24.777 777 777 777 877 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 877(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
24.777 777 777 777 877(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: 0.777 777 777 777 877.

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) 0.777 777 777 777 877 × 2 = 1 + 0.555 555 555 555 754;
  • 2) 0.555 555 555 555 754 × 2 = 1 + 0.111 111 111 111 508;
  • 3) 0.111 111 111 111 508 × 2 = 0 + 0.222 222 222 223 016;
  • 4) 0.222 222 222 223 016 × 2 = 0 + 0.444 444 444 446 032;
  • 5) 0.444 444 444 446 032 × 2 = 0 + 0.888 888 888 892 064;
  • 6) 0.888 888 888 892 064 × 2 = 1 + 0.777 777 777 784 128;
  • 7) 0.777 777 777 784 128 × 2 = 1 + 0.555 555 555 568 256;
  • 8) 0.555 555 555 568 256 × 2 = 1 + 0.111 111 111 136 512;
  • 9) 0.111 111 111 136 512 × 2 = 0 + 0.222 222 222 273 024;
  • 10) 0.222 222 222 273 024 × 2 = 0 + 0.444 444 444 546 048;
  • 11) 0.444 444 444 546 048 × 2 = 0 + 0.888 888 889 092 096;
  • 12) 0.888 888 889 092 096 × 2 = 1 + 0.777 777 778 184 192;
  • 13) 0.777 777 778 184 192 × 2 = 1 + 0.555 555 556 368 384;
  • 14) 0.555 555 556 368 384 × 2 = 1 + 0.111 111 112 736 768;
  • 15) 0.111 111 112 736 768 × 2 = 0 + 0.222 222 225 473 536;
  • 16) 0.222 222 225 473 536 × 2 = 0 + 0.444 444 450 947 072;
  • 17) 0.444 444 450 947 072 × 2 = 0 + 0.888 888 901 894 144;
  • 18) 0.888 888 901 894 144 × 2 = 1 + 0.777 777 803 788 288;
  • 19) 0.777 777 803 788 288 × 2 = 1 + 0.555 555 607 576 576;
  • 20) 0.555 555 607 576 576 × 2 = 1 + 0.111 111 215 153 152;
  • 21) 0.111 111 215 153 152 × 2 = 0 + 0.222 222 430 306 304;
  • 22) 0.222 222 430 306 304 × 2 = 0 + 0.444 444 860 612 608;
  • 23) 0.444 444 860 612 608 × 2 = 0 + 0.888 889 721 225 216;
  • 24) 0.888 889 721 225 216 × 2 = 1 + 0.777 779 442 450 432;
  • 25) 0.777 779 442 450 432 × 2 = 1 + 0.555 558 884 900 864;
  • 26) 0.555 558 884 900 864 × 2 = 1 + 0.111 117 769 801 728;
  • 27) 0.111 117 769 801 728 × 2 = 0 + 0.222 235 539 603 456;
  • 28) 0.222 235 539 603 456 × 2 = 0 + 0.444 471 079 206 912;
  • 29) 0.444 471 079 206 912 × 2 = 0 + 0.888 942 158 413 824;
  • 30) 0.888 942 158 413 824 × 2 = 1 + 0.777 884 316 827 648;
  • 31) 0.777 884 316 827 648 × 2 = 1 + 0.555 768 633 655 296;
  • 32) 0.555 768 633 655 296 × 2 = 1 + 0.111 537 267 310 592;
  • 33) 0.111 537 267 310 592 × 2 = 0 + 0.223 074 534 621 184;
  • 34) 0.223 074 534 621 184 × 2 = 0 + 0.446 149 069 242 368;
  • 35) 0.446 149 069 242 368 × 2 = 0 + 0.892 298 138 484 736;
  • 36) 0.892 298 138 484 736 × 2 = 1 + 0.784 596 276 969 472;
  • 37) 0.784 596 276 969 472 × 2 = 1 + 0.569 192 553 938 944;
  • 38) 0.569 192 553 938 944 × 2 = 1 + 0.138 385 107 877 888;
  • 39) 0.138 385 107 877 888 × 2 = 0 + 0.276 770 215 755 776;
  • 40) 0.276 770 215 755 776 × 2 = 0 + 0.553 540 431 511 552;
  • 41) 0.553 540 431 511 552 × 2 = 1 + 0.107 080 863 023 104;
  • 42) 0.107 080 863 023 104 × 2 = 0 + 0.214 161 726 046 208;
  • 43) 0.214 161 726 046 208 × 2 = 0 + 0.428 323 452 092 416;
  • 44) 0.428 323 452 092 416 × 2 = 0 + 0.856 646 904 184 832;
  • 45) 0.856 646 904 184 832 × 2 = 1 + 0.713 293 808 369 664;
  • 46) 0.713 293 808 369 664 × 2 = 1 + 0.426 587 616 739 328;
  • 47) 0.426 587 616 739 328 × 2 = 0 + 0.853 175 233 478 656;
  • 48) 0.853 175 233 478 656 × 2 = 1 + 0.706 350 466 957 312;
  • 49) 0.706 350 466 957 312 × 2 = 1 + 0.412 700 933 914 624;
  • 50) 0.412 700 933 914 624 × 2 = 0 + 0.825 401 867 829 248;
  • 51) 0.825 401 867 829 248 × 2 = 1 + 0.650 803 735 658 496;
  • 52) 0.650 803 735 658 496 × 2 = 1 + 0.301 607 471 316 992;
  • 53) 0.301 607 471 316 992 × 2 = 0 + 0.603 214 942 633 984;

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:


0.777 777 777 777 877(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1101 1011 0(2)

5. Positive number before normalization:

24.777 777 777 777 877(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1101 1011 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:


24.777 777 777 777 877(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1101 1011 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1101 1011 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1101 1011 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 1000 1101 1011 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 1000 1101 1 0110 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1101


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 1000 1101


Decimal number 24.777 777 777 777 877 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 1000 1101

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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