24.777 777 777 777 777 783 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 783 99(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
24.777 777 777 777 777 783 99(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: 0.777 777 777 777 777 783 99.

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) 0.777 777 777 777 777 783 99 × 2 = 1 + 0.555 555 555 555 555 567 98;
  • 2) 0.555 555 555 555 555 567 98 × 2 = 1 + 0.111 111 111 111 111 135 96;
  • 3) 0.111 111 111 111 111 135 96 × 2 = 0 + 0.222 222 222 222 222 271 92;
  • 4) 0.222 222 222 222 222 271 92 × 2 = 0 + 0.444 444 444 444 444 543 84;
  • 5) 0.444 444 444 444 444 543 84 × 2 = 0 + 0.888 888 888 888 889 087 68;
  • 6) 0.888 888 888 888 889 087 68 × 2 = 1 + 0.777 777 777 777 778 175 36;
  • 7) 0.777 777 777 777 778 175 36 × 2 = 1 + 0.555 555 555 555 556 350 72;
  • 8) 0.555 555 555 555 556 350 72 × 2 = 1 + 0.111 111 111 111 112 701 44;
  • 9) 0.111 111 111 111 112 701 44 × 2 = 0 + 0.222 222 222 222 225 402 88;
  • 10) 0.222 222 222 222 225 402 88 × 2 = 0 + 0.444 444 444 444 450 805 76;
  • 11) 0.444 444 444 444 450 805 76 × 2 = 0 + 0.888 888 888 888 901 611 52;
  • 12) 0.888 888 888 888 901 611 52 × 2 = 1 + 0.777 777 777 777 803 223 04;
  • 13) 0.777 777 777 777 803 223 04 × 2 = 1 + 0.555 555 555 555 606 446 08;
  • 14) 0.555 555 555 555 606 446 08 × 2 = 1 + 0.111 111 111 111 212 892 16;
  • 15) 0.111 111 111 111 212 892 16 × 2 = 0 + 0.222 222 222 222 425 784 32;
  • 16) 0.222 222 222 222 425 784 32 × 2 = 0 + 0.444 444 444 444 851 568 64;
  • 17) 0.444 444 444 444 851 568 64 × 2 = 0 + 0.888 888 888 889 703 137 28;
  • 18) 0.888 888 888 889 703 137 28 × 2 = 1 + 0.777 777 777 779 406 274 56;
  • 19) 0.777 777 777 779 406 274 56 × 2 = 1 + 0.555 555 555 558 812 549 12;
  • 20) 0.555 555 555 558 812 549 12 × 2 = 1 + 0.111 111 111 117 625 098 24;
  • 21) 0.111 111 111 117 625 098 24 × 2 = 0 + 0.222 222 222 235 250 196 48;
  • 22) 0.222 222 222 235 250 196 48 × 2 = 0 + 0.444 444 444 470 500 392 96;
  • 23) 0.444 444 444 470 500 392 96 × 2 = 0 + 0.888 888 888 941 000 785 92;
  • 24) 0.888 888 888 941 000 785 92 × 2 = 1 + 0.777 777 777 882 001 571 84;
  • 25) 0.777 777 777 882 001 571 84 × 2 = 1 + 0.555 555 555 764 003 143 68;
  • 26) 0.555 555 555 764 003 143 68 × 2 = 1 + 0.111 111 111 528 006 287 36;
  • 27) 0.111 111 111 528 006 287 36 × 2 = 0 + 0.222 222 223 056 012 574 72;
  • 28) 0.222 222 223 056 012 574 72 × 2 = 0 + 0.444 444 446 112 025 149 44;
  • 29) 0.444 444 446 112 025 149 44 × 2 = 0 + 0.888 888 892 224 050 298 88;
  • 30) 0.888 888 892 224 050 298 88 × 2 = 1 + 0.777 777 784 448 100 597 76;
  • 31) 0.777 777 784 448 100 597 76 × 2 = 1 + 0.555 555 568 896 201 195 52;
  • 32) 0.555 555 568 896 201 195 52 × 2 = 1 + 0.111 111 137 792 402 391 04;
  • 33) 0.111 111 137 792 402 391 04 × 2 = 0 + 0.222 222 275 584 804 782 08;
  • 34) 0.222 222 275 584 804 782 08 × 2 = 0 + 0.444 444 551 169 609 564 16;
  • 35) 0.444 444 551 169 609 564 16 × 2 = 0 + 0.888 889 102 339 219 128 32;
  • 36) 0.888 889 102 339 219 128 32 × 2 = 1 + 0.777 778 204 678 438 256 64;
  • 37) 0.777 778 204 678 438 256 64 × 2 = 1 + 0.555 556 409 356 876 513 28;
  • 38) 0.555 556 409 356 876 513 28 × 2 = 1 + 0.111 112 818 713 753 026 56;
  • 39) 0.111 112 818 713 753 026 56 × 2 = 0 + 0.222 225 637 427 506 053 12;
  • 40) 0.222 225 637 427 506 053 12 × 2 = 0 + 0.444 451 274 855 012 106 24;
  • 41) 0.444 451 274 855 012 106 24 × 2 = 0 + 0.888 902 549 710 024 212 48;
  • 42) 0.888 902 549 710 024 212 48 × 2 = 1 + 0.777 805 099 420 048 424 96;
  • 43) 0.777 805 099 420 048 424 96 × 2 = 1 + 0.555 610 198 840 096 849 92;
  • 44) 0.555 610 198 840 096 849 92 × 2 = 1 + 0.111 220 397 680 193 699 84;
  • 45) 0.111 220 397 680 193 699 84 × 2 = 0 + 0.222 440 795 360 387 399 68;
  • 46) 0.222 440 795 360 387 399 68 × 2 = 0 + 0.444 881 590 720 774 799 36;
  • 47) 0.444 881 590 720 774 799 36 × 2 = 0 + 0.889 763 181 441 549 598 72;
  • 48) 0.889 763 181 441 549 598 72 × 2 = 1 + 0.779 526 362 883 099 197 44;
  • 49) 0.779 526 362 883 099 197 44 × 2 = 1 + 0.559 052 725 766 198 394 88;
  • 50) 0.559 052 725 766 198 394 88 × 2 = 1 + 0.118 105 451 532 396 789 76;
  • 51) 0.118 105 451 532 396 789 76 × 2 = 0 + 0.236 210 903 064 793 579 52;
  • 52) 0.236 210 903 064 793 579 52 × 2 = 0 + 0.472 421 806 129 587 159 04;
  • 53) 0.472 421 806 129 587 159 04 × 2 = 0 + 0.944 843 612 259 174 318 08;

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:


0.777 777 777 777 777 783 99(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 783 99(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:


24.777 777 777 777 777 783 99(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 24.777 777 777 777 777 783 99 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