24.777 777 777 777 777 761 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 761 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
24.777 777 777 777 777 761 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: 0.777 777 777 777 777 761 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) 0.777 777 777 777 777 761 3 × 2 = 1 + 0.555 555 555 555 555 522 6;
  • 2) 0.555 555 555 555 555 522 6 × 2 = 1 + 0.111 111 111 111 111 045 2;
  • 3) 0.111 111 111 111 111 045 2 × 2 = 0 + 0.222 222 222 222 222 090 4;
  • 4) 0.222 222 222 222 222 090 4 × 2 = 0 + 0.444 444 444 444 444 180 8;
  • 5) 0.444 444 444 444 444 180 8 × 2 = 0 + 0.888 888 888 888 888 361 6;
  • 6) 0.888 888 888 888 888 361 6 × 2 = 1 + 0.777 777 777 777 776 723 2;
  • 7) 0.777 777 777 777 776 723 2 × 2 = 1 + 0.555 555 555 555 553 446 4;
  • 8) 0.555 555 555 555 553 446 4 × 2 = 1 + 0.111 111 111 111 106 892 8;
  • 9) 0.111 111 111 111 106 892 8 × 2 = 0 + 0.222 222 222 222 213 785 6;
  • 10) 0.222 222 222 222 213 785 6 × 2 = 0 + 0.444 444 444 444 427 571 2;
  • 11) 0.444 444 444 444 427 571 2 × 2 = 0 + 0.888 888 888 888 855 142 4;
  • 12) 0.888 888 888 888 855 142 4 × 2 = 1 + 0.777 777 777 777 710 284 8;
  • 13) 0.777 777 777 777 710 284 8 × 2 = 1 + 0.555 555 555 555 420 569 6;
  • 14) 0.555 555 555 555 420 569 6 × 2 = 1 + 0.111 111 111 110 841 139 2;
  • 15) 0.111 111 111 110 841 139 2 × 2 = 0 + 0.222 222 222 221 682 278 4;
  • 16) 0.222 222 222 221 682 278 4 × 2 = 0 + 0.444 444 444 443 364 556 8;
  • 17) 0.444 444 444 443 364 556 8 × 2 = 0 + 0.888 888 888 886 729 113 6;
  • 18) 0.888 888 888 886 729 113 6 × 2 = 1 + 0.777 777 777 773 458 227 2;
  • 19) 0.777 777 777 773 458 227 2 × 2 = 1 + 0.555 555 555 546 916 454 4;
  • 20) 0.555 555 555 546 916 454 4 × 2 = 1 + 0.111 111 111 093 832 908 8;
  • 21) 0.111 111 111 093 832 908 8 × 2 = 0 + 0.222 222 222 187 665 817 6;
  • 22) 0.222 222 222 187 665 817 6 × 2 = 0 + 0.444 444 444 375 331 635 2;
  • 23) 0.444 444 444 375 331 635 2 × 2 = 0 + 0.888 888 888 750 663 270 4;
  • 24) 0.888 888 888 750 663 270 4 × 2 = 1 + 0.777 777 777 501 326 540 8;
  • 25) 0.777 777 777 501 326 540 8 × 2 = 1 + 0.555 555 555 002 653 081 6;
  • 26) 0.555 555 555 002 653 081 6 × 2 = 1 + 0.111 111 110 005 306 163 2;
  • 27) 0.111 111 110 005 306 163 2 × 2 = 0 + 0.222 222 220 010 612 326 4;
  • 28) 0.222 222 220 010 612 326 4 × 2 = 0 + 0.444 444 440 021 224 652 8;
  • 29) 0.444 444 440 021 224 652 8 × 2 = 0 + 0.888 888 880 042 449 305 6;
  • 30) 0.888 888 880 042 449 305 6 × 2 = 1 + 0.777 777 760 084 898 611 2;
  • 31) 0.777 777 760 084 898 611 2 × 2 = 1 + 0.555 555 520 169 797 222 4;
  • 32) 0.555 555 520 169 797 222 4 × 2 = 1 + 0.111 111 040 339 594 444 8;
  • 33) 0.111 111 040 339 594 444 8 × 2 = 0 + 0.222 222 080 679 188 889 6;
  • 34) 0.222 222 080 679 188 889 6 × 2 = 0 + 0.444 444 161 358 377 779 2;
  • 35) 0.444 444 161 358 377 779 2 × 2 = 0 + 0.888 888 322 716 755 558 4;
  • 36) 0.888 888 322 716 755 558 4 × 2 = 1 + 0.777 776 645 433 511 116 8;
  • 37) 0.777 776 645 433 511 116 8 × 2 = 1 + 0.555 553 290 867 022 233 6;
  • 38) 0.555 553 290 867 022 233 6 × 2 = 1 + 0.111 106 581 734 044 467 2;
  • 39) 0.111 106 581 734 044 467 2 × 2 = 0 + 0.222 213 163 468 088 934 4;
  • 40) 0.222 213 163 468 088 934 4 × 2 = 0 + 0.444 426 326 936 177 868 8;
  • 41) 0.444 426 326 936 177 868 8 × 2 = 0 + 0.888 852 653 872 355 737 6;
  • 42) 0.888 852 653 872 355 737 6 × 2 = 1 + 0.777 705 307 744 711 475 2;
  • 43) 0.777 705 307 744 711 475 2 × 2 = 1 + 0.555 410 615 489 422 950 4;
  • 44) 0.555 410 615 489 422 950 4 × 2 = 1 + 0.110 821 230 978 845 900 8;
  • 45) 0.110 821 230 978 845 900 8 × 2 = 0 + 0.221 642 461 957 691 801 6;
  • 46) 0.221 642 461 957 691 801 6 × 2 = 0 + 0.443 284 923 915 383 603 2;
  • 47) 0.443 284 923 915 383 603 2 × 2 = 0 + 0.886 569 847 830 767 206 4;
  • 48) 0.886 569 847 830 767 206 4 × 2 = 1 + 0.773 139 695 661 534 412 8;
  • 49) 0.773 139 695 661 534 412 8 × 2 = 1 + 0.546 279 391 323 068 825 6;
  • 50) 0.546 279 391 323 068 825 6 × 2 = 1 + 0.092 558 782 646 137 651 2;
  • 51) 0.092 558 782 646 137 651 2 × 2 = 0 + 0.185 117 565 292 275 302 4;
  • 52) 0.185 117 565 292 275 302 4 × 2 = 0 + 0.370 235 130 584 550 604 8;
  • 53) 0.370 235 130 584 550 604 8 × 2 = 0 + 0.740 470 261 169 101 209 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:


0.777 777 777 777 777 761 3(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 761 3(10) =

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


24.777 777 777 777 777 761 3(10) =

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

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

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 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 0111 0001 1100 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 0111 0001 1 1000 =

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 24.777 777 777 777 777 761 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