Decimal to 64 Bit IEEE 754 Binary: Convert Number 922 337 203 685 477 548 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 922 337 203 685 477 548(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;
  • 922 337 203 685 477 548 ÷ 2 = 461 168 601 842 738 774 + 0;
  • 461 168 601 842 738 774 ÷ 2 = 230 584 300 921 369 387 + 0;
  • 230 584 300 921 369 387 ÷ 2 = 115 292 150 460 684 693 + 1;
  • 115 292 150 460 684 693 ÷ 2 = 57 646 075 230 342 346 + 1;
  • 57 646 075 230 342 346 ÷ 2 = 28 823 037 615 171 173 + 0;
  • 28 823 037 615 171 173 ÷ 2 = 14 411 518 807 585 586 + 1;
  • 14 411 518 807 585 586 ÷ 2 = 7 205 759 403 792 793 + 0;
  • 7 205 759 403 792 793 ÷ 2 = 3 602 879 701 896 396 + 1;
  • 3 602 879 701 896 396 ÷ 2 = 1 801 439 850 948 198 + 0;
  • 1 801 439 850 948 198 ÷ 2 = 900 719 925 474 099 + 0;
  • 900 719 925 474 099 ÷ 2 = 450 359 962 737 049 + 1;
  • 450 359 962 737 049 ÷ 2 = 225 179 981 368 524 + 1;
  • 225 179 981 368 524 ÷ 2 = 112 589 990 684 262 + 0;
  • 112 589 990 684 262 ÷ 2 = 56 294 995 342 131 + 0;
  • 56 294 995 342 131 ÷ 2 = 28 147 497 671 065 + 1;
  • 28 147 497 671 065 ÷ 2 = 14 073 748 835 532 + 1;
  • 14 073 748 835 532 ÷ 2 = 7 036 874 417 766 + 0;
  • 7 036 874 417 766 ÷ 2 = 3 518 437 208 883 + 0;
  • 3 518 437 208 883 ÷ 2 = 1 759 218 604 441 + 1;
  • 1 759 218 604 441 ÷ 2 = 879 609 302 220 + 1;
  • 879 609 302 220 ÷ 2 = 439 804 651 110 + 0;
  • 439 804 651 110 ÷ 2 = 219 902 325 555 + 0;
  • 219 902 325 555 ÷ 2 = 109 951 162 777 + 1;
  • 109 951 162 777 ÷ 2 = 54 975 581 388 + 1;
  • 54 975 581 388 ÷ 2 = 27 487 790 694 + 0;
  • 27 487 790 694 ÷ 2 = 13 743 895 347 + 0;
  • 13 743 895 347 ÷ 2 = 6 871 947 673 + 1;
  • 6 871 947 673 ÷ 2 = 3 435 973 836 + 1;
  • 3 435 973 836 ÷ 2 = 1 717 986 918 + 0;
  • 1 717 986 918 ÷ 2 = 858 993 459 + 0;
  • 858 993 459 ÷ 2 = 429 496 729 + 1;
  • 429 496 729 ÷ 2 = 214 748 364 + 1;
  • 214 748 364 ÷ 2 = 107 374 182 + 0;
  • 107 374 182 ÷ 2 = 53 687 091 + 0;
  • 53 687 091 ÷ 2 = 26 843 545 + 1;
  • 26 843 545 ÷ 2 = 13 421 772 + 1;
  • 13 421 772 ÷ 2 = 6 710 886 + 0;
  • 6 710 886 ÷ 2 = 3 355 443 + 0;
  • 3 355 443 ÷ 2 = 1 677 721 + 1;
  • 1 677 721 ÷ 2 = 838 860 + 1;
  • 838 860 ÷ 2 = 419 430 + 0;
  • 419 430 ÷ 2 = 209 715 + 0;
  • 209 715 ÷ 2 = 104 857 + 1;
  • 104 857 ÷ 2 = 52 428 + 1;
  • 52 428 ÷ 2 = 26 214 + 0;
  • 26 214 ÷ 2 = 13 107 + 0;
  • 13 107 ÷ 2 = 6 553 + 1;
  • 6 553 ÷ 2 = 3 276 + 1;
  • 3 276 ÷ 2 = 1 638 + 0;
  • 1 638 ÷ 2 = 819 + 0;
  • 819 ÷ 2 = 409 + 1;
  • 409 ÷ 2 = 204 + 1;
  • 204 ÷ 2 = 102 + 0;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 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.

922 337 203 685 477 548(10) =


1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1010 1100(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 59 positions to the left, so that only one non zero digit remains to the left of it:


922 337 203 685 477 548(10) =


1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1010 1100(2) =


1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1010 1100(2) × 20 =


1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0101 100(2) × 259


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): 59


Mantissa (not normalized):
1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0101 100


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


59 + 2(11-1) - 1 =


(59 + 1 023)(10) =


1 082(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;
  • 1 082 ÷ 2 = 541 + 0;
  • 541 ÷ 2 = 270 + 1;
  • 270 ÷ 2 = 135 + 0;
  • 135 ÷ 2 = 67 + 1;
  • 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) =


1082(10) =


100 0011 1010(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 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 010 1100 =


1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001


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 1010


Mantissa (52 bits) =
1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001


The base ten decimal number 922 337 203 685 477 548 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0011 1010 - 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001

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