24.777 777 777 777 777 783 35 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 783 35(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
24.777 777 777 777 777 783 35(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: 0.777 777 777 777 777 783 35.

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) 0.777 777 777 777 777 783 35 × 2 = 1 + 0.555 555 555 555 555 566 7;
  • 2) 0.555 555 555 555 555 566 7 × 2 = 1 + 0.111 111 111 111 111 133 4;
  • 3) 0.111 111 111 111 111 133 4 × 2 = 0 + 0.222 222 222 222 222 266 8;
  • 4) 0.222 222 222 222 222 266 8 × 2 = 0 + 0.444 444 444 444 444 533 6;
  • 5) 0.444 444 444 444 444 533 6 × 2 = 0 + 0.888 888 888 888 889 067 2;
  • 6) 0.888 888 888 888 889 067 2 × 2 = 1 + 0.777 777 777 777 778 134 4;
  • 7) 0.777 777 777 777 778 134 4 × 2 = 1 + 0.555 555 555 555 556 268 8;
  • 8) 0.555 555 555 555 556 268 8 × 2 = 1 + 0.111 111 111 111 112 537 6;
  • 9) 0.111 111 111 111 112 537 6 × 2 = 0 + 0.222 222 222 222 225 075 2;
  • 10) 0.222 222 222 222 225 075 2 × 2 = 0 + 0.444 444 444 444 450 150 4;
  • 11) 0.444 444 444 444 450 150 4 × 2 = 0 + 0.888 888 888 888 900 300 8;
  • 12) 0.888 888 888 888 900 300 8 × 2 = 1 + 0.777 777 777 777 800 601 6;
  • 13) 0.777 777 777 777 800 601 6 × 2 = 1 + 0.555 555 555 555 601 203 2;
  • 14) 0.555 555 555 555 601 203 2 × 2 = 1 + 0.111 111 111 111 202 406 4;
  • 15) 0.111 111 111 111 202 406 4 × 2 = 0 + 0.222 222 222 222 404 812 8;
  • 16) 0.222 222 222 222 404 812 8 × 2 = 0 + 0.444 444 444 444 809 625 6;
  • 17) 0.444 444 444 444 809 625 6 × 2 = 0 + 0.888 888 888 889 619 251 2;
  • 18) 0.888 888 888 889 619 251 2 × 2 = 1 + 0.777 777 777 779 238 502 4;
  • 19) 0.777 777 777 779 238 502 4 × 2 = 1 + 0.555 555 555 558 477 004 8;
  • 20) 0.555 555 555 558 477 004 8 × 2 = 1 + 0.111 111 111 116 954 009 6;
  • 21) 0.111 111 111 116 954 009 6 × 2 = 0 + 0.222 222 222 233 908 019 2;
  • 22) 0.222 222 222 233 908 019 2 × 2 = 0 + 0.444 444 444 467 816 038 4;
  • 23) 0.444 444 444 467 816 038 4 × 2 = 0 + 0.888 888 888 935 632 076 8;
  • 24) 0.888 888 888 935 632 076 8 × 2 = 1 + 0.777 777 777 871 264 153 6;
  • 25) 0.777 777 777 871 264 153 6 × 2 = 1 + 0.555 555 555 742 528 307 2;
  • 26) 0.555 555 555 742 528 307 2 × 2 = 1 + 0.111 111 111 485 056 614 4;
  • 27) 0.111 111 111 485 056 614 4 × 2 = 0 + 0.222 222 222 970 113 228 8;
  • 28) 0.222 222 222 970 113 228 8 × 2 = 0 + 0.444 444 445 940 226 457 6;
  • 29) 0.444 444 445 940 226 457 6 × 2 = 0 + 0.888 888 891 880 452 915 2;
  • 30) 0.888 888 891 880 452 915 2 × 2 = 1 + 0.777 777 783 760 905 830 4;
  • 31) 0.777 777 783 760 905 830 4 × 2 = 1 + 0.555 555 567 521 811 660 8;
  • 32) 0.555 555 567 521 811 660 8 × 2 = 1 + 0.111 111 135 043 623 321 6;
  • 33) 0.111 111 135 043 623 321 6 × 2 = 0 + 0.222 222 270 087 246 643 2;
  • 34) 0.222 222 270 087 246 643 2 × 2 = 0 + 0.444 444 540 174 493 286 4;
  • 35) 0.444 444 540 174 493 286 4 × 2 = 0 + 0.888 889 080 348 986 572 8;
  • 36) 0.888 889 080 348 986 572 8 × 2 = 1 + 0.777 778 160 697 973 145 6;
  • 37) 0.777 778 160 697 973 145 6 × 2 = 1 + 0.555 556 321 395 946 291 2;
  • 38) 0.555 556 321 395 946 291 2 × 2 = 1 + 0.111 112 642 791 892 582 4;
  • 39) 0.111 112 642 791 892 582 4 × 2 = 0 + 0.222 225 285 583 785 164 8;
  • 40) 0.222 225 285 583 785 164 8 × 2 = 0 + 0.444 450 571 167 570 329 6;
  • 41) 0.444 450 571 167 570 329 6 × 2 = 0 + 0.888 901 142 335 140 659 2;
  • 42) 0.888 901 142 335 140 659 2 × 2 = 1 + 0.777 802 284 670 281 318 4;
  • 43) 0.777 802 284 670 281 318 4 × 2 = 1 + 0.555 604 569 340 562 636 8;
  • 44) 0.555 604 569 340 562 636 8 × 2 = 1 + 0.111 209 138 681 125 273 6;
  • 45) 0.111 209 138 681 125 273 6 × 2 = 0 + 0.222 418 277 362 250 547 2;
  • 46) 0.222 418 277 362 250 547 2 × 2 = 0 + 0.444 836 554 724 501 094 4;
  • 47) 0.444 836 554 724 501 094 4 × 2 = 0 + 0.889 673 109 449 002 188 8;
  • 48) 0.889 673 109 449 002 188 8 × 2 = 1 + 0.779 346 218 898 004 377 6;
  • 49) 0.779 346 218 898 004 377 6 × 2 = 1 + 0.558 692 437 796 008 755 2;
  • 50) 0.558 692 437 796 008 755 2 × 2 = 1 + 0.117 384 875 592 017 510 4;
  • 51) 0.117 384 875 592 017 510 4 × 2 = 0 + 0.234 769 751 184 035 020 8;
  • 52) 0.234 769 751 184 035 020 8 × 2 = 0 + 0.469 539 502 368 070 041 6;
  • 53) 0.469 539 502 368 070 041 6 × 2 = 0 + 0.939 079 004 736 140 083 2;

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:


0.777 777 777 777 777 783 35(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 783 35(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:


24.777 777 777 777 777 783 35(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 24.777 777 777 777 777 783 35 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