24.777 777 777 777 777 768 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 768 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
24.777 777 777 777 777 768 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: 0.777 777 777 777 777 768 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) 0.777 777 777 777 777 768 7 × 2 = 1 + 0.555 555 555 555 555 537 4;
  • 2) 0.555 555 555 555 555 537 4 × 2 = 1 + 0.111 111 111 111 111 074 8;
  • 3) 0.111 111 111 111 111 074 8 × 2 = 0 + 0.222 222 222 222 222 149 6;
  • 4) 0.222 222 222 222 222 149 6 × 2 = 0 + 0.444 444 444 444 444 299 2;
  • 5) 0.444 444 444 444 444 299 2 × 2 = 0 + 0.888 888 888 888 888 598 4;
  • 6) 0.888 888 888 888 888 598 4 × 2 = 1 + 0.777 777 777 777 777 196 8;
  • 7) 0.777 777 777 777 777 196 8 × 2 = 1 + 0.555 555 555 555 554 393 6;
  • 8) 0.555 555 555 555 554 393 6 × 2 = 1 + 0.111 111 111 111 108 787 2;
  • 9) 0.111 111 111 111 108 787 2 × 2 = 0 + 0.222 222 222 222 217 574 4;
  • 10) 0.222 222 222 222 217 574 4 × 2 = 0 + 0.444 444 444 444 435 148 8;
  • 11) 0.444 444 444 444 435 148 8 × 2 = 0 + 0.888 888 888 888 870 297 6;
  • 12) 0.888 888 888 888 870 297 6 × 2 = 1 + 0.777 777 777 777 740 595 2;
  • 13) 0.777 777 777 777 740 595 2 × 2 = 1 + 0.555 555 555 555 481 190 4;
  • 14) 0.555 555 555 555 481 190 4 × 2 = 1 + 0.111 111 111 110 962 380 8;
  • 15) 0.111 111 111 110 962 380 8 × 2 = 0 + 0.222 222 222 221 924 761 6;
  • 16) 0.222 222 222 221 924 761 6 × 2 = 0 + 0.444 444 444 443 849 523 2;
  • 17) 0.444 444 444 443 849 523 2 × 2 = 0 + 0.888 888 888 887 699 046 4;
  • 18) 0.888 888 888 887 699 046 4 × 2 = 1 + 0.777 777 777 775 398 092 8;
  • 19) 0.777 777 777 775 398 092 8 × 2 = 1 + 0.555 555 555 550 796 185 6;
  • 20) 0.555 555 555 550 796 185 6 × 2 = 1 + 0.111 111 111 101 592 371 2;
  • 21) 0.111 111 111 101 592 371 2 × 2 = 0 + 0.222 222 222 203 184 742 4;
  • 22) 0.222 222 222 203 184 742 4 × 2 = 0 + 0.444 444 444 406 369 484 8;
  • 23) 0.444 444 444 406 369 484 8 × 2 = 0 + 0.888 888 888 812 738 969 6;
  • 24) 0.888 888 888 812 738 969 6 × 2 = 1 + 0.777 777 777 625 477 939 2;
  • 25) 0.777 777 777 625 477 939 2 × 2 = 1 + 0.555 555 555 250 955 878 4;
  • 26) 0.555 555 555 250 955 878 4 × 2 = 1 + 0.111 111 110 501 911 756 8;
  • 27) 0.111 111 110 501 911 756 8 × 2 = 0 + 0.222 222 221 003 823 513 6;
  • 28) 0.222 222 221 003 823 513 6 × 2 = 0 + 0.444 444 442 007 647 027 2;
  • 29) 0.444 444 442 007 647 027 2 × 2 = 0 + 0.888 888 884 015 294 054 4;
  • 30) 0.888 888 884 015 294 054 4 × 2 = 1 + 0.777 777 768 030 588 108 8;
  • 31) 0.777 777 768 030 588 108 8 × 2 = 1 + 0.555 555 536 061 176 217 6;
  • 32) 0.555 555 536 061 176 217 6 × 2 = 1 + 0.111 111 072 122 352 435 2;
  • 33) 0.111 111 072 122 352 435 2 × 2 = 0 + 0.222 222 144 244 704 870 4;
  • 34) 0.222 222 144 244 704 870 4 × 2 = 0 + 0.444 444 288 489 409 740 8;
  • 35) 0.444 444 288 489 409 740 8 × 2 = 0 + 0.888 888 576 978 819 481 6;
  • 36) 0.888 888 576 978 819 481 6 × 2 = 1 + 0.777 777 153 957 638 963 2;
  • 37) 0.777 777 153 957 638 963 2 × 2 = 1 + 0.555 554 307 915 277 926 4;
  • 38) 0.555 554 307 915 277 926 4 × 2 = 1 + 0.111 108 615 830 555 852 8;
  • 39) 0.111 108 615 830 555 852 8 × 2 = 0 + 0.222 217 231 661 111 705 6;
  • 40) 0.222 217 231 661 111 705 6 × 2 = 0 + 0.444 434 463 322 223 411 2;
  • 41) 0.444 434 463 322 223 411 2 × 2 = 0 + 0.888 868 926 644 446 822 4;
  • 42) 0.888 868 926 644 446 822 4 × 2 = 1 + 0.777 737 853 288 893 644 8;
  • 43) 0.777 737 853 288 893 644 8 × 2 = 1 + 0.555 475 706 577 787 289 6;
  • 44) 0.555 475 706 577 787 289 6 × 2 = 1 + 0.110 951 413 155 574 579 2;
  • 45) 0.110 951 413 155 574 579 2 × 2 = 0 + 0.221 902 826 311 149 158 4;
  • 46) 0.221 902 826 311 149 158 4 × 2 = 0 + 0.443 805 652 622 298 316 8;
  • 47) 0.443 805 652 622 298 316 8 × 2 = 0 + 0.887 611 305 244 596 633 6;
  • 48) 0.887 611 305 244 596 633 6 × 2 = 1 + 0.775 222 610 489 193 267 2;
  • 49) 0.775 222 610 489 193 267 2 × 2 = 1 + 0.550 445 220 978 386 534 4;
  • 50) 0.550 445 220 978 386 534 4 × 2 = 1 + 0.100 890 441 956 773 068 8;
  • 51) 0.100 890 441 956 773 068 8 × 2 = 0 + 0.201 780 883 913 546 137 6;
  • 52) 0.201 780 883 913 546 137 6 × 2 = 0 + 0.403 561 767 827 092 275 2;
  • 53) 0.403 561 767 827 092 275 2 × 2 = 0 + 0.807 123 535 654 184 550 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:


0.777 777 777 777 777 768 7(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 768 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:


24.777 777 777 777 777 768 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 24.777 777 777 777 777 768 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

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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