24.777 777 777 777 891 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 891(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
24.777 777 777 777 891(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: 0.777 777 777 777 891.

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) 0.777 777 777 777 891 × 2 = 1 + 0.555 555 555 555 782;
  • 2) 0.555 555 555 555 782 × 2 = 1 + 0.111 111 111 111 564;
  • 3) 0.111 111 111 111 564 × 2 = 0 + 0.222 222 222 223 128;
  • 4) 0.222 222 222 223 128 × 2 = 0 + 0.444 444 444 446 256;
  • 5) 0.444 444 444 446 256 × 2 = 0 + 0.888 888 888 892 512;
  • 6) 0.888 888 888 892 512 × 2 = 1 + 0.777 777 777 785 024;
  • 7) 0.777 777 777 785 024 × 2 = 1 + 0.555 555 555 570 048;
  • 8) 0.555 555 555 570 048 × 2 = 1 + 0.111 111 111 140 096;
  • 9) 0.111 111 111 140 096 × 2 = 0 + 0.222 222 222 280 192;
  • 10) 0.222 222 222 280 192 × 2 = 0 + 0.444 444 444 560 384;
  • 11) 0.444 444 444 560 384 × 2 = 0 + 0.888 888 889 120 768;
  • 12) 0.888 888 889 120 768 × 2 = 1 + 0.777 777 778 241 536;
  • 13) 0.777 777 778 241 536 × 2 = 1 + 0.555 555 556 483 072;
  • 14) 0.555 555 556 483 072 × 2 = 1 + 0.111 111 112 966 144;
  • 15) 0.111 111 112 966 144 × 2 = 0 + 0.222 222 225 932 288;
  • 16) 0.222 222 225 932 288 × 2 = 0 + 0.444 444 451 864 576;
  • 17) 0.444 444 451 864 576 × 2 = 0 + 0.888 888 903 729 152;
  • 18) 0.888 888 903 729 152 × 2 = 1 + 0.777 777 807 458 304;
  • 19) 0.777 777 807 458 304 × 2 = 1 + 0.555 555 614 916 608;
  • 20) 0.555 555 614 916 608 × 2 = 1 + 0.111 111 229 833 216;
  • 21) 0.111 111 229 833 216 × 2 = 0 + 0.222 222 459 666 432;
  • 22) 0.222 222 459 666 432 × 2 = 0 + 0.444 444 919 332 864;
  • 23) 0.444 444 919 332 864 × 2 = 0 + 0.888 889 838 665 728;
  • 24) 0.888 889 838 665 728 × 2 = 1 + 0.777 779 677 331 456;
  • 25) 0.777 779 677 331 456 × 2 = 1 + 0.555 559 354 662 912;
  • 26) 0.555 559 354 662 912 × 2 = 1 + 0.111 118 709 325 824;
  • 27) 0.111 118 709 325 824 × 2 = 0 + 0.222 237 418 651 648;
  • 28) 0.222 237 418 651 648 × 2 = 0 + 0.444 474 837 303 296;
  • 29) 0.444 474 837 303 296 × 2 = 0 + 0.888 949 674 606 592;
  • 30) 0.888 949 674 606 592 × 2 = 1 + 0.777 899 349 213 184;
  • 31) 0.777 899 349 213 184 × 2 = 1 + 0.555 798 698 426 368;
  • 32) 0.555 798 698 426 368 × 2 = 1 + 0.111 597 396 852 736;
  • 33) 0.111 597 396 852 736 × 2 = 0 + 0.223 194 793 705 472;
  • 34) 0.223 194 793 705 472 × 2 = 0 + 0.446 389 587 410 944;
  • 35) 0.446 389 587 410 944 × 2 = 0 + 0.892 779 174 821 888;
  • 36) 0.892 779 174 821 888 × 2 = 1 + 0.785 558 349 643 776;
  • 37) 0.785 558 349 643 776 × 2 = 1 + 0.571 116 699 287 552;
  • 38) 0.571 116 699 287 552 × 2 = 1 + 0.142 233 398 575 104;
  • 39) 0.142 233 398 575 104 × 2 = 0 + 0.284 466 797 150 208;
  • 40) 0.284 466 797 150 208 × 2 = 0 + 0.568 933 594 300 416;
  • 41) 0.568 933 594 300 416 × 2 = 1 + 0.137 867 188 600 832;
  • 42) 0.137 867 188 600 832 × 2 = 0 + 0.275 734 377 201 664;
  • 43) 0.275 734 377 201 664 × 2 = 0 + 0.551 468 754 403 328;
  • 44) 0.551 468 754 403 328 × 2 = 1 + 0.102 937 508 806 656;
  • 45) 0.102 937 508 806 656 × 2 = 0 + 0.205 875 017 613 312;
  • 46) 0.205 875 017 613 312 × 2 = 0 + 0.411 750 035 226 624;
  • 47) 0.411 750 035 226 624 × 2 = 0 + 0.823 500 070 453 248;
  • 48) 0.823 500 070 453 248 × 2 = 1 + 0.647 000 140 906 496;
  • 49) 0.647 000 140 906 496 × 2 = 1 + 0.294 000 281 812 992;
  • 50) 0.294 000 281 812 992 × 2 = 0 + 0.588 000 563 625 984;
  • 51) 0.588 000 563 625 984 × 2 = 1 + 0.176 001 127 251 968;
  • 52) 0.176 001 127 251 968 × 2 = 0 + 0.352 002 254 503 936;
  • 53) 0.352 002 254 503 936 × 2 = 0 + 0.704 004 509 007 872;

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:


0.777 777 777 777 891(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1010 0(2)

5. Positive number before normalization:

24.777 777 777 777 891(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1010 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:


24.777 777 777 777 891(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1010 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1010 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1010 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 1001 0001 1010 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 1001 0001 1 0100 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 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 1001 0001


Decimal number 24.777 777 777 777 891 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 1001 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