24.777 777 777 777 691 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 691(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
24.777 777 777 777 691(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: 0.777 777 777 777 691.

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) 0.777 777 777 777 691 × 2 = 1 + 0.555 555 555 555 382;
  • 2) 0.555 555 555 555 382 × 2 = 1 + 0.111 111 111 110 764;
  • 3) 0.111 111 111 110 764 × 2 = 0 + 0.222 222 222 221 528;
  • 4) 0.222 222 222 221 528 × 2 = 0 + 0.444 444 444 443 056;
  • 5) 0.444 444 444 443 056 × 2 = 0 + 0.888 888 888 886 112;
  • 6) 0.888 888 888 886 112 × 2 = 1 + 0.777 777 777 772 224;
  • 7) 0.777 777 777 772 224 × 2 = 1 + 0.555 555 555 544 448;
  • 8) 0.555 555 555 544 448 × 2 = 1 + 0.111 111 111 088 896;
  • 9) 0.111 111 111 088 896 × 2 = 0 + 0.222 222 222 177 792;
  • 10) 0.222 222 222 177 792 × 2 = 0 + 0.444 444 444 355 584;
  • 11) 0.444 444 444 355 584 × 2 = 0 + 0.888 888 888 711 168;
  • 12) 0.888 888 888 711 168 × 2 = 1 + 0.777 777 777 422 336;
  • 13) 0.777 777 777 422 336 × 2 = 1 + 0.555 555 554 844 672;
  • 14) 0.555 555 554 844 672 × 2 = 1 + 0.111 111 109 689 344;
  • 15) 0.111 111 109 689 344 × 2 = 0 + 0.222 222 219 378 688;
  • 16) 0.222 222 219 378 688 × 2 = 0 + 0.444 444 438 757 376;
  • 17) 0.444 444 438 757 376 × 2 = 0 + 0.888 888 877 514 752;
  • 18) 0.888 888 877 514 752 × 2 = 1 + 0.777 777 755 029 504;
  • 19) 0.777 777 755 029 504 × 2 = 1 + 0.555 555 510 059 008;
  • 20) 0.555 555 510 059 008 × 2 = 1 + 0.111 111 020 118 016;
  • 21) 0.111 111 020 118 016 × 2 = 0 + 0.222 222 040 236 032;
  • 22) 0.222 222 040 236 032 × 2 = 0 + 0.444 444 080 472 064;
  • 23) 0.444 444 080 472 064 × 2 = 0 + 0.888 888 160 944 128;
  • 24) 0.888 888 160 944 128 × 2 = 1 + 0.777 776 321 888 256;
  • 25) 0.777 776 321 888 256 × 2 = 1 + 0.555 552 643 776 512;
  • 26) 0.555 552 643 776 512 × 2 = 1 + 0.111 105 287 553 024;
  • 27) 0.111 105 287 553 024 × 2 = 0 + 0.222 210 575 106 048;
  • 28) 0.222 210 575 106 048 × 2 = 0 + 0.444 421 150 212 096;
  • 29) 0.444 421 150 212 096 × 2 = 0 + 0.888 842 300 424 192;
  • 30) 0.888 842 300 424 192 × 2 = 1 + 0.777 684 600 848 384;
  • 31) 0.777 684 600 848 384 × 2 = 1 + 0.555 369 201 696 768;
  • 32) 0.555 369 201 696 768 × 2 = 1 + 0.110 738 403 393 536;
  • 33) 0.110 738 403 393 536 × 2 = 0 + 0.221 476 806 787 072;
  • 34) 0.221 476 806 787 072 × 2 = 0 + 0.442 953 613 574 144;
  • 35) 0.442 953 613 574 144 × 2 = 0 + 0.885 907 227 148 288;
  • 36) 0.885 907 227 148 288 × 2 = 1 + 0.771 814 454 296 576;
  • 37) 0.771 814 454 296 576 × 2 = 1 + 0.543 628 908 593 152;
  • 38) 0.543 628 908 593 152 × 2 = 1 + 0.087 257 817 186 304;
  • 39) 0.087 257 817 186 304 × 2 = 0 + 0.174 515 634 372 608;
  • 40) 0.174 515 634 372 608 × 2 = 0 + 0.349 031 268 745 216;
  • 41) 0.349 031 268 745 216 × 2 = 0 + 0.698 062 537 490 432;
  • 42) 0.698 062 537 490 432 × 2 = 1 + 0.396 125 074 980 864;
  • 43) 0.396 125 074 980 864 × 2 = 0 + 0.792 250 149 961 728;
  • 44) 0.792 250 149 961 728 × 2 = 1 + 0.584 500 299 923 456;
  • 45) 0.584 500 299 923 456 × 2 = 1 + 0.169 000 599 846 912;
  • 46) 0.169 000 599 846 912 × 2 = 0 + 0.338 001 199 693 824;
  • 47) 0.338 001 199 693 824 × 2 = 0 + 0.676 002 399 387 648;
  • 48) 0.676 002 399 387 648 × 2 = 1 + 0.352 004 798 775 296;
  • 49) 0.352 004 798 775 296 × 2 = 0 + 0.704 009 597 550 592;
  • 50) 0.704 009 597 550 592 × 2 = 1 + 0.408 019 195 101 184;
  • 51) 0.408 019 195 101 184 × 2 = 0 + 0.816 038 390 202 368;
  • 52) 0.816 038 390 202 368 × 2 = 1 + 0.632 076 780 404 736;
  • 53) 0.632 076 780 404 736 × 2 = 1 + 0.264 153 560 809 472;

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:


0.777 777 777 777 691(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1001 0101 1(2)

5. Positive number before normalization:

24.777 777 777 777 691(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1001 0101 1(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:


24.777 777 777 777 691(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1001 0101 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1001 0101 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1001 0101 1(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 0101 1001 0101 1


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 0101 1001 0 1011 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1001


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 0101 1001


Decimal number 24.777 777 777 777 691 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 0101 1001

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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