24.777 777 777 777 777 777 777 918 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 777 918 1(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
24.777 777 777 777 777 777 777 918 1(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: 0.777 777 777 777 777 777 777 918 1.

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) 0.777 777 777 777 777 777 777 918 1 × 2 = 1 + 0.555 555 555 555 555 555 555 836 2;
  • 2) 0.555 555 555 555 555 555 555 836 2 × 2 = 1 + 0.111 111 111 111 111 111 111 672 4;
  • 3) 0.111 111 111 111 111 111 111 672 4 × 2 = 0 + 0.222 222 222 222 222 222 223 344 8;
  • 4) 0.222 222 222 222 222 222 223 344 8 × 2 = 0 + 0.444 444 444 444 444 444 446 689 6;
  • 5) 0.444 444 444 444 444 444 446 689 6 × 2 = 0 + 0.888 888 888 888 888 888 893 379 2;
  • 6) 0.888 888 888 888 888 888 893 379 2 × 2 = 1 + 0.777 777 777 777 777 777 786 758 4;
  • 7) 0.777 777 777 777 777 777 786 758 4 × 2 = 1 + 0.555 555 555 555 555 555 573 516 8;
  • 8) 0.555 555 555 555 555 555 573 516 8 × 2 = 1 + 0.111 111 111 111 111 111 147 033 6;
  • 9) 0.111 111 111 111 111 111 147 033 6 × 2 = 0 + 0.222 222 222 222 222 222 294 067 2;
  • 10) 0.222 222 222 222 222 222 294 067 2 × 2 = 0 + 0.444 444 444 444 444 444 588 134 4;
  • 11) 0.444 444 444 444 444 444 588 134 4 × 2 = 0 + 0.888 888 888 888 888 889 176 268 8;
  • 12) 0.888 888 888 888 888 889 176 268 8 × 2 = 1 + 0.777 777 777 777 777 778 352 537 6;
  • 13) 0.777 777 777 777 777 778 352 537 6 × 2 = 1 + 0.555 555 555 555 555 556 705 075 2;
  • 14) 0.555 555 555 555 555 556 705 075 2 × 2 = 1 + 0.111 111 111 111 111 113 410 150 4;
  • 15) 0.111 111 111 111 111 113 410 150 4 × 2 = 0 + 0.222 222 222 222 222 226 820 300 8;
  • 16) 0.222 222 222 222 222 226 820 300 8 × 2 = 0 + 0.444 444 444 444 444 453 640 601 6;
  • 17) 0.444 444 444 444 444 453 640 601 6 × 2 = 0 + 0.888 888 888 888 888 907 281 203 2;
  • 18) 0.888 888 888 888 888 907 281 203 2 × 2 = 1 + 0.777 777 777 777 777 814 562 406 4;
  • 19) 0.777 777 777 777 777 814 562 406 4 × 2 = 1 + 0.555 555 555 555 555 629 124 812 8;
  • 20) 0.555 555 555 555 555 629 124 812 8 × 2 = 1 + 0.111 111 111 111 111 258 249 625 6;
  • 21) 0.111 111 111 111 111 258 249 625 6 × 2 = 0 + 0.222 222 222 222 222 516 499 251 2;
  • 22) 0.222 222 222 222 222 516 499 251 2 × 2 = 0 + 0.444 444 444 444 445 032 998 502 4;
  • 23) 0.444 444 444 444 445 032 998 502 4 × 2 = 0 + 0.888 888 888 888 890 065 997 004 8;
  • 24) 0.888 888 888 888 890 065 997 004 8 × 2 = 1 + 0.777 777 777 777 780 131 994 009 6;
  • 25) 0.777 777 777 777 780 131 994 009 6 × 2 = 1 + 0.555 555 555 555 560 263 988 019 2;
  • 26) 0.555 555 555 555 560 263 988 019 2 × 2 = 1 + 0.111 111 111 111 120 527 976 038 4;
  • 27) 0.111 111 111 111 120 527 976 038 4 × 2 = 0 + 0.222 222 222 222 241 055 952 076 8;
  • 28) 0.222 222 222 222 241 055 952 076 8 × 2 = 0 + 0.444 444 444 444 482 111 904 153 6;
  • 29) 0.444 444 444 444 482 111 904 153 6 × 2 = 0 + 0.888 888 888 888 964 223 808 307 2;
  • 30) 0.888 888 888 888 964 223 808 307 2 × 2 = 1 + 0.777 777 777 777 928 447 616 614 4;
  • 31) 0.777 777 777 777 928 447 616 614 4 × 2 = 1 + 0.555 555 555 555 856 895 233 228 8;
  • 32) 0.555 555 555 555 856 895 233 228 8 × 2 = 1 + 0.111 111 111 111 713 790 466 457 6;
  • 33) 0.111 111 111 111 713 790 466 457 6 × 2 = 0 + 0.222 222 222 223 427 580 932 915 2;
  • 34) 0.222 222 222 223 427 580 932 915 2 × 2 = 0 + 0.444 444 444 446 855 161 865 830 4;
  • 35) 0.444 444 444 446 855 161 865 830 4 × 2 = 0 + 0.888 888 888 893 710 323 731 660 8;
  • 36) 0.888 888 888 893 710 323 731 660 8 × 2 = 1 + 0.777 777 777 787 420 647 463 321 6;
  • 37) 0.777 777 777 787 420 647 463 321 6 × 2 = 1 + 0.555 555 555 574 841 294 926 643 2;
  • 38) 0.555 555 555 574 841 294 926 643 2 × 2 = 1 + 0.111 111 111 149 682 589 853 286 4;
  • 39) 0.111 111 111 149 682 589 853 286 4 × 2 = 0 + 0.222 222 222 299 365 179 706 572 8;
  • 40) 0.222 222 222 299 365 179 706 572 8 × 2 = 0 + 0.444 444 444 598 730 359 413 145 6;
  • 41) 0.444 444 444 598 730 359 413 145 6 × 2 = 0 + 0.888 888 889 197 460 718 826 291 2;
  • 42) 0.888 888 889 197 460 718 826 291 2 × 2 = 1 + 0.777 777 778 394 921 437 652 582 4;
  • 43) 0.777 777 778 394 921 437 652 582 4 × 2 = 1 + 0.555 555 556 789 842 875 305 164 8;
  • 44) 0.555 555 556 789 842 875 305 164 8 × 2 = 1 + 0.111 111 113 579 685 750 610 329 6;
  • 45) 0.111 111 113 579 685 750 610 329 6 × 2 = 0 + 0.222 222 227 159 371 501 220 659 2;
  • 46) 0.222 222 227 159 371 501 220 659 2 × 2 = 0 + 0.444 444 454 318 743 002 441 318 4;
  • 47) 0.444 444 454 318 743 002 441 318 4 × 2 = 0 + 0.888 888 908 637 486 004 882 636 8;
  • 48) 0.888 888 908 637 486 004 882 636 8 × 2 = 1 + 0.777 777 817 274 972 009 765 273 6;
  • 49) 0.777 777 817 274 972 009 765 273 6 × 2 = 1 + 0.555 555 634 549 944 019 530 547 2;
  • 50) 0.555 555 634 549 944 019 530 547 2 × 2 = 1 + 0.111 111 269 099 888 039 061 094 4;
  • 51) 0.111 111 269 099 888 039 061 094 4 × 2 = 0 + 0.222 222 538 199 776 078 122 188 8;
  • 52) 0.222 222 538 199 776 078 122 188 8 × 2 = 0 + 0.444 445 076 399 552 156 244 377 6;
  • 53) 0.444 445 076 399 552 156 244 377 6 × 2 = 0 + 0.888 890 152 799 104 312 488 755 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:


0.777 777 777 777 777 777 777 918 1(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 777 918 1(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:


24.777 777 777 777 777 777 777 918 1(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 24.777 777 777 777 777 777 777 918 1 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