24.777 777 777 777 777 777 777 944 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 777 944(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
24.777 777 777 777 777 777 777 944(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: 0.777 777 777 777 777 777 777 944.

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) 0.777 777 777 777 777 777 777 944 × 2 = 1 + 0.555 555 555 555 555 555 555 888;
  • 2) 0.555 555 555 555 555 555 555 888 × 2 = 1 + 0.111 111 111 111 111 111 111 776;
  • 3) 0.111 111 111 111 111 111 111 776 × 2 = 0 + 0.222 222 222 222 222 222 223 552;
  • 4) 0.222 222 222 222 222 222 223 552 × 2 = 0 + 0.444 444 444 444 444 444 447 104;
  • 5) 0.444 444 444 444 444 444 447 104 × 2 = 0 + 0.888 888 888 888 888 888 894 208;
  • 6) 0.888 888 888 888 888 888 894 208 × 2 = 1 + 0.777 777 777 777 777 777 788 416;
  • 7) 0.777 777 777 777 777 777 788 416 × 2 = 1 + 0.555 555 555 555 555 555 576 832;
  • 8) 0.555 555 555 555 555 555 576 832 × 2 = 1 + 0.111 111 111 111 111 111 153 664;
  • 9) 0.111 111 111 111 111 111 153 664 × 2 = 0 + 0.222 222 222 222 222 222 307 328;
  • 10) 0.222 222 222 222 222 222 307 328 × 2 = 0 + 0.444 444 444 444 444 444 614 656;
  • 11) 0.444 444 444 444 444 444 614 656 × 2 = 0 + 0.888 888 888 888 888 889 229 312;
  • 12) 0.888 888 888 888 888 889 229 312 × 2 = 1 + 0.777 777 777 777 777 778 458 624;
  • 13) 0.777 777 777 777 777 778 458 624 × 2 = 1 + 0.555 555 555 555 555 556 917 248;
  • 14) 0.555 555 555 555 555 556 917 248 × 2 = 1 + 0.111 111 111 111 111 113 834 496;
  • 15) 0.111 111 111 111 111 113 834 496 × 2 = 0 + 0.222 222 222 222 222 227 668 992;
  • 16) 0.222 222 222 222 222 227 668 992 × 2 = 0 + 0.444 444 444 444 444 455 337 984;
  • 17) 0.444 444 444 444 444 455 337 984 × 2 = 0 + 0.888 888 888 888 888 910 675 968;
  • 18) 0.888 888 888 888 888 910 675 968 × 2 = 1 + 0.777 777 777 777 777 821 351 936;
  • 19) 0.777 777 777 777 777 821 351 936 × 2 = 1 + 0.555 555 555 555 555 642 703 872;
  • 20) 0.555 555 555 555 555 642 703 872 × 2 = 1 + 0.111 111 111 111 111 285 407 744;
  • 21) 0.111 111 111 111 111 285 407 744 × 2 = 0 + 0.222 222 222 222 222 570 815 488;
  • 22) 0.222 222 222 222 222 570 815 488 × 2 = 0 + 0.444 444 444 444 445 141 630 976;
  • 23) 0.444 444 444 444 445 141 630 976 × 2 = 0 + 0.888 888 888 888 890 283 261 952;
  • 24) 0.888 888 888 888 890 283 261 952 × 2 = 1 + 0.777 777 777 777 780 566 523 904;
  • 25) 0.777 777 777 777 780 566 523 904 × 2 = 1 + 0.555 555 555 555 561 133 047 808;
  • 26) 0.555 555 555 555 561 133 047 808 × 2 = 1 + 0.111 111 111 111 122 266 095 616;
  • 27) 0.111 111 111 111 122 266 095 616 × 2 = 0 + 0.222 222 222 222 244 532 191 232;
  • 28) 0.222 222 222 222 244 532 191 232 × 2 = 0 + 0.444 444 444 444 489 064 382 464;
  • 29) 0.444 444 444 444 489 064 382 464 × 2 = 0 + 0.888 888 888 888 978 128 764 928;
  • 30) 0.888 888 888 888 978 128 764 928 × 2 = 1 + 0.777 777 777 777 956 257 529 856;
  • 31) 0.777 777 777 777 956 257 529 856 × 2 = 1 + 0.555 555 555 555 912 515 059 712;
  • 32) 0.555 555 555 555 912 515 059 712 × 2 = 1 + 0.111 111 111 111 825 030 119 424;
  • 33) 0.111 111 111 111 825 030 119 424 × 2 = 0 + 0.222 222 222 223 650 060 238 848;
  • 34) 0.222 222 222 223 650 060 238 848 × 2 = 0 + 0.444 444 444 447 300 120 477 696;
  • 35) 0.444 444 444 447 300 120 477 696 × 2 = 0 + 0.888 888 888 894 600 240 955 392;
  • 36) 0.888 888 888 894 600 240 955 392 × 2 = 1 + 0.777 777 777 789 200 481 910 784;
  • 37) 0.777 777 777 789 200 481 910 784 × 2 = 1 + 0.555 555 555 578 400 963 821 568;
  • 38) 0.555 555 555 578 400 963 821 568 × 2 = 1 + 0.111 111 111 156 801 927 643 136;
  • 39) 0.111 111 111 156 801 927 643 136 × 2 = 0 + 0.222 222 222 313 603 855 286 272;
  • 40) 0.222 222 222 313 603 855 286 272 × 2 = 0 + 0.444 444 444 627 207 710 572 544;
  • 41) 0.444 444 444 627 207 710 572 544 × 2 = 0 + 0.888 888 889 254 415 421 145 088;
  • 42) 0.888 888 889 254 415 421 145 088 × 2 = 1 + 0.777 777 778 508 830 842 290 176;
  • 43) 0.777 777 778 508 830 842 290 176 × 2 = 1 + 0.555 555 557 017 661 684 580 352;
  • 44) 0.555 555 557 017 661 684 580 352 × 2 = 1 + 0.111 111 114 035 323 369 160 704;
  • 45) 0.111 111 114 035 323 369 160 704 × 2 = 0 + 0.222 222 228 070 646 738 321 408;
  • 46) 0.222 222 228 070 646 738 321 408 × 2 = 0 + 0.444 444 456 141 293 476 642 816;
  • 47) 0.444 444 456 141 293 476 642 816 × 2 = 0 + 0.888 888 912 282 586 953 285 632;
  • 48) 0.888 888 912 282 586 953 285 632 × 2 = 1 + 0.777 777 824 565 173 906 571 264;
  • 49) 0.777 777 824 565 173 906 571 264 × 2 = 1 + 0.555 555 649 130 347 813 142 528;
  • 50) 0.555 555 649 130 347 813 142 528 × 2 = 1 + 0.111 111 298 260 695 626 285 056;
  • 51) 0.111 111 298 260 695 626 285 056 × 2 = 0 + 0.222 222 596 521 391 252 570 112;
  • 52) 0.222 222 596 521 391 252 570 112 × 2 = 0 + 0.444 445 193 042 782 505 140 224;
  • 53) 0.444 445 193 042 782 505 140 224 × 2 = 0 + 0.888 890 386 085 565 010 280 448;

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:


0.777 777 777 777 777 777 777 944(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 777 944(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:


24.777 777 777 777 777 777 777 944(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 24.777 777 777 777 777 777 777 944 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