24.777 777 777 777 777 784 41 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 784 41(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
24.777 777 777 777 777 784 41(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: 0.777 777 777 777 777 784 41.

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) 0.777 777 777 777 777 784 41 × 2 = 1 + 0.555 555 555 555 555 568 82;
  • 2) 0.555 555 555 555 555 568 82 × 2 = 1 + 0.111 111 111 111 111 137 64;
  • 3) 0.111 111 111 111 111 137 64 × 2 = 0 + 0.222 222 222 222 222 275 28;
  • 4) 0.222 222 222 222 222 275 28 × 2 = 0 + 0.444 444 444 444 444 550 56;
  • 5) 0.444 444 444 444 444 550 56 × 2 = 0 + 0.888 888 888 888 889 101 12;
  • 6) 0.888 888 888 888 889 101 12 × 2 = 1 + 0.777 777 777 777 778 202 24;
  • 7) 0.777 777 777 777 778 202 24 × 2 = 1 + 0.555 555 555 555 556 404 48;
  • 8) 0.555 555 555 555 556 404 48 × 2 = 1 + 0.111 111 111 111 112 808 96;
  • 9) 0.111 111 111 111 112 808 96 × 2 = 0 + 0.222 222 222 222 225 617 92;
  • 10) 0.222 222 222 222 225 617 92 × 2 = 0 + 0.444 444 444 444 451 235 84;
  • 11) 0.444 444 444 444 451 235 84 × 2 = 0 + 0.888 888 888 888 902 471 68;
  • 12) 0.888 888 888 888 902 471 68 × 2 = 1 + 0.777 777 777 777 804 943 36;
  • 13) 0.777 777 777 777 804 943 36 × 2 = 1 + 0.555 555 555 555 609 886 72;
  • 14) 0.555 555 555 555 609 886 72 × 2 = 1 + 0.111 111 111 111 219 773 44;
  • 15) 0.111 111 111 111 219 773 44 × 2 = 0 + 0.222 222 222 222 439 546 88;
  • 16) 0.222 222 222 222 439 546 88 × 2 = 0 + 0.444 444 444 444 879 093 76;
  • 17) 0.444 444 444 444 879 093 76 × 2 = 0 + 0.888 888 888 889 758 187 52;
  • 18) 0.888 888 888 889 758 187 52 × 2 = 1 + 0.777 777 777 779 516 375 04;
  • 19) 0.777 777 777 779 516 375 04 × 2 = 1 + 0.555 555 555 559 032 750 08;
  • 20) 0.555 555 555 559 032 750 08 × 2 = 1 + 0.111 111 111 118 065 500 16;
  • 21) 0.111 111 111 118 065 500 16 × 2 = 0 + 0.222 222 222 236 131 000 32;
  • 22) 0.222 222 222 236 131 000 32 × 2 = 0 + 0.444 444 444 472 262 000 64;
  • 23) 0.444 444 444 472 262 000 64 × 2 = 0 + 0.888 888 888 944 524 001 28;
  • 24) 0.888 888 888 944 524 001 28 × 2 = 1 + 0.777 777 777 889 048 002 56;
  • 25) 0.777 777 777 889 048 002 56 × 2 = 1 + 0.555 555 555 778 096 005 12;
  • 26) 0.555 555 555 778 096 005 12 × 2 = 1 + 0.111 111 111 556 192 010 24;
  • 27) 0.111 111 111 556 192 010 24 × 2 = 0 + 0.222 222 223 112 384 020 48;
  • 28) 0.222 222 223 112 384 020 48 × 2 = 0 + 0.444 444 446 224 768 040 96;
  • 29) 0.444 444 446 224 768 040 96 × 2 = 0 + 0.888 888 892 449 536 081 92;
  • 30) 0.888 888 892 449 536 081 92 × 2 = 1 + 0.777 777 784 899 072 163 84;
  • 31) 0.777 777 784 899 072 163 84 × 2 = 1 + 0.555 555 569 798 144 327 68;
  • 32) 0.555 555 569 798 144 327 68 × 2 = 1 + 0.111 111 139 596 288 655 36;
  • 33) 0.111 111 139 596 288 655 36 × 2 = 0 + 0.222 222 279 192 577 310 72;
  • 34) 0.222 222 279 192 577 310 72 × 2 = 0 + 0.444 444 558 385 154 621 44;
  • 35) 0.444 444 558 385 154 621 44 × 2 = 0 + 0.888 889 116 770 309 242 88;
  • 36) 0.888 889 116 770 309 242 88 × 2 = 1 + 0.777 778 233 540 618 485 76;
  • 37) 0.777 778 233 540 618 485 76 × 2 = 1 + 0.555 556 467 081 236 971 52;
  • 38) 0.555 556 467 081 236 971 52 × 2 = 1 + 0.111 112 934 162 473 943 04;
  • 39) 0.111 112 934 162 473 943 04 × 2 = 0 + 0.222 225 868 324 947 886 08;
  • 40) 0.222 225 868 324 947 886 08 × 2 = 0 + 0.444 451 736 649 895 772 16;
  • 41) 0.444 451 736 649 895 772 16 × 2 = 0 + 0.888 903 473 299 791 544 32;
  • 42) 0.888 903 473 299 791 544 32 × 2 = 1 + 0.777 806 946 599 583 088 64;
  • 43) 0.777 806 946 599 583 088 64 × 2 = 1 + 0.555 613 893 199 166 177 28;
  • 44) 0.555 613 893 199 166 177 28 × 2 = 1 + 0.111 227 786 398 332 354 56;
  • 45) 0.111 227 786 398 332 354 56 × 2 = 0 + 0.222 455 572 796 664 709 12;
  • 46) 0.222 455 572 796 664 709 12 × 2 = 0 + 0.444 911 145 593 329 418 24;
  • 47) 0.444 911 145 593 329 418 24 × 2 = 0 + 0.889 822 291 186 658 836 48;
  • 48) 0.889 822 291 186 658 836 48 × 2 = 1 + 0.779 644 582 373 317 672 96;
  • 49) 0.779 644 582 373 317 672 96 × 2 = 1 + 0.559 289 164 746 635 345 92;
  • 50) 0.559 289 164 746 635 345 92 × 2 = 1 + 0.118 578 329 493 270 691 84;
  • 51) 0.118 578 329 493 270 691 84 × 2 = 0 + 0.237 156 658 986 541 383 68;
  • 52) 0.237 156 658 986 541 383 68 × 2 = 0 + 0.474 313 317 973 082 767 36;
  • 53) 0.474 313 317 973 082 767 36 × 2 = 0 + 0.948 626 635 946 165 534 72;

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:


0.777 777 777 777 777 784 41(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 784 41(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:


24.777 777 777 777 777 784 41(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 24.777 777 777 777 777 784 41 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