24.777 777 777 777 777 777 777 926 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 777 926 6(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
24.777 777 777 777 777 777 777 926 6(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: 0.777 777 777 777 777 777 777 926 6.

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) 0.777 777 777 777 777 777 777 926 6 × 2 = 1 + 0.555 555 555 555 555 555 555 853 2;
  • 2) 0.555 555 555 555 555 555 555 853 2 × 2 = 1 + 0.111 111 111 111 111 111 111 706 4;
  • 3) 0.111 111 111 111 111 111 111 706 4 × 2 = 0 + 0.222 222 222 222 222 222 223 412 8;
  • 4) 0.222 222 222 222 222 222 223 412 8 × 2 = 0 + 0.444 444 444 444 444 444 446 825 6;
  • 5) 0.444 444 444 444 444 444 446 825 6 × 2 = 0 + 0.888 888 888 888 888 888 893 651 2;
  • 6) 0.888 888 888 888 888 888 893 651 2 × 2 = 1 + 0.777 777 777 777 777 777 787 302 4;
  • 7) 0.777 777 777 777 777 777 787 302 4 × 2 = 1 + 0.555 555 555 555 555 555 574 604 8;
  • 8) 0.555 555 555 555 555 555 574 604 8 × 2 = 1 + 0.111 111 111 111 111 111 149 209 6;
  • 9) 0.111 111 111 111 111 111 149 209 6 × 2 = 0 + 0.222 222 222 222 222 222 298 419 2;
  • 10) 0.222 222 222 222 222 222 298 419 2 × 2 = 0 + 0.444 444 444 444 444 444 596 838 4;
  • 11) 0.444 444 444 444 444 444 596 838 4 × 2 = 0 + 0.888 888 888 888 888 889 193 676 8;
  • 12) 0.888 888 888 888 888 889 193 676 8 × 2 = 1 + 0.777 777 777 777 777 778 387 353 6;
  • 13) 0.777 777 777 777 777 778 387 353 6 × 2 = 1 + 0.555 555 555 555 555 556 774 707 2;
  • 14) 0.555 555 555 555 555 556 774 707 2 × 2 = 1 + 0.111 111 111 111 111 113 549 414 4;
  • 15) 0.111 111 111 111 111 113 549 414 4 × 2 = 0 + 0.222 222 222 222 222 227 098 828 8;
  • 16) 0.222 222 222 222 222 227 098 828 8 × 2 = 0 + 0.444 444 444 444 444 454 197 657 6;
  • 17) 0.444 444 444 444 444 454 197 657 6 × 2 = 0 + 0.888 888 888 888 888 908 395 315 2;
  • 18) 0.888 888 888 888 888 908 395 315 2 × 2 = 1 + 0.777 777 777 777 777 816 790 630 4;
  • 19) 0.777 777 777 777 777 816 790 630 4 × 2 = 1 + 0.555 555 555 555 555 633 581 260 8;
  • 20) 0.555 555 555 555 555 633 581 260 8 × 2 = 1 + 0.111 111 111 111 111 267 162 521 6;
  • 21) 0.111 111 111 111 111 267 162 521 6 × 2 = 0 + 0.222 222 222 222 222 534 325 043 2;
  • 22) 0.222 222 222 222 222 534 325 043 2 × 2 = 0 + 0.444 444 444 444 445 068 650 086 4;
  • 23) 0.444 444 444 444 445 068 650 086 4 × 2 = 0 + 0.888 888 888 888 890 137 300 172 8;
  • 24) 0.888 888 888 888 890 137 300 172 8 × 2 = 1 + 0.777 777 777 777 780 274 600 345 6;
  • 25) 0.777 777 777 777 780 274 600 345 6 × 2 = 1 + 0.555 555 555 555 560 549 200 691 2;
  • 26) 0.555 555 555 555 560 549 200 691 2 × 2 = 1 + 0.111 111 111 111 121 098 401 382 4;
  • 27) 0.111 111 111 111 121 098 401 382 4 × 2 = 0 + 0.222 222 222 222 242 196 802 764 8;
  • 28) 0.222 222 222 222 242 196 802 764 8 × 2 = 0 + 0.444 444 444 444 484 393 605 529 6;
  • 29) 0.444 444 444 444 484 393 605 529 6 × 2 = 0 + 0.888 888 888 888 968 787 211 059 2;
  • 30) 0.888 888 888 888 968 787 211 059 2 × 2 = 1 + 0.777 777 777 777 937 574 422 118 4;
  • 31) 0.777 777 777 777 937 574 422 118 4 × 2 = 1 + 0.555 555 555 555 875 148 844 236 8;
  • 32) 0.555 555 555 555 875 148 844 236 8 × 2 = 1 + 0.111 111 111 111 750 297 688 473 6;
  • 33) 0.111 111 111 111 750 297 688 473 6 × 2 = 0 + 0.222 222 222 223 500 595 376 947 2;
  • 34) 0.222 222 222 223 500 595 376 947 2 × 2 = 0 + 0.444 444 444 447 001 190 753 894 4;
  • 35) 0.444 444 444 447 001 190 753 894 4 × 2 = 0 + 0.888 888 888 894 002 381 507 788 8;
  • 36) 0.888 888 888 894 002 381 507 788 8 × 2 = 1 + 0.777 777 777 788 004 763 015 577 6;
  • 37) 0.777 777 777 788 004 763 015 577 6 × 2 = 1 + 0.555 555 555 576 009 526 031 155 2;
  • 38) 0.555 555 555 576 009 526 031 155 2 × 2 = 1 + 0.111 111 111 152 019 052 062 310 4;
  • 39) 0.111 111 111 152 019 052 062 310 4 × 2 = 0 + 0.222 222 222 304 038 104 124 620 8;
  • 40) 0.222 222 222 304 038 104 124 620 8 × 2 = 0 + 0.444 444 444 608 076 208 249 241 6;
  • 41) 0.444 444 444 608 076 208 249 241 6 × 2 = 0 + 0.888 888 889 216 152 416 498 483 2;
  • 42) 0.888 888 889 216 152 416 498 483 2 × 2 = 1 + 0.777 777 778 432 304 832 996 966 4;
  • 43) 0.777 777 778 432 304 832 996 966 4 × 2 = 1 + 0.555 555 556 864 609 665 993 932 8;
  • 44) 0.555 555 556 864 609 665 993 932 8 × 2 = 1 + 0.111 111 113 729 219 331 987 865 6;
  • 45) 0.111 111 113 729 219 331 987 865 6 × 2 = 0 + 0.222 222 227 458 438 663 975 731 2;
  • 46) 0.222 222 227 458 438 663 975 731 2 × 2 = 0 + 0.444 444 454 916 877 327 951 462 4;
  • 47) 0.444 444 454 916 877 327 951 462 4 × 2 = 0 + 0.888 888 909 833 754 655 902 924 8;
  • 48) 0.888 888 909 833 754 655 902 924 8 × 2 = 1 + 0.777 777 819 667 509 311 805 849 6;
  • 49) 0.777 777 819 667 509 311 805 849 6 × 2 = 1 + 0.555 555 639 335 018 623 611 699 2;
  • 50) 0.555 555 639 335 018 623 611 699 2 × 2 = 1 + 0.111 111 278 670 037 247 223 398 4;
  • 51) 0.111 111 278 670 037 247 223 398 4 × 2 = 0 + 0.222 222 557 340 074 494 446 796 8;
  • 52) 0.222 222 557 340 074 494 446 796 8 × 2 = 0 + 0.444 445 114 680 148 988 893 593 6;
  • 53) 0.444 445 114 680 148 988 893 593 6 × 2 = 0 + 0.888 890 229 360 297 977 787 187 2;

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:


0.777 777 777 777 777 777 777 926 6(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 777 926 6(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:


24.777 777 777 777 777 777 777 926 6(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 24.777 777 777 777 777 777 777 926 6 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