24.777 777 777 777 777 777 557 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 557(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
24.777 777 777 777 777 777 557(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: 0.777 777 777 777 777 777 557.

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) 0.777 777 777 777 777 777 557 × 2 = 1 + 0.555 555 555 555 555 555 114;
  • 2) 0.555 555 555 555 555 555 114 × 2 = 1 + 0.111 111 111 111 111 110 228;
  • 3) 0.111 111 111 111 111 110 228 × 2 = 0 + 0.222 222 222 222 222 220 456;
  • 4) 0.222 222 222 222 222 220 456 × 2 = 0 + 0.444 444 444 444 444 440 912;
  • 5) 0.444 444 444 444 444 440 912 × 2 = 0 + 0.888 888 888 888 888 881 824;
  • 6) 0.888 888 888 888 888 881 824 × 2 = 1 + 0.777 777 777 777 777 763 648;
  • 7) 0.777 777 777 777 777 763 648 × 2 = 1 + 0.555 555 555 555 555 527 296;
  • 8) 0.555 555 555 555 555 527 296 × 2 = 1 + 0.111 111 111 111 111 054 592;
  • 9) 0.111 111 111 111 111 054 592 × 2 = 0 + 0.222 222 222 222 222 109 184;
  • 10) 0.222 222 222 222 222 109 184 × 2 = 0 + 0.444 444 444 444 444 218 368;
  • 11) 0.444 444 444 444 444 218 368 × 2 = 0 + 0.888 888 888 888 888 436 736;
  • 12) 0.888 888 888 888 888 436 736 × 2 = 1 + 0.777 777 777 777 776 873 472;
  • 13) 0.777 777 777 777 776 873 472 × 2 = 1 + 0.555 555 555 555 553 746 944;
  • 14) 0.555 555 555 555 553 746 944 × 2 = 1 + 0.111 111 111 111 107 493 888;
  • 15) 0.111 111 111 111 107 493 888 × 2 = 0 + 0.222 222 222 222 214 987 776;
  • 16) 0.222 222 222 222 214 987 776 × 2 = 0 + 0.444 444 444 444 429 975 552;
  • 17) 0.444 444 444 444 429 975 552 × 2 = 0 + 0.888 888 888 888 859 951 104;
  • 18) 0.888 888 888 888 859 951 104 × 2 = 1 + 0.777 777 777 777 719 902 208;
  • 19) 0.777 777 777 777 719 902 208 × 2 = 1 + 0.555 555 555 555 439 804 416;
  • 20) 0.555 555 555 555 439 804 416 × 2 = 1 + 0.111 111 111 110 879 608 832;
  • 21) 0.111 111 111 110 879 608 832 × 2 = 0 + 0.222 222 222 221 759 217 664;
  • 22) 0.222 222 222 221 759 217 664 × 2 = 0 + 0.444 444 444 443 518 435 328;
  • 23) 0.444 444 444 443 518 435 328 × 2 = 0 + 0.888 888 888 887 036 870 656;
  • 24) 0.888 888 888 887 036 870 656 × 2 = 1 + 0.777 777 777 774 073 741 312;
  • 25) 0.777 777 777 774 073 741 312 × 2 = 1 + 0.555 555 555 548 147 482 624;
  • 26) 0.555 555 555 548 147 482 624 × 2 = 1 + 0.111 111 111 096 294 965 248;
  • 27) 0.111 111 111 096 294 965 248 × 2 = 0 + 0.222 222 222 192 589 930 496;
  • 28) 0.222 222 222 192 589 930 496 × 2 = 0 + 0.444 444 444 385 179 860 992;
  • 29) 0.444 444 444 385 179 860 992 × 2 = 0 + 0.888 888 888 770 359 721 984;
  • 30) 0.888 888 888 770 359 721 984 × 2 = 1 + 0.777 777 777 540 719 443 968;
  • 31) 0.777 777 777 540 719 443 968 × 2 = 1 + 0.555 555 555 081 438 887 936;
  • 32) 0.555 555 555 081 438 887 936 × 2 = 1 + 0.111 111 110 162 877 775 872;
  • 33) 0.111 111 110 162 877 775 872 × 2 = 0 + 0.222 222 220 325 755 551 744;
  • 34) 0.222 222 220 325 755 551 744 × 2 = 0 + 0.444 444 440 651 511 103 488;
  • 35) 0.444 444 440 651 511 103 488 × 2 = 0 + 0.888 888 881 303 022 206 976;
  • 36) 0.888 888 881 303 022 206 976 × 2 = 1 + 0.777 777 762 606 044 413 952;
  • 37) 0.777 777 762 606 044 413 952 × 2 = 1 + 0.555 555 525 212 088 827 904;
  • 38) 0.555 555 525 212 088 827 904 × 2 = 1 + 0.111 111 050 424 177 655 808;
  • 39) 0.111 111 050 424 177 655 808 × 2 = 0 + 0.222 222 100 848 355 311 616;
  • 40) 0.222 222 100 848 355 311 616 × 2 = 0 + 0.444 444 201 696 710 623 232;
  • 41) 0.444 444 201 696 710 623 232 × 2 = 0 + 0.888 888 403 393 421 246 464;
  • 42) 0.888 888 403 393 421 246 464 × 2 = 1 + 0.777 776 806 786 842 492 928;
  • 43) 0.777 776 806 786 842 492 928 × 2 = 1 + 0.555 553 613 573 684 985 856;
  • 44) 0.555 553 613 573 684 985 856 × 2 = 1 + 0.111 107 227 147 369 971 712;
  • 45) 0.111 107 227 147 369 971 712 × 2 = 0 + 0.222 214 454 294 739 943 424;
  • 46) 0.222 214 454 294 739 943 424 × 2 = 0 + 0.444 428 908 589 479 886 848;
  • 47) 0.444 428 908 589 479 886 848 × 2 = 0 + 0.888 857 817 178 959 773 696;
  • 48) 0.888 857 817 178 959 773 696 × 2 = 1 + 0.777 715 634 357 919 547 392;
  • 49) 0.777 715 634 357 919 547 392 × 2 = 1 + 0.555 431 268 715 839 094 784;
  • 50) 0.555 431 268 715 839 094 784 × 2 = 1 + 0.110 862 537 431 678 189 568;
  • 51) 0.110 862 537 431 678 189 568 × 2 = 0 + 0.221 725 074 863 356 379 136;
  • 52) 0.221 725 074 863 356 379 136 × 2 = 0 + 0.443 450 149 726 712 758 272;
  • 53) 0.443 450 149 726 712 758 272 × 2 = 0 + 0.886 900 299 453 425 516 544;

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:


0.777 777 777 777 777 777 557(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 557(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:


24.777 777 777 777 777 777 557(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 24.777 777 777 777 777 777 557 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