Decimal to 64 Bit IEEE 754 Binary: Convert Number 112 704 614 771 458 041 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 112 704 614 771 458 041(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. 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;
  • 112 704 614 771 458 041 ÷ 2 = 56 352 307 385 729 020 + 1;
  • 56 352 307 385 729 020 ÷ 2 = 28 176 153 692 864 510 + 0;
  • 28 176 153 692 864 510 ÷ 2 = 14 088 076 846 432 255 + 0;
  • 14 088 076 846 432 255 ÷ 2 = 7 044 038 423 216 127 + 1;
  • 7 044 038 423 216 127 ÷ 2 = 3 522 019 211 608 063 + 1;
  • 3 522 019 211 608 063 ÷ 2 = 1 761 009 605 804 031 + 1;
  • 1 761 009 605 804 031 ÷ 2 = 880 504 802 902 015 + 1;
  • 880 504 802 902 015 ÷ 2 = 440 252 401 451 007 + 1;
  • 440 252 401 451 007 ÷ 2 = 220 126 200 725 503 + 1;
  • 220 126 200 725 503 ÷ 2 = 110 063 100 362 751 + 1;
  • 110 063 100 362 751 ÷ 2 = 55 031 550 181 375 + 1;
  • 55 031 550 181 375 ÷ 2 = 27 515 775 090 687 + 1;
  • 27 515 775 090 687 ÷ 2 = 13 757 887 545 343 + 1;
  • 13 757 887 545 343 ÷ 2 = 6 878 943 772 671 + 1;
  • 6 878 943 772 671 ÷ 2 = 3 439 471 886 335 + 1;
  • 3 439 471 886 335 ÷ 2 = 1 719 735 943 167 + 1;
  • 1 719 735 943 167 ÷ 2 = 859 867 971 583 + 1;
  • 859 867 971 583 ÷ 2 = 429 933 985 791 + 1;
  • 429 933 985 791 ÷ 2 = 214 966 992 895 + 1;
  • 214 966 992 895 ÷ 2 = 107 483 496 447 + 1;
  • 107 483 496 447 ÷ 2 = 53 741 748 223 + 1;
  • 53 741 748 223 ÷ 2 = 26 870 874 111 + 1;
  • 26 870 874 111 ÷ 2 = 13 435 437 055 + 1;
  • 13 435 437 055 ÷ 2 = 6 717 718 527 + 1;
  • 6 717 718 527 ÷ 2 = 3 358 859 263 + 1;
  • 3 358 859 263 ÷ 2 = 1 679 429 631 + 1;
  • 1 679 429 631 ÷ 2 = 839 714 815 + 1;
  • 839 714 815 ÷ 2 = 419 857 407 + 1;
  • 419 857 407 ÷ 2 = 209 928 703 + 1;
  • 209 928 703 ÷ 2 = 104 964 351 + 1;
  • 104 964 351 ÷ 2 = 52 482 175 + 1;
  • 52 482 175 ÷ 2 = 26 241 087 + 1;
  • 26 241 087 ÷ 2 = 13 120 543 + 1;
  • 13 120 543 ÷ 2 = 6 560 271 + 1;
  • 6 560 271 ÷ 2 = 3 280 135 + 1;
  • 3 280 135 ÷ 2 = 1 640 067 + 1;
  • 1 640 067 ÷ 2 = 820 033 + 1;
  • 820 033 ÷ 2 = 410 016 + 1;
  • 410 016 ÷ 2 = 205 008 + 0;
  • 205 008 ÷ 2 = 102 504 + 0;
  • 102 504 ÷ 2 = 51 252 + 0;
  • 51 252 ÷ 2 = 25 626 + 0;
  • 25 626 ÷ 2 = 12 813 + 0;
  • 12 813 ÷ 2 = 6 406 + 1;
  • 6 406 ÷ 2 = 3 203 + 0;
  • 3 203 ÷ 2 = 1 601 + 1;
  • 1 601 ÷ 2 = 800 + 1;
  • 800 ÷ 2 = 400 + 0;
  • 400 ÷ 2 = 200 + 0;
  • 200 ÷ 2 = 100 + 0;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

Take all the remainders starting from the bottom of the list constructed above.

112 704 614 771 458 041(10) =


1 1001 0000 0110 1000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1001(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 56 positions to the left, so that only one non zero digit remains to the left of it:


112 704 614 771 458 041(10) =


1 1001 0000 0110 1000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1001(2) =


1 1001 0000 0110 1000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1001(2) × 20 =


1.1001 0000 0110 1000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1001(2) × 256


4. 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): 56


Mantissa (not normalized):
1.1001 0000 0110 1000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1001


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


56 + 2(11-1) - 1 =


(56 + 1 023)(10) =


1 079(10)


6. 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;
  • 1 079 ÷ 2 = 539 + 1;
  • 539 ÷ 2 = 269 + 1;
  • 269 ÷ 2 = 134 + 1;
  • 134 ÷ 2 = 67 + 0;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

7. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1079(10) =


100 0011 0111(2)


8. 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. 1001 0000 0110 1000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1001 =


1001 0000 0110 1000 0011 1111 1111 1111 1111 1111 1111 1111 1111


9. 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 0011 0111


Mantissa (52 bits) =
1001 0000 0110 1000 0011 1111 1111 1111 1111 1111 1111 1111 1111


The base ten decimal number 112 704 614 771 458 041 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0011 0111 - 1001 0000 0110 1000 0011 1111 1111 1111 1111 1111 1111 1111 1111

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