 0.888 888 888 888 916 582 4;
  • 12) 0.888 888 888 888 916 582 4 × 2 = 1 + 0.777 777 777 777 833 164 8;
  • 13) 0.777 777 777 777 833 164 8 × 2 = 1 + 0.555 555 555 555 666 329 6;
  • 14) 0.555 555 555 555 666 329 6 × 2 = 1 + 0.111 111 111 111 332 659 2;
  • 15) 0.111 111 111 111 332 659 2 × 2 = 0 + 0.222 222 222 222 665 318 4;
  • 16) 0.222 222 222 222 665 318 4 × 2 = 0 + 0.444 444 444 445 330 636 8;
  • 17) 0.444 444 444 445 330 636 8 × 2 = 0 + 0.888 888 888 890 661 273 6;
  • 18) 0.888 888 888 890 661 273 6 × 2 = 1 + 0.777 777 777 781 322 547 2;
  • 19) 0.777 777 777 781 322 547 2 × 2 = 1 + 0.555 555 555 562 645 094 4;
  • 20) 0.555 555 555 562 645 094 4 × 2 = 1 + 0.111 111 111 125 290 188 8;
  • 21) 0.111 111 111 125 290 188 8 × 2 = 0 + 0.222 222 222 250 580 377 6;
  • 22) 0.222 222 222 250 580 377 6 × 2 = 0 + 0.444 444 444 501 160 755 2;
  • 23) 0.444 444 444 501 160 755 2 × 2 = 0 + 0.888 888 889 002 321 510 4;
  • 24) 0.888 888 889 002 321 510 4 × 2 = 1 + 0.777 777 778 004 643 020 8;
  • 25) 0.777 777 778 004 643 020 8 × 2 = 1 + 0.555 555 556 009 286 041 6;
  • 26) 0.555 555 556 009 286 041 6 × 2 = 1 + 0.111 111 112 018 572 083 2;
  • 27) 0.111 111 112 018 572 083 2 × 2 = 0 + 0.222 222 224 037 144 166 4;
  • 28) 0.222 222 224 037 144 166 4 × 2 = 0 + 0.444 444 448 074 288 332 8;
  • 29) 0.444 444 448 074 288 332 8 × 2 = 0 + 0.888 888 896 148 576 665 6;
  • 30) 0.888 888 896 148 576 665 6 × 2 = 1 + 0.777 777 792 297 153 331 2;
  • 31) 0.777 777 792 297 153 331 2 × 2 = 1 + 0.555 555 584 594 306 662 4;
  • 32) 0.555 555 584 594 306 662 4 × 2 = 1 + 0.111 111 169 188 613 324 8;
  • 33) 0.111 111 169 188 613 324 8 × 2 = 0 + 0.222 222 338 377 226 649 6;
  • 34) 0.222 222 338 377 226 649 6 × 2 = 0 + 0.444 444 676 754 453 299 2;
  • 35) 0.444 444 676 754 453 299 2 × 2 = 0 + 0.888 889 353 508 906 598 4;
  • 36) 0.888 889 353 508 906 598 4 × 2 = 1 + 0.777 778 707 017 813 196 8;
  • 37) 0.777 778 707 017 813 196 8 × 2 = 1 + 0.555 557 414 035 626 393 6;
  • 38) 0.555 557 414 035 626 393 6 × 2 = 1 + 0.111 114 828 071 252 787 2;
  • 39) 0.111 114 828 071 252 787 2 × 2 = 0 + 0.222 229 656 142 505 574 4;
  • 40) 0.222 229 656 142 505 574 4 × 2 = 0 + 0.444 459 312 285 011 148 8;
  • 41) 0.444 459 312 285 011 148 8 × 2 = 0 + 0.888 918 624 570 022 297 6;
  • 42) 0.888 918 624 570 022 297 6 × 2 = 1 + 0.777 837 249 140 044 595 2;
  • 43) 0.777 837 249 140 044 595 2 × 2 = 1 + 0.555 674 498 280 089 190 4;
  • 44) 0.555 674 498 280 089 190 4 × 2 = 1 + 0.111 348 996 560 178 380 8;
  • 45) 0.111 348 996 560 178 380 8 × 2 = 0 + 0.222 697 993 120 356 761 6;
  • 46) 0.222 697 993 120 356 761 6 × 2 = 0 + 0.445 395 986 240 713 523 2;
  • 47) 0.445 395 986 240 713 523 2 × 2 = 0 + 0.890 791 972 481 427 046 4;
  • 48) 0.890 791 972 481 427 046 4 × 2 = 1 + 0.781 583 944 962 854 092 8;
  • 49) 0.781 583 944 962 854 092 8 × 2 = 1 + 0.563 167 889 925 708 185 6;
  • 50) 0.563 167 889 925 708 185 6 × 2 = 1 + 0.126 335 779 851 416 371 2;
  • 51) 0.126 335 779 851 416 371 2 × 2 = 0 + 0.252 671 559 702 832 742 4;
  • 52) 0.252 671 559 702 832 742 4 × 2 = 0 + 0.505 343 119 405 665 484 8;
  • 53) 0.505 343 119 405 665 484 8 × 2 = 1 + 0.010 686 238 811 330 969 6;

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:


0.777 777 777 777 777 791 3(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1(2)

5. Positive number before normalization:

24.777 777 777 777 777 791 3(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 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:


24.777 777 777 777 777 791 3(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 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 0111 0001 1100 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 0111 0001 1 1001 =

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 24.777 777 777 777 777 791 3 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