24.777 777 777 777 777 777 777 930 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 777 930 7(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
24.777 777 777 777 777 777 777 930 7(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: 0.777 777 777 777 777 777 777 930 7.

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) 0.777 777 777 777 777 777 777 930 7 × 2 = 1 + 0.555 555 555 555 555 555 555 861 4;
  • 2) 0.555 555 555 555 555 555 555 861 4 × 2 = 1 + 0.111 111 111 111 111 111 111 722 8;
  • 3) 0.111 111 111 111 111 111 111 722 8 × 2 = 0 + 0.222 222 222 222 222 222 223 445 6;
  • 4) 0.222 222 222 222 222 222 223 445 6 × 2 = 0 + 0.444 444 444 444 444 444 446 891 2;
  • 5) 0.444 444 444 444 444 444 446 891 2 × 2 = 0 + 0.888 888 888 888 888 888 893 782 4;
  • 6) 0.888 888 888 888 888 888 893 782 4 × 2 = 1 + 0.777 777 777 777 777 777 787 564 8;
  • 7) 0.777 777 777 777 777 777 787 564 8 × 2 = 1 + 0.555 555 555 555 555 555 575 129 6;
  • 8) 0.555 555 555 555 555 555 575 129 6 × 2 = 1 + 0.111 111 111 111 111 111 150 259 2;
  • 9) 0.111 111 111 111 111 111 150 259 2 × 2 = 0 + 0.222 222 222 222 222 222 300 518 4;
  • 10) 0.222 222 222 222 222 222 300 518 4 × 2 = 0 + 0.444 444 444 444 444 444 601 036 8;
  • 11) 0.444 444 444 444 444 444 601 036 8 × 2 = 0 + 0.888 888 888 888 888 889 202 073 6;
  • 12) 0.888 888 888 888 888 889 202 073 6 × 2 = 1 + 0.777 777 777 777 777 778 404 147 2;
  • 13) 0.777 777 777 777 777 778 404 147 2 × 2 = 1 + 0.555 555 555 555 555 556 808 294 4;
  • 14) 0.555 555 555 555 555 556 808 294 4 × 2 = 1 + 0.111 111 111 111 111 113 616 588 8;
  • 15) 0.111 111 111 111 111 113 616 588 8 × 2 = 0 + 0.222 222 222 222 222 227 233 177 6;
  • 16) 0.222 222 222 222 222 227 233 177 6 × 2 = 0 + 0.444 444 444 444 444 454 466 355 2;
  • 17) 0.444 444 444 444 444 454 466 355 2 × 2 = 0 + 0.888 888 888 888 888 908 932 710 4;
  • 18) 0.888 888 888 888 888 908 932 710 4 × 2 = 1 + 0.777 777 777 777 777 817 865 420 8;
  • 19) 0.777 777 777 777 777 817 865 420 8 × 2 = 1 + 0.555 555 555 555 555 635 730 841 6;
  • 20) 0.555 555 555 555 555 635 730 841 6 × 2 = 1 + 0.111 111 111 111 111 271 461 683 2;
  • 21) 0.111 111 111 111 111 271 461 683 2 × 2 = 0 + 0.222 222 222 222 222 542 923 366 4;
  • 22) 0.222 222 222 222 222 542 923 366 4 × 2 = 0 + 0.444 444 444 444 445 085 846 732 8;
  • 23) 0.444 444 444 444 445 085 846 732 8 × 2 = 0 + 0.888 888 888 888 890 171 693 465 6;
  • 24) 0.888 888 888 888 890 171 693 465 6 × 2 = 1 + 0.777 777 777 777 780 343 386 931 2;
  • 25) 0.777 777 777 777 780 343 386 931 2 × 2 = 1 + 0.555 555 555 555 560 686 773 862 4;
  • 26) 0.555 555 555 555 560 686 773 862 4 × 2 = 1 + 0.111 111 111 111 121 373 547 724 8;
  • 27) 0.111 111 111 111 121 373 547 724 8 × 2 = 0 + 0.222 222 222 222 242 747 095 449 6;
  • 28) 0.222 222 222 222 242 747 095 449 6 × 2 = 0 + 0.444 444 444 444 485 494 190 899 2;
  • 29) 0.444 444 444 444 485 494 190 899 2 × 2 = 0 + 0.888 888 888 888 970 988 381 798 4;
  • 30) 0.888 888 888 888 970 988 381 798 4 × 2 = 1 + 0.777 777 777 777 941 976 763 596 8;
  • 31) 0.777 777 777 777 941 976 763 596 8 × 2 = 1 + 0.555 555 555 555 883 953 527 193 6;
  • 32) 0.555 555 555 555 883 953 527 193 6 × 2 = 1 + 0.111 111 111 111 767 907 054 387 2;
  • 33) 0.111 111 111 111 767 907 054 387 2 × 2 = 0 + 0.222 222 222 223 535 814 108 774 4;
  • 34) 0.222 222 222 223 535 814 108 774 4 × 2 = 0 + 0.444 444 444 447 071 628 217 548 8;
  • 35) 0.444 444 444 447 071 628 217 548 8 × 2 = 0 + 0.888 888 888 894 143 256 435 097 6;
  • 36) 0.888 888 888 894 143 256 435 097 6 × 2 = 1 + 0.777 777 777 788 286 512 870 195 2;
  • 37) 0.777 777 777 788 286 512 870 195 2 × 2 = 1 + 0.555 555 555 576 573 025 740 390 4;
  • 38) 0.555 555 555 576 573 025 740 390 4 × 2 = 1 + 0.111 111 111 153 146 051 480 780 8;
  • 39) 0.111 111 111 153 146 051 480 780 8 × 2 = 0 + 0.222 222 222 306 292 102 961 561 6;
  • 40) 0.222 222 222 306 292 102 961 561 6 × 2 = 0 + 0.444 444 444 612 584 205 923 123 2;
  • 41) 0.444 444 444 612 584 205 923 123 2 × 2 = 0 + 0.888 888 889 225 168 411 846 246 4;
  • 42) 0.888 888 889 225 168 411 846 246 4 × 2 = 1 + 0.777 777 778 450 336 823 692 492 8;
  • 43) 0.777 777 778 450 336 823 692 492 8 × 2 = 1 + 0.555 555 556 900 673 647 384 985 6;
  • 44) 0.555 555 556 900 673 647 384 985 6 × 2 = 1 + 0.111 111 113 801 347 294 769 971 2;
  • 45) 0.111 111 113 801 347 294 769 971 2 × 2 = 0 + 0.222 222 227 602 694 589 539 942 4;
  • 46) 0.222 222 227 602 694 589 539 942 4 × 2 = 0 + 0.444 444 455 205 389 179 079 884 8;
  • 47) 0.444 444 455 205 389 179 079 884 8 × 2 = 0 + 0.888 888 910 410 778 358 159 769 6;
  • 48) 0.888 888 910 410 778 358 159 769 6 × 2 = 1 + 0.777 777 820 821 556 716 319 539 2;
  • 49) 0.777 777 820 821 556 716 319 539 2 × 2 = 1 + 0.555 555 641 643 113 432 639 078 4;
  • 50) 0.555 555 641 643 113 432 639 078 4 × 2 = 1 + 0.111 111 283 286 226 865 278 156 8;
  • 51) 0.111 111 283 286 226 865 278 156 8 × 2 = 0 + 0.222 222 566 572 453 730 556 313 6;
  • 52) 0.222 222 566 572 453 730 556 313 6 × 2 = 0 + 0.444 445 133 144 907 461 112 627 2;
  • 53) 0.444 445 133 144 907 461 112 627 2 × 2 = 0 + 0.888 890 266 289 814 922 225 254 4;

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:


0.777 777 777 777 777 777 777 930 7(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 777 930 7(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:


24.777 777 777 777 777 777 777 930 7(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 24.777 777 777 777 777 777 777 930 7 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

Share

» Convert decimal 24.777 777 777 777 777 777 777 936 6 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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